{"id":537,"date":"2017-10-27T16:30:10","date_gmt":"2017-10-27T16:30:10","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/chapter\/rotational-kinetic-energy-work-and-energy-revisited\/"},"modified":"2017-11-08T03:24:49","modified_gmt":"2017-11-08T03:24:49","slug":"rotational-kinetic-energy-work-and-energy-revisited","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/chapter\/rotational-kinetic-energy-work-and-energy-revisited\/","title":{"raw":"Rotational Kinetic Energy: Work and Energy Revisited","rendered":"Rotational Kinetic Energy: Work and Energy Revisited"},"content":{"raw":"\n<div class=\"textbox learning-objectives\">\n<h3 itemprop=\"educationalUse\">Learning Objectives<\/h3>\n<ul>\n<li>Derive the equation for rotational work.<\/li>\n<li>Calculate rotational kinetic energy.<\/li>\n<li>Demonstrate the Law of Conservation of Energy.<\/li>\n<\/ul>\n<\/div>\n<p id=\"import-auto-id3035380\">In this module, we will learn about work and energy associated with rotational motion. <a href=\"#import-auto-id3229349\" class=\"autogenerated-content\">(Figure)<\/a> shows a worker using an electric grindstone propelled by a motor. Sparks are flying, and noise and vibration are created as layers of steel are pared from the pole. The stone continues to turn even after the motor is turned off, but it is eventually brought to a stop by friction. Clearly, the motor had to work to get the stone spinning. This work went into heat, light, sound, vibration, and considerable <span data-type=\"term\" id=\"import-auto-id2669979\">rotational kinetic energy<\/span>.<\/p>\n<div class=\"bc-figure figure\" id=\"import-auto-id3229349\">\n<div class=\"bc-figcaption figcaption\">The motor works in spinning the grindstone, giving it rotational kinetic energy. That energy is then converted to heat, light, sound, and vibration. (credit: U.S. Navy photo by Mass Communication Specialist Seaman Zachary David Bell)<\/div>\n<p><span data-type=\"media\" id=\"import-auto-id2630228\" data-alt=\"The figure shows a mechanic cutting metal with a metal grinder. The sparks are emerging from the point of contact and jumping off tangentially from the cutter.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/Figure_11_04_01a.jpg\" data-media-type=\"image\/png\" alt=\"The figure shows a mechanic cutting metal with a metal grinder. The sparks are emerging from the point of contact and jumping off tangentially from the cutter.\" width=\"250\"><\/span><\/p><\/div>\n<p id=\"import-auto-id2612157\">Work must be done to rotate objects such as grindstones or merry-go-rounds. Work was defined in <a href=\"\/contents\/3ef5dfb6-0a8d-433e-9c8f-b8c860a3903b@2\">Uniform Circular Motion and Gravitation<\/a> for translational motion, and we can build on that knowledge when considering work done in rotational motion. The simplest rotational situation is one in which the net force is exerted perpendicular to the radius of a disk (as shown in <a href=\"#import-auto-id2009471\" class=\"autogenerated-content\">(Figure)<\/a>) and remains perpendicular as the disk starts to rotate. The force is parallel to the displacement, and so the net work done is the product of the force times the arc length traveled:<\/p>\n<div data-type=\"equation\" class=\"equation\">[latex]\\text{net}\\phantom{\\rule{0.25em}{0ex}}W=\\left(\\text{net}\\phantom{\\rule{0.25em}{0ex}}F\\right)\\text{\u0394}s.[\/latex]<\/div>\n<p id=\"import-auto-id1587900\">To get torque and other rotational quantities into the equation, we multiply and divide the right-hand side of the equation by [latex]r[\/latex], and gather terms:<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-909\">[latex]\\text{net}\\phantom{\\rule{0.25em}{0ex}}W=\\left(r\\phantom{\\rule{0.25em}{0ex}}\\text{net}\\phantom{\\rule{0.25em}{0ex}}F\\right)\\frac{\\text{\u0394}s}{r}.[\/latex]<\/div>\n<p id=\"import-auto-id2448795\">We recognize that [latex]r\\phantom{\\rule{0.25em}{0ex}}\\text{net}\\phantom{\\rule{0.25em}{0ex}}F=\\text{net \u03c4}[\/latex] and [latex]\\Delta s\/r=\\theta [\/latex], so that<\/p>\n<div data-type=\"equation\" class=\"equation\">[latex]\\text{net}\\phantom{\\rule{0.25em}{0ex}}W=\\left(\\text{net \u03c4}\\right)\\theta .[\/latex]<\/div>\n<p id=\"import-auto-id2009328\">This equation is the expression for rotational work. It is very similar to the familiar definition of translational work as force multiplied by distance. Here, torque is analogous to force, and angle is analogous to distance. The equation [latex]\\text{net}\\phantom{\\rule{0.25em}{0ex}}W=\\left(\\text{net \u03c4}\\right)\\theta [\/latex] is valid in general, even though it was derived for a special case.<\/p>\n<p id=\"import-auto-id3026626\">To get an expression for rotational kinetic energy, we must again perform some algebraic manipulations. The first step is to note that [latex]\\text{net \u03c4}=\\mathrm{I\\alpha }[\/latex], so that<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-404\">[latex]\\text{net}\\phantom{\\rule{0.25em}{0ex}}W=I\\text{\u03b1\u03b8}.[\/latex]<\/div>\n<div class=\"bc-figure figure\" id=\"import-auto-id2009471\">\n<div class=\"bc-figcaption figcaption\">The net force on this disk is kept perpendicular to its radius as the force causes the disk to rotate. The net work done is thus [latex]\\left(\\text{net}\\phantom{\\rule{0.25em}{0ex}}F\\right)\\Delta s[\/latex]. The net work goes into rotational kinetic energy.<\/div>\n<p><span data-type=\"media\" id=\"import-auto-id2613010\" data-alt=\"The figure shows a circular disc of radius r. A net force F is applied perpendicular to the radius, rotating the disc in an anti-clockwise direction and producing a displacement equal to delta S, in a direction parallel to the direction of the force applied. The angle covered is theta.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/Figure_11_04_02a.jpg\" data-media-type=\"image\/jpg\" alt=\"The figure shows a circular disc of radius r. A net force F is applied perpendicular to the radius, rotating the disc in an anti-clockwise direction and producing a displacement equal to delta S, in a direction parallel to the direction of the force applied. The angle covered is theta.\" width=\"275\"><\/span><\/p><\/div>\n<div data-type=\"note\" class=\"note\" data-has-label=\"true\" id=\"fs-id3191720\" data-label=\"\">\n<div data-type=\"title\" class=\"title\">Making Connections<\/div>\n<p id=\"import-auto-id3372728\">Work and energy in rotational motion are completely analogous to work and energy in translational motion, first presented in <a href=\"\/contents\/3ef5dfb6-0a8d-433e-9c8f-b8c860a3903b@2\">Uniform Circular Motion and Gravitation<\/a>.<\/p>\n<\/div>\n<p id=\"import-auto-id3418348\">Now, we solve one of the rotational kinematics equations for [latex]\\text{\u03b1\u03b8}[\/latex]. We start with the equation<\/p>\n<div data-type=\"equation\" class=\"equation\">[latex]{{\\omega }_{}}^{2}={{\\omega }_{\\text{0}}}^{2}+2\\text{\u03b1\u03b8}.[\/latex]<\/div>\n<p id=\"import-auto-id3154919\">Next, we solve for [latex]\\text{\u03b1\u03b8}[\/latex]:<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-241\">[latex]\\text{\u03b1\u03b8}=\\frac{{\\omega }^{2}-{{\\omega }_{\\text{0}}}^{2}}{2}.[\/latex]<\/div>\n<p id=\"import-auto-id2041695\">Substituting this into the equation for net [latex]W[\/latex] and gathering terms yields<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-789\">[latex]\\text{net}\\phantom{\\rule{0.25em}{0ex}}W=\\frac{1}{2}{\\mathrm{I\\omega }}^{2}-\\frac{1}{2}I{{\\omega }_{\\text{0}}}^{2}.[\/latex]<\/div>\n<p id=\"import-auto-id1373539\">This equation is the <span data-type=\"term\" id=\"import-auto-id2409820\">work-energy theorem<\/span> for rotational motion only. As you may recall, net work changes the kinetic energy of a system. Through an analogy with translational motion, we define the term [latex]\\left(\\frac{1}{2}\\right){\\mathrm{I\\omega }}^{2}[\/latex] to be <span data-type=\"term\" id=\"import-auto-id3012447\">rotational kinetic energy<\/span> [latex]{\\text{KE}}_{\\text{rot}}[\/latex] for an object with a moment of inertia [latex]I[\/latex] and an angular velocity [latex]\\omega [\/latex]:<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-682\">[latex]{\\text{KE}}_{\\text{rot}}=\\frac{1}{2}{\\mathrm{I\\omega }}^{2}.[\/latex]<\/div>\n<p id=\"import-auto-id1434731\">The expression for rotational kinetic energy is exactly analogous to translational kinetic energy, with [latex]I[\/latex] being analogous to [latex]m[\/latex] and [latex]\\omega [\/latex] to [latex]v[\/latex]. Rotational kinetic energy has important effects. Flywheels, for example, can be used to store large amounts of rotational kinetic energy in a vehicle, as seen in <a href=\"#import-auto-id1614457\" class=\"autogenerated-content\">(Figure)<\/a>.<\/p>\n<div class=\"bc-figure figure\" id=\"import-auto-id1614457\">\n<div class=\"bc-figcaption figcaption\">Experimental vehicles, such as this bus, have been constructed in which rotational kinetic energy is stored in a large flywheel. When the bus goes down a hill, its transmission converts its gravitational potential energy into [latex]{\\text{KE}}_{\\text{rot}}[\/latex]. It can also convert translational kinetic energy, when the bus stops, into [latex]{\\text{KE}}_{\\text{rot}}[\/latex]. The flywheel\u2019s energy can then be used to accelerate, to go up another hill, or to keep the bus from going against friction.<\/div>\n<p><span data-type=\"media\" id=\"import-auto-id1817378\" data-alt=\"The figure shows a bus carrying a large flywheel on its board in which rotational kinetic energy is stored.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/Figure_11_04_03a.jpg\" data-media-type=\"image\/jpg\" alt=\"The figure shows a bus carrying a large flywheel on its board in which rotational kinetic energy is stored.\" width=\"250\"><\/span><\/p><\/div>\n<div data-type=\"example\" class=\"textbox examples\" id=\"fs-id3354618\">\n<div data-type=\"title\" class=\"title\">Calculating the Work and Energy for Spinning a Grindstone<\/div>\n<p id=\"import-auto-id2600755\">Consider a person who spins a large grindstone by placing her hand on its edge and exerting a force through part of a revolution as shown in <a href=\"#import-auto-id2674234\" class=\"autogenerated-content\">(Figure)<\/a>. In this example, we verify that the work done by the torque she exerts equals the change in rotational energy. (a) How much work is done if she exerts a force of 200 N through a rotation of [latex]\\text{1.00 rad}\\left(57.3\u00ba\\right)[\/latex]? The force is kept perpendicular to the grindstone\u2019s 0.320-m radius at the point of application, and the effects of friction are negligible. (b) What is the final angular velocity if the grindstone has a mass of 85.0 kg? (c) What is the final rotational kinetic energy? (It should equal the work.)<\/p>\n<p id=\"import-auto-id2421131\"><strong>Strategy<\/strong><\/p>\n<p id=\"fs-id2422273\">To find the work, we can use the equation [latex]\\text{net}\\phantom{\\rule{0.25em}{0ex}}W=\\left(\\text{net \u03c4}\\right)\\theta [\/latex]. We have enough information to calculate the torque and are given the rotation angle. In the second part, we can find the final angular velocity using one of the kinematic relationships. In the last part, we can calculate the rotational kinetic energy from its expression in [latex]{\\text{KE}}_{\\text{rot}}=\\frac{1}{2}{\\mathrm{I\\omega }}^{2}[\/latex].<\/p>\n<p id=\"import-auto-id3181145\"><strong>Solution for (a)<\/strong><\/p>\n<p id=\"fs-id1485088\">The net work is expressed in the equation<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-117\">[latex]\\text{net}\\phantom{\\rule{0.25em}{0ex}}W=\\left(\\text{net \u03c4}\\right)\\theta ,[\/latex]<\/div>\n<p id=\"import-auto-id1828262\">where net [latex]\\tau [\/latex] is the applied force multiplied by the radius [latex]\\left(\\text{rF}\\right)[\/latex] because there is no retarding friction, and the force is perpendicular to [latex]r[\/latex]. The angle [latex]\\theta [\/latex] is given. Substituting the given values in the equation above yields<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-348\">[latex]\\begin{array}{lll}\\text{net}\\phantom{\\rule{0.25em}{0ex}}W&amp; =&amp; \\text{rF}\\theta =\\left(\\text{0.320 m}\\right)\\left(\\text{200 N}\\right)\\left(\\text{1.00 rad}\\right)\\\\ &amp; =&amp; \\text{64.0 N}\\cdot \\text{m.}\\end{array}[\/latex]<\/div>\n<p id=\"import-auto-id2406953\">Noting that [latex]1 N\u00b7\\text{m}=1 J[\/latex],<\/p>\n<div data-type=\"equation\" class=\"equation\">[latex]\\text{net}\\phantom{\\rule{0.25em}{0ex}}W=\\text{64.0 J}.[\/latex]<\/div>\n<div class=\"bc-figure figure\" id=\"import-auto-id2674234\">\n<div class=\"bc-figcaption figcaption\">A large grindstone is given a spin by a person grasping its outer edge.<\/div>\n<p><span data-type=\"media\" id=\"import-auto-id3402470\" data-alt=\"The figure shows a large grindstone of radius r which is being given a spin by applying a force F in a counterclockwise direction, as indicated by the arrows.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/Figure_11_04_04a.jpg\" data-media-type=\"image\/jpg\" alt=\"The figure shows a large grindstone of radius r which is being given a spin by applying a force F in a counterclockwise direction, as indicated by the arrows.\" height=\"225\"><\/span><\/p><\/div>\n<p id=\"import-auto-id1468054\"><strong>Solution for (b)<\/strong><\/p>\n<p id=\"fs-id1276292\">To find [latex]\\omega [\/latex] from the given information requires more than one step. We start with the kinematic relationship in the equation<\/p>\n<div data-type=\"equation\" class=\"equation\">[latex]{\\omega }^{2}={{\\omega }_{\\text{0}}}^{2}+2\\text{\u03b1\u03b8}.[\/latex]<\/div>\n<p id=\"import-auto-id1946811\">Note that [latex]{\\omega }_{0}=0[\/latex] because we start from rest. Taking the square root of the resulting equation gives<\/p>\n<div data-type=\"equation\" class=\"equation\">[latex]\\omega ={\\left(2\\text{\u03b1\u03b8}\\right)}^{1\/2}.[\/latex]<\/div>\n<p id=\"import-auto-id2435786\">Now we need to find [latex]\\alpha [\/latex]. One possibility is<\/p>\n<div data-type=\"equation\" class=\"equation\">[latex]\\alpha =\\frac{\\text{net \u03c4}}{I},[\/latex]<\/div>\n<p id=\"import-auto-id1151420\">where the torque is<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-643\">[latex]\\text{net \u03c4}=\\text{rF}=\\left(\\text{0.320 m}\\right)\\left(\\text{200 N}\\right)=\\text{64.0 N}\\cdot \\text{m}.[\/latex]<\/div>\n<p id=\"import-auto-id2588420\">The formula for the moment of inertia for a disk is found in <a href=\"\/contents\/db59f656-e708-4094-a742-1e5560fe97c9@6#fs-id1838666\" class=\"autogenerated-content\">(Figure)<\/a>:<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-119\">[latex]I=\\frac{1}{2}{\\text{MR}}^{2}=0.5\\left(\\text{85.0 kg}\\right){\\left(\\text{0.320 m}\\right)}^{2}=\\text{4.352 kg}\\cdot {\\text{m}}^{2}.[\/latex]<\/div>\n<p id=\"import-auto-id2670890\">Substituting the values of torque and moment of inertia into the expression for [latex]\\alpha [\/latex], we obtain<\/p>\n<div data-type=\"equation\" class=\"equation\">[latex]\\alpha =\\frac{\\text{64}\\text{.}\\text{0 N}\\cdot \\text{m}}{\\text{4.352 kg}\\cdot {\\text{m}}^{2}}=\\text{14.7}\\frac{\\text{rad}}{{\\text{s}}^{2}}.[\/latex]<\/div>\n<p id=\"import-auto-id3400808\">Now, substitute this value and the given value for [latex]\\theta [\/latex] into the above expression for [latex]\\omega [\/latex]:<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-116\">[latex]\\omega ={\\left(2\\text{\u03b1\u03b8}\\right)}^{1\/2}={\\left[2\\left(\\text{14.7}\\frac{\\text{rad}}{{\\text{s}}^{2}}\\right)\\left(\\text{1.00 rad}\\right)\\right]}^{1\/2}=\\text{5.42}\\frac{\\text{rad}}{\\text{s}}.[\/latex]<\/div>\n<p id=\"import-auto-id2408772\"><strong>Solution for (c)<\/strong><\/p>\n<p>The final rotational kinetic energy is<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-291\">[latex]{\\text{KE}}_{\\text{rot}}=\\frac{1}{2}{\\mathrm{I\\omega }}^{2}.[\/latex]<\/div>\n<p id=\"import-auto-id3077647\">Both [latex]I[\/latex] and [latex]\\omega [\/latex] were found above. Thus,<\/p>\n<div data-type=\"equation\" class=\"equation\">[latex]{\\text{KE}}_{\\text{rot}}=\\left(0.5\\right)\\left(\\text{4.352 kg}\\cdot {\\text{m}}^{2}\\right){\\left(\\text{5.42 rad\/s}\\right)}^{2}=\\text{64.0 J}.[\/latex]<\/div>\n<p id=\"import-auto-id1427676\"><strong>Discussion<\/strong><\/p>\n<p id=\"eip-id1935871\">The final rotational kinetic energy equals the work done by the torque, which confirms that the work done went into rotational kinetic energy. We could, in fact, have used an expression for energy instead of a kinematic relation to solve part (b). We will do this in later examples.<\/p>\n<\/div>\n<p id=\"import-auto-id3162150\">Helicopter pilots are quite familiar with rotational kinetic energy. They know, for example, that a point of no return will be reached if they allow their blades to slow below a critical angular velocity during flight. The blades lose lift, and it is impossible to immediately get the blades spinning fast enough to regain it. Rotational kinetic energy must be supplied to the blades to get them to rotate faster, and enough energy cannot be supplied in time to avoid a crash. Because of weight limitations, helicopter engines are too small to supply both the energy needed for lift and to replenish the rotational kinetic energy of the blades once they have slowed down. The rotational kinetic energy is put into them before takeoff and must not be allowed to drop below this crucial level. One possible way to avoid a crash is to use the gravitational potential energy of the helicopter to replenish the rotational kinetic energy of the blades by losing altitude and aligning the blades so that the helicopter is spun up in the descent. Of course, if the helicopter\u2019s altitude is too low, then there is insufficient time for the blade to regain lift before reaching the ground.<\/p>\n<div data-type=\"note\" class=\"note\" data-has-label=\"true\" id=\"fs-id1986333\" data-label=\"\">\n<div data-type=\"title\" class=\"title\">Problem-Solving Strategy for Rotational Energy<\/div>\n<ol id=\"fs-id2438401\" data-number-style=\"arabic\">\n<li id=\"import-auto-id3054599\"><em data-effect=\"italics\">Determine that energy or work is involved in the rotation<\/em>.<\/li>\n<li id=\"import-auto-id1389478\"><em data-effect=\"italics\">Determine the system of interest<\/em>. A sketch usually helps.<\/li>\n<li id=\"import-auto-id2930250\"><em data-effect=\"italics\">Analyze the situation to determine the types of work and energy involved<\/em>.<\/li>\n<li id=\"import-auto-id2937257\"><em data-effect=\"italics\">For closed systems, mechanical energy is conserved<\/em>. That is, [latex]{\\text{KE}}_{\\text{i}}+{\\text{PE}}_{\\text{i}}={\\text{KE}}_{\\text{f}}+{\\text{PE}}_{\\text{f}}.[\/latex] Note that [latex]{\\text{KE}}_{\\text{i}}[\/latex] and [latex]{\\text{KE}}_{\\text{f}}[\/latex] may each include translational and rotational contributions.<\/li>\n<li id=\"import-auto-id1439188\"><em data-effect=\"italics\">For open systems<\/em>, mechanical energy may not be conserved, and other forms of energy (referred to previously as [latex]\\text{OE}[\/latex]), such as heat transfer, may enter or leave the system. Determine what they are, and calculate them as necessary.<\/li>\n<li id=\"import-auto-id1429438\"><em data-effect=\"italics\">Eliminate terms wherever possible to simplify the algebra<\/em>.<\/li>\n<li id=\"import-auto-id3386776\"><em data-effect=\"italics\">Check the answer to see if it is reasonable<\/em>.<\/li>\n<\/ol>\n<\/div>\n<div data-type=\"example\" class=\"textbox examples\" id=\"fs-id3173123\">\n<div data-type=\"title\" class=\"title\">Calculating Helicopter Energies<\/div>\n<p>A typical small rescue helicopter, similar to the one in <a href=\"#import-auto-id2420248\" class=\"autogenerated-content\">(Figure)<\/a>, has four blades, each is 4.00 m long and has a mass of 50.0 kg. The blades can be approximated as thin rods that rotate about one end of an axis perpendicular to their length. The helicopter has a total loaded mass of 1000 kg. (a) Calculate the rotational kinetic energy in the blades when they rotate at 300 rpm. (b) Calculate the translational kinetic energy of the helicopter when it flies at 20.0 m\/s, and compare it with the rotational energy in the blades. (c) To what height could the helicopter be raised if all of the rotational kinetic energy could be used to lift it?<\/p>\n<p id=\"import-auto-id1389334\"><strong>Strategy<\/strong><\/p>\n<p id=\"fs-id1519185\">Rotational and translational kinetic energies can be calculated from their definitions. The last part of the problem relates to the idea that energy can change form, in this case from rotational kinetic energy to gravitational potential energy.<\/p>\n<p id=\"import-auto-id3430265\"><strong>Solution for (a)<\/strong><\/p>\n<p id=\"fs-id1569357\">The rotational kinetic energy is <\/p>\n<div data-type=\"equation\" class=\"equation\">[latex]{\\text{KE}}_{\\text{rot}}=\\frac{1}{2}{\\mathrm{I\\omega }}^{2}.[\/latex]<\/div>\n<p id=\"import-auto-id1426438\">We must convert the angular velocity to radians per second and calculate the moment of inertia before we can find [latex]{\\text{KE}}_{\\text{rot}}[\/latex]. The angular velocity [latex]\\omega [\/latex] is<\/p>\n<div data-type=\"equation\" class=\"equation\">[latex]\\omega =\\frac{\\text{300 rev}}{\\text{1.00 min}}\\cdot \\frac{\\text{2\u03c0 rad}}{\\text{1 rev}}\\cdot \\frac{\\text{1.00 min}}{\\text{60.0 s}}=\\text{31.4}\\frac{\\text{rad}}{\\text{s}}.[\/latex]<\/div>\n<p id=\"import-auto-id2598986\">The moment of inertia of one blade will be that of a thin rod rotated about its end, found in <a href=\"\/contents\/db59f656-e708-4094-a742-1e5560fe97c9@6#fs-id1838666\" class=\"autogenerated-content\">(Figure)<\/a>. The total [latex]I[\/latex] is four times this moment of inertia, because there are four blades. Thus,<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-897\">[latex]I=4\\frac{{\\mathrm{M\\ell }}^{2}}{3}=4\u00d7\\frac{\\left(\\text{50.0 kg}\\right){\\left(\\text{4.00 m}\\right)}^{2}}{3}=\\text{1067 kg}\\cdot {\\text{m}}^{2}.[\/latex]<\/div>\n<p id=\"import-auto-id2673012\">Entering [latex]\\omega [\/latex] and [latex]I[\/latex] into the expression for rotational kinetic energy gives<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-913\">[latex]\\begin{array}{lll}{\\text{KE}}_{\\text{rot}}&amp; =&amp; 0.5\\left(\\text{1067 kg}\\cdot {\\text{m}}^{2}\\right){\\left(\\text{31.4 rad\/s}\\right)}^{2}\\\\ &amp; =&amp; 5.26\u00d7{\\text{10}}^{5}\\phantom{\\rule{0.25em}{0ex}}\\text{J}\\end{array}[\/latex]<\/div>\n<p id=\"import-auto-id2616110\"><strong>Solution for (b)<\/strong><\/p>\n<p>Translational kinetic energy was defined in <a href=\"\/contents\/3ef5dfb6-0a8d-433e-9c8f-b8c860a3903b@2\">Uniform Circular Motion and Gravitation<\/a>. Entering the given values of mass and velocity, we obtain<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-2\">[latex]{\\text{KE}}_{\\text{trans}}=\\frac{1}{2}{\\mathit{mv}}^{2}=\\left(0.5\\right)\\left(\\text{1000 kg}\\right){\\left(\\text{20.0 m\/s}\\right)}^{2}=2\\text{.}\\text{00}\u00d7{\\text{10}}^{5}\\phantom{\\rule{0.25em}{0ex}}\\text{J}.[\/latex]<\/div>\n<p id=\"import-auto-id1587080\">To compare kinetic energies, we take the ratio of translational kinetic energy to rotational kinetic energy. This ratio is<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-973\">[latex]\\frac{2\\text{.}\\text{00}\u00d7{\\text{10}}^{5}\\phantom{\\rule{0.25em}{0ex}}\\text{J}}{5\\text{.}\\text{26}\u00d7{\\text{10}}^{5}\\phantom{\\rule{0.25em}{0ex}}\\text{J}}=0.380.[\/latex]<\/div>\n<p id=\"import-auto-id1848503\"><strong>Solution for (c)<\/strong><\/p>\n<p id=\"fs-id2407243\">At the maximum height, all rotational kinetic energy will have been converted to gravitational energy. To find this height, we equate those two energies:<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-487\">[latex]{\\text{KE}}_{\\text{rot}}={\\text{PE}}_{\\text{grav}}[\/latex]<\/div>\n<p id=\"import-auto-id3148882\">or<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-154\">[latex]\\frac{1}{2}{\\mathrm{I\\omega }}^{2}=\\text{mgh}.[\/latex]<\/div>\n<p id=\"import-auto-id1920250\">We now solve for [latex]h[\/latex] and substitute known values into the resulting equation<\/p>\n<div data-type=\"equation\" class=\"equation\">[latex]h=\\frac{{\\frac{1}{2}\\mathrm{I\\omega }}^{2}}{\\text{mg}}=\\frac{5.26\u00d7{\\text{10}}^{5}\\phantom{\\rule{0.25em}{0ex}}\\text{J}}{\\left(\\text{1000 kg}\\right)\\left(9.80\\phantom{\\rule{0.25em}{0ex}}{\\text{m\/s}}^{2}\\right)}=\\text{53.7 m}.[\/latex]<\/div>\n<p id=\"import-auto-id3068968\"><strong>Discussion<\/strong><\/p>\n<p id=\"fs-id2449340\">The ratio of translational energy to rotational kinetic energy is only 0.380. This ratio tells us that most of the kinetic energy of the helicopter is in its spinning blades\u2014something you probably would not suspect. The 53.7 m height to which the helicopter could be raised with the rotational kinetic energy is also impressive, again emphasizing the amount of rotational kinetic energy in the blades.<\/p>\n<\/div>\n<div class=\"bc-figure figure\" id=\"import-auto-id2420248\">\n<div class=\"bc-figcaption figcaption\">The first image shows how helicopters store large amounts of rotational kinetic energy in their blades. This energy must be put into the blades before takeoff and maintained until the end of the flight. The engines do not have enough power to simultaneously provide lift and put significant rotational energy into the blades. The second image shows a helicopter from the Auckland Westpac Rescue Helicopter Service. Over 50,000 lives have been saved since its operations beginning in 1973. Here, a water rescue operation is shown. (credit: 111 Emergency, Flickr)<\/div>\n<p><span data-type=\"media\" id=\"import-auto-id2930076\" data-alt=\"The given figure here shows a helicopter from the Auckland Westpac Rescue Helicopter Service over a sea. A rescue diver is shown holding the iron stand bar at the bottom of the helicopter, clutching a person. In the other image just above this, the blades of the helicopter are shown with their anti-clockwise rotation direction shown with an arrow and the length of one blade is given as four meters.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/Figure_11_04_05a.jpg\" data-media-type=\"image\/jpg\" alt=\"The given figure here shows a helicopter from the Auckland Westpac Rescue Helicopter Service over a sea. A rescue diver is shown holding the iron stand bar at the bottom of the helicopter, clutching a person. In the other image just above this, the blades of the helicopter are shown with their anti-clockwise rotation direction shown with an arrow and the length of one blade is given as four meters.\" height=\"300\"><\/span><\/p><\/div>\n<div data-type=\"note\" class=\"note\" data-has-label=\"true\" id=\"fs-id3421186\" data-label=\"\">\n<div data-type=\"title\" class=\"title\">Making Connections<\/div>\n<p id=\"import-auto-id1418758\">Conservation of energy includes rotational motion, because rotational kinetic energy is another form of [latex]\\text{KE}[\/latex]<br>\n    . <a href=\"\/contents\/3ef5dfb6-0a8d-433e-9c8f-b8c860a3903b@2\">Uniform Circular Motion and Gravitation<\/a> has a detailed treatment of conservation of energy.<\/p>\n<\/div>\n<div class=\"bc-section section\" data-depth=\"1\" id=\"fs-id2616556\">\n<h1 data-type=\"title\">How Thick Is the Soup? Or Why Don\u2019t All Objects Roll Downhill at the Same Rate?<\/h1>\n<p id=\"import-auto-id1970058\">One of the quality controls in a tomato soup factory consists of rolling filled cans down a ramp. If they roll too fast, the soup is too thin. Why should cans of identical size and mass roll down an incline at different rates? And why should the thickest soup roll the slowest?<\/p>\n<p id=\"import-auto-id2969200\">The easiest way to answer these questions is to consider energy. Suppose each can starts down the ramp from rest. Each can starting from rest means each starts with the same gravitational potential energy [latex]{\\text{PE}}_{\\text{grav}}[\/latex], which is converted entirely to [latex]\\text{KE}[\/latex], provided each rolls without slipping. [latex]\\text{KE}[\/latex], however, can take the form of [latex]{\\text{KE}}_{\\text{trans}}[\/latex] or [latex]{\\text{KE}}_{\\text{rot}}[\/latex], and total [latex]\\text{KE}[\/latex] is the sum of the two. If a can rolls down a ramp, it puts part of its energy into rotation, leaving less for translation. Thus, the can goes slower than it would if it slid down. Furthermore, the thin soup does not rotate, whereas the thick soup does, because it sticks to the can. The thick soup thus puts more of the can\u2019s original gravitational potential energy into rotation than the thin soup, and the can rolls more slowly, as seen in <a href=\"#import-auto-id3105621\" class=\"autogenerated-content\">(Figure)<\/a>.<\/p>\n<div class=\"bc-figure figure\" id=\"import-auto-id3105621\">\n<div class=\"bc-figcaption figcaption\">Three cans of soup with identical masses race down an incline. The first can has a low friction coating and does not roll but just slides down the incline. It wins because it converts its entire PE into translational KE. The second and third cans both roll down the incline without slipping. The second can contains thin soup and comes in second because part of its initial PE goes into rotating the can (but not the thin soup). The third can contains thick soup. It comes in third because the soup rotates along with the can, taking even more of the initial PE for rotational KE, leaving less for translational KE.<\/div>\n<p><span data-type=\"media\" id=\"import-auto-id3026629\" data-alt=\"The figure shows a flat surface inclined at a height of h from the surface level, with three cans of soup of different densities numbered as one, two, and three rolling along it.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/Figure_11_04_06a.jpg\" data-media-type=\"image\/jpg\" alt=\"The figure shows a flat surface inclined at a height of h from the surface level, with three cans of soup of different densities numbered as one, two, and three rolling along it.\" width=\"300\"><\/span><\/p><\/div>\n<p id=\"import-auto-id2679189\">Assuming no losses due to friction, there is only one force doing work\u2014gravity. Therefore the total work done is the change in kinetic energy. As the cans start moving, the potential energy is changing into kinetic energy. Conservation of energy gives<\/p>\n<div data-type=\"equation\" class=\"equation\">[latex]{\\text{PE}}_{\\text{i}}={\\text{KE}}_{\\text{f}}.[\/latex]<\/div>\n<p id=\"import-auto-id3018171\">More specifically,<\/p>\n<div data-type=\"equation\" class=\"equation\">[latex]{\\text{PE}}_{\\text{grav}}={\\text{KE}}_{\\text{trans}}+{\\text{KE}}_{\\text{rot}}[\/latex]<\/div>\n<p id=\"import-auto-id1431966\">or<\/p>\n<div data-type=\"equation\" class=\"equation\">[latex]\\text{mgh}=\\frac{1}{2}{\\text{mv}}^{2}+\\frac{1}{2}{\\mathrm{I\\omega }}^{2}.[\/latex]<\/div>\n<p id=\"import-auto-id1868893\">So, the initial [latex]\\text{mgh}[\/latex] is divided between translational kinetic energy and rotational kinetic energy; and the greater [latex]I[\/latex] is, the less energy goes into translation. If the can slides down without friction, then [latex]\\omega =0[\/latex] and all the energy goes into translation; thus, the can goes faster.<\/p>\n<div data-type=\"note\" class=\"note\" data-has-label=\"true\" id=\"fs-id2398409\" data-label=\"\">\n<div data-type=\"title\" class=\"title\">Take-Home Experiment<\/div>\n<p id=\"import-auto-id3246662\">Locate several cans each containing different types of food. First, predict which can will win the race down an inclined plane and explain why. See if your prediction is correct. You could also do this experiment by collecting several empty cylindrical containers of the same size and filling them with different materials such as wet or dry sand.<\/p>\n<\/div>\n<div data-type=\"example\" class=\"textbox examples\" id=\"fs-id3073422\">\n<div data-type=\"title\" class=\"title\">Calculating the Speed of a Cylinder Rolling Down an Incline<\/div>\n<p id=\"import-auto-id1818016\">Calculate the final speed of a solid cylinder that rolls down a 2.00-m-high incline. The cylinder starts from rest, has a mass of 0.750 kg, and has a radius of 4.00 cm.<\/p>\n<p id=\"import-auto-id1429770\"><strong>Strategy<\/strong><\/p>\n<p id=\"fs-id2445196\">We can solve for the final velocity using conservation of energy, but we must first express rotational quantities in terms of translational quantities to end up with [latex]v[\/latex] as the only unknown.<\/p>\n<p id=\"import-auto-id2052619\"><strong>Solution<\/strong><\/p>\n<p id=\"fs-id3158522\">Conservation of energy for this situation is written as described above:<\/p>\n<div data-type=\"equation\" class=\"equation\">[latex]\\text{mgh}=\\frac{1}{2}{\\mathit{mv}}^{2}+\\frac{1}{2}{\\mathrm{I\\omega }}^{2}.[\/latex]<\/div>\n<p id=\"import-auto-id3385654\">Before we can solve for [latex]v[\/latex] , we must get an expression for [latex]I[\/latex] from <a href=\"\/contents\/db59f656-e708-4094-a742-1e5560fe97c9@6#fs-id1838666\" class=\"autogenerated-content\">(Figure)<\/a>. Because [latex]v[\/latex] and [latex]\\omega [\/latex] are related (note here that the cylinder is rolling without slipping), we must also substitute the relationship [latex]\\omega =v\/R[\/latex] into the expression. These substitutions yield<\/p>\n<div data-type=\"equation\" class=\"equation\">[latex]\\text{mgh}=\\frac{1}{2}{\\mathit{mv}}^{2}+\\frac{1}{2}\\left(\\frac{1}{2}{\\mathit{mR}}^{2}\\right)\\left(\\frac{{v}^{2}}{{R}^{2}}\\right).[\/latex]<\/div>\n<p id=\"import-auto-id1951682\">Interestingly, the cylinder\u2019s radius [latex]R[\/latex] and mass [latex]m[\/latex] cancel, yielding<\/p>\n<div data-type=\"equation\" class=\"equation\">[latex]\\text{gh}=\\frac{1}{2}{v}^{2}+\\frac{1}{4}{v}^{2}=\\frac{3}{4}{v}^{2}.[\/latex]<\/div>\n<p id=\"import-auto-id2956864\">Solving algebraically, the equation for the final velocity [latex]v[\/latex] gives<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-320\">[latex]v={\\left(\\frac{4\\text{gh}}{3}\\right)}^{1\/2}.[\/latex]<\/div>\n<p id=\"import-auto-id2618589\">Substituting known values into the resulting expression yields<\/p>\n<div data-type=\"equation\" class=\"equation\">[latex]v={\\left[\\frac{4\\left(9.80\\phantom{\\rule{0.25em}{0ex}}{\\text{m\/s}}^{2}\\right)\\left(\\text{2.00 m}\\right)}{3}\\right]}^{1\/2}=\\text{5.11 m\/s}.[\/latex]<\/div>\n<p id=\"import-auto-id1362701\"><strong>Discussion<\/strong><\/p>\n<p id=\"fs-id3007358\">Because [latex]m[\/latex] and [latex]R[\/latex] cancel, the result [latex]v={\\left(\\frac{4}{3}\\text{gh}\\right)}^{1\/2}[\/latex] is valid for any solid cylinder, implying that all solid cylinders will roll down an incline at the same rate independent of their masses and sizes. (Rolling cylinders down inclines is what Galileo actually did to show that objects fall at the same rate independent of mass.) Note that if the cylinder slid without friction down the incline without rolling, then the entire gravitational potential energy would go into translational kinetic energy. Thus, [latex]\\frac{1}{2}{\\text{mv}}^{2}=\\text{mgh}[\/latex] and [latex]v=\\left(2\\text{gh}{\\right)}^{1\/2}[\/latex], which is 22% greater than [latex]\\left(4\\text{gh}\/3{\\right)}^{1\/2}[\/latex]. That is, the cylinder would go faster at the bottom.<\/p>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id2931518\" data-element-type=\"check-understanding\" data-label=\"\">\n<div data-type=\"title\">Check Your Understanding<\/div>\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id1613668\">\n<p id=\"import-auto-id3043952\"><span data-type=\"title\">Analogy of Rotational and Translational Kinetic Energy<\/span>Is rotational kinetic energy completely analogous to translational kinetic energy? What, if any, are their differences? Give an example of each type of kinetic energy.<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"fs-id1412799\">\n<p id=\"import-auto-id3081539\">Yes, rotational and translational kinetic energy are exact analogs. They both are the energy of motion involved with the coordinated (non-random) movement of mass relative to some reference frame. The only difference between rotational and translational kinetic energy is that translational is straight line motion while rotational is not. An example of both kinetic and translational kinetic energy is found in a bike tire while being ridden down a bike path. The rotational motion of the tire means it has rotational kinetic energy while the movement of the bike along the path means the tire also has translational kinetic energy. If you were to lift the front wheel of the bike and spin it while the bike is stationary, then the wheel would have only rotational kinetic energy relative to the Earth.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div data-type=\"note\" class=\"note\" data-has-label=\"true\" data-label=\"\">\n<div data-type=\"title\" class=\"title\">PhET Explorations: My Solar System<\/div>\n<p id=\"eip-id1169738171827\">Build your own system of heavenly bodies and watch the gravitational ballet. With this orbit simulator, you can set initial positions, velocities, and masses of 2, 3, or 4 bodies, and then see them orbit each other.<\/p>\n<div class=\"bc-figure figure\" id=\"eip-id2625118\">\n<div class=\"bc-figcaption figcaption\"><a href=\"\/resources\/e7460a4a0c31d5af5265b49a67e8778040ce3630\/my-solar-system_en.jar\">My Solar System<\/a><\/div>\n<p><span data-type=\"media\" id=\"Phet_module_11.4\" data-alt=\"\"><a href=\"\/resources\/e7460a4a0c31d5af5265b49a67e8778040ce3630\/my-solar-system_en.jar\" data-type=\"image\"><img src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/PhET_Icon.png\" data-media-type=\"image\/png\" alt=\"\" data-print=\"false\" width=\"450\"><\/a><span data-media-type=\"image\/png\" data-print=\"true\" data-src=\"\/resources\/075500ad9f71890a85fe3f7a4137ac08e2b7907c\/PhET_Icon.png\" data-type=\"image\"><\/span><\/span><\/p><\/div>\n<\/div>\n<div class=\"section-summary\" data-depth=\"1\" id=\"fs-id3176558\">\n<h1 data-type=\"title\">Section Summary<\/h1>\n<ul id=\"fs-id1917137\">\n<li id=\"import-auto-id1360816\">The rotational kinetic energy [latex]{\\text{KE}}_{\\text{rot}}[\/latex] for an object with a moment of inertia [latex]I[\/latex] and an angular velocity [latex]\\omega [\/latex] is given by\n<div data-type=\"equation\" class=\"equation\">[latex]{\\text{KE}}_{\\text{rot}}=\\frac{1}{2}{\\mathrm{I\\omega }}^{2}.[\/latex]<\/div>\n<\/li>\n<li id=\"import-auto-id1972580\">Helicopters store large amounts of rotational kinetic energy in their blades. This energy must be put into the blades before takeoff and maintained until the end of the flight. The engines do not have enough power to simultaneously provide lift and put significant rotational energy into the blades.<\/li>\n<li id=\"import-auto-id3224025\">Work and energy in rotational motion are completely analogous to work and energy in translational motion.<\/li>\n<li id=\"import-auto-id2453212\">The equation for the <span data-type=\"term\">work-energy theorem<\/span> for rotational motion is,\n<div data-type=\"equation\" class=\"equation\">[latex]\\text{net}\\phantom{\\rule{0.25em}{0ex}}W=\\frac{1}{2}{\\mathrm{I\\omega }}^{2}-\\frac{1}{2}I{{\\omega }_{\\text{0}}}^{2}.[\/latex]<\/div>\n<\/li>\n<\/ul>\n<\/div>\n<div class=\"conceptual-questions\" data-depth=\"1\" id=\"fs-id1578001\" data-element-type=\"conceptual-questions\">\n<h1 data-type=\"title\">Conceptual Questions<\/h1>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id3026004\" data-element-type=\"conceptual-questions\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id3025680\">\n<p id=\"import-auto-id3012901\">Describe the energy transformations involved when a yo-yo is thrown downward and then climbs back up its string to be caught in the user\u2019s hand.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id1428194\" data-element-type=\"conceptual-questions\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id1995300\">\n<p id=\"import-auto-id1413725\">What energy transformations are involved when a dragster engine is revved, its clutch let out rapidly, its tires spun, and it starts to accelerate forward? Describe the source and transformation of energy at each step.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id2640555\" data-element-type=\"conceptual-questions\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id3192098\">\n<p id=\"eip-id2695457\">The Earth has more rotational kinetic energy now than did the cloud of gas and dust from which it formed. Where did this energy come from?<\/p>\n<div class=\"bc-figure figure\" id=\"import-auto-id2615448\">\n<div class=\"bc-figcaption figcaption\">An immense cloud of rotating gas and dust contracted under the influence of gravity to form the Earth and in the process rotational kinetic energy increased. (credit: NASA)<\/div>\n<p><span data-type=\"media\" id=\"import-auto-id3045924\" data-alt=\"The figure shows a closed view of a red planet in the sky, with a sun like object seen at the far right and the planet shown here being surrounded by circles of gas and dust.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/Figure_11_04_07a.jpg\" data-media-type=\"image\/png\" alt=\"The figure shows a closed view of a red planet in the sky, with a sun like object seen at the far right and the planet shown here being surrounded by circles of gas and dust.\" width=\"250\"><\/span><\/p><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"problems-exercises\" data-depth=\"1\" id=\"fs-id1864335\" data-element-type=\"problems-exercises\">\n<h1 data-type=\"title\">Problems &amp; Exercises<\/h1>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id2402678\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id3255931\">\n<p id=\"import-auto-id2963400\">This problem considers energy and work aspects of <a href=\"\/contents\/db59f656-e708-4094-a742-1e5560fe97c9@6#fs-id1468671\" class=\"autogenerated-content\">(Figure)<\/a>\u2014use data from that example as needed. (a) Calculate the rotational kinetic energy in the merry-go-round plus child when they have an angular velocity of 20.0 rpm. (b) Using energy considerations, find the number of revolutions the father will have to push to achieve this angular velocity starting from rest. (c) Again, using energy considerations, calculate the force the father must exert to stop the merry-go-round in two revolutions<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"fs-id2991773\">\n<p id=\"import-auto-id3384991\">(a) 185 J<\/p>\n<p id=\"import-auto-id3055432\">(b) 0.0785 rev<\/p>\n<p id=\"import-auto-id3358608\">(c) [latex]W=9\\text{.}\\text{81 N}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id2601323\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id2640866\">\n<p id=\"import-auto-id1431802\">What is the final velocity of a hoop that rolls without slipping down a 5.00-m-high hill, starting from rest?<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id3017926\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id1461913\">\n<p id=\"import-auto-id3091855\">(a) Calculate the rotational kinetic energy of Earth on its axis. (b) What is the rotational kinetic energy of Earth in its orbit around the Sun?<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"fs-id2382524\">\n<p id=\"import-auto-id1326234\">(a) [latex]2.57\u00d7{\\text{10}}^{\\text{29}}\\phantom{\\rule{0.25em}{0ex}}\\text{J}[\/latex]<\/p>\n<p id=\"import-auto-id3036677\">(b) [latex]{\\text{KE}}_{\\text{rot}}=2\\text{.}\\text{65}\u00d7{\\text{10}}^{\\text{33}}\\phantom{\\rule{0.25em}{0ex}}\\text{J}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id1596687\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id1577764\">\n<p id=\"import-auto-id2041602\">Calculate the rotational kinetic energy in the motorcycle wheel (<a href=\"\/contents\/db59f656-e708-4094-a742-1e5560fe97c9@6#import-auto-id3370574\" class=\"autogenerated-content\">(Figure)<\/a>) if its angular velocity is 120 rad\/s. Assume M = 12.0 kg, R<sub>1<\/sub> = 0.280 m, and R<sub>2<\/sub> = 0.330 m.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id1580820\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id3007150\">\n<p id=\"import-auto-id2403290\">A baseball pitcher throws the ball in a motion where there is rotation of the forearm about the elbow joint as well as other movements. If the linear velocity of the ball relative to the elbow joint is 20.0 m\/s at a distance of 0.480 m from the joint and the moment of inertia of the forearm is [latex]\\text{0.500 kg}\\cdot {\\text{m}}^{2}[\/latex], what is the rotational kinetic energy of the forearm?<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"fs-id3028612\">\n<div data-type=\"equation\" class=\"equation\" id=\"eip-id1583692\">[latex]{\\text{KE}}_{\\text{rot}}=\\text{434 J}[\/latex]<\/div>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id2604037\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id1818585\">\n<p id=\"fs-id3053480\">While punting a football, a kicker rotates his leg about the hip joint. The moment of inertia of the leg is [latex]\\text{3.75 kg}\\cdot {\\text{m}}^{2}[\/latex] and its rotational kinetic energy is 175 J. (a) What is the angular velocity of the leg? (b) What is the velocity of tip of the punter\u2019s shoe if it is 1.05 m from the hip joint? (c) Explain how the football can be given a velocity greater than the tip of the shoe (necessary for a decent kick distance).<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id2662255\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id2679107\">\n<p id=\"fs-id2589915\">A bus contains a 1500 kg flywheel (a disk that has a 0.600 m radius) and has a total mass of 10,000 kg. (a) Calculate the angular velocity the flywheel must have to contain enough energy to take the bus from rest to a speed of 20.0 m\/s, assuming 90.0% of the rotational kinetic energy can be transformed into translational energy. (b) How high a hill can the bus climb with this stored energy and still have a speed of 3.00 m\/s at the top of the hill? Explicitly show how you follow the steps in the <a href=\"#fs-id1986333\">Problem-Solving Strategy for Rotational Energy<\/a>.<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"fs-id1917152\">\n<p id=\"fs-id3450208\">(a) [latex]\\text{128 rad\/s}[\/latex]<\/p>\n<p id=\"import-auto-id1514659\">(b) [latex]\\text{19.9 m}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id3250372\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id1922322\">\n<p id=\"import-auto-id1942736\">A ball with an initial velocity of 8.00 m\/s rolls up a hill without slipping. Treating the ball as a spherical shell, calculate the vertical height it reaches. (b) Repeat the calculation for the same ball if it slides up the hill without rolling.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id2583778\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id3254741\">\n<p id=\"import-auto-id3161408\">While exercising in a fitness center, a man lies face down on a bench and lifts a weight with one lower leg by contacting the muscles in the back of the upper leg. (a) Find the angular acceleration produced given the mass lifted is 10.0 kg at a distance of 28.0 cm from the knee joint, the moment of inertia of the lower leg is [latex]\\text{0.900 kg}\\cdot {\\text{m}}^{2}[\/latex], the muscle force is 1500 N, and its effective perpendicular lever arm is 3.00 cm. (b) How much work is done if the leg rotates through an angle of [latex]\\text{20.0\u00ba}[\/latex] with a constant force exerted by the muscle?<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"fs-id3115463\">\n<p id=\"import-auto-id1517396\">(a) [latex]\\text{10.}{\\text{4 rad\/s}}^{2}[\/latex]<\/p>\n<p id=\"import-auto-id3051710\">(b) [latex]\\text{net}\\phantom{\\rule{0.25em}{0ex}}W=6.\\text{11 J}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id3199856\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id3008300\">\n<p id=\"import-auto-id1930100\">To develop muscle tone, a woman lifts a 2.00-kg weight held in her hand. She uses her biceps muscle to flex the lower arm through an angle of [latex]\\text{60.0\u00ba}[\/latex]. (a) What is the angular acceleration if the weight is 24.0 cm from the elbow joint, her forearm has a moment of inertia of [latex]\\text{0.250 kg}\\cdot {\\text{m}}^{2}[\/latex], and the net force she exerts is 750 N at an effective perpendicular lever arm of 2.00 cm? (b) How much work does she do?<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id3245199\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id2963718\">\n<p id=\"import-auto-id3080929\">Consider two cylinders that start down identical inclines from rest except that one is frictionless. Thus one cylinder rolls without slipping, while the other slides frictionlessly without rolling. They both travel a short distance at the bottom and then start up another incline. (a) Show that they both reach the same height on the other incline, and that this height is equal to their original height. (b) Find the ratio of the time the rolling cylinder takes to reach the height on the second incline to the time the sliding cylinder takes to reach the height on the second incline. (c) Explain why the time for the rolling motion is greater than that for the sliding motion.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id2402928\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id3176898\">\n<p id=\"import-auto-id3398559\">What is the moment of inertia of an object that rolls without slipping down a 2.00-m-high incline starting from rest, and has a final velocity of 6.00 m\/s? Express the moment of inertia as a multiple of [latex]{\\mathit{MR}}^{2}[\/latex], where [latex]M[\/latex] is the mass of the object and [latex]R[\/latex] is its radius.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id2406116\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id2679003\">\n<p id=\"import-auto-id3232936\">Suppose a 200-kg motorcycle has two wheels like, <a href=\"\/contents\/db59f656-e708-4094-a742-1e5560fe97c9@6#import-auto-id3370574\">the one described in Problem 10.15<\/a> and is heading toward a hill at a speed of 30.0 m\/s. (a) How high can it coast up the hill, if you neglect friction? (b) How much energy is lost to friction if the motorcycle only gains an altitude of 35.0 m before coming to rest?<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id3399194\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id2583176\">\n<p id=\"import-auto-id2950423\">In softball, the pitcher throws with the arm fully extended (straight at the elbow). In a fast pitch the ball leaves the hand with a speed of 139 km\/h. (a) Find the rotational kinetic energy of the pitcher\u2019s arm given its moment of inertia is [latex]\\text{0.720 kg}\\cdot {\\text{m}}^{2}[\/latex] and the ball leaves the hand at a distance of 0.600 m from the pivot at the shoulder. (b) What force did the muscles exert to cause the arm to rotate if their effective perpendicular lever arm is 4.00 cm and the ball is 0.156 kg?<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"fs-id2616505\">\n<p id=\"import-auto-id3013127\">(a) 1.49 kJ<\/p>\n<p id=\"import-auto-id1561424\">(b) [latex]2.52\u00d7{\\text{10}}^{4}\\phantom{\\rule{0.25em}{0ex}}\\text{N}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id3073542\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id3158726\">\n<p id=\"import-auto-id3104836\"><strong>Construct Your Own Problem<\/strong><\/p>\n<p id=\"eip-id1169762600538\">Consider the work done by a spinning skater pulling her arms in to increase her rate of spin. Construct a problem in which you calculate the work done with a \u201cforce multiplied by distance\u201d calculation and compare it to the skater\u2019s increase in kinetic energy.\n    <\/p>\n<\/div>\n<\/div>\n<\/div>\n<div data-type=\"glossary\" class=\"textbox shaded\">\n<h2 data-type=\"glossary-title\">Glossary<\/h2>\n<dl class=\"definition\" id=\"import-auto-id3007462\">\n<dt>work-energy theorem<\/dt>\n<dd id=\"fs-id2070291\">if one or more external forces act upon a rigid object, causing its kinetic energy to change from [latex]{\\text{KE}}_{\\text{1}}[\/latex] to [latex]{\\text{KE}}_{\\text{2}}[\/latex], then the work [latex]W[\/latex] done by the net force is equal to the change in kinetic energy<\/dd>\n<\/dl>\n<dl class=\"definition\" id=\"import-auto-id1771451\">\n<dt>rotational kinetic energy<\/dt>\n<dd id=\"fs-id2383989\">the kinetic energy due to the rotation of an object. This is part of its total kinetic energy<\/dd>\n<\/dl>\n<\/div>\n\n","rendered":"<div class=\"textbox learning-objectives\">\n<h3 itemprop=\"educationalUse\">Learning Objectives<\/h3>\n<ul>\n<li>Derive the equation for rotational work.<\/li>\n<li>Calculate rotational kinetic energy.<\/li>\n<li>Demonstrate the Law of Conservation of Energy.<\/li>\n<\/ul>\n<\/div>\n<p id=\"import-auto-id3035380\">In this module, we will learn about work and energy associated with rotational motion. <a href=\"#import-auto-id3229349\" class=\"autogenerated-content\">(Figure)<\/a> shows a worker using an electric grindstone propelled by a motor. Sparks are flying, and noise and vibration are created as layers of steel are pared from the pole. The stone continues to turn even after the motor is turned off, but it is eventually brought to a stop by friction. Clearly, the motor had to work to get the stone spinning. This work went into heat, light, sound, vibration, and considerable <span data-type=\"term\" id=\"import-auto-id2669979\">rotational kinetic energy<\/span>.<\/p>\n<div class=\"bc-figure figure\" id=\"import-auto-id3229349\">\n<div class=\"bc-figcaption figcaption\">The motor works in spinning the grindstone, giving it rotational kinetic energy. That energy is then converted to heat, light, sound, and vibration. (credit: U.S. Navy photo by Mass Communication Specialist Seaman Zachary David Bell)<\/div>\n<p><span data-type=\"media\" id=\"import-auto-id2630228\" data-alt=\"The figure shows a mechanic cutting metal with a metal grinder. The sparks are emerging from the point of contact and jumping off tangentially from the cutter.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/Figure_11_04_01a.jpg\" data-media-type=\"image\/png\" alt=\"The figure shows a mechanic cutting metal with a metal grinder. The sparks are emerging from the point of contact and jumping off tangentially from the cutter.\" width=\"250\" \/><\/span><\/p>\n<\/div>\n<p id=\"import-auto-id2612157\">Work must be done to rotate objects such as grindstones or merry-go-rounds. Work was defined in <a href=\"\/contents\/3ef5dfb6-0a8d-433e-9c8f-b8c860a3903b@2\">Uniform Circular Motion and Gravitation<\/a> for translational motion, and we can build on that knowledge when considering work done in rotational motion. The simplest rotational situation is one in which the net force is exerted perpendicular to the radius of a disk (as shown in <a href=\"#import-auto-id2009471\" class=\"autogenerated-content\">(Figure)<\/a>) and remains perpendicular as the disk starts to rotate. The force is parallel to the displacement, and so the net work done is the product of the force times the arc length traveled:<\/p>\n<div data-type=\"equation\" class=\"equation\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-3be863efc49cb518df512cd340609e5b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#53;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#87;&#61;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#53;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#70;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#92;&#116;&#101;&#120;&#116;&#123;&Delta;&#125;&#115;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"144\" style=\"vertical-align: -4px;\" \/><\/div>\n<p id=\"import-auto-id1587900\">To get torque and other rotational quantities into the equation, we multiply and divide the right-hand side of the equation by <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-c409433a9e2dfcdb83360a974d243f18_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#114;\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"8\" style=\"vertical-align: 0px;\" \/>, and gather terms:<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-909\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-f5c5d2c12c4ccdcf1925551da9266e9c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#53;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#87;&#61;&#92;&#108;&#101;&#102;&#116;&#40;&#114;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#53;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#53;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#70;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#92;&#102;&#114;&#97;&#99;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&Delta;&#125;&#115;&#125;&#123;&#114;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"159\" style=\"vertical-align: -6px;\" \/><\/div>\n<p id=\"import-auto-id2448795\">We recognize that <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-f97f3a4ac6506a6fe5edd3725e849fdc_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#114;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#53;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#53;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#70;&#61;&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#32;&tau;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"103\" style=\"vertical-align: 0px;\" \/> and <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-8feade05fd39b187a34242da9224a598_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#68;&#101;&#108;&#116;&#97;&#32;&#115;&#47;&#114;&#61;&#92;&#116;&#104;&#101;&#116;&#97;&#32;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"73\" style=\"vertical-align: -5px;\" \/>, so that<\/p>\n<div data-type=\"equation\" class=\"equation\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-96d1ca9b0b87521a3e41256d8ce86d8a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#53;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#87;&#61;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#32;&tau;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#92;&#116;&#104;&#101;&#116;&#97;&#32;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"132\" style=\"vertical-align: -4px;\" \/><\/div>\n<p id=\"import-auto-id2009328\">This equation is the expression for rotational work. It is very similar to the familiar definition of translational work as force multiplied by distance. Here, torque is analogous to force, and angle is analogous to distance. The equation <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-ec354f61651b7c13194214ec0972ba90_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#53;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#87;&#61;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#32;&tau;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#92;&#116;&#104;&#101;&#116;&#97;&#32;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"128\" style=\"vertical-align: -4px;\" \/> is valid in general, even though it was derived for a special case.<\/p>\n<p id=\"import-auto-id3026626\">To get an expression for rotational kinetic energy, we must again perform some algebraic manipulations. The first step is to note that <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-b57f46319869a6fcb472c04dff56ab2e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#32;&tau;&#125;&#61;&#92;&#109;&#97;&#116;&#104;&#114;&#109;&#123;&#73;&#92;&#97;&#108;&#112;&#104;&#97;&#32;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"14\" width=\"71\" style=\"vertical-align: -1px;\" \/>, so that<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-404\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-54a4a362877a77be61398baca441c39e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#53;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#87;&#61;&#73;&#92;&#116;&#101;&#120;&#116;&#123;&alpha;&theta;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"85\" style=\"vertical-align: 0px;\" \/><\/div>\n<div class=\"bc-figure figure\" id=\"import-auto-id2009471\">\n<div class=\"bc-figcaption figcaption\">The net force on this disk is kept perpendicular to its radius as the force causes the disk to rotate. The net work done is thus <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-12b9673ff28a9ba6d3c4eba2a370f351_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#53;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#70;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#92;&#68;&#101;&#108;&#116;&#97;&#32;&#115;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"82\" style=\"vertical-align: -4px;\" \/>. The net work goes into rotational kinetic energy.<\/div>\n<p><span data-type=\"media\" id=\"import-auto-id2613010\" data-alt=\"The figure shows a circular disc of radius r. A net force F is applied perpendicular to the radius, rotating the disc in an anti-clockwise direction and producing a displacement equal to delta S, in a direction parallel to the direction of the force applied. The angle covered is theta.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/Figure_11_04_02a.jpg\" data-media-type=\"image\/jpg\" alt=\"The figure shows a circular disc of radius r. A net force F is applied perpendicular to the radius, rotating the disc in an anti-clockwise direction and producing a displacement equal to delta S, in a direction parallel to the direction of the force applied. The angle covered is theta.\" width=\"275\" \/><\/span><\/p>\n<\/div>\n<div data-type=\"note\" class=\"note\" data-has-label=\"true\" id=\"fs-id3191720\" data-label=\"\">\n<div data-type=\"title\" class=\"title\">Making Connections<\/div>\n<p id=\"import-auto-id3372728\">Work and energy in rotational motion are completely analogous to work and energy in translational motion, first presented in <a href=\"\/contents\/3ef5dfb6-0a8d-433e-9c8f-b8c860a3903b@2\">Uniform Circular Motion and Gravitation<\/a>.<\/p>\n<\/div>\n<p id=\"import-auto-id3418348\">Now, we solve one of the rotational kinematics equations for <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-87ebfb942ddbb3627be1bc97df129b9d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&alpha;&theta;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"1\" width=\"1\" style=\"vertical-align: 0px;\" \/>. We start with the equation<\/p>\n<div data-type=\"equation\" class=\"equation\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-aa3421548d55eb9d9e60308faacb7e4b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#123;&#92;&#111;&#109;&#101;&#103;&#97;&#32;&#125;&#95;&#123;&#125;&#125;&#94;&#123;&#50;&#125;&#61;&#123;&#123;&#92;&#111;&#109;&#101;&#103;&#97;&#32;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#48;&#125;&#125;&#125;&#94;&#123;&#50;&#125;&#43;&#50;&#92;&#116;&#101;&#120;&#116;&#123;&alpha;&theta;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"104\" style=\"vertical-align: -3px;\" \/><\/div>\n<p id=\"import-auto-id3154919\">Next, we solve for <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-87ebfb942ddbb3627be1bc97df129b9d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&alpha;&theta;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"1\" width=\"1\" style=\"vertical-align: 0px;\" \/>:<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-241\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-ab57b72074524c153f8e93fe989b2711_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&alpha;&theta;&#125;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#123;&#92;&#111;&#109;&#101;&#103;&#97;&#32;&#125;&#94;&#123;&#50;&#125;&#45;&#123;&#123;&#92;&#111;&#109;&#101;&#103;&#97;&#32;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#48;&#125;&#125;&#125;&#94;&#123;&#50;&#125;&#125;&#123;&#50;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"24\" width=\"75\" style=\"vertical-align: -6px;\" \/><\/div>\n<p id=\"import-auto-id2041695\">Substituting this into the equation for net <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-4caed22919a1780df1b6310b338b904e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#87;\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"19\" style=\"vertical-align: 0px;\" \/> and gathering terms yields<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-789\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-1c4f8ad681292f506a6caa3a37479861_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#53;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#87;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#49;&#125;&#123;&#50;&#125;&#123;&#92;&#109;&#97;&#116;&#104;&#114;&#109;&#123;&#73;&#92;&#111;&#109;&#101;&#103;&#97;&#32;&#125;&#125;&#94;&#123;&#50;&#125;&#45;&#92;&#102;&#114;&#97;&#99;&#123;&#49;&#125;&#123;&#50;&#125;&#73;&#123;&#123;&#92;&#111;&#109;&#101;&#103;&#97;&#32;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#48;&#125;&#125;&#125;&#94;&#123;&#50;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"22\" width=\"180\" style=\"vertical-align: -6px;\" \/><\/div>\n<p id=\"import-auto-id1373539\">This equation is the <span data-type=\"term\" id=\"import-auto-id2409820\">work-energy theorem<\/span> for rotational motion only. As you may recall, net work changes the kinetic energy of a system. Through an analogy with translational motion, we define the term <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-3949164c5a46e4c730ff002bd9bb6987_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#102;&#114;&#97;&#99;&#123;&#49;&#125;&#123;&#50;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#123;&#92;&#109;&#97;&#116;&#104;&#114;&#109;&#123;&#73;&#92;&#111;&#109;&#101;&#103;&#97;&#32;&#125;&#125;&#94;&#123;&#50;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"23\" width=\"53\" style=\"vertical-align: -7px;\" \/> to be <span data-type=\"term\" id=\"import-auto-id3012447\">rotational kinetic energy<\/span> <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-66178ae9cd7eefcecb69a17e362e498b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#75;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#114;&#111;&#116;&#125;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"43\" style=\"vertical-align: -4px;\" \/> for an object with a moment of inertia <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-18b5e45cb4a1ee02e81b9a980f828db8_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#73;\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"9\" style=\"vertical-align: 0px;\" \/> and an angular velocity <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-8ffb415af81ab9c23c1d2e7ec67d29c6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#111;&#109;&#101;&#103;&#97;&#32;\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"11\" style=\"vertical-align: 0px;\" \/>:<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-682\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-9621d67ae6ff2929c3ffcec377429438_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#75;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#114;&#111;&#116;&#125;&#125;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#49;&#125;&#123;&#50;&#125;&#123;&#92;&#109;&#97;&#116;&#104;&#114;&#109;&#123;&#73;&#92;&#111;&#109;&#101;&#103;&#97;&#32;&#125;&#125;&#94;&#123;&#50;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"22\" width=\"108\" style=\"vertical-align: -6px;\" \/><\/div>\n<p id=\"import-auto-id1434731\">The expression for rotational kinetic energy is exactly analogous to translational kinetic energy, with <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-18b5e45cb4a1ee02e81b9a980f828db8_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#73;\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"9\" style=\"vertical-align: 0px;\" \/> being analogous to <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-6b41df788161942c6f98604d37de8098_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#109;\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"15\" style=\"vertical-align: 0px;\" \/> and <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-8ffb415af81ab9c23c1d2e7ec67d29c6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#111;&#109;&#101;&#103;&#97;&#32;\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"11\" style=\"vertical-align: 0px;\" \/> to <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-ef71511c70f0e4b25cc6bd69f3bc20c2_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#118;\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"9\" style=\"vertical-align: 0px;\" \/>. Rotational kinetic energy has important effects. Flywheels, for example, can be used to store large amounts of rotational kinetic energy in a vehicle, as seen in <a href=\"#import-auto-id1614457\" class=\"autogenerated-content\">(Figure)<\/a>.<\/p>\n<div class=\"bc-figure figure\" id=\"import-auto-id1614457\">\n<div class=\"bc-figcaption figcaption\">Experimental vehicles, such as this bus, have been constructed in which rotational kinetic energy is stored in a large flywheel. When the bus goes down a hill, its transmission converts its gravitational potential energy into <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-66178ae9cd7eefcecb69a17e362e498b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#75;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#114;&#111;&#116;&#125;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"43\" style=\"vertical-align: -4px;\" \/>. It can also convert translational kinetic energy, when the bus stops, into <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-66178ae9cd7eefcecb69a17e362e498b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#75;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#114;&#111;&#116;&#125;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"43\" style=\"vertical-align: -4px;\" \/>. The flywheel\u2019s energy can then be used to accelerate, to go up another hill, or to keep the bus from going against friction.<\/div>\n<p><span data-type=\"media\" id=\"import-auto-id1817378\" data-alt=\"The figure shows a bus carrying a large flywheel on its board in which rotational kinetic energy is stored.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/Figure_11_04_03a.jpg\" data-media-type=\"image\/jpg\" alt=\"The figure shows a bus carrying a large flywheel on its board in which rotational kinetic energy is stored.\" width=\"250\" \/><\/span><\/p>\n<\/div>\n<div data-type=\"example\" class=\"textbox examples\" id=\"fs-id3354618\">\n<div data-type=\"title\" class=\"title\">Calculating the Work and Energy for Spinning a Grindstone<\/div>\n<p id=\"import-auto-id2600755\">Consider a person who spins a large grindstone by placing her hand on its edge and exerting a force through part of a revolution as shown in <a href=\"#import-auto-id2674234\" class=\"autogenerated-content\">(Figure)<\/a>. In this example, we verify that the work done by the torque she exerts equals the change in rotational energy. (a) How much work is done if she exerts a force of 200 N through a rotation of <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-b13f6eead74654932e1c3f4cfd8f11c6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#49;&#46;&#48;&#48;&#32;&#114;&#97;&#100;&#125;&#92;&#108;&#101;&#102;&#116;&#40;&#53;&#55;&#46;&#51;&ordm;&#92;&#114;&#105;&#103;&#104;&#116;&#41;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"110\" style=\"vertical-align: -4px;\" \/>? The force is kept perpendicular to the grindstone\u2019s 0.320-m radius at the point of application, and the effects of friction are negligible. (b) What is the final angular velocity if the grindstone has a mass of 85.0 kg? (c) What is the final rotational kinetic energy? (It should equal the work.)<\/p>\n<p id=\"import-auto-id2421131\"><strong>Strategy<\/strong><\/p>\n<p id=\"fs-id2422273\">To find the work, we can use the equation <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-ec354f61651b7c13194214ec0972ba90_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#53;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#87;&#61;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#32;&tau;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#92;&#116;&#104;&#101;&#116;&#97;&#32;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"128\" style=\"vertical-align: -4px;\" \/>. We have enough information to calculate the torque and are given the rotation angle. In the second part, we can find the final angular velocity using one of the kinematic relationships. In the last part, we can calculate the rotational kinetic energy from its expression in <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-630e98629052469857a3c8cd9662113e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#75;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#114;&#111;&#116;&#125;&#125;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#49;&#125;&#123;&#50;&#125;&#123;&#92;&#109;&#97;&#116;&#104;&#114;&#109;&#123;&#73;&#92;&#111;&#109;&#101;&#103;&#97;&#32;&#125;&#125;&#94;&#123;&#50;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"22\" width=\"104\" style=\"vertical-align: -6px;\" \/>.<\/p>\n<p id=\"import-auto-id3181145\"><strong>Solution for (a)<\/strong><\/p>\n<p id=\"fs-id1485088\">The net work is expressed in the equation<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-117\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-86f95260fe56304d2112107fda470963_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#53;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#87;&#61;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#32;&tau;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#92;&#116;&#104;&#101;&#116;&#97;&#32;&#44;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"132\" style=\"vertical-align: -4px;\" \/><\/div>\n<p id=\"import-auto-id1828262\">where net <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-ac03dc06d394ae61cf07e44856527651_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#97;&#117;&#32;\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"10\" style=\"vertical-align: 0px;\" \/> is the applied force multiplied by the radius <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-058ba4549a4b7165bb7694429c44682b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#116;&#101;&#120;&#116;&#123;&#114;&#70;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"30\" style=\"vertical-align: -4px;\" \/> because there is no retarding friction, and the force is perpendicular to <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-c409433a9e2dfcdb83360a974d243f18_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#114;\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"8\" style=\"vertical-align: 0px;\" \/>. The angle <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-761998727948942ceb1b5763e45f01e4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#104;&#101;&#116;&#97;&#32;\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"9\" style=\"vertical-align: 0px;\" \/> is given. Substituting the given values in the equation above yields<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-348\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-099d18e967e2136dbcbe192451355423_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#98;&#101;&#103;&#105;&#110;&#123;&#97;&#114;&#114;&#97;&#121;&#125;&#123;&#108;&#108;&#108;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#53;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#87;&#38;&#32;&#61;&#38;&#32;&#92;&#116;&#101;&#120;&#116;&#123;&#114;&#70;&#125;&#92;&#116;&#104;&#101;&#116;&#97;&#32;&#61;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#116;&#101;&#120;&#116;&#123;&#48;&#46;&#51;&#50;&#48;&#32;&#109;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#116;&#101;&#120;&#116;&#123;&#50;&#48;&#48;&#32;&#78;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#116;&#101;&#120;&#116;&#123;&#49;&#46;&#48;&#48;&#32;&#114;&#97;&#100;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#92;&#92;&#32;&#38;&#32;&#61;&#38;&#32;&#92;&#116;&#101;&#120;&#116;&#123;&#54;&#52;&#46;&#48;&#32;&#78;&#125;&#92;&#99;&#100;&#111;&#116;&#32;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#46;&#125;&#92;&#101;&#110;&#100;&#123;&#97;&#114;&#114;&#97;&#121;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"37\" width=\"362\" style=\"vertical-align: -12px;\" \/><\/div>\n<p id=\"import-auto-id2406953\">Noting that <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-0f62c22a4980ca186e7d73354faa959c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#49;&#32;&#78;&middot;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#125;&#61;&#49;&#32;&#74;\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"83\" style=\"vertical-align: -1px;\" \/>,<\/p>\n<div data-type=\"equation\" class=\"equation\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-8e7a3fedd1d69ec4899b1d0347b5306b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#53;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#87;&#61;&#92;&#116;&#101;&#120;&#116;&#123;&#54;&#52;&#46;&#48;&#32;&#74;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"14\" width=\"122\" style=\"vertical-align: -1px;\" \/><\/div>\n<div class=\"bc-figure figure\" id=\"import-auto-id2674234\">\n<div class=\"bc-figcaption figcaption\">A large grindstone is given a spin by a person grasping its outer edge.<\/div>\n<p><span data-type=\"media\" id=\"import-auto-id3402470\" data-alt=\"The figure shows a large grindstone of radius r which is being given a spin by applying a force F in a counterclockwise direction, as indicated by the arrows.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/Figure_11_04_04a.jpg\" data-media-type=\"image\/jpg\" alt=\"The figure shows a large grindstone of radius r which is being given a spin by applying a force F in a counterclockwise direction, as indicated by the arrows.\" height=\"225\" \/><\/span><\/p>\n<\/div>\n<p id=\"import-auto-id1468054\"><strong>Solution for (b)<\/strong><\/p>\n<p id=\"fs-id1276292\">To find <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-8ffb415af81ab9c23c1d2e7ec67d29c6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#111;&#109;&#101;&#103;&#97;&#32;\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"11\" style=\"vertical-align: 0px;\" \/> from the given information requires more than one step. We start with the kinematic relationship in the equation<\/p>\n<div data-type=\"equation\" class=\"equation\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-fc47bba8c28fd3be164f9336a2405528_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#111;&#109;&#101;&#103;&#97;&#32;&#125;&#94;&#123;&#50;&#125;&#61;&#123;&#123;&#92;&#111;&#109;&#101;&#103;&#97;&#32;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#48;&#125;&#125;&#125;&#94;&#123;&#50;&#125;&#43;&#50;&#92;&#116;&#101;&#120;&#116;&#123;&alpha;&theta;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"104\" style=\"vertical-align: -3px;\" \/><\/div>\n<p id=\"import-auto-id1946811\">Note that <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-4e725f49672e9c389c2e661120540991_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#111;&#109;&#101;&#103;&#97;&#32;&#125;&#95;&#123;&#48;&#125;&#61;&#48;\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"51\" style=\"vertical-align: -3px;\" \/> because we start from rest. Taking the square root of the resulting equation gives<\/p>\n<div data-type=\"equation\" class=\"equation\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-f53ce9eb21c328e007edafbcd68ac4e2_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#111;&#109;&#101;&#103;&#97;&#32;&#61;&#123;&#92;&#108;&#101;&#102;&#116;&#40;&#50;&#92;&#116;&#101;&#120;&#116;&#123;&alpha;&theta;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#125;&#94;&#123;&#49;&#47;&#50;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"23\" width=\"83\" style=\"vertical-align: -4px;\" \/><\/div>\n<p id=\"import-auto-id2435786\">Now we need to find <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-946f8144d4e3d460c8621773145884d3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#97;&#108;&#112;&#104;&#97;&#32;\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"11\" style=\"vertical-align: 0px;\" \/>. One possibility is<\/p>\n<div data-type=\"equation\" class=\"equation\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-27cebbad5507cf57133289f6be5f6003_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#97;&#108;&#112;&#104;&#97;&#32;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#32;&tau;&#125;&#125;&#123;&#73;&#125;&#44;\" title=\"Rendered by QuickLaTeX.com\" height=\"22\" width=\"67\" style=\"vertical-align: -6px;\" \/><\/div>\n<p id=\"import-auto-id1151420\">where the torque is<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-643\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-4f98aa07901fc24f3530d5cb7a174964_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#32;&tau;&#125;&#61;&#92;&#116;&#101;&#120;&#116;&#123;&#114;&#70;&#125;&#61;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#116;&#101;&#120;&#116;&#123;&#48;&#46;&#51;&#50;&#48;&#32;&#109;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#116;&#101;&#120;&#116;&#123;&#50;&#48;&#48;&#32;&#78;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#61;&#92;&#116;&#101;&#120;&#116;&#123;&#54;&#52;&#46;&#48;&#32;&#78;&#125;&#92;&#99;&#100;&#111;&#116;&#32;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"340\" style=\"vertical-align: -4px;\" \/><\/div>\n<p id=\"import-auto-id2588420\">The formula for the moment of inertia for a disk is found in <a href=\"\/contents\/db59f656-e708-4094-a742-1e5560fe97c9@6#fs-id1838666\" class=\"autogenerated-content\">(Figure)<\/a>:<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-119\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-d60e5a17f7eab251eed8ef73553ab4ff_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#73;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#49;&#125;&#123;&#50;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#77;&#82;&#125;&#125;&#94;&#123;&#50;&#125;&#61;&#48;&#46;&#53;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#116;&#101;&#120;&#116;&#123;&#56;&#53;&#46;&#48;&#32;&#107;&#103;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#123;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#116;&#101;&#120;&#116;&#123;&#48;&#46;&#51;&#50;&#48;&#32;&#109;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#125;&#94;&#123;&#50;&#125;&#61;&#92;&#116;&#101;&#120;&#116;&#123;&#52;&#46;&#51;&#53;&#50;&#32;&#107;&#103;&#125;&#92;&#99;&#100;&#111;&#116;&#32;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#125;&#125;&#94;&#123;&#50;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"23\" width=\"412\" style=\"vertical-align: -6px;\" \/><\/div>\n<p id=\"import-auto-id2670890\">Substituting the values of torque and moment of inertia into the expression for <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-946f8144d4e3d460c8621773145884d3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#97;&#108;&#112;&#104;&#97;&#32;\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"11\" style=\"vertical-align: 0px;\" \/>, we obtain<\/p>\n<div data-type=\"equation\" class=\"equation\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-c2ebd0a5d8184c9a7461e216e21810a9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#97;&#108;&#112;&#104;&#97;&#32;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#54;&#52;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#46;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#48;&#32;&#78;&#125;&#92;&#99;&#100;&#111;&#116;&#32;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#125;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#52;&#46;&#51;&#53;&#50;&#32;&#107;&#103;&#125;&#92;&#99;&#100;&#111;&#116;&#32;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#125;&#125;&#94;&#123;&#50;&#125;&#125;&#61;&#92;&#116;&#101;&#120;&#116;&#123;&#49;&#52;&#46;&#55;&#125;&#92;&#102;&#114;&#97;&#99;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#114;&#97;&#100;&#125;&#125;&#123;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#115;&#125;&#125;&#94;&#123;&#50;&#125;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"27\" width=\"194\" style=\"vertical-align: -10px;\" \/><\/div>\n<p id=\"import-auto-id3400808\">Now, substitute this value and the given value for <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-761998727948942ceb1b5763e45f01e4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#104;&#101;&#116;&#97;&#32;\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"9\" style=\"vertical-align: 0px;\" \/> into the above expression for <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-8ffb415af81ab9c23c1d2e7ec67d29c6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#111;&#109;&#101;&#103;&#97;&#32;\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"11\" style=\"vertical-align: 0px;\" \/>:<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-116\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-abf2abe8bece511d974762195e9bcb7d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#111;&#109;&#101;&#103;&#97;&#32;&#61;&#123;&#92;&#108;&#101;&#102;&#116;&#40;&#50;&#92;&#116;&#101;&#120;&#116;&#123;&alpha;&theta;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#125;&#94;&#123;&#49;&#47;&#50;&#125;&#61;&#123;&#92;&#108;&#101;&#102;&#116;&#91;&#50;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#116;&#101;&#120;&#116;&#123;&#49;&#52;&#46;&#55;&#125;&#92;&#102;&#114;&#97;&#99;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#114;&#97;&#100;&#125;&#125;&#123;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#115;&#125;&#125;&#94;&#123;&#50;&#125;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#116;&#101;&#120;&#116;&#123;&#49;&#46;&#48;&#48;&#32;&#114;&#97;&#100;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#92;&#114;&#105;&#103;&#104;&#116;&#93;&#125;&#94;&#123;&#49;&#47;&#50;&#125;&#61;&#92;&#116;&#101;&#120;&#116;&#123;&#53;&#46;&#52;&#50;&#125;&#92;&#102;&#114;&#97;&#99;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#114;&#97;&#100;&#125;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#115;&#125;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"29\" width=\"385\" style=\"vertical-align: -7px;\" \/><\/div>\n<p id=\"import-auto-id2408772\"><strong>Solution for (c)<\/strong><\/p>\n<p>The final rotational kinetic energy is<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-291\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-9621d67ae6ff2929c3ffcec377429438_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#75;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#114;&#111;&#116;&#125;&#125;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#49;&#125;&#123;&#50;&#125;&#123;&#92;&#109;&#97;&#116;&#104;&#114;&#109;&#123;&#73;&#92;&#111;&#109;&#101;&#103;&#97;&#32;&#125;&#125;&#94;&#123;&#50;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"22\" width=\"108\" style=\"vertical-align: -6px;\" \/><\/div>\n<p id=\"import-auto-id3077647\">Both <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-18b5e45cb4a1ee02e81b9a980f828db8_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#73;\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"9\" style=\"vertical-align: 0px;\" \/> and <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-8ffb415af81ab9c23c1d2e7ec67d29c6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#111;&#109;&#101;&#103;&#97;&#32;\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"11\" style=\"vertical-align: 0px;\" \/> were found above. Thus,<\/p>\n<div data-type=\"equation\" class=\"equation\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-3e3d1c6aba580296b7c09d07b4bbba67_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#75;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#114;&#111;&#116;&#125;&#125;&#61;&#92;&#108;&#101;&#102;&#116;&#40;&#48;&#46;&#53;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#116;&#101;&#120;&#116;&#123;&#52;&#46;&#51;&#53;&#50;&#32;&#107;&#103;&#125;&#92;&#99;&#100;&#111;&#116;&#32;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#125;&#125;&#94;&#123;&#50;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#123;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#116;&#101;&#120;&#116;&#123;&#53;&#46;&#52;&#50;&#32;&#114;&#97;&#100;&#47;&#115;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#125;&#94;&#123;&#50;&#125;&#61;&#92;&#116;&#101;&#120;&#116;&#123;&#54;&#52;&#46;&#48;&#32;&#74;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"24\" width=\"401\" style=\"vertical-align: -7px;\" \/><\/div>\n<p id=\"import-auto-id1427676\"><strong>Discussion<\/strong><\/p>\n<p id=\"eip-id1935871\">The final rotational kinetic energy equals the work done by the torque, which confirms that the work done went into rotational kinetic energy. We could, in fact, have used an expression for energy instead of a kinematic relation to solve part (b). We will do this in later examples.<\/p>\n<\/div>\n<p id=\"import-auto-id3162150\">Helicopter pilots are quite familiar with rotational kinetic energy. They know, for example, that a point of no return will be reached if they allow their blades to slow below a critical angular velocity during flight. The blades lose lift, and it is impossible to immediately get the blades spinning fast enough to regain it. Rotational kinetic energy must be supplied to the blades to get them to rotate faster, and enough energy cannot be supplied in time to avoid a crash. Because of weight limitations, helicopter engines are too small to supply both the energy needed for lift and to replenish the rotational kinetic energy of the blades once they have slowed down. The rotational kinetic energy is put into them before takeoff and must not be allowed to drop below this crucial level. One possible way to avoid a crash is to use the gravitational potential energy of the helicopter to replenish the rotational kinetic energy of the blades by losing altitude and aligning the blades so that the helicopter is spun up in the descent. Of course, if the helicopter\u2019s altitude is too low, then there is insufficient time for the blade to regain lift before reaching the ground.<\/p>\n<div data-type=\"note\" class=\"note\" data-has-label=\"true\" id=\"fs-id1986333\" data-label=\"\">\n<div data-type=\"title\" class=\"title\">Problem-Solving Strategy for Rotational Energy<\/div>\n<ol id=\"fs-id2438401\" data-number-style=\"arabic\">\n<li id=\"import-auto-id3054599\"><em data-effect=\"italics\">Determine that energy or work is involved in the rotation<\/em>.<\/li>\n<li id=\"import-auto-id1389478\"><em data-effect=\"italics\">Determine the system of interest<\/em>. A sketch usually helps.<\/li>\n<li id=\"import-auto-id2930250\"><em data-effect=\"italics\">Analyze the situation to determine the types of work and energy involved<\/em>.<\/li>\n<li id=\"import-auto-id2937257\"><em data-effect=\"italics\">For closed systems, mechanical energy is conserved<\/em>. That is, <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-d17ceaff4bc85a0e7ead12046d69e5e5_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#75;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#105;&#125;&#125;&#43;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#80;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#105;&#125;&#125;&#61;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#75;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#102;&#125;&#125;&#43;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#80;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#102;&#125;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"190\" style=\"vertical-align: -4px;\" \/> Note that <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-3414ecd35b8bb4ec5bfe9cba398b6944_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#75;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#105;&#125;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"30\" style=\"vertical-align: -4px;\" \/> and <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-7e5c86d808d8e41532b1f8d1bdff09fa_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#75;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#102;&#125;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"31\" style=\"vertical-align: -4px;\" \/> may each include translational and rotational contributions.<\/li>\n<li id=\"import-auto-id1439188\"><em data-effect=\"italics\">For open systems<\/em>, mechanical energy may not be conserved, and other forms of energy (referred to previously as <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-52eedfe51fc2f3e3d977c38d91c931b7_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#79;&#69;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"26\" style=\"vertical-align: 0px;\" \/>), such as heat transfer, may enter or leave the system. Determine what they are, and calculate them as necessary.<\/li>\n<li id=\"import-auto-id1429438\"><em data-effect=\"italics\">Eliminate terms wherever possible to simplify the algebra<\/em>.<\/li>\n<li id=\"import-auto-id3386776\"><em data-effect=\"italics\">Check the answer to see if it is reasonable<\/em>.<\/li>\n<\/ol>\n<\/div>\n<div data-type=\"example\" class=\"textbox examples\" id=\"fs-id3173123\">\n<div data-type=\"title\" class=\"title\">Calculating Helicopter Energies<\/div>\n<p>A typical small rescue helicopter, similar to the one in <a href=\"#import-auto-id2420248\" class=\"autogenerated-content\">(Figure)<\/a>, has four blades, each is 4.00 m long and has a mass of 50.0 kg. The blades can be approximated as thin rods that rotate about one end of an axis perpendicular to their length. The helicopter has a total loaded mass of 1000 kg. (a) Calculate the rotational kinetic energy in the blades when they rotate at 300 rpm. (b) Calculate the translational kinetic energy of the helicopter when it flies at 20.0 m\/s, and compare it with the rotational energy in the blades. (c) To what height could the helicopter be raised if all of the rotational kinetic energy could be used to lift it?<\/p>\n<p id=\"import-auto-id1389334\"><strong>Strategy<\/strong><\/p>\n<p id=\"fs-id1519185\">Rotational and translational kinetic energies can be calculated from their definitions. The last part of the problem relates to the idea that energy can change form, in this case from rotational kinetic energy to gravitational potential energy.<\/p>\n<p id=\"import-auto-id3430265\"><strong>Solution for (a)<\/strong><\/p>\n<p id=\"fs-id1569357\">The rotational kinetic energy is <\/p>\n<div data-type=\"equation\" class=\"equation\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-9621d67ae6ff2929c3ffcec377429438_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#75;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#114;&#111;&#116;&#125;&#125;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#49;&#125;&#123;&#50;&#125;&#123;&#92;&#109;&#97;&#116;&#104;&#114;&#109;&#123;&#73;&#92;&#111;&#109;&#101;&#103;&#97;&#32;&#125;&#125;&#94;&#123;&#50;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"22\" width=\"108\" style=\"vertical-align: -6px;\" \/><\/div>\n<p id=\"import-auto-id1426438\">We must convert the angular velocity to radians per second and calculate the moment of inertia before we can find <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-66178ae9cd7eefcecb69a17e362e498b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#75;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#114;&#111;&#116;&#125;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"43\" style=\"vertical-align: -4px;\" \/>. The angular velocity <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-8ffb415af81ab9c23c1d2e7ec67d29c6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#111;&#109;&#101;&#103;&#97;&#32;\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"11\" style=\"vertical-align: 0px;\" \/> is<\/p>\n<div data-type=\"equation\" class=\"equation\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-9c27a4482666e65ec244b5ad3a5634d8_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#111;&#109;&#101;&#103;&#97;&#32;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#51;&#48;&#48;&#32;&#114;&#101;&#118;&#125;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#49;&#46;&#48;&#48;&#32;&#109;&#105;&#110;&#125;&#125;&#92;&#99;&#100;&#111;&#116;&#32;&#92;&#102;&#114;&#97;&#99;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#50;&pi;&#32;&#114;&#97;&#100;&#125;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#49;&#32;&#114;&#101;&#118;&#125;&#125;&#92;&#99;&#100;&#111;&#116;&#32;&#92;&#102;&#114;&#97;&#99;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#49;&#46;&#48;&#48;&#32;&#109;&#105;&#110;&#125;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#54;&#48;&#46;&#48;&#32;&#115;&#125;&#125;&#61;&#92;&#116;&#101;&#120;&#116;&#123;&#51;&#49;&#46;&#52;&#125;&#92;&#102;&#114;&#97;&#99;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#114;&#97;&#100;&#125;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#115;&#125;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"24\" width=\"291\" style=\"vertical-align: -7px;\" \/><\/div>\n<p id=\"import-auto-id2598986\">The moment of inertia of one blade will be that of a thin rod rotated about its end, found in <a href=\"\/contents\/db59f656-e708-4094-a742-1e5560fe97c9@6#fs-id1838666\" class=\"autogenerated-content\">(Figure)<\/a>. The total <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-18b5e45cb4a1ee02e81b9a980f828db8_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#73;\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"9\" style=\"vertical-align: 0px;\" \/> is four times this moment of inertia, because there are four blades. Thus,<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-897\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-fceeed7b278cf969f716fa517d71adac_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#73;&#61;&#52;&#92;&#102;&#114;&#97;&#99;&#123;&#123;&#92;&#109;&#97;&#116;&#104;&#114;&#109;&#123;&#77;&#92;&#101;&#108;&#108;&#32;&#125;&#125;&#94;&#123;&#50;&#125;&#125;&#123;&#51;&#125;&#61;&#52;&times;&#92;&#102;&#114;&#97;&#99;&#123;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#116;&#101;&#120;&#116;&#123;&#53;&#48;&#46;&#48;&#32;&#107;&#103;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#123;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#116;&#101;&#120;&#116;&#123;&#52;&#46;&#48;&#48;&#32;&#109;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#125;&#94;&#123;&#50;&#125;&#125;&#123;&#51;&#125;&#61;&#92;&#116;&#101;&#120;&#116;&#123;&#49;&#48;&#54;&#55;&#32;&#107;&#103;&#125;&#92;&#99;&#100;&#111;&#116;&#32;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#125;&#125;&#94;&#123;&#50;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"26\" width=\"341\" style=\"vertical-align: -6px;\" \/><\/div>\n<p id=\"import-auto-id2673012\">Entering <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-8ffb415af81ab9c23c1d2e7ec67d29c6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#111;&#109;&#101;&#103;&#97;&#32;\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"11\" style=\"vertical-align: 0px;\" \/> and <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-18b5e45cb4a1ee02e81b9a980f828db8_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#73;\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"9\" style=\"vertical-align: 0px;\" \/> into the expression for rotational kinetic energy gives<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-913\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-53a7bdb6c8addf242c817314a3763884_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#98;&#101;&#103;&#105;&#110;&#123;&#97;&#114;&#114;&#97;&#121;&#125;&#123;&#108;&#108;&#108;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#75;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#114;&#111;&#116;&#125;&#125;&#38;&#32;&#61;&#38;&#32;&#48;&#46;&#53;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#116;&#101;&#120;&#116;&#123;&#49;&#48;&#54;&#55;&#32;&#107;&#103;&#125;&#92;&#99;&#100;&#111;&#116;&#32;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#125;&#125;&#94;&#123;&#50;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#123;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#116;&#101;&#120;&#116;&#123;&#51;&#49;&#46;&#52;&#32;&#114;&#97;&#100;&#47;&#115;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#125;&#94;&#123;&#50;&#125;&#92;&#92;&#32;&#38;&#32;&#61;&#38;&#32;&#53;&#46;&#50;&#54;&times;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#49;&#48;&#125;&#125;&#94;&#123;&#53;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#53;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#74;&#125;&#92;&#101;&#110;&#100;&#123;&#97;&#114;&#114;&#97;&#121;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"41\" width=\"330\" style=\"vertical-align: -13px;\" \/><\/div>\n<p id=\"import-auto-id2616110\"><strong>Solution for (b)<\/strong><\/p>\n<p>Translational kinetic energy was defined in <a href=\"\/contents\/3ef5dfb6-0a8d-433e-9c8f-b8c860a3903b@2\">Uniform Circular Motion and Gravitation<\/a>. Entering the given values of mass and velocity, we obtain<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-2\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-342543fb0443a08fd4349b4f4506af6e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#75;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#116;&#114;&#97;&#110;&#115;&#125;&#125;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#49;&#125;&#123;&#50;&#125;&#123;&#92;&#109;&#97;&#116;&#104;&#105;&#116;&#123;&#109;&#118;&#125;&#125;&#94;&#123;&#50;&#125;&#61;&#92;&#108;&#101;&#102;&#116;&#40;&#48;&#46;&#53;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#116;&#101;&#120;&#116;&#123;&#49;&#48;&#48;&#48;&#32;&#107;&#103;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#123;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#116;&#101;&#120;&#116;&#123;&#50;&#48;&#46;&#48;&#32;&#109;&#47;&#115;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#125;&#94;&#123;&#50;&#125;&#61;&#50;&#92;&#116;&#101;&#120;&#116;&#123;&#46;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#48;&#48;&#125;&times;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#49;&#48;&#125;&#125;&#94;&#123;&#53;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#53;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#74;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"23\" width=\"451\" style=\"vertical-align: -6px;\" \/><\/div>\n<p id=\"import-auto-id1587080\">To compare kinetic energies, we take the ratio of translational kinetic energy to rotational kinetic energy. This ratio is<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-973\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-8a6fd3d2207ac589cff6493a074b3882_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#102;&#114;&#97;&#99;&#123;&#50;&#92;&#116;&#101;&#120;&#116;&#123;&#46;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#48;&#48;&#125;&times;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#49;&#48;&#125;&#125;&#94;&#123;&#53;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#53;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#74;&#125;&#125;&#123;&#53;&#92;&#116;&#101;&#120;&#116;&#123;&#46;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#50;&#54;&#125;&times;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#49;&#48;&#125;&#125;&#94;&#123;&#53;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#53;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#74;&#125;&#125;&#61;&#48;&#46;&#51;&#56;&#48;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"27\" width=\"127\" style=\"vertical-align: -8px;\" \/><\/div>\n<p id=\"import-auto-id1848503\"><strong>Solution for (c)<\/strong><\/p>\n<p id=\"fs-id2407243\">At the maximum height, all rotational kinetic energy will have been converted to gravitational energy. To find this height, we equate those two energies:<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-487\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-3b4f960b659dded2b18d88b8f592ab53_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#75;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#114;&#111;&#116;&#125;&#125;&#61;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#80;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#103;&#114;&#97;&#118;&#125;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"118\" style=\"vertical-align: -6px;\" \/><\/div>\n<p id=\"import-auto-id3148882\">or<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-154\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-f2ae37b88bc2e07b516e1f7292beb202_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#102;&#114;&#97;&#99;&#123;&#49;&#125;&#123;&#50;&#125;&#123;&#92;&#109;&#97;&#116;&#104;&#114;&#109;&#123;&#73;&#92;&#111;&#109;&#101;&#103;&#97;&#32;&#125;&#125;&#94;&#123;&#50;&#125;&#61;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#103;&#104;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"22\" width=\"96\" style=\"vertical-align: -6px;\" \/><\/div>\n<p id=\"import-auto-id1920250\">We now solve for <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-14b463d0ecd5b350ced6cf1d6a12eef3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#104;\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"10\" style=\"vertical-align: 0px;\" \/> and substitute known values into the resulting equation<\/p>\n<div data-type=\"equation\" class=\"equation\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-d5829d635cc47b39c927c19c2e42f227_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#104;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#123;&#92;&#102;&#114;&#97;&#99;&#123;&#49;&#125;&#123;&#50;&#125;&#92;&#109;&#97;&#116;&#104;&#114;&#109;&#123;&#73;&#92;&#111;&#109;&#101;&#103;&#97;&#32;&#125;&#125;&#94;&#123;&#50;&#125;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#103;&#125;&#125;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#53;&#46;&#50;&#54;&times;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#49;&#48;&#125;&#125;&#94;&#123;&#53;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#53;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#74;&#125;&#125;&#123;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#116;&#101;&#120;&#116;&#123;&#49;&#48;&#48;&#48;&#32;&#107;&#103;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#92;&#108;&#101;&#102;&#116;&#40;&#57;&#46;&#56;&#48;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#53;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#47;&#115;&#125;&#125;&#94;&#123;&#50;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#125;&#61;&#92;&#116;&#101;&#120;&#116;&#123;&#53;&#51;&#46;&#55;&#32;&#109;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"306\" style=\"vertical-align: -14px;\" \/><\/div>\n<p id=\"import-auto-id3068968\"><strong>Discussion<\/strong><\/p>\n<p id=\"fs-id2449340\">The ratio of translational energy to rotational kinetic energy is only 0.380. This ratio tells us that most of the kinetic energy of the helicopter is in its spinning blades\u2014something you probably would not suspect. The 53.7 m height to which the helicopter could be raised with the rotational kinetic energy is also impressive, again emphasizing the amount of rotational kinetic energy in the blades.<\/p>\n<\/div>\n<div class=\"bc-figure figure\" id=\"import-auto-id2420248\">\n<div class=\"bc-figcaption figcaption\">The first image shows how helicopters store large amounts of rotational kinetic energy in their blades. This energy must be put into the blades before takeoff and maintained until the end of the flight. The engines do not have enough power to simultaneously provide lift and put significant rotational energy into the blades. The second image shows a helicopter from the Auckland Westpac Rescue Helicopter Service. Over 50,000 lives have been saved since its operations beginning in 1973. Here, a water rescue operation is shown. (credit: 111 Emergency, Flickr)<\/div>\n<p><span data-type=\"media\" id=\"import-auto-id2930076\" data-alt=\"The given figure here shows a helicopter from the Auckland Westpac Rescue Helicopter Service over a sea. A rescue diver is shown holding the iron stand bar at the bottom of the helicopter, clutching a person. In the other image just above this, the blades of the helicopter are shown with their anti-clockwise rotation direction shown with an arrow and the length of one blade is given as four meters.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/Figure_11_04_05a.jpg\" data-media-type=\"image\/jpg\" alt=\"The given figure here shows a helicopter from the Auckland Westpac Rescue Helicopter Service over a sea. A rescue diver is shown holding the iron stand bar at the bottom of the helicopter, clutching a person. In the other image just above this, the blades of the helicopter are shown with their anti-clockwise rotation direction shown with an arrow and the length of one blade is given as four meters.\" height=\"300\" \/><\/span><\/p>\n<\/div>\n<div data-type=\"note\" class=\"note\" data-has-label=\"true\" id=\"fs-id3421186\" data-label=\"\">\n<div data-type=\"title\" class=\"title\">Making Connections<\/div>\n<p id=\"import-auto-id1418758\">Conservation of energy includes rotational motion, because rotational kinetic energy is another form of <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-909db8dd14ec527e98eef010c3baba6e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#75;&#69;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"14\" width=\"26\" style=\"vertical-align: -1px;\" \/><br \/>\n    . <a href=\"\/contents\/3ef5dfb6-0a8d-433e-9c8f-b8c860a3903b@2\">Uniform Circular Motion and Gravitation<\/a> has a detailed treatment of conservation of energy.<\/p>\n<\/div>\n<div class=\"bc-section section\" data-depth=\"1\" id=\"fs-id2616556\">\n<h1 data-type=\"title\">How Thick Is the Soup? Or Why Don\u2019t All Objects Roll Downhill at the Same Rate?<\/h1>\n<p id=\"import-auto-id1970058\">One of the quality controls in a tomato soup factory consists of rolling filled cans down a ramp. If they roll too fast, the soup is too thin. Why should cans of identical size and mass roll down an incline at different rates? And why should the thickest soup roll the slowest?<\/p>\n<p id=\"import-auto-id2969200\">The easiest way to answer these questions is to consider energy. Suppose each can starts down the ramp from rest. Each can starting from rest means each starts with the same gravitational potential energy <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-2c4392fef0201e4e66f08799e2ae55d0_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#80;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#103;&#114;&#97;&#118;&#125;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"50\" style=\"vertical-align: -6px;\" \/>, which is converted entirely to <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-909db8dd14ec527e98eef010c3baba6e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#75;&#69;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"14\" width=\"26\" style=\"vertical-align: -1px;\" \/>, provided each rolls without slipping. <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-909db8dd14ec527e98eef010c3baba6e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#75;&#69;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"14\" width=\"26\" style=\"vertical-align: -1px;\" \/>, however, can take the form of <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-f5d796dde74f13a1b570426b45c013fc_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#75;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#116;&#114;&#97;&#110;&#115;&#125;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"56\" style=\"vertical-align: -4px;\" \/> or <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-66178ae9cd7eefcecb69a17e362e498b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#75;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#114;&#111;&#116;&#125;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"43\" style=\"vertical-align: -4px;\" \/>, and total <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-909db8dd14ec527e98eef010c3baba6e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#75;&#69;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"14\" width=\"26\" style=\"vertical-align: -1px;\" \/> is the sum of the two. If a can rolls down a ramp, it puts part of its energy into rotation, leaving less for translation. Thus, the can goes slower than it would if it slid down. Furthermore, the thin soup does not rotate, whereas the thick soup does, because it sticks to the can. The thick soup thus puts more of the can\u2019s original gravitational potential energy into rotation than the thin soup, and the can rolls more slowly, as seen in <a href=\"#import-auto-id3105621\" class=\"autogenerated-content\">(Figure)<\/a>.<\/p>\n<div class=\"bc-figure figure\" id=\"import-auto-id3105621\">\n<div class=\"bc-figcaption figcaption\">Three cans of soup with identical masses race down an incline. The first can has a low friction coating and does not roll but just slides down the incline. It wins because it converts its entire PE into translational KE. The second and third cans both roll down the incline without slipping. The second can contains thin soup and comes in second because part of its initial PE goes into rotating the can (but not the thin soup). The third can contains thick soup. It comes in third because the soup rotates along with the can, taking even more of the initial PE for rotational KE, leaving less for translational KE.<\/div>\n<p><span data-type=\"media\" id=\"import-auto-id3026629\" data-alt=\"The figure shows a flat surface inclined at a height of h from the surface level, with three cans of soup of different densities numbered as one, two, and three rolling along it.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/Figure_11_04_06a.jpg\" data-media-type=\"image\/jpg\" alt=\"The figure shows a flat surface inclined at a height of h from the surface level, with three cans of soup of different densities numbered as one, two, and three rolling along it.\" width=\"300\" \/><\/span><\/p>\n<\/div>\n<p id=\"import-auto-id2679189\">Assuming no losses due to friction, there is only one force doing work\u2014gravity. Therefore the total work done is the change in kinetic energy. As the cans start moving, the potential energy is changing into kinetic energy. Conservation of energy gives<\/p>\n<div data-type=\"equation\" class=\"equation\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-8c8c0d7d7fa45ea24dabeac1853cc301_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#80;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#105;&#125;&#125;&#61;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#75;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#102;&#125;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"87\" style=\"vertical-align: -4px;\" \/><\/div>\n<p id=\"import-auto-id3018171\">More specifically,<\/p>\n<div data-type=\"equation\" class=\"equation\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-c587d766a5ed648324197399bbf10611_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#80;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#103;&#114;&#97;&#118;&#125;&#125;&#61;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#75;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#116;&#114;&#97;&#110;&#115;&#125;&#125;&#43;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#75;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#114;&#111;&#116;&#125;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"196\" style=\"vertical-align: -6px;\" \/><\/div>\n<p id=\"import-auto-id1431966\">or<\/p>\n<div data-type=\"equation\" class=\"equation\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-213755f176447945fb41b107670e5e39_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#103;&#104;&#125;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#49;&#125;&#123;&#50;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#118;&#125;&#125;&#94;&#123;&#50;&#125;&#43;&#92;&#102;&#114;&#97;&#99;&#123;&#49;&#125;&#123;&#50;&#125;&#123;&#92;&#109;&#97;&#116;&#104;&#114;&#109;&#123;&#73;&#92;&#111;&#109;&#101;&#103;&#97;&#32;&#125;&#125;&#94;&#123;&#50;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"22\" width=\"161\" style=\"vertical-align: -6px;\" \/><\/div>\n<p id=\"import-auto-id1868893\">So, the initial <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-c6103d60cd471f934368e423f05e3741_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#103;&#104;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"34\" style=\"vertical-align: -3px;\" \/> is divided between translational kinetic energy and rotational kinetic energy; and the greater <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-18b5e45cb4a1ee02e81b9a980f828db8_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#73;\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"9\" style=\"vertical-align: 0px;\" \/> is, the less energy goes into translation. If the can slides down without friction, then <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-f787d1a62f0180301830a38231fe6fa5_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#111;&#109;&#101;&#103;&#97;&#32;&#61;&#48;\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"44\" style=\"vertical-align: 0px;\" \/> and all the energy goes into translation; thus, the can goes faster.<\/p>\n<div data-type=\"note\" class=\"note\" data-has-label=\"true\" id=\"fs-id2398409\" data-label=\"\">\n<div data-type=\"title\" class=\"title\">Take-Home Experiment<\/div>\n<p id=\"import-auto-id3246662\">Locate several cans each containing different types of food. First, predict which can will win the race down an inclined plane and explain why. See if your prediction is correct. You could also do this experiment by collecting several empty cylindrical containers of the same size and filling them with different materials such as wet or dry sand.<\/p>\n<\/div>\n<div data-type=\"example\" class=\"textbox examples\" id=\"fs-id3073422\">\n<div data-type=\"title\" class=\"title\">Calculating the Speed of a Cylinder Rolling Down an Incline<\/div>\n<p id=\"import-auto-id1818016\">Calculate the final speed of a solid cylinder that rolls down a 2.00-m-high incline. The cylinder starts from rest, has a mass of 0.750 kg, and has a radius of 4.00 cm.<\/p>\n<p id=\"import-auto-id1429770\"><strong>Strategy<\/strong><\/p>\n<p id=\"fs-id2445196\">We can solve for the final velocity using conservation of energy, but we must first express rotational quantities in terms of translational quantities to end up with <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-ef71511c70f0e4b25cc6bd69f3bc20c2_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#118;\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"9\" style=\"vertical-align: 0px;\" \/> as the only unknown.<\/p>\n<p id=\"import-auto-id2052619\"><strong>Solution<\/strong><\/p>\n<p id=\"fs-id3158522\">Conservation of energy for this situation is written as described above:<\/p>\n<div data-type=\"equation\" class=\"equation\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-c7250ff89340b94c85c0d6b73ca2b87d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#103;&#104;&#125;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#49;&#125;&#123;&#50;&#125;&#123;&#92;&#109;&#97;&#116;&#104;&#105;&#116;&#123;&#109;&#118;&#125;&#125;&#94;&#123;&#50;&#125;&#43;&#92;&#102;&#114;&#97;&#99;&#123;&#49;&#125;&#123;&#50;&#125;&#123;&#92;&#109;&#97;&#116;&#104;&#114;&#109;&#123;&#73;&#92;&#111;&#109;&#101;&#103;&#97;&#32;&#125;&#125;&#94;&#123;&#50;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"22\" width=\"162\" style=\"vertical-align: -6px;\" \/><\/div>\n<p id=\"import-auto-id3385654\">Before we can solve for <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-ef71511c70f0e4b25cc6bd69f3bc20c2_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#118;\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"9\" style=\"vertical-align: 0px;\" \/> , we must get an expression for <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-18b5e45cb4a1ee02e81b9a980f828db8_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#73;\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"9\" style=\"vertical-align: 0px;\" \/> from <a href=\"\/contents\/db59f656-e708-4094-a742-1e5560fe97c9@6#fs-id1838666\" class=\"autogenerated-content\">(Figure)<\/a>. Because <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-ef71511c70f0e4b25cc6bd69f3bc20c2_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#118;\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"9\" style=\"vertical-align: 0px;\" \/> and <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-8ffb415af81ab9c23c1d2e7ec67d29c6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#111;&#109;&#101;&#103;&#97;&#32;\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"11\" style=\"vertical-align: 0px;\" \/> are related (note here that the cylinder is rolling without slipping), we must also substitute the relationship <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-b9745b93d94f9d47aab510e008008853_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#111;&#109;&#101;&#103;&#97;&#32;&#61;&#118;&#47;&#82;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"68\" style=\"vertical-align: -5px;\" \/> into the expression. These substitutions yield<\/p>\n<div data-type=\"equation\" class=\"equation\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-2144972a4cf69939a538389060b60139_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#103;&#104;&#125;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#49;&#125;&#123;&#50;&#125;&#123;&#92;&#109;&#97;&#116;&#104;&#105;&#116;&#123;&#109;&#118;&#125;&#125;&#94;&#123;&#50;&#125;&#43;&#92;&#102;&#114;&#97;&#99;&#123;&#49;&#125;&#123;&#50;&#125;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#102;&#114;&#97;&#99;&#123;&#49;&#125;&#123;&#50;&#125;&#123;&#92;&#109;&#97;&#116;&#104;&#105;&#116;&#123;&#109;&#82;&#125;&#125;&#94;&#123;&#50;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#102;&#114;&#97;&#99;&#123;&#123;&#118;&#125;&#94;&#123;&#50;&#125;&#125;&#123;&#123;&#82;&#125;&#94;&#123;&#50;&#125;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"33\" width=\"251\" style=\"vertical-align: -12px;\" \/><\/div>\n<p id=\"import-auto-id1951682\">Interestingly, the cylinder\u2019s radius <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-dae6bae3dcdac4629730754352c5e329_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#82;\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"14\" style=\"vertical-align: 0px;\" \/> and mass <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-6b41df788161942c6f98604d37de8098_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#109;\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"15\" style=\"vertical-align: 0px;\" \/> cancel, yielding<\/p>\n<div data-type=\"equation\" class=\"equation\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-10e97f7a12216a32e6cb7cc7b16ea4cd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#103;&#104;&#125;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#49;&#125;&#123;&#50;&#125;&#123;&#118;&#125;&#94;&#123;&#50;&#125;&#43;&#92;&#102;&#114;&#97;&#99;&#123;&#49;&#125;&#123;&#52;&#125;&#123;&#118;&#125;&#94;&#123;&#50;&#125;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#51;&#125;&#123;&#52;&#125;&#123;&#118;&#125;&#94;&#123;&#50;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"22\" width=\"175\" style=\"vertical-align: -6px;\" \/><\/div>\n<p id=\"import-auto-id2956864\">Solving algebraically, the equation for the final velocity <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-ef71511c70f0e4b25cc6bd69f3bc20c2_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#118;\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"9\" style=\"vertical-align: 0px;\" \/> gives<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-320\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-30ef927d282c29c92ae6644977f505ad_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#118;&#61;&#123;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#102;&#114;&#97;&#99;&#123;&#52;&#92;&#116;&#101;&#120;&#116;&#123;&#103;&#104;&#125;&#125;&#123;&#51;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#125;&#94;&#123;&#49;&#47;&#50;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"105\" style=\"vertical-align: -12px;\" \/><\/div>\n<p id=\"import-auto-id2618589\">Substituting known values into the resulting expression yields<\/p>\n<div data-type=\"equation\" class=\"equation\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-eb0947466aaf47ddeeebf060e2755ee6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#118;&#61;&#123;&#92;&#108;&#101;&#102;&#116;&#91;&#92;&#102;&#114;&#97;&#99;&#123;&#52;&#92;&#108;&#101;&#102;&#116;&#40;&#57;&#46;&#56;&#48;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#53;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#47;&#115;&#125;&#125;&#94;&#123;&#50;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#116;&#101;&#120;&#116;&#123;&#50;&#46;&#48;&#48;&#32;&#109;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#125;&#123;&#51;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#93;&#125;&#94;&#123;&#49;&#47;&#50;&#125;&#61;&#92;&#116;&#101;&#120;&#116;&#123;&#53;&#46;&#49;&#49;&#32;&#109;&#47;&#115;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"49\" width=\"304\" style=\"vertical-align: -17px;\" \/><\/div>\n<p id=\"import-auto-id1362701\"><strong>Discussion<\/strong><\/p>\n<p id=\"fs-id3007358\">Because <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-6b41df788161942c6f98604d37de8098_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#109;\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"15\" style=\"vertical-align: 0px;\" \/> and <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-dae6bae3dcdac4629730754352c5e329_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#82;\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"14\" style=\"vertical-align: 0px;\" \/> cancel, the result <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-c7823416856bba2dced523407f185426_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#118;&#61;&#123;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#102;&#114;&#97;&#99;&#123;&#52;&#125;&#123;&#51;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#103;&#104;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#125;&#94;&#123;&#49;&#47;&#50;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"28\" width=\"100\" style=\"vertical-align: -7px;\" \/> is valid for any solid cylinder, implying that all solid cylinders will roll down an incline at the same rate independent of their masses and sizes. (Rolling cylinders down inclines is what Galileo actually did to show that objects fall at the same rate independent of mass.) Note that if the cylinder slid without friction down the incline without rolling, then the entire gravitational potential energy would go into translational kinetic energy. Thus, <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-8ead15815f225936a01774a4b5712372_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#102;&#114;&#97;&#99;&#123;&#49;&#125;&#123;&#50;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#118;&#125;&#125;&#94;&#123;&#50;&#125;&#61;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#103;&#104;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"22\" width=\"98\" style=\"vertical-align: -6px;\" \/> and <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-61c88153004c2b8f41530ec3c1be13fe_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#118;&#61;&#92;&#108;&#101;&#102;&#116;&#40;&#50;&#92;&#116;&#101;&#120;&#116;&#123;&#103;&#104;&#125;&#123;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#125;&#94;&#123;&#49;&#47;&#50;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"23\" width=\"95\" style=\"vertical-align: -4px;\" \/>, which is 22% greater than <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-12c9718113e9ade5166eaa67aa8751bb_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#108;&#101;&#102;&#116;&#40;&#52;&#92;&#116;&#101;&#120;&#116;&#123;&#103;&#104;&#125;&#47;&#51;&#123;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#125;&#94;&#123;&#49;&#47;&#50;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"24\" width=\"79\" style=\"vertical-align: -5px;\" \/>. That is, the cylinder would go faster at the bottom.<\/p>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id2931518\" data-element-type=\"check-understanding\" data-label=\"\">\n<div data-type=\"title\">Check Your Understanding<\/div>\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id1613668\">\n<p id=\"import-auto-id3043952\"><span data-type=\"title\">Analogy of Rotational and Translational Kinetic Energy<\/span>Is rotational kinetic energy completely analogous to translational kinetic energy? What, if any, are their differences? Give an example of each type of kinetic energy.<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"fs-id1412799\">\n<p id=\"import-auto-id3081539\">Yes, rotational and translational kinetic energy are exact analogs. They both are the energy of motion involved with the coordinated (non-random) movement of mass relative to some reference frame. The only difference between rotational and translational kinetic energy is that translational is straight line motion while rotational is not. An example of both kinetic and translational kinetic energy is found in a bike tire while being ridden down a bike path. The rotational motion of the tire means it has rotational kinetic energy while the movement of the bike along the path means the tire also has translational kinetic energy. If you were to lift the front wheel of the bike and spin it while the bike is stationary, then the wheel would have only rotational kinetic energy relative to the Earth.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div data-type=\"note\" class=\"note\" data-has-label=\"true\" data-label=\"\">\n<div data-type=\"title\" class=\"title\">PhET Explorations: My Solar System<\/div>\n<p id=\"eip-id1169738171827\">Build your own system of heavenly bodies and watch the gravitational ballet. With this orbit simulator, you can set initial positions, velocities, and masses of 2, 3, or 4 bodies, and then see them orbit each other.<\/p>\n<div class=\"bc-figure figure\" id=\"eip-id2625118\">\n<div class=\"bc-figcaption figcaption\"><a href=\"\/resources\/e7460a4a0c31d5af5265b49a67e8778040ce3630\/my-solar-system_en.jar\">My Solar System<\/a><\/div>\n<p><span data-type=\"media\" id=\"Phet_module_11.4\" data-alt=\"\"><a href=\"\/resources\/e7460a4a0c31d5af5265b49a67e8778040ce3630\/my-solar-system_en.jar\" data-type=\"image\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/PhET_Icon.png\" data-media-type=\"image\/png\" alt=\"\" data-print=\"false\" width=\"450\" \/><\/a><span data-media-type=\"image\/png\" data-print=\"true\" data-src=\"\/resources\/075500ad9f71890a85fe3f7a4137ac08e2b7907c\/PhET_Icon.png\" data-type=\"image\"><\/span><\/span><\/p>\n<\/div>\n<\/div>\n<div class=\"section-summary\" data-depth=\"1\" id=\"fs-id3176558\">\n<h1 data-type=\"title\">Section Summary<\/h1>\n<ul id=\"fs-id1917137\">\n<li id=\"import-auto-id1360816\">The rotational kinetic energy <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-66178ae9cd7eefcecb69a17e362e498b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#75;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#114;&#111;&#116;&#125;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"43\" style=\"vertical-align: -4px;\" \/> for an object with a moment of inertia <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-18b5e45cb4a1ee02e81b9a980f828db8_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#73;\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"9\" style=\"vertical-align: 0px;\" \/> and an angular velocity <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-8ffb415af81ab9c23c1d2e7ec67d29c6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#111;&#109;&#101;&#103;&#97;&#32;\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"11\" style=\"vertical-align: 0px;\" \/> is given by\n<div data-type=\"equation\" class=\"equation\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-9621d67ae6ff2929c3ffcec377429438_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#75;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#114;&#111;&#116;&#125;&#125;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#49;&#125;&#123;&#50;&#125;&#123;&#92;&#109;&#97;&#116;&#104;&#114;&#109;&#123;&#73;&#92;&#111;&#109;&#101;&#103;&#97;&#32;&#125;&#125;&#94;&#123;&#50;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"22\" width=\"108\" style=\"vertical-align: -6px;\" \/><\/div>\n<\/li>\n<li id=\"import-auto-id1972580\">Helicopters store large amounts of rotational kinetic energy in their blades. This energy must be put into the blades before takeoff and maintained until the end of the flight. The engines do not have enough power to simultaneously provide lift and put significant rotational energy into the blades.<\/li>\n<li id=\"import-auto-id3224025\">Work and energy in rotational motion are completely analogous to work and energy in translational motion.<\/li>\n<li id=\"import-auto-id2453212\">The equation for the <span data-type=\"term\">work-energy theorem<\/span> for rotational motion is,\n<div data-type=\"equation\" class=\"equation\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-1c4f8ad681292f506a6caa3a37479861_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#53;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#87;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#49;&#125;&#123;&#50;&#125;&#123;&#92;&#109;&#97;&#116;&#104;&#114;&#109;&#123;&#73;&#92;&#111;&#109;&#101;&#103;&#97;&#32;&#125;&#125;&#94;&#123;&#50;&#125;&#45;&#92;&#102;&#114;&#97;&#99;&#123;&#49;&#125;&#123;&#50;&#125;&#73;&#123;&#123;&#92;&#111;&#109;&#101;&#103;&#97;&#32;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#48;&#125;&#125;&#125;&#94;&#123;&#50;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"22\" width=\"180\" style=\"vertical-align: -6px;\" \/><\/div>\n<\/li>\n<\/ul>\n<\/div>\n<div class=\"conceptual-questions\" data-depth=\"1\" id=\"fs-id1578001\" data-element-type=\"conceptual-questions\">\n<h1 data-type=\"title\">Conceptual Questions<\/h1>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id3026004\" data-element-type=\"conceptual-questions\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id3025680\">\n<p id=\"import-auto-id3012901\">Describe the energy transformations involved when a yo-yo is thrown downward and then climbs back up its string to be caught in the user\u2019s hand.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id1428194\" data-element-type=\"conceptual-questions\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id1995300\">\n<p id=\"import-auto-id1413725\">What energy transformations are involved when a dragster engine is revved, its clutch let out rapidly, its tires spun, and it starts to accelerate forward? Describe the source and transformation of energy at each step.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id2640555\" data-element-type=\"conceptual-questions\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id3192098\">\n<p id=\"eip-id2695457\">The Earth has more rotational kinetic energy now than did the cloud of gas and dust from which it formed. Where did this energy come from?<\/p>\n<div class=\"bc-figure figure\" id=\"import-auto-id2615448\">\n<div class=\"bc-figcaption figcaption\">An immense cloud of rotating gas and dust contracted under the influence of gravity to form the Earth and in the process rotational kinetic energy increased. (credit: NASA)<\/div>\n<p><span data-type=\"media\" id=\"import-auto-id3045924\" data-alt=\"The figure shows a closed view of a red planet in the sky, with a sun like object seen at the far right and the planet shown here being surrounded by circles of gas and dust.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/Figure_11_04_07a.jpg\" data-media-type=\"image\/png\" alt=\"The figure shows a closed view of a red planet in the sky, with a sun like object seen at the far right and the planet shown here being surrounded by circles of gas and dust.\" width=\"250\" \/><\/span><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"problems-exercises\" data-depth=\"1\" id=\"fs-id1864335\" data-element-type=\"problems-exercises\">\n<h1 data-type=\"title\">Problems &amp; Exercises<\/h1>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id2402678\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id3255931\">\n<p id=\"import-auto-id2963400\">This problem considers energy and work aspects of <a href=\"\/contents\/db59f656-e708-4094-a742-1e5560fe97c9@6#fs-id1468671\" class=\"autogenerated-content\">(Figure)<\/a>\u2014use data from that example as needed. (a) Calculate the rotational kinetic energy in the merry-go-round plus child when they have an angular velocity of 20.0 rpm. (b) Using energy considerations, find the number of revolutions the father will have to push to achieve this angular velocity starting from rest. (c) Again, using energy considerations, calculate the force the father must exert to stop the merry-go-round in two revolutions<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"fs-id2991773\">\n<p id=\"import-auto-id3384991\">(a) 185 J<\/p>\n<p id=\"import-auto-id3055432\">(b) 0.0785 rev<\/p>\n<p id=\"import-auto-id3358608\">(c) <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-5ef210021aa6c31831e6da0bb9527784_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#87;&#61;&#57;&#92;&#116;&#101;&#120;&#116;&#123;&#46;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#56;&#49;&#32;&#78;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"93\" style=\"vertical-align: -1px;\" \/><\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id2601323\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id2640866\">\n<p id=\"import-auto-id1431802\">What is the final velocity of a hoop that rolls without slipping down a 5.00-m-high hill, starting from rest?<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id3017926\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id1461913\">\n<p id=\"import-auto-id3091855\">(a) Calculate the rotational kinetic energy of Earth on its axis. (b) What is the rotational kinetic energy of Earth in its orbit around the Sun?<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"fs-id2382524\">\n<p id=\"import-auto-id1326234\">(a) <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-54568b793c6fd6e164395b30f1fdadf4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#50;&#46;&#53;&#55;&times;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#49;&#48;&#125;&#125;&#94;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#50;&#57;&#125;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#53;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#74;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"77\" style=\"vertical-align: -1px;\" \/><\/p>\n<p id=\"import-auto-id3036677\">(b) <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-4235495cc878d035a0994aa4ac32acea_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#75;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#114;&#111;&#116;&#125;&#125;&#61;&#50;&#92;&#116;&#101;&#120;&#116;&#123;&#46;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#54;&#53;&#125;&times;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#49;&#48;&#125;&#125;&#94;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#51;&#51;&#125;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#53;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#74;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"145\" style=\"vertical-align: -4px;\" \/><\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id1596687\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id1577764\">\n<p id=\"import-auto-id2041602\">Calculate the rotational kinetic energy in the motorcycle wheel (<a href=\"\/contents\/db59f656-e708-4094-a742-1e5560fe97c9@6#import-auto-id3370574\" class=\"autogenerated-content\">(Figure)<\/a>) if its angular velocity is 120 rad\/s. Assume M = 12.0 kg, R<sub>1<\/sub> = 0.280 m, and R<sub>2<\/sub> = 0.330 m.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id1580820\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id3007150\">\n<p id=\"import-auto-id2403290\">A baseball pitcher throws the ball in a motion where there is rotation of the forearm about the elbow joint as well as other movements. If the linear velocity of the ball relative to the elbow joint is 20.0 m\/s at a distance of 0.480 m from the joint and the moment of inertia of the forearm is <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-0ebc4ba6d8a54bfe459aa8f4c59849fd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#48;&#46;&#53;&#48;&#48;&#32;&#107;&#103;&#125;&#92;&#99;&#100;&#111;&#116;&#32;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#125;&#125;&#94;&#123;&#50;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"99\" style=\"vertical-align: -3px;\" \/>, what is the rotational kinetic energy of the forearm?<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"fs-id3028612\">\n<div data-type=\"equation\" class=\"equation\" id=\"eip-id1583692\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-0a10300c7cb896845d13e1c0f58e6dd6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#75;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#114;&#111;&#116;&#125;&#125;&#61;&#92;&#116;&#101;&#120;&#116;&#123;&#52;&#51;&#52;&#32;&#74;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"109\" style=\"vertical-align: -4px;\" \/><\/div>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id2604037\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id1818585\">\n<p id=\"fs-id3053480\">While punting a football, a kicker rotates his leg about the hip joint. The moment of inertia of the leg is <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-51ea2f61ec744a10579efe9e7c5afdf5_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#51;&#46;&#55;&#53;&#32;&#107;&#103;&#125;&#92;&#99;&#100;&#111;&#116;&#32;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#125;&#125;&#94;&#123;&#50;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"90\" style=\"vertical-align: -3px;\" \/> and its rotational kinetic energy is 175 J. (a) What is the angular velocity of the leg? (b) What is the velocity of tip of the punter\u2019s shoe if it is 1.05 m from the hip joint? (c) Explain how the football can be given a velocity greater than the tip of the shoe (necessary for a decent kick distance).<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id2662255\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id2679107\">\n<p id=\"fs-id2589915\">A bus contains a 1500 kg flywheel (a disk that has a 0.600 m radius) and has a total mass of 10,000 kg. (a) Calculate the angular velocity the flywheel must have to contain enough energy to take the bus from rest to a speed of 20.0 m\/s, assuming 90.0% of the rotational kinetic energy can be transformed into translational energy. (b) How high a hill can the bus climb with this stored energy and still have a speed of 3.00 m\/s at the top of the hill? Explicitly show how you follow the steps in the <a href=\"#fs-id1986333\">Problem-Solving Strategy for Rotational Energy<\/a>.<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"fs-id1917152\">\n<p id=\"fs-id3450208\">(a) <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-e878468547d4bd6079216ffb2c0c2b0b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#49;&#50;&#56;&#32;&#114;&#97;&#100;&#47;&#115;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"74\" style=\"vertical-align: -4px;\" \/><\/p>\n<p id=\"import-auto-id1514659\">(b) <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-51c6e97762dbb9863293effd57e736e9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#49;&#57;&#46;&#57;&#32;&#109;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"51\" style=\"vertical-align: -1px;\" \/><\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id3250372\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id1922322\">\n<p id=\"import-auto-id1942736\">A ball with an initial velocity of 8.00 m\/s rolls up a hill without slipping. Treating the ball as a spherical shell, calculate the vertical height it reaches. (b) Repeat the calculation for the same ball if it slides up the hill without rolling.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id2583778\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id3254741\">\n<p id=\"import-auto-id3161408\">While exercising in a fitness center, a man lies face down on a bench and lifts a weight with one lower leg by contacting the muscles in the back of the upper leg. (a) Find the angular acceleration produced given the mass lifted is 10.0 kg at a distance of 28.0 cm from the knee joint, the moment of inertia of the lower leg is <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-1ca672ace42dbbb02fba9986aeb1b91b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#48;&#46;&#57;&#48;&#48;&#32;&#107;&#103;&#125;&#92;&#99;&#100;&#111;&#116;&#32;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#125;&#125;&#94;&#123;&#50;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"99\" style=\"vertical-align: -3px;\" \/>, the muscle force is 1500 N, and its effective perpendicular lever arm is 3.00 cm. (b) How much work is done if the leg rotates through an angle of <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-c918e80bba36ea1d2e565858d4a55186_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#50;&#48;&#46;&#48;&ordm;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"32\" style=\"vertical-align: 0px;\" \/> with a constant force exerted by the muscle?<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"fs-id3115463\">\n<p id=\"import-auto-id1517396\">(a) <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-30edbc7b0898f7aa46ff42246bf5e64e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#49;&#48;&#46;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#52;&#32;&#114;&#97;&#100;&#47;&#115;&#125;&#125;&#94;&#123;&#50;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"21\" width=\"85\" style=\"vertical-align: -4px;\" \/><\/p>\n<p id=\"import-auto-id3051710\">(b) <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-01b872ec4ab8208d0453eb3d613d2093_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#101;&#116;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#53;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#87;&#61;&#54;&#46;&#92;&#116;&#101;&#120;&#116;&#123;&#49;&#49;&#32;&#74;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"14\" width=\"118\" style=\"vertical-align: -1px;\" \/><\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id3199856\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id3008300\">\n<p id=\"import-auto-id1930100\">To develop muscle tone, a woman lifts a 2.00-kg weight held in her hand. She uses her biceps muscle to flex the lower arm through an angle of <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-c9b21c457af08380c1d3f91224476771_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#54;&#48;&#46;&#48;&ordm;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"32\" style=\"vertical-align: 0px;\" \/>. (a) What is the angular acceleration if the weight is 24.0 cm from the elbow joint, her forearm has a moment of inertia of <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-33cfabd7141769fb1542ef6f8930dd55_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#48;&#46;&#50;&#53;&#48;&#32;&#107;&#103;&#125;&#92;&#99;&#100;&#111;&#116;&#32;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#125;&#125;&#94;&#123;&#50;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"99\" style=\"vertical-align: -3px;\" \/>, and the net force she exerts is 750 N at an effective perpendicular lever arm of 2.00 cm? (b) How much work does she do?<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id3245199\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id2963718\">\n<p id=\"import-auto-id3080929\">Consider two cylinders that start down identical inclines from rest except that one is frictionless. Thus one cylinder rolls without slipping, while the other slides frictionlessly without rolling. They both travel a short distance at the bottom and then start up another incline. (a) Show that they both reach the same height on the other incline, and that this height is equal to their original height. (b) Find the ratio of the time the rolling cylinder takes to reach the height on the second incline to the time the sliding cylinder takes to reach the height on the second incline. (c) Explain why the time for the rolling motion is greater than that for the sliding motion.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id2402928\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id3176898\">\n<p id=\"import-auto-id3398559\">What is the moment of inertia of an object that rolls without slipping down a 2.00-m-high incline starting from rest, and has a final velocity of 6.00 m\/s? Express the moment of inertia as a multiple of <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-9b0d23c4f2a4593686b555ee91c04508_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#109;&#97;&#116;&#104;&#105;&#116;&#123;&#77;&#82;&#125;&#125;&#94;&#123;&#50;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"36\" style=\"vertical-align: 0px;\" \/>, where <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-10ebb71bad275c1815a8f2a8c5dea0be_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#77;\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"19\" style=\"vertical-align: 0px;\" \/> is the mass of the object and <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-dae6bae3dcdac4629730754352c5e329_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#82;\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"14\" style=\"vertical-align: 0px;\" \/> is its radius.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id2406116\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id2679003\">\n<p id=\"import-auto-id3232936\">Suppose a 200-kg motorcycle has two wheels like, <a href=\"\/contents\/db59f656-e708-4094-a742-1e5560fe97c9@6#import-auto-id3370574\">the one described in Problem 10.15<\/a> and is heading toward a hill at a speed of 30.0 m\/s. (a) How high can it coast up the hill, if you neglect friction? (b) How much energy is lost to friction if the motorcycle only gains an altitude of 35.0 m before coming to rest?<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id3399194\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id2583176\">\n<p id=\"import-auto-id2950423\">In softball, the pitcher throws with the arm fully extended (straight at the elbow). In a fast pitch the ball leaves the hand with a speed of 139 km\/h. (a) Find the rotational kinetic energy of the pitcher\u2019s arm given its moment of inertia is <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-f789002a394675f2551d2fba12f4a947_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#48;&#46;&#55;&#50;&#48;&#32;&#107;&#103;&#125;&#92;&#99;&#100;&#111;&#116;&#32;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#125;&#125;&#94;&#123;&#50;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"99\" style=\"vertical-align: -3px;\" \/> and the ball leaves the hand at a distance of 0.600 m from the pivot at the shoulder. (b) What force did the muscles exert to cause the arm to rotate if their effective perpendicular lever arm is 4.00 cm and the ball is 0.156 kg?<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"fs-id2616505\">\n<p id=\"import-auto-id3013127\">(a) 1.49 kJ<\/p>\n<p id=\"import-auto-id1561424\">(b) <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-3923a569237737d49ebd5901b381f356_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#50;&#46;&#53;&#50;&times;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#49;&#48;&#125;&#125;&#94;&#123;&#52;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#53;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#78;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"74\" style=\"vertical-align: -1px;\" \/><\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id3073542\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id3158726\">\n<p id=\"import-auto-id3104836\"><strong>Construct Your Own Problem<\/strong><\/p>\n<p id=\"eip-id1169762600538\">Consider the work done by a spinning skater pulling her arms in to increase her rate of spin. Construct a problem in which you calculate the work done with a \u201cforce multiplied by distance\u201d calculation and compare it to the skater\u2019s increase in kinetic energy.\n    <\/p>\n<\/div>\n<\/div>\n<\/div>\n<div data-type=\"glossary\" class=\"textbox shaded\">\n<h2 data-type=\"glossary-title\">Glossary<\/h2>\n<dl class=\"definition\" id=\"import-auto-id3007462\">\n<dt>work-energy theorem<\/dt>\n<dd id=\"fs-id2070291\">if one or more external forces act upon a rigid object, causing its kinetic energy to change from <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-439c72ac8c4ae88dfc66fb9fdf3380bc_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#75;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#49;&#125;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"32\" style=\"vertical-align: -4px;\" \/> to <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-bfc6fc199832e695fea1ff25ae51dd76_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#75;&#69;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#50;&#125;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"33\" style=\"vertical-align: -3px;\" \/>, then the work <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-4caed22919a1780df1b6310b338b904e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#87;\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"19\" style=\"vertical-align: 0px;\" \/> done by the net force is equal to the change in kinetic energy<\/dd>\n<\/dl>\n<dl class=\"definition\" id=\"import-auto-id1771451\">\n<dt>rotational kinetic energy<\/dt>\n<dd id=\"fs-id2383989\">the kinetic energy due to the rotation of an object. This is part of its total kinetic energy<\/dd>\n<\/dl>\n<\/div>\n","protected":false},"author":211,"menu_order":1,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":"all-rights-reserved"},"chapter-type":[],"contributor":[],"license":[56],"class_list":["post-537","chapter","type-chapter","status-publish","hentry","license-all-rights-reserved"],"part":506,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-json\/pressbooks\/v2\/chapters\/537","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-json\/wp\/v2\/users\/211"}],"version-history":[{"count":1,"href":"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-json\/pressbooks\/v2\/chapters\/537\/revisions"}],"predecessor-version":[{"id":538,"href":"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-json\/pressbooks\/v2\/chapters\/537\/revisions\/538"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-json\/pressbooks\/v2\/parts\/506"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-json\/pressbooks\/v2\/chapters\/537\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-json\/wp\/v2\/media?parent=537"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-json\/pressbooks\/v2\/chapter-type?post=537"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-json\/wp\/v2\/contributor?post=537"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-json\/wp\/v2\/license?post=537"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}