{"id":328,"date":"2017-10-27T16:29:30","date_gmt":"2017-10-27T16:29:30","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/chapter\/centripetal-force\/"},"modified":"2017-11-08T03:24:18","modified_gmt":"2017-11-08T03:24:18","slug":"centripetal-force","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/chapter\/centripetal-force\/","title":{"raw":"Centripetal Force","rendered":"Centripetal Force"},"content":{"raw":"\n<div class=\"textbox learning-objectives\">\n<h3 itemprop=\"educationalUse\">Learning Objectives<\/h3>\n<ul>\n<li>Calculate coefficient of friction on a car tire.<\/li>\n<li>Calculate ideal speed and angle of a car on a turn.<\/li>\n<\/ul>\n<\/div>\n<p id=\"import-auto-id2589409\">Any force or combination of forces can cause a centripetal or radial acceleration. Just a few examples are the tension in the rope on a tether ball, the force of Earth\u2019s gravity on the Moon, friction between roller skates and a rink floor, a banked roadway\u2019s force on a car, and forces on the tube of a spinning centrifuge.<\/p>\n<p id=\"import-auto-id3025916\">Any net force causing uniform circular motion is called a <span data-type=\"term\" id=\"import-auto-id3450370\">centripetal force<\/span>. The direction of a centripetal force is toward the center of curvature, the same as the direction of centripetal acceleration. According to Newton\u2019s second law of motion, net force is mass times acceleration: net [latex]\\text{F}=\\text{ma}[\/latex]. For uniform circular motion, the acceleration is the centripetal acceleration\u2014 [latex]a={a}_{c}[\/latex]. Thus, the magnitude of centripetal force [latex]{\\text{F}}_{\\text{c}}[\/latex] is<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-696\">[latex]{\\text{F}}_{\\text{c}}={m\\text{a}}_{\\text{c}}.[\/latex]<\/div>\n<p id=\"import-auto-id3397174\">By using the expressions for centripetal acceleration [latex]{a}_{c}[\/latex] from [latex]{a}_{c}=\\frac{{v}^{2}}{r};\\phantom{\\rule{0.25em}{0ex}}{a}_{c}={\\mathrm{r\\omega }}^{2}[\/latex], we get two expressions for the centripetal force [latex]{\\text{F}}_{\\text{c}}[\/latex] in terms of mass, velocity, angular velocity, and radius of curvature:<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-562\">[latex]{F}_{c}=m\\frac{{v}^{2}}{r};\\phantom{\\rule{0.25em}{0ex}}{F}_{c}=\\text{mr}{\\omega }^{2}.[\/latex]<\/div>\n<p id=\"import-auto-id2616689\">You may use whichever expression for centripetal force is more convenient. Centripetal force [latex]{F}_{\\text{c}}[\/latex] is always perpendicular to the path and pointing to the center of curvature, because [latex]{\\mathbf{a}}_{c}[\/latex] is perpendicular to the velocity and pointing to the center of curvature.<\/p>\n<p id=\"import-auto-id3105986\">Note that if you solve the first expression for [latex]r[\/latex], you get<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-58\">[latex]r=\\frac{{\\mathrm{mv}}^{2}}{{F}_{c}}\\text{.}[\/latex]<\/div>\n<p id=\"import-auto-id1410994\">This implies that for a given mass and velocity, a large centripetal force causes a small radius of curvature\u2014that is, a tight curve.<\/p>\n<div class=\"bc-figure figure\" id=\"import-auto-id2601650\">\n<div class=\"bc-figcaption figcaption\">The frictional force supplies the centripetal force and is numerically equal to it. Centripetal force is perpendicular to velocity and causes uniform circular motion. The larger the [latex]{\\text{F}}_{\\text{c}}[\/latex], the smaller the radius of curvature [latex]r[\/latex] and the sharper the curve. The second curve has the same [latex]v[\/latex], but a larger [latex]{\\text{F}}_{\\text{c}}[\/latex] produces a smaller [latex]r\\prime [\/latex].<\/div>\n<p><span data-type=\"media\" id=\"import-auto-id1581195\" data-alt=\"The given figure consists of two semicircles, one over the other. The top semicircle is bigger and the one below is smaller. In both the figures, the direction of the path is given along the semicircle in the counter-clockwise direction. A point is shown on the path, where the radius from the circle, r, is shown with an arrow from the center of the circle. At the same point, the centripetal force is shown in the opposite direction to that of radius arrow. The velocity, v, is shown along this point in the left upward direction and is perpendicular to the force. In both the figures, the velocity is same, but the radius is smaller and centripetal force is larger in the lower figure.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/Figure_07_03_01a.jpg\" data-media-type=\"image\/jpg\" alt=\"The given figure consists of two semicircles, one over the other. The top semicircle is bigger and the one below is smaller. In both the figures, the direction of the path is given along the semicircle in the counter-clockwise direction. A point is shown on the path, where the radius from the circle, r, is shown with an arrow from the center of the circle. At the same point, the centripetal force is shown in the opposite direction to that of radius arrow. The velocity, v, is shown along this point in the left upward direction and is perpendicular to the force. In both the figures, the velocity is same, but the radius is smaller and centripetal force is larger in the lower figure.\" width=\"200\"><\/span><\/p><\/div>\n<div data-type=\"example\" class=\"textbox examples\" id=\"fs-id3130274\">\n<div data-type=\"title\" class=\"title\">What Coefficient of Friction Do Car Tires Need on a Flat Curve?<\/div>\n<p id=\"import-auto-id1849101\">(a) Calculate the centripetal force exerted on a 900 kg car that negotiates a 500 m radius curve at 25.0 m\/s. <\/p>\n<p id=\"import-auto-id1409542\">(b) Assuming an unbanked curve, find the minimum static coefficient of friction, between the tires and the road, static friction being the reason that keeps the car from slipping (see <a href=\"#import-auto-id3184285\" class=\"autogenerated-content\">(Figure)<\/a>).<\/p>\n<p id=\"import-auto-id3205483\"><strong>Strategy and Solution for (a)<\/strong><\/p>\n<p id=\"import-auto-id1418337\">We know that<br>\n[latex]{F}_{\\text{c}}=\\frac{{\\mathrm{mv}}^{\\text{2}}}{r}[\/latex]. Thus,<\/p>\n<div data-type=\"equation\" class=\"equation\">[latex]{F}_{\\text{c}}=\\frac{{\\mathrm{mv}}^{\\text{2}}}{r}=\\frac{\\left(\\text{900 kg}\\right)\\left(\\text{25.0 m\/s}{\\right)}^{\\text{2}}}{\\left(\\text{500 m}\\right)}=\\text{1125 N.}[\/latex]<\/div>\n<p id=\"import-auto-id3102728\"><strong>Strategy for (b)<\/strong><\/p>\n<p id=\"fs-id3078905\"><a href=\"#import-auto-id3184285\" class=\"autogenerated-content\">(Figure)<\/a> shows the forces acting on the car on an unbanked (level ground) curve. Friction is to the left, keeping the car from slipping, and because it is the only horizontal force acting on the car, the friction is the centripetal force in this case. We know that the maximum static friction (at which the tires roll but do not slip) is [latex]{\\mu }_{\\text{s}}N[\/latex], where [latex]{\\mu }_{\\text{s}}[\/latex] is the static coefficient of friction and N is the normal force. The normal force equals the car\u2019s weight on level ground, so that <\/p>\n<p>[latex]N=\\mathit{mg}[\/latex]. Thus the centripetal force in this situation is<\/p>\n<div data-type=\"equation\" class=\"equation\">[latex]{F}_{\\text{c}}=f={\\mu }_{\\text{s}}N={\\mu }_{\\text{s}}\\text{mg}\\text{.}[\/latex]<\/div>\n<p id=\"import-auto-id3098784\">Now we have a relationship between centripetal force and the coefficient of friction. Using the first expression for [latex]{F}_{\\text{c}}[\/latex] from the equation<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-735\">[latex]\\begin{array}{c}{F}_{\\text{c}}=m\\frac{{v}^{2}}{r}\\\\ {F}_{\\text{c}}=\\text{mr}{\\omega }^{2}\\end{array}\\right\\},[\/latex]<\/div>\n<div data-type=\"equation\" class=\"equation\">[latex]m\\frac{{v}^{2}}{r}={\\mu }_{\\text{s}}\\text{mg}\\text{.}[\/latex]<\/div>\n<p id=\"import-auto-id1467842\">We solve this for [latex]{\\mu }_{\\text{s}}[\/latex], noting that mass cancels, and obtain<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-99\">[latex]{\\mu }_{\\text{s}}=\\frac{{v}^{2}}{\\text{rg}}\\text{.}[\/latex]<\/div>\n<p id=\"import-auto-id937261\"><strong>Solution for (b)<\/strong><\/p>\n<p id=\"fs-id1916407\">Substituting the knowns,<\/p>\n<div data-type=\"equation\" class=\"equation\">[latex]{\\mu }_{\\text{s}}=\\frac{\\left(\\text{25.0 m\/s}{\\right)}^{2}}{\\left(\\text{500 m}\\right)\\left(9\\text{.}{\\text{80 m\/s}}^{2}\\right)}=0\\text{.}\\text{13}\\text{.}[\/latex]<\/div>\n<p id=\"eip-508\">(Because coefficients of friction are approximate, the answer is given to only two digits.)<\/p>\n<p id=\"import-auto-id1348664\"><strong>Discussion<\/strong><\/p>\n<p id=\"fs-id2397853\">We could also solve part (a) using the first expression in [latex]\\begin{array}{c}{F}_{\\text{c}}=m\\frac{{v}^{2}}{r}\\\\ {F}_{\\text{c}}=\\text{mr}{\\omega }^{2}\\end{array}\\right\\},[\/latex] because [latex]m,[\/latex][latex]v,[\/latex] and [latex]r[\/latex] are given. The coefficient of friction found in part (b) is much smaller than is typically found between tires and roads. The car will still negotiate the curve if the coefficient is greater than 0.13, because static friction is a responsive force, being able to assume a value less than but no more than [latex]{\\mu }_{\\text{s}}N[\/latex]. A higher coefficient would also allow the car to negotiate the curve at a higher speed, but if the coefficient of friction is less, the safe speed would be less than 25 m\/s. Note that mass cancels, implying that in this example, it does not matter how heavily loaded the car is to negotiate the turn. Mass cancels because friction is assumed proportional to the normal force, which in turn is proportional to mass. If the surface of the road were banked, the normal force would be less as will be discussed below.<\/p>\n<\/div>\n<p id=\"import-auto-id2602826\">\n<\/p><div class=\"bc-figure figure\" id=\"import-auto-id3184285\">\n<div class=\"bc-figcaption figcaption\">This car on level ground is moving away and turning to the left. The centripetal force causing the car to turn in a circular path is due to friction between the tires and the road. A minimum coefficient of friction is needed, or the car will move in a larger-radius curve and leave the roadway.<\/div>\n<p><span data-type=\"media\" id=\"import-auto-id3017370\" data-alt=\"In the given figure, a car is shown from the back, which is turning to the left. The weight, w, of the car is shown with a down arrow and N with an up arrow at the back of the car. At the right rear wheel, centripetal force is shown along with its equation formula in a leftward horizontal arrow. The free-body diagram shows three vectors, one upward, depicting N, one downward, depicting w, and one leftward, depicting centripetal force.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/Figure_07_03_02a.jpg\" data-media-type=\"image\/jpg\" alt=\"In the given figure, a car is shown from the back, which is turning to the left. The weight, w, of the car is shown with a down arrow and N with an up arrow at the back of the car. At the right rear wheel, centripetal force is shown along with its equation formula in a leftward horizontal arrow. The free-body diagram shows three vectors, one upward, depicting N, one downward, depicting w, and one leftward, depicting centripetal force.\" width=\"300\"><\/span><\/p><\/div>\n<p id=\"import-auto-id3200990\">Let us now consider <span data-type=\"term\" id=\"import-auto-id2403430\">banked curves<\/span>, where the slope of the road helps you negotiate the curve. See <a href=\"#import-auto-id1571739\" class=\"autogenerated-content\">(Figure)<\/a>. The greater the angle [latex]\\theta [\/latex], the faster you can take the curve. Race tracks for bikes as well as cars, for example, often have steeply banked curves. In an \u201cideally banked curve,\u201d the angle [latex]\\theta [\/latex] is such that you can negotiate the curve at a certain speed without the aid of friction between the tires and the road. We will derive an expression for [latex]\\theta [\/latex] for an ideally banked curve and consider an example related to it.<\/p>\n<p id=\"import-auto-id2441377\">For <span data-type=\"term\" id=\"import-auto-id1402139\">ideal banking<\/span>, the net external force equals the horizontal centripetal force in the absence of friction. The components of the normal force N in the horizontal and vertical directions must equal the centripetal force and the weight of the car, respectively. In cases in which forces are not parallel, it is most convenient to consider components along perpendicular axes\u2014in this case, the vertical and horizontal directions.<\/p>\n<p id=\"import-auto-id1580876\"><a href=\"#import-auto-id1571739\" class=\"autogenerated-content\">(Figure)<\/a> shows a free body diagram for a car on a frictionless banked curve. If the angle [latex]\\theta [\/latex] is ideal for the speed and radius, then the net external force will equal the necessary centripetal force. The only two external forces acting on the car are its weight [latex]\\mathbf{w}[\/latex] and the normal force of the road [latex]\\mathbf{N}[\/latex]. (A frictionless surface can only exert a force perpendicular to the surface\u2014that is, a normal force.) These two forces must add to give a net external force that is horizontal toward the center of curvature and has magnitude [latex]{\\text{mv}}^{2}\\text{\/r}[\/latex]. Because this is the crucial force and it is horizontal, we use a coordinate system with vertical and horizontal axes. Only the normal force has a horizontal component, and so this must equal the centripetal force\u2014that is,<\/p>\n<div data-type=\"equation\" class=\"equation\">[latex]N\\phantom{\\rule{0.25em}{0ex}}\\text{sin}\\phantom{\\rule{0.25em}{0ex}}\\theta =\\frac{{\\mathrm{mv}}^{2}}{r}\\text{.}[\/latex]<\/div>\n<p id=\"import-auto-id3305945\">Because the car does not leave the surface of the road, the net vertical force must be zero, meaning that the vertical components of the two external forces must be equal in magnitude and opposite in direction. From the figure, we see that the vertical component of the normal force is [latex]N\\phantom{\\rule{0.25em}{0ex}}\\text{cos}\\phantom{\\rule{0.25em}{0ex}}\\theta [\/latex], and the only other vertical force is the car\u2019s weight. These must be equal in magnitude; thus,<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-125\">[latex]N\\phantom{\\rule{0.25em}{0ex}}\\text{cos}\\phantom{\\rule{0.25em}{0ex}}\\theta =\\text{mg}\\text{.}[\/latex]<\/div>\n<p id=\"import-auto-id2598289\">Now we can combine the last two equations to eliminate [latex]N[\/latex] and get an expression for [latex]\\theta [\/latex], as desired. Solving the second equation for [latex]N=\\text{mg}\/\\left(\\text{cos}\\phantom{\\rule{0.25em}{0ex}}\\theta \\right)[\/latex], and substituting this into the first yields<\/p>\n<div data-type=\"equation\" class=\"equation\">[latex]\\text{mg}\\frac{\\text{sin}\\phantom{\\rule{0.25em}{0ex}}\\theta }{\\text{cos}\\phantom{\\rule{0.25em}{0ex}}\\theta }=\\frac{{\\text{mv}}^{2}}{r}[\/latex]<\/div>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-285\">[latex]\\begin{array}{rrr}\\text{mg}\\phantom{\\rule{0.25em}{0ex}}\\text{tan}\\left(\\theta \\right)&amp; =&amp; \\frac{{\\mathrm{mv}}^{2}}{r}\\\\ \\text{tan}\\phantom{\\rule{0.25em}{0ex}}\\theta &amp; =&amp; \\frac{{v}^{2}}{\\text{rg.}}\\end{array}[\/latex]<\/div>\n<p id=\"import-auto-id1562443\">Taking the inverse tangent gives<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-239\">[latex]\\theta ={\\text{tan}}^{-1}\\left(\\frac{{v}^{2}}{\\text{rg}}\\right)\\phantom{\\rule{0.25em}{0ex}}\\text{(ideally banked curve, no friction).}[\/latex]<\/div>\n<p id=\"import-auto-id2667857\">This expression can be understood by considering how [latex]\\theta [\/latex] depends on [latex]v[\/latex] and [latex]r[\/latex]. A large [latex]\\theta [\/latex] will be obtained for a large [latex]v[\/latex] and a small [latex]r[\/latex]. That is, roads must be steeply banked for high speeds and sharp curves. Friction helps, because it allows you to take the curve at greater or lower speed than if the curve is frictionless. Note that [latex]\\theta [\/latex] does not depend on the mass of the vehicle.<\/p>\n<div class=\"bc-figure figure\" id=\"import-auto-id1571739\">\n<div class=\"bc-figcaption figcaption\">The car on this banked curve is moving away and turning to the left.<\/div>\n<p><span data-type=\"media\" id=\"import-auto-id2456078\" data-alt=\"In this figure, a car from the backside is shown, turning to the left, on a slope angling downward to the left. A point in the middle of the back of the car is shown which shows one downward vector depicting weight, w, and an upward arrow depicting force N, which is a linear line along the car and is at an angle theta with the straight up arrow. The slope is at an angle theta with the horizontal surface below the slope. The force values, N multipliy sine theta equals to centripetal force, the net force on the car and N cosine theta equal to w are given below the car.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/Figure_07_03_03a.jpg\" data-media-type=\"image\/jpg\" alt=\"In this figure, a car from the backside is shown, turning to the left, on a slope angling downward to the left. A point in the middle of the back of the car is shown which shows one downward vector depicting weight, w, and an upward arrow depicting force N, which is a linear line along the car and is at an angle theta with the straight up arrow. The slope is at an angle theta with the horizontal surface below the slope. The force values, N multipliy sine theta equals to centripetal force, the net force on the car and N cosine theta equal to w are given below the car.\" width=\"300\"><\/span><\/p><\/div>\n<div data-type=\"example\" class=\"textbox examples\" id=\"fs-id982366\">\n<div data-type=\"title\" class=\"title\">What Is the Ideal Speed to Take a Steeply Banked Tight Curve?<\/div>\n<p id=\"import-auto-id3062348\">Curves on some test tracks and race courses, such as the Daytona International Speedway in Florida, are very steeply banked. This banking, with the aid of tire friction and very stable car configurations, allows the curves to be taken at very high speed. To illustrate, calculate the speed at which a 100 m radius curve banked at 65.0\u00b0 should be driven if the road is frictionless.<\/p>\n<p id=\"import-auto-id1927008\"><strong>Strategy<\/strong><\/p>\n<p id=\"import-auto-id3123411\">We first note that all terms in the expression for the ideal angle of a banked curve except for speed are known; thus, we need only rearrange it so that speed appears on the left-hand side and then substitute known quantities.<\/p>\n<p id=\"import-auto-id3484355\"><strong>Solution<\/strong><\/p>\n<p id=\"import-auto-id2392096\">Starting with<\/p>\n<div data-type=\"equation\" class=\"equation\">[latex]\\text{tan}\\phantom{\\rule{0.25em}{0ex}}\\theta =\\frac{{v}^{2}}{\\text{rg}}[\/latex]<\/div>\n<p id=\"import-auto-id2397201\">we get<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-6\">[latex]v=\\left(\\text{rg}\\phantom{\\rule{0.25em}{0ex}}\\text{tan}\\phantom{\\rule{0.25em}{0ex}}\\theta {\\right)}^{1\/2}\\text{.}[\/latex]<\/div>\n<p id=\"import-auto-id3375232\">Noting that tan 65.0\u00ba = 2.14, we obtain<\/p>\n<div data-type=\"equation\" class=\"equation\">[latex]\\begin{array}{lll}v&amp; =&amp; {\\left[\\left(\\text{100 m}\\right)\\left(9.80 m{\\text{\/s}}^{2}\\right)\\left(2\\text{.}\\text{14}\\right)\\right]}^{1\/2}\\\\ &amp; =&amp; \\text{45.8 m\/s.}\\end{array}[\/latex]<\/div>\n<p id=\"import-auto-id1995411\"><strong>Discussion<\/strong><\/p>\n<p id=\"fs-id1967548\">This is just about 165 km\/h, consistent with a very steeply banked and rather sharp curve. Tire friction enables a vehicle to take the curve at significantly higher speeds.<\/p>\n<p id=\"import-auto-id3007417\">Calculations similar to those in the preceding examples can be performed for a host of interesting situations in which centripetal force is involved\u2014a number of these are presented in this chapter\u2019s Problems and Exercises.<\/p>\n<\/div>\n<div data-type=\"note\" class=\"note\" data-has-label=\"true\" id=\"fs-id1398255\" data-label=\"\">\n<div data-type=\"title\" class=\"title\">Take-Home Experiment<\/div>\n<p id=\"import-auto-id1446432\">Ask a friend or relative to swing a golf club or a tennis racquet. Take appropriate measurements to estimate the centripetal acceleration of the end of the club or racquet. You may choose to do this in slow motion.<\/p>\n<\/div>\n<div data-type=\"note\" class=\"note\" data-has-label=\"true\" id=\"eip-52\" data-label=\"\">\n<div data-type=\"title\" class=\"title\">PhET Explorations: Gravity and Orbits<\/div>\n<p id=\"eip-id1169737830697\">Move the sun, earth, moon and space station to see how it affects their gravitational forces and orbital paths. Visualize the sizes and distances between different heavenly bodies, and turn off gravity to see what would happen without it!<\/p>\n<div class=\"bc-figure figure\" id=\"eip-id1751967\">\n<div class=\"bc-figcaption figcaption\"><a href=\"\/resources\/70d6df9fd891a2e8ed242506e51efa0b65c52ea8\/gravity-and-orbits_en.jar\">Gravity and Orbits<\/a><\/div>\n<p><span data-type=\"media\" id=\"Phet_module_7.3\" data-alt=\"\"><a href=\"\/resources\/70d6df9fd891a2e8ed242506e51efa0b65c52ea8\/gravity-and-orbits_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-id2604525\">\n<h1 data-type=\"title\">Section Summary<\/h1>\n<ul id=\"fs-id1416963\">\n<li id=\"import-auto-id1381455\">Centripetal force [latex]{\\text{F}}_{\\text{c}}[\/latex] is any force causing uniform circular motion. It is a \u201ccenter-seeking\u201d force that always points toward the center of rotation. It is perpendicular to linear velocity [latex]v[\/latex]<em data-effect=\"italics\"> and has magnitude<br>\n    [latex]\\phantom{\\rule{0.25em}{0ex}}{F}_{\\text{c}}={\\text{ma}}_{\\text{c}}\\text{,}[\/latex]<\/em><\/li>\n<\/ul>\n<\/div>\n<p id=\"import-auto-id2950190\">which can also be expressed as<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-967\">[latex]\\begin{array}{c}{F}_{\\text{c}}=m\\frac{{v}^{2}}{r}\\\\ \\begin{array}{}\\text{or}\\\\ {F}_{\\text{c}}=\\text{mr}{\\omega }^{2}\\end{array}\\end{array},}[\/latex]<\/div>\n<div class=\"conceptual-questions\" data-depth=\"1\" id=\"fs-id1977486\" data-element-type=\"conceptual-questions\">\n<h1 data-type=\"title\">Conceptual Questions<\/h1>\n<div data-type=\"exercise\" class=\"exercise\" data-element-type=\"conceptual-questions\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id2405338\">\n<p id=\"fs-id1903962\">If you wish to reduce the stress (which is related to centripetal force) on high-speed tires, would you use large- or small-diameter tires? Explain.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id3055510\" data-element-type=\"conceptual-questions\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id2449871\">\n<p id=\"fs-id1452850\">Define centripetal force. Can any type of force (for example, tension, gravitational force, friction, and so on) be a centripetal force? Can any combination of forces be a centripetal force?<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id3035023\" data-element-type=\"conceptual-questions\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id2447888\">\n<p id=\"fs-id1332111\">If centripetal force is directed toward the center, why do you feel that you are \u2018thrown\u2019 away from the center as a car goes around a curve? Explain.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id3286004\" data-element-type=\"conceptual-questions\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id2402563\">\n<p id=\"import-auto-id2655220\">Race car drivers routinely cut corners as shown in <a href=\"#import-auto-id2673863\" class=\"autogenerated-content\">(Figure)<\/a>. Explain how this allows the curve to be taken at the greatest speed.<\/p>\n<div class=\"bc-figure figure\" id=\"import-auto-id2673863\">\n<div class=\"bc-figcaption figcaption\">Two paths around a race track curve are shown. Race car drivers will take the inside path (called cutting the corner) whenever possible because it allows them to take the curve at the highest speed.<\/div>\n<p><span data-type=\"media\" id=\"import-auto-id2588311\" data-alt=\"In the figure, two paths are shown inside a race track through a steep curve, approximately equal to ninety degrees. Two cars are shown. One car is on the path one, which is the inside path along the track. The path of this car is shown with an arrow through the inside path. The second car is shown overtaking the first car, while taking a left turn, showing it to be crossing into the inside path from the second path. The path of this car is also shown with an arrow throughout.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/Figure_07_03_04a.jpg\" data-media-type=\"image\/jpg\" alt=\"In the figure, two paths are shown inside a race track through a steep curve, approximately equal to ninety degrees. Two cars are shown. One car is on the path one, which is the inside path along the track. The path of this car is shown with an arrow through the inside path. The second car is shown overtaking the first car, while taking a left turn, showing it to be crossing into the inside path from the second path. The path of this car is also shown with an arrow throughout.\" width=\"200\"><\/span><\/p><\/div>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id2405334\" data-element-type=\"conceptual-questions\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id2684442\">\n<p id=\"import-auto-id1824608\">A number of amusement parks have rides that make vertical loops like the one shown in <a href=\"#import-auto-id2452691\" class=\"autogenerated-content\">(Figure)<\/a>. For safety, the cars are attached to the rails in such a way that they cannot fall off. If the car goes over the top at just the right speed, gravity alone will supply the centripetal force. What other force acts and what is its direction if:<\/p>\n<p id=\"fs-id2382029\">(a) The car goes over the top at faster than this speed?<\/p>\n<p id=\"fs-id3064329\">(b)The car goes over the top at slower than this speed?<\/p>\n<div class=\"bc-figure figure\" id=\"import-auto-id2452691\">\n<div class=\"bc-figcaption figcaption\">Amusement rides with a vertical loop are an example of a form of curved motion.<\/div>\n<p><span data-type=\"media\" id=\"import-auto-id3026136\" data-alt=\"In the given line diagram, a circular amusement ride is shown from the front with a boat having four people seated in it going upward from the left to the right. The ride starts from the left in a horizontal direction, then goes upward, then turns sideways to the left, comes down from the right and moves horizontal to the right and then ends. It looks like a single knot of a thread, viewed from sideways. Some square iron blocks are also shown below the ride path.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/Figure_07_03_05a.jpg\" data-media-type=\"image\/jpg\" alt=\"In the given line diagram, a circular amusement ride is shown from the front with a boat having four people seated in it going upward from the left to the right. The ride starts from the left in a horizontal direction, then goes upward, then turns sideways to the left, comes down from the right and moves horizontal to the right and then ends. It looks like a single knot of a thread, viewed from sideways. Some square iron blocks are also shown below the ride path.\" width=\"300\"><\/span><\/p><\/div>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id1186294\" data-element-type=\"conceptual-questions\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id1845036\">\n<p id=\"fs-id2654826\">What is the direction of the force exerted by the car on the passenger as the car goes over the top of the amusement ride pictured in <a href=\"#import-auto-id2452691\" class=\"autogenerated-content\">(Figure)<\/a> under the following circumstances:<\/p>\n<p id=\"fs-id3016908\">(a) The car goes over the top at such a speed that the gravitational force is the only force acting?<\/p>\n<p id=\"fs-id2669590\">(b) The car goes over the top faster than this speed?<\/p>\n<p id=\"fs-id982124\">(c) The car goes over the top slower than this speed?<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id3011795\" data-element-type=\"conceptual-questions\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id1276085\">\n<p id=\"fs-id3037465\">As a skater forms a circle, what force is responsible for making her turn? Use a free body diagram in your answer.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" data-element-type=\"conceptual-questions\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id3064151\">\n<p id=\"fs-id2392046\">Suppose a child is riding on a merry-go-round at a distance about halfway between its center and edge. She has a lunch box resting on wax paper, so that there is very little friction between it and the merry-go-round. Which path shown in <a href=\"#import-auto-id2990767\" class=\"autogenerated-content\">(Figure)<\/a> will the lunch box take when she lets go? The lunch box leaves a trail in the dust on the merry-go-round. Is that trail straight, curved to the left, or curved to the right? Explain your answer.<\/p>\n<div class=\"bc-figure figure\" id=\"import-auto-id2990767\">\n<div class=\"bc-figcaption figcaption\">A child riding on a merry-go-round releases her lunch box at point P. This is a view from above the clockwise rotation. Assuming it slides with negligible friction, will it follow path A, B, or C, as viewed from Earth\u2019s frame of reference? What will be the shape of the path it leaves in the dust on the merry-go-round?<\/div>\n<p><span data-type=\"media\" id=\"import-auto-id2658360\" data-alt=\"The given figure shows the circular base of a merry-go-round, whose angular velocity is clockwise, shown here with an arrow. A single horse is shown on whom a child is sitting, with a vertical line shown passed through her, which goes from the bottom of the merry-go-round to the top of it. A point P is shown alongside the horse, through which three arrows in downward three directions are shown which depicts the three possible path of the fall of the lunch box.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/Figure_07_03_06a.jpg\" data-media-type=\"image\/jpg\" alt=\"The given figure shows the circular base of a merry-go-round, whose angular velocity is clockwise, shown here with an arrow. A single horse is shown on whom a child is sitting, with a vertical line shown passed through her, which goes from the bottom of the merry-go-round to the top of it. A point P is shown alongside the horse, through which three arrows in downward three directions are shown which depicts the three possible path of the fall of the lunch box.\" width=\"300\"><\/span><\/p><\/div>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id2404351\" data-element-type=\"conceptual-questions\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id2660428\">\n<p id=\"import-auto-id693773\">Do you feel yourself thrown to either side when you negotiate a curve that is ideally banked for your car\u2019s speed? What is the direction of the force exerted on you by the car seat?<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id3340905\" data-element-type=\"conceptual-questions\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id2530716\">\n<p id=\"import-auto-id3224639\">Suppose a mass is moving in a circular path on a frictionless table as shown in figure. In the Earth\u2019s frame of reference, there is no centrifugal force pulling the mass away from the centre of rotation, yet there is a very real force stretching the string attaching the mass to the nail. Using concepts related to centripetal force and Newton\u2019s third law, explain what force stretches the string, identifying its physical origin.<\/p>\n<div class=\"bc-figure figure\" id=\"import-auto-id2435986\">\n<div class=\"bc-figcaption figcaption\">A mass attached to a nail on a frictionless table moves in a circular path. The force stretching the string is real and not fictional. What is the physical origin of the force on the string?<\/div>\n<p><span data-type=\"media\" id=\"import-auto-id2671045\" data-alt=\"In the figure a table is shown. On the table a mass is attached to a nail at the center with the help of a string. The mass is moving on a circular path in counterclockwise direction.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/Figure_07_03_07a.jpg\" data-media-type=\"image\/jpg\" alt=\"In the figure a table is shown. On the table a mass is attached to a nail at the center with the help of a string. The mass is moving on a circular path in counterclockwise direction.\" width=\"250\"><\/span><\/p><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"problems-exercises\" data-depth=\"1\" id=\"fs-id2687518\">\n<h1 data-type=\"title\">Problems Exercise<\/h1>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id3245705\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id3035351\">\n<p id=\"import-auto-id2424366\">(a) A 22.0 kg child is riding a playground merry-go-round that is rotating at 40.0 rev\/min. What centripetal force must she exert to stay on if she is 1.25 m from its center? <\/p>\n<p id=\"import-auto-id789255\">(b) What centripetal force does she need to stay on an amusement park merry-go-round that rotates at 3.00 rev\/min if she is 8.00 m from its center? <\/p>\n<p id=\"import-auto-id2659195\">(c) Compare each force with her weight.<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"fs-id3025606\" data-element-type=\"problems-exercises\">\n<p id=\"import-auto-id2950375\">a) 483 N<\/p>\n<p id=\"import-auto-id2590108\">b) 17.4 N<\/p>\n<p id=\"import-auto-id1355232\">c) 2.24 times her weight, 0.0807 times her weight<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id1586925\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id3006657\">\n<p id=\"import-auto-id2437110\">Calculate the centripetal force on the end of a 100 m (radius) wind turbine blade that is rotating at 0.5 rev\/s. Assume the mass is 4 kg.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id1864125\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id3451983\">\n<p id=\"import-auto-id1900542\">What is the ideal banking angle for a gentle turn of 1.20 km radius on a highway with a 105 km\/h speed limit (about 65 mi\/h), assuming everyone travels at the limit?<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"fs-id3192173\">\n<p id=\"fs-id1920896\">[latex]4\\text{.}\\text{14\u00ba}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id2009808\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id2662619\">\n<p id=\"fs-id1849074\">What is the ideal speed to take a 100 m radius curve banked at a 20.0\u00b0 angle?<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"eip-id1167458029802\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"eip-id1167457015961\">\n<p id=\"fs-id2054959\">(a) What is the radius of a bobsled turn banked at 75.0\u00b0 and taken at 30.0 m\/s, assuming it is ideally banked?<\/p>\n<p id=\"fs-id2663295\">(b) Calculate the centripetal acceleration.<\/p>\n<p id=\"fs-id2682824\">(c) Does this acceleration seem large to you?<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"fs-id2053936\">\n<p id=\"import-auto-id1921665\">a) 24.6 m<\/p>\n<p id=\"import-auto-id3110349\">b) [latex]36.6 m\/{\\text{s}}^{2}[\/latex]<\/p>\n<p id=\"import-auto-id1917809\">c) [latex]{a}_{\\text{c}}=3.73\\phantom{\\rule{0.25em}{0ex}}\\mathrm{g.}[\/latex]<br>\nThis does not seem too large, but it is clear that bobsledders feel a lot of force on them going through sharply banked turns.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id2992610\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id2953850\">\n<p id=\"import-auto-id1272129\">Part of riding a bicycle involves leaning at the correct angle when making a turn, as seen in <a href=\"#import-auto-id2682310\" class=\"autogenerated-content\">(Figure)<\/a>. To be stable, the force exerted by the ground must be on a line going through the center of gravity. The force on the bicycle wheel can be resolved into two perpendicular components\u2014friction parallel to the road (this must supply the centripetal force), and the vertical normal force (which must equal the system\u2019s weight).<\/p>\n<p id=\"fs-id2032385\">(a) Show that [latex]\\theta [\/latex] (as defined in the figure) is related to the speed [latex]v[\/latex] and radius of curvature [latex]r[\/latex] of the turn in the same way as for an ideally banked roadway\u2014that is,<br>\n[latex]\\theta ={\\text{tan}}^{\\text{\u20131}}{v}^{2}\/\\mathrm{rg}[\/latex]<\/p>\n<p id=\"import-auto-id2973190\">(b) Calculate [latex]\\theta [\/latex] for a 12.0 m\/s turn of radius 30.0 m (as in a race).<\/p>\n<div class=\"bc-figure figure\" id=\"import-auto-id2682310\">\n<div class=\"bc-figcaption figcaption\">A bicyclist negotiating a turn on level ground must lean at the correct angle\u2014the ability to do this becomes instinctive. The force of the ground on the wheel needs to be on a line through the center of gravity. The net external force on the system is the centripetal force. The vertical component of the force on the wheel cancels the weight of the system while its horizontal component must supply the centripetal force. This process produces a relationship among the angle [latex]\\theta [\/latex], the speed [latex]v[\/latex], and the radius of curvature [latex]r[\/latex] of the turn similar to that for the ideal banking of roadways.<\/div>\n<p><span data-type=\"media\" id=\"import-auto-id3259740\" data-alt=\"The given figure shows a boy riding a bicycle, from the front. The boy is sliding leftward to his left. Three vectors are shown. One is from the bottom the front cycle wheel to the right depicting centripetal force, another one is from the same point drawn vertically upward showing the force N, making an angle theta with the slope of the front cycle wheel. The third vector is drawn from the chest of the boy to vertically downward to the bottom showing his weight, w. An arrow from the bottom of the wheel to the chest point of the boy is also shown depicting the slope of the bicycle with force F exerting on it. A free-body diagram is also given alongside the figure showing the direction of weight and force vectors. And the values of net F equals to sum of N and centripetal force, and N equals to weight W also given alongside on the right.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/Figure_07_03_08a.jpg\" data-media-type=\"image\/jpg\" alt=\"The given figure shows a boy riding a bicycle, from the front. The boy is sliding leftward to his left. Three vectors are shown. One is from the bottom the front cycle wheel to the right depicting centripetal force, another one is from the same point drawn vertically upward showing the force N, making an angle theta with the slope of the front cycle wheel. The third vector is drawn from the chest of the boy to vertically downward to the bottom showing his weight, w. An arrow from the bottom of the wheel to the chest point of the boy is also shown depicting the slope of the bicycle with force F exerting on it. A free-body diagram is also given alongside the figure showing the direction of weight and force vectors. And the values of net F equals to sum of N and centripetal force, and N equals to weight W also given alongside on the right.\" width=\"300\"><\/span><\/p><\/div>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id3424499\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id1282324\">\n<p id=\"import-auto-id3253194\">A large centrifuge, like the one shown in <a href=\"#import-auto-id3099006\" class=\"autogenerated-content\">(Figure)<\/a>(a), is used to expose aspiring astronauts to accelerations similar to those experienced in rocket launches and atmospheric reentries.<\/p>\n<p id=\"fs-id2929873\">(a) At what angular velocity is the centripetal acceleration <\/p>\n[latex]\\mathrm{10}\\phantom{\\rule{0.25em}{0ex}}g[\/latex]\n<p>if the rider is 15.0 m from the center of rotation?<\/p>\n<p id=\"fs-id1431970\">(b) The rider\u2019s cage hangs on a pivot at the end of the arm, allowing it to swing outward during rotation as shown in <a href=\"#import-auto-id3099006\" class=\"autogenerated-content\">(Figure)<\/a>(b). At what angle [latex]\\theta [\/latex] below the horizontal will the cage hang when the centripetal acceleration is [latex]\\mathrm{10}\\phantom{\\rule{0.25em}{0ex}}g[\/latex]  ? (Hint: The arm supplies centripetal force and supports the weight of the cage. Draw a free body diagram of the forces to see what the angle [latex]\\theta [\/latex] should be.)<\/p>\n<p id=\"import-auto-id3163736\">\n<\/p><div class=\"bc-figure figure\" id=\"import-auto-id3099006\">\n<div class=\"bc-figcaption figcaption\">(a) NASA centrifuge used to subject trainees to accelerations similar to those experienced in rocket launches and reentries. (credit: NASA) (b) Rider in cage showing how the cage pivots outward during rotation. This allows the total force exerted on the rider by the cage to be along its axis at all times.<\/div>\n<p><span data-type=\"media\" id=\"import-auto-id1932760\" data-alt=\"Figure a shows a NASA centrifuge n a big hall. In figure b, there is a girl sitting in the cage of the centrifuge. The centripetal force on the cage is directed toward left. The direction of the weight of the cage is downward and the force on the arm is directed in north-west direction.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/Figure_07_03_09a.jpg\" data-media-type=\"image\/jpg\" alt=\"Figure a shows a NASA centrifuge n a big hall. In figure b, there is a girl sitting in the cage of the centrifuge. The centripetal force on the cage is directed toward left. The direction of the weight of the cage is downward and the force on the arm is directed in north-west direction.\" width=\"300\"><\/span><\/p><\/div>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"fs-id1344086\" data-element-type=\"problems-exercises\">\n<p id=\"import-auto-id2659571\">a) 2.56 rad\/s<\/p>\n<p id=\"import-auto-id2658565\">b) [latex]5.71\u00ba[\/latex]<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id2963096\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id2677993\">\n<p id=\"import-auto-id1572171\"><strong>Integrated Concepts<\/strong><\/p>\n<p id=\"import-auto-id3006132\">If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of the curve (a real problem on icy mountain roads). (a) Calculate the ideal speed to take a 100 m radius curve banked at 15.0\u00ba. (b) What is the minimum coefficient of friction needed for a frightened driver to take the same curve at 20.0 km\/h?<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"fs-id1870875\" data-element-type=\"problems-exercises\">\n<p id=\"import-auto-id1986122\">a) 16.2 m\/s<\/p>\n<p id=\"import-auto-id2669620\">b) 0.234<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id1933865\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id2673689\">\n<p id=\"import-auto-id1916232\">Modern roller coasters have vertical loops like the one shown in <a href=\"#import-auto-id2514816\" class=\"autogenerated-content\">(Figure)<\/a>. The radius of curvature is smaller at the top than on the sides so that the downward centripetal acceleration at the top will be greater than the acceleration due to gravity, keeping the passengers pressed firmly into their seats. What is the speed of the roller coaster at the top of the loop if the radius of curvature there is 15.0 m and the downward acceleration of the car is 1.50 g?<\/p>\n<div class=\"bc-figure figure\" id=\"import-auto-id2514816\">\n<div class=\"bc-figcaption figcaption\">Teardrop-shaped loops are used in the latest roller coasters so that the radius of curvature gradually decreases to a minimum at the top. This means that the centripetal acceleration builds from zero to a maximum at the top and gradually decreases again. A circular loop would cause a jolting change in acceleration at entry, a disadvantage discovered long ago in railroad curve design. With a small radius of curvature at the top, the centripetal acceleration can more easily be kept greater than [latex]g[\/latex]  so that the passengers do not lose contact with their seats nor do they need seat belts to keep them in place.<\/div>\n<p><span data-type=\"media\" id=\"import-auto-id2681688\" data-alt=\"A teardrop shaped loop of a roller coaster is shown. The car of the roller coaster starts from the point A near the right of the base and covers the teardrop portion of the roller coaster and move to a point D at the left of base. Near the top of tear drop portion an upward arrow is shown labeled as r-minimum. Also at a point near the base toward A there is a label called r-maximum. The wire frame of the base is also shown.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/Figure_07_03_10a.jpg\" data-media-type=\"image\/jpg\" alt=\"A teardrop shaped loop of a roller coaster is shown. The car of the roller coaster starts from the point A near the right of the base and covers the teardrop portion of the roller coaster and move to a point D at the left of base. Near the top of tear drop portion an upward arrow is shown labeled as r-minimum. Also at a point near the base toward A there is a label called r-maximum. The wire frame of the base is also shown.\" width=\"250\"><\/span><\/p><\/div>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id2667652\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id1863783\">\n<p id=\"import-auto-id3258600\"><strong>Unreasonable Results<\/strong><\/p>\n<p id=\"import-auto-id2403623\">(a) Calculate the minimum coefficient of friction needed for a car to negotiate an unbanked 50.0 m radius curve at 30.0 m\/s. <\/p>\n<p id=\"import-auto-id1771358\">(b) What is unreasonable about the result? <\/p>\n<p id=\"import-auto-id2663006\">(c) Which premises are unreasonable or inconsistent?<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"fs-id1607940\">\n<p id=\"import-auto-id2677862\">a) 1.84<\/p>\n<p id=\"import-auto-id2661282\">b) A coefficient of friction this much greater than 1 is unreasonable .<\/p>\n<p id=\"import-auto-id3229848\">c) The assumed speed is too great for the tight curve.<\/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-id2658751\">\n<dt>centripetal force<\/dt>\n<dd id=\"fs-id2671027\">any net force causing uniform circular motion<\/dd>\n<\/dl>\n<dl class=\"definition\" id=\"import-auto-id969602\">\n<dt>ideal banking<\/dt>\n<dd id=\"fs-id2673973\">the sloping of a curve in a road, where the angle of the slope allows the vehicle to  negotiate the curve at a certain speed without the aid of friction between the tires and the road; the net external force on the vehicle equals the horizontal centripetal force in the absence of friction<\/dd>\n<\/dl>\n<dl class=\"definition\" id=\"import-auto-id2384610\">\n<dt>ideal speed<\/dt>\n<dd id=\"fs-id1594023\">the maximum safe speed at which a vehicle can turn on a curve without the aid of friction between the tire and the road<\/dd>\n<\/dl>\n<dl class=\"definition\" id=\"import-auto-id3149885\">\n<dt>ideal angle<\/dt>\n<dd id=\"fs-id3225958\">the angle at which a car can turn safely on a steep curve, which is in proportion to the ideal speed<\/dd>\n<\/dl>\n<dl class=\"definition\" id=\"import-auto-id3013678\">\n<dt>banked curve<\/dt>\n<dd id=\"fs-id3148861\"> the curve in a road that is sloping in a manner that helps a vehicle negotiate the curve<\/dd>\n<\/dl>\n<\/div>\n\n","rendered":"<div class=\"textbox learning-objectives\">\n<h3 itemprop=\"educationalUse\">Learning Objectives<\/h3>\n<ul>\n<li>Calculate coefficient of friction on a car tire.<\/li>\n<li>Calculate ideal speed and angle of a car on a turn.<\/li>\n<\/ul>\n<\/div>\n<p id=\"import-auto-id2589409\">Any force or combination of forces can cause a centripetal or radial acceleration. Just a few examples are the tension in the rope on a tether ball, the force of Earth\u2019s gravity on the Moon, friction between roller skates and a rink floor, a banked roadway\u2019s force on a car, and forces on the tube of a spinning centrifuge.<\/p>\n<p id=\"import-auto-id3025916\">Any net force causing uniform circular motion is called a <span data-type=\"term\" id=\"import-auto-id3450370\">centripetal force<\/span>. The direction of a centripetal force is toward the center of curvature, the same as the direction of centripetal acceleration. According to Newton\u2019s second law of motion, net force is mass times acceleration: net <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-96b2771b0a1356454b961795069eec69_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#70;&#125;&#61;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#97;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"59\" style=\"vertical-align: -1px;\" \/>. For uniform circular motion, the acceleration is the centripetal acceleration\u2014 <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-ae5d440bf38a469768d66136841983cd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#97;&#61;&#123;&#97;&#125;&#95;&#123;&#99;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"11\" width=\"48\" style=\"vertical-align: -3px;\" \/>. Thus, the magnitude of centripetal force <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-bafe46f2d283f128c3c9f354652f0ecd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#70;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#99;&#125;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"18\" style=\"vertical-align: -3px;\" \/> is<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-696\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-14686f442c32d41cc8f5b963d812db01_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#70;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#99;&#125;&#125;&#61;&#123;&#109;&#92;&#116;&#101;&#120;&#116;&#123;&#97;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#99;&#125;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"77\" style=\"vertical-align: -3px;\" \/><\/div>\n<p id=\"import-auto-id3397174\">By using the expressions for centripetal acceleration <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-f4bcdfb66caba0f995b123987e74d778_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#97;&#125;&#95;&#123;&#99;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"11\" width=\"15\" style=\"vertical-align: -3px;\" \/> from <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-0af14d2a66ca6561ed479f30ec322990_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#97;&#125;&#95;&#123;&#99;&#125;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#123;&#118;&#125;&#94;&#123;&#50;&#125;&#125;&#123;&#114;&#125;&#59;&#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;&#97;&#125;&#95;&#123;&#99;&#125;&#61;&#123;&#92;&#109;&#97;&#116;&#104;&#114;&#109;&#123;&#114;&#92;&#111;&#109;&#101;&#103;&#97;&#32;&#125;&#125;&#94;&#123;&#50;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"24\" width=\"135\" style=\"vertical-align: -6px;\" \/>, we get two expressions for the centripetal force <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-bafe46f2d283f128c3c9f354652f0ecd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#70;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#99;&#125;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"18\" style=\"vertical-align: -3px;\" \/> in terms of mass, velocity, angular velocity, and radius of curvature:<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-562\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-3da0d0a592ba3e711375fa7f7848a2b9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#70;&#125;&#95;&#123;&#99;&#125;&#61;&#109;&#92;&#102;&#114;&#97;&#99;&#123;&#123;&#118;&#125;&#94;&#123;&#50;&#125;&#125;&#123;&#114;&#125;&#59;&#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;&#70;&#125;&#95;&#123;&#99;&#125;&#61;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#114;&#125;&#123;&#92;&#111;&#109;&#101;&#103;&#97;&#32;&#125;&#94;&#123;&#50;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"24\" width=\"174\" style=\"vertical-align: -6px;\" \/><\/div>\n<p id=\"import-auto-id2616689\">You may use whichever expression for centripetal force is more convenient. Centripetal force <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-a6fdfe842a80199b5eac17ad242db0d4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#70;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#99;&#125;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"17\" style=\"vertical-align: -3px;\" \/> is always perpendicular to the path and pointing to the center of curvature, because <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-ffd935b21805af3e1447ffc5f6ccb3a4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#109;&#97;&#116;&#104;&#98;&#102;&#123;&#97;&#125;&#125;&#95;&#123;&#99;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"11\" width=\"16\" style=\"vertical-align: -3px;\" \/> is perpendicular to the velocity and pointing to the center of curvature.<\/p>\n<p id=\"import-auto-id3105986\">Note that if you solve the first expression for <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;\" \/>, you get<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-58\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-edfe42c18d0bd0d1c1c60a063b925295_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#114;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#123;&#92;&#109;&#97;&#116;&#104;&#114;&#109;&#123;&#109;&#118;&#125;&#125;&#94;&#123;&#50;&#125;&#125;&#123;&#123;&#70;&#125;&#95;&#123;&#99;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#46;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"26\" width=\"65\" style=\"vertical-align: -8px;\" \/><\/div>\n<p id=\"import-auto-id1410994\">This implies that for a given mass and velocity, a large centripetal force causes a small radius of curvature\u2014that is, a tight curve.<\/p>\n<div class=\"bc-figure figure\" id=\"import-auto-id2601650\">\n<div class=\"bc-figcaption figcaption\">The frictional force supplies the centripetal force and is numerically equal to it. Centripetal force is perpendicular to velocity and causes uniform circular motion. The larger the <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-bafe46f2d283f128c3c9f354652f0ecd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#70;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#99;&#125;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"18\" style=\"vertical-align: -3px;\" \/>, the smaller the radius of curvature <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 the sharper the curve. The second curve has the same <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;\" \/>, but a larger <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-bafe46f2d283f128c3c9f354652f0ecd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#70;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#99;&#125;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"18\" style=\"vertical-align: -3px;\" \/> produces a smaller <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-bfb95134c6ae22035355e84586640522_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#114;&#92;&#112;&#114;&#105;&#109;&#101;&#32;\" title=\"Rendered by QuickLaTeX.com\" height=\"10\" width=\"13\" style=\"vertical-align: 0px;\" \/>.<\/div>\n<p><span data-type=\"media\" id=\"import-auto-id1581195\" data-alt=\"The given figure consists of two semicircles, one over the other. The top semicircle is bigger and the one below is smaller. In both the figures, the direction of the path is given along the semicircle in the counter-clockwise direction. A point is shown on the path, where the radius from the circle, r, is shown with an arrow from the center of the circle. At the same point, the centripetal force is shown in the opposite direction to that of radius arrow. The velocity, v, is shown along this point in the left upward direction and is perpendicular to the force. In both the figures, the velocity is same, but the radius is smaller and centripetal force is larger in the lower figure.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/Figure_07_03_01a.jpg\" data-media-type=\"image\/jpg\" alt=\"The given figure consists of two semicircles, one over the other. The top semicircle is bigger and the one below is smaller. In both the figures, the direction of the path is given along the semicircle in the counter-clockwise direction. A point is shown on the path, where the radius from the circle, r, is shown with an arrow from the center of the circle. At the same point, the centripetal force is shown in the opposite direction to that of radius arrow. The velocity, v, is shown along this point in the left upward direction and is perpendicular to the force. In both the figures, the velocity is same, but the radius is smaller and centripetal force is larger in the lower figure.\" width=\"200\" \/><\/span><\/p>\n<\/div>\n<div data-type=\"example\" class=\"textbox examples\" id=\"fs-id3130274\">\n<div data-type=\"title\" class=\"title\">What Coefficient of Friction Do Car Tires Need on a Flat Curve?<\/div>\n<p id=\"import-auto-id1849101\">(a) Calculate the centripetal force exerted on a 900 kg car that negotiates a 500 m radius curve at 25.0 m\/s. <\/p>\n<p id=\"import-auto-id1409542\">(b) Assuming an unbanked curve, find the minimum static coefficient of friction, between the tires and the road, static friction being the reason that keeps the car from slipping (see <a href=\"#import-auto-id3184285\" class=\"autogenerated-content\">(Figure)<\/a>).<\/p>\n<p id=\"import-auto-id3205483\"><strong>Strategy and Solution for (a)<\/strong><\/p>\n<p id=\"import-auto-id1418337\">We know that<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-3c32237309f5671a7412fcdba00f9cfb_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#70;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#99;&#125;&#125;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#123;&#92;&#109;&#97;&#116;&#104;&#114;&#109;&#123;&#109;&#118;&#125;&#125;&#94;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#50;&#125;&#125;&#125;&#123;&#114;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"24\" width=\"70\" style=\"vertical-align: -6px;\" \/>. 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-ad8ee8a74cddfaf4337cbc1072d52fb3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#70;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#99;&#125;&#125;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#123;&#92;&#109;&#97;&#116;&#104;&#114;&#109;&#123;&#109;&#118;&#125;&#125;&#94;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#50;&#125;&#125;&#125;&#123;&#114;&#125;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#116;&#101;&#120;&#116;&#123;&#57;&#48;&#48;&#32;&#107;&#103;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#116;&#101;&#120;&#116;&#123;&#50;&#53;&#46;&#48;&#32;&#109;&#47;&#115;&#125;&#123;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#125;&#94;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#50;&#125;&#125;&#125;&#123;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#116;&#101;&#120;&#116;&#123;&#53;&#48;&#48;&#32;&#109;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#125;&#61;&#92;&#116;&#101;&#120;&#116;&#123;&#49;&#49;&#50;&#53;&#32;&#78;&#46;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"29\" width=\"301\" style=\"vertical-align: -9px;\" \/><\/div>\n<p id=\"import-auto-id3102728\"><strong>Strategy for (b)<\/strong><\/p>\n<p id=\"fs-id3078905\"><a href=\"#import-auto-id3184285\" class=\"autogenerated-content\">(Figure)<\/a> shows the forces acting on the car on an unbanked (level ground) curve. Friction is to the left, keeping the car from slipping, and because it is the only horizontal force acting on the car, the friction is the centripetal force in this case. We know that the maximum static friction (at which the tires roll but do not slip) is <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-eeef174b4de604a04dba4443b8cbdfff_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#109;&#117;&#32;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#115;&#125;&#125;&#78;\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"33\" style=\"vertical-align: -4px;\" \/>, where <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-cf50b75d4684151cd2ce6adb8041fa14_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#109;&#117;&#32;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#115;&#125;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"16\" style=\"vertical-align: -4px;\" \/> is the static coefficient of friction and N is the normal force. The normal force equals the car\u2019s weight on level ground, so that <\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-ed00da850ef10d4363a3a7bb35701d04_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#78;&#61;&#92;&#109;&#97;&#116;&#104;&#105;&#116;&#123;&#109;&#103;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"63\" style=\"vertical-align: -4px;\" \/>. Thus the centripetal force in this situation 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-bd3594cc2ba109aabb76baa00324e291_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#70;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#99;&#125;&#125;&#61;&#102;&#61;&#123;&#92;&#109;&#117;&#32;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#115;&#125;&#125;&#78;&#61;&#123;&#92;&#109;&#117;&#32;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#115;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#103;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#46;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"177\" style=\"vertical-align: -4px;\" \/><\/div>\n<p id=\"import-auto-id3098784\">Now we have a relationship between centripetal force and the coefficient of friction. Using the first expression for <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-a6fdfe842a80199b5eac17ad242db0d4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#70;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#99;&#125;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"17\" style=\"vertical-align: -3px;\" \/> from the equation<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-735\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-ae62332179e54e6491854cd617b27945_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;&#99;&#125;&#123;&#70;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#99;&#125;&#125;&#61;&#109;&#92;&#102;&#114;&#97;&#99;&#123;&#123;&#118;&#125;&#94;&#123;&#50;&#125;&#125;&#123;&#114;&#125;&#92;&#92;&#32;&#123;&#70;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#99;&#125;&#125;&#61;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#114;&#125;&#123;&#92;&#111;&#109;&#101;&#103;&#97;&#32;&#125;&#94;&#123;&#50;&#125;&#92;&#101;&#110;&#100;&#123;&#97;&#114;&#114;&#97;&#121;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#92;&#125;&#44;\" title=\"Rendered by QuickLaTeX.com\" height=\"42\" width=\"95\" style=\"vertical-align: -15px;\" \/><\/div>\n<div data-type=\"equation\" class=\"equation\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-c6295e20fb29b02ca68473e93e730713_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#109;&#92;&#102;&#114;&#97;&#99;&#123;&#123;&#118;&#125;&#94;&#123;&#50;&#125;&#125;&#123;&#114;&#125;&#61;&#123;&#92;&#109;&#117;&#32;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#115;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#103;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#46;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"24\" width=\"102\" style=\"vertical-align: -6px;\" \/><\/div>\n<p id=\"import-auto-id1467842\">We solve this for <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-cf50b75d4684151cd2ce6adb8041fa14_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#109;&#117;&#32;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#115;&#125;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"16\" style=\"vertical-align: -4px;\" \/>, noting that mass cancels, and obtain<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-99\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-ff51c93e8156a6fad52647152467449f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#109;&#117;&#32;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#115;&#125;&#125;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#123;&#118;&#125;&#94;&#123;&#50;&#125;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#114;&#103;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#46;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"27\" width=\"62\" style=\"vertical-align: -9px;\" \/><\/div>\n<p id=\"import-auto-id937261\"><strong>Solution for (b)<\/strong><\/p>\n<p id=\"fs-id1916407\">Substituting the knowns,<\/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-c90bdb7e57448e2216a25348f20a5e8a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#109;&#117;&#32;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#115;&#125;&#125;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#116;&#101;&#120;&#116;&#123;&#50;&#53;&#46;&#48;&#32;&#109;&#47;&#115;&#125;&#123;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#125;&#94;&#123;&#50;&#125;&#125;&#123;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#116;&#101;&#120;&#116;&#123;&#53;&#48;&#48;&#32;&#109;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#92;&#108;&#101;&#102;&#116;&#40;&#57;&#92;&#116;&#101;&#120;&#116;&#123;&#46;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#56;&#48;&#32;&#109;&#47;&#115;&#125;&#125;&#94;&#123;&#50;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#125;&#61;&#48;&#92;&#116;&#101;&#120;&#116;&#123;&#46;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#49;&#51;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#46;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"34\" width=\"224\" style=\"vertical-align: -14px;\" \/><\/div>\n<p id=\"eip-508\">(Because coefficients of friction are approximate, the answer is given to only two digits.)<\/p>\n<p id=\"import-auto-id1348664\"><strong>Discussion<\/strong><\/p>\n<p id=\"fs-id2397853\">We could also solve part (a) using the first expression in <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-ae62332179e54e6491854cd617b27945_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;&#99;&#125;&#123;&#70;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#99;&#125;&#125;&#61;&#109;&#92;&#102;&#114;&#97;&#99;&#123;&#123;&#118;&#125;&#94;&#123;&#50;&#125;&#125;&#123;&#114;&#125;&#92;&#92;&#32;&#123;&#70;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#99;&#125;&#125;&#61;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#114;&#125;&#123;&#92;&#111;&#109;&#101;&#103;&#97;&#32;&#125;&#94;&#123;&#50;&#125;&#92;&#101;&#110;&#100;&#123;&#97;&#114;&#114;&#97;&#121;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#92;&#125;&#44;\" title=\"Rendered by QuickLaTeX.com\" height=\"42\" width=\"95\" style=\"vertical-align: -15px;\" \/> because <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-77c771aa52eccc9f4f8878d25d110b84_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#109;&#44;\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"20\" style=\"vertical-align: -4px;\" \/>[latex]v,[\/latex] and <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;\" \/> are given. The coefficient of friction found in part (b) is much smaller than is typically found between tires and roads. The car will still negotiate the curve if the coefficient is greater than 0.13, because static friction is a responsive force, being able to assume a value less than but no more than <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-eeef174b4de604a04dba4443b8cbdfff_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#109;&#117;&#32;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#115;&#125;&#125;&#78;\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"33\" style=\"vertical-align: -4px;\" \/>. A higher coefficient would also allow the car to negotiate the curve at a higher speed, but if the coefficient of friction is less, the safe speed would be less than 25 m\/s. Note that mass cancels, implying that in this example, it does not matter how heavily loaded the car is to negotiate the turn. Mass cancels because friction is assumed proportional to the normal force, which in turn is proportional to mass. If the surface of the road were banked, the normal force would be less as will be discussed below.<\/p>\n<\/div>\n<p id=\"import-auto-id2602826\">\n<div class=\"bc-figure figure\" id=\"import-auto-id3184285\">\n<div class=\"bc-figcaption figcaption\">This car on level ground is moving away and turning to the left. The centripetal force causing the car to turn in a circular path is due to friction between the tires and the road. A minimum coefficient of friction is needed, or the car will move in a larger-radius curve and leave the roadway.<\/div>\n<p><span data-type=\"media\" id=\"import-auto-id3017370\" data-alt=\"In the given figure, a car is shown from the back, which is turning to the left. The weight, w, of the car is shown with a down arrow and N with an up arrow at the back of the car. At the right rear wheel, centripetal force is shown along with its equation formula in a leftward horizontal arrow. The free-body diagram shows three vectors, one upward, depicting N, one downward, depicting w, and one leftward, depicting centripetal force.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/Figure_07_03_02a.jpg\" data-media-type=\"image\/jpg\" alt=\"In the given figure, a car is shown from the back, which is turning to the left. The weight, w, of the car is shown with a down arrow and N with an up arrow at the back of the car. At the right rear wheel, centripetal force is shown along with its equation formula in a leftward horizontal arrow. The free-body diagram shows three vectors, one upward, depicting N, one downward, depicting w, and one leftward, depicting centripetal force.\" width=\"300\" \/><\/span><\/p>\n<\/div>\n<p id=\"import-auto-id3200990\">Let us now consider <span data-type=\"term\" id=\"import-auto-id2403430\">banked curves<\/span>, where the slope of the road helps you negotiate the curve. See <a href=\"#import-auto-id1571739\" class=\"autogenerated-content\">(Figure)<\/a>. The greater 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;\" \/>, the faster you can take the curve. Race tracks for bikes as well as cars, for example, often have steeply banked curves. In an \u201cideally banked curve,\u201d 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 such that you can negotiate the curve at a certain speed without the aid of friction between the tires and the road. We will derive an expression 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;\" \/> for an ideally banked curve and consider an example related to it.<\/p>\n<p id=\"import-auto-id2441377\">For <span data-type=\"term\" id=\"import-auto-id1402139\">ideal banking<\/span>, the net external force equals the horizontal centripetal force in the absence of friction. The components of the normal force N in the horizontal and vertical directions must equal the centripetal force and the weight of the car, respectively. In cases in which forces are not parallel, it is most convenient to consider components along perpendicular axes\u2014in this case, the vertical and horizontal directions.<\/p>\n<p id=\"import-auto-id1580876\"><a href=\"#import-auto-id1571739\" class=\"autogenerated-content\">(Figure)<\/a> shows a free body diagram for a car on a frictionless banked curve. If 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 ideal for the speed and radius, then the net external force will equal the necessary centripetal force. The only two external forces acting on the car are its weight <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-098fe4fc91886b0f8da407e07e59a15f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#109;&#97;&#116;&#104;&#98;&#102;&#123;&#119;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"9\" width=\"15\" style=\"vertical-align: 0px;\" \/> and the normal force of the road <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-df519733808cec3d8d8df3ad2682a59c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#109;&#97;&#116;&#104;&#98;&#102;&#123;&#78;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"14\" width=\"16\" style=\"vertical-align: -1px;\" \/>. (A frictionless surface can only exert a force perpendicular to the surface\u2014that is, a normal force.) These two forces must add to give a net external force that is horizontal toward the center of curvature and has magnitude <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-2d034e694aa1c58b191aef8ff734c5b0_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#118;&#125;&#125;&#94;&#123;&#50;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#47;&#114;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"47\" style=\"vertical-align: -4px;\" \/>. Because this is the crucial force and it is horizontal, we use a coordinate system with vertical and horizontal axes. Only the normal force has a horizontal component, and so this must equal the centripetal force\u2014that 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-6903934596a89ed6d5e0891820edeb92_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#78;&#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;&#115;&#105;&#110;&#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;&#104;&#101;&#116;&#97;&#32;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#123;&#92;&#109;&#97;&#116;&#104;&#114;&#109;&#123;&#109;&#118;&#125;&#125;&#94;&#123;&#50;&#125;&#125;&#123;&#114;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#46;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"24\" width=\"112\" style=\"vertical-align: -6px;\" \/><\/div>\n<p id=\"import-auto-id3305945\">Because the car does not leave the surface of the road, the net vertical force must be zero, meaning that the vertical components of the two external forces must be equal in magnitude and opposite in direction. From the figure, we see that the vertical component of the normal force is <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-55d64ab6967c1a5d1367a8daf0b7b599_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#78;&#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;&#99;&#111;&#115;&#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;&#104;&#101;&#116;&#97;&#32;\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"58\" style=\"vertical-align: 0px;\" \/>, and the only other vertical force is the car\u2019s weight. These must be equal in magnitude; thus,<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-125\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-3f0fd7f5faa2219f116c2b1585ef4d39_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#78;&#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;&#99;&#111;&#115;&#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;&#104;&#101;&#116;&#97;&#32;&#61;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#103;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#46;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"109\" style=\"vertical-align: -3px;\" \/><\/div>\n<p id=\"import-auto-id2598289\">Now we can combine the last two equations to eliminate <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-5793832f979c2268e3694c246d53b1bb_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#78;\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"16\" style=\"vertical-align: 0px;\" \/> and get an expression 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;\" \/>, as desired. Solving the second equation for <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-db5eba7466f60cca7e3988a1b0f945d2_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#78;&#61;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#103;&#125;&#47;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#116;&#101;&#120;&#116;&#123;&#99;&#111;&#115;&#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;&#104;&#101;&#116;&#97;&#32;&#92;&#114;&#105;&#103;&#104;&#116;&#41;\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"125\" style=\"vertical-align: -5px;\" \/>, and substituting this into the first 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-a7fbfe0c83613a5f74fdc752ee31c4b0_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#103;&#125;&#92;&#102;&#114;&#97;&#99;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#115;&#105;&#110;&#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;&#104;&#101;&#116;&#97;&#32;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#99;&#111;&#115;&#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;&#104;&#101;&#116;&#97;&#32;&#125;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#118;&#125;&#125;&#94;&#123;&#50;&#125;&#125;&#123;&#114;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"24\" width=\"109\" style=\"vertical-align: -6px;\" \/><\/div>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-285\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-50c2265a98d93fa9865522494dfa953b_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;&#114;&#114;&#114;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#103;&#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;&#116;&#97;&#110;&#125;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#116;&#104;&#101;&#116;&#97;&#32;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#38;&#32;&#61;&#38;&#32;&#92;&#102;&#114;&#97;&#99;&#123;&#123;&#92;&#109;&#97;&#116;&#104;&#114;&#109;&#123;&#109;&#118;&#125;&#125;&#94;&#123;&#50;&#125;&#125;&#123;&#114;&#125;&#92;&#92;&#32;&#92;&#116;&#101;&#120;&#116;&#123;&#116;&#97;&#110;&#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;&#104;&#101;&#116;&#97;&#32;&#38;&#32;&#61;&#38;&#32;&#92;&#102;&#114;&#97;&#99;&#123;&#123;&#118;&#125;&#94;&#123;&#50;&#125;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#114;&#103;&#46;&#125;&#125;&#92;&#101;&#110;&#100;&#123;&#97;&#114;&#114;&#97;&#121;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"51\" width=\"154\" style=\"vertical-align: -21px;\" \/><\/div>\n<p id=\"import-auto-id1562443\">Taking the inverse tangent gives<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-239\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-066992cc09940d247de4ecee22a91ed5_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#104;&#101;&#116;&#97;&#32;&#61;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#116;&#97;&#110;&#125;&#125;&#94;&#123;&#45;&#49;&#125;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#102;&#114;&#97;&#99;&#123;&#123;&#118;&#125;&#94;&#123;&#50;&#125;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#114;&#103;&#125;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#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;&#40;&#105;&#100;&#101;&#97;&#108;&#108;&#121;&#32;&#98;&#97;&#110;&#107;&#101;&#100;&#32;&#99;&#117;&#114;&#118;&#101;&#44;&#32;&#110;&#111;&#32;&#102;&#114;&#105;&#99;&#116;&#105;&#111;&#110;&#41;&#46;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"33\" width=\"394\" style=\"vertical-align: -12px;\" \/><\/div>\n<p id=\"import-auto-id2667857\">This expression can be understood by considering how <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;\" \/> depends on <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-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;\" \/>. A large <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;\" \/> will be obtained for a large <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 a small <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;\" \/>. That is, roads must be steeply banked for high speeds and sharp curves. Friction helps, because it allows you to take the curve at greater or lower speed than if the curve is frictionless. Note that <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;\" \/> does not depend on the mass of the vehicle.<\/p>\n<div class=\"bc-figure figure\" id=\"import-auto-id1571739\">\n<div class=\"bc-figcaption figcaption\">The car on this banked curve is moving away and turning to the left.<\/div>\n<p><span data-type=\"media\" id=\"import-auto-id2456078\" data-alt=\"In this figure, a car from the backside is shown, turning to the left, on a slope angling downward to the left. A point in the middle of the back of the car is shown which shows one downward vector depicting weight, w, and an upward arrow depicting force N, which is a linear line along the car and is at an angle theta with the straight up arrow. The slope is at an angle theta with the horizontal surface below the slope. The force values, N multipliy sine theta equals to centripetal force, the net force on the car and N cosine theta equal to w are given below the car.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/Figure_07_03_03a.jpg\" data-media-type=\"image\/jpg\" alt=\"In this figure, a car from the backside is shown, turning to the left, on a slope angling downward to the left. A point in the middle of the back of the car is shown which shows one downward vector depicting weight, w, and an upward arrow depicting force N, which is a linear line along the car and is at an angle theta with the straight up arrow. The slope is at an angle theta with the horizontal surface below the slope. The force values, N multipliy sine theta equals to centripetal force, the net force on the car and N cosine theta equal to w are given below the car.\" width=\"300\" \/><\/span><\/p>\n<\/div>\n<div data-type=\"example\" class=\"textbox examples\" id=\"fs-id982366\">\n<div data-type=\"title\" class=\"title\">What Is the Ideal Speed to Take a Steeply Banked Tight Curve?<\/div>\n<p id=\"import-auto-id3062348\">Curves on some test tracks and race courses, such as the Daytona International Speedway in Florida, are very steeply banked. This banking, with the aid of tire friction and very stable car configurations, allows the curves to be taken at very high speed. To illustrate, calculate the speed at which a 100 m radius curve banked at 65.0\u00b0 should be driven if the road is frictionless.<\/p>\n<p id=\"import-auto-id1927008\"><strong>Strategy<\/strong><\/p>\n<p id=\"import-auto-id3123411\">We first note that all terms in the expression for the ideal angle of a banked curve except for speed are known; thus, we need only rearrange it so that speed appears on the left-hand side and then substitute known quantities.<\/p>\n<p id=\"import-auto-id3484355\"><strong>Solution<\/strong><\/p>\n<p id=\"import-auto-id2392096\">Starting with<\/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-2d319a97083ef5ba5b7fdd39538f88ae_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#101;&#120;&#116;&#123;&#116;&#97;&#110;&#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;&#104;&#101;&#116;&#97;&#32;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#123;&#118;&#125;&#94;&#123;&#50;&#125;&#125;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#114;&#103;&#125;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"27\" width=\"78\" style=\"vertical-align: -9px;\" \/><\/div>\n<p id=\"import-auto-id2397201\">we get<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-6\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-01b6339aaa2f1697053eefebb5c36232_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#118;&#61;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#116;&#101;&#120;&#116;&#123;&#114;&#103;&#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;&#116;&#97;&#110;&#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;&#104;&#101;&#116;&#97;&#32;&#123;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#125;&#94;&#123;&#49;&#47;&#50;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#46;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"23\" width=\"134\" style=\"vertical-align: -4px;\" \/><\/div>\n<p id=\"import-auto-id3375232\">Noting that tan 65.0\u00ba = 2.14, 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-2b1d26c937e0d3a0d3afa08d188205db_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;&#118;&#38;&#32;&#61;&#38;&#32;&#123;&#92;&#108;&#101;&#102;&#116;&#91;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#116;&#101;&#120;&#116;&#123;&#49;&#48;&#48;&#32;&#109;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#92;&#108;&#101;&#102;&#116;&#40;&#57;&#46;&#56;&#48;&#32;&#109;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#47;&#115;&#125;&#125;&#94;&#123;&#50;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#92;&#108;&#101;&#102;&#116;&#40;&#50;&#92;&#116;&#101;&#120;&#116;&#123;&#46;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#49;&#52;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#92;&#114;&#105;&#103;&#104;&#116;&#93;&#125;&#94;&#123;&#49;&#47;&#50;&#125;&#92;&#92;&#32;&#38;&#32;&#61;&#38;&#32;&#92;&#116;&#101;&#120;&#116;&#123;&#52;&#53;&#46;&#56;&#32;&#109;&#47;&#115;&#46;&#125;&#92;&#101;&#110;&#100;&#123;&#97;&#114;&#114;&#97;&#121;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"57\" width=\"297\" style=\"vertical-align: -22px;\" \/><\/div>\n<p id=\"import-auto-id1995411\"><strong>Discussion<\/strong><\/p>\n<p id=\"fs-id1967548\">This is just about 165 km\/h, consistent with a very steeply banked and rather sharp curve. Tire friction enables a vehicle to take the curve at significantly higher speeds.<\/p>\n<p id=\"import-auto-id3007417\">Calculations similar to those in the preceding examples can be performed for a host of interesting situations in which centripetal force is involved\u2014a number of these are presented in this chapter\u2019s Problems and Exercises.<\/p>\n<\/div>\n<div data-type=\"note\" class=\"note\" data-has-label=\"true\" id=\"fs-id1398255\" data-label=\"\">\n<div data-type=\"title\" class=\"title\">Take-Home Experiment<\/div>\n<p id=\"import-auto-id1446432\">Ask a friend or relative to swing a golf club or a tennis racquet. Take appropriate measurements to estimate the centripetal acceleration of the end of the club or racquet. You may choose to do this in slow motion.<\/p>\n<\/div>\n<div data-type=\"note\" class=\"note\" data-has-label=\"true\" id=\"eip-52\" data-label=\"\">\n<div data-type=\"title\" class=\"title\">PhET Explorations: Gravity and Orbits<\/div>\n<p id=\"eip-id1169737830697\">Move the sun, earth, moon and space station to see how it affects their gravitational forces and orbital paths. Visualize the sizes and distances between different heavenly bodies, and turn off gravity to see what would happen without it!<\/p>\n<div class=\"bc-figure figure\" id=\"eip-id1751967\">\n<div class=\"bc-figcaption figcaption\"><a href=\"\/resources\/70d6df9fd891a2e8ed242506e51efa0b65c52ea8\/gravity-and-orbits_en.jar\">Gravity and Orbits<\/a><\/div>\n<p><span data-type=\"media\" id=\"Phet_module_7.3\" data-alt=\"\"><a href=\"\/resources\/70d6df9fd891a2e8ed242506e51efa0b65c52ea8\/gravity-and-orbits_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-id2604525\">\n<h1 data-type=\"title\">Section Summary<\/h1>\n<ul id=\"fs-id1416963\">\n<li id=\"import-auto-id1381455\">Centripetal force <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-bafe46f2d283f128c3c9f354652f0ecd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#70;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#99;&#125;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"18\" style=\"vertical-align: -3px;\" \/> is any force causing uniform circular motion. It is a \u201ccenter-seeking\u201d force that always points toward the center of rotation. It is perpendicular to linear 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;\" \/><em data-effect=\"italics\"> and has magnitude<br \/>\n    <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-653380ebb3b5b2fa1ed351957864cec3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#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;&#70;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#99;&#125;&#125;&#61;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#97;&#125;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#99;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#44;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"77\" style=\"vertical-align: -3px;\" \/><\/em><\/li>\n<\/ul>\n<\/div>\n<p id=\"import-auto-id2950190\">which can also be expressed as<\/p>\n<div data-type=\"equation\" class=\"equation\" id=\"eip-967\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-fc5ae3588864f0cabd206d4e2025280a_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;&#99;&#125;&#123;&#70;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#99;&#125;&#125;&#61;&#109;&#92;&#102;&#114;&#97;&#99;&#123;&#123;&#118;&#125;&#94;&#123;&#50;&#125;&#125;&#123;&#114;&#125;&#92;&#92;&#32;&#92;&#98;&#101;&#103;&#105;&#110;&#123;&#97;&#114;&#114;&#97;&#121;&#125;&#123;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#111;&#114;&#125;&#92;&#92;&#32;&#123;&#70;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#99;&#125;&#125;&#61;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#114;&#125;&#123;&#92;&#111;&#109;&#101;&#103;&#97;&#32;&#125;&#94;&#123;&#50;&#125;&#92;&#101;&#110;&#100;&#123;&#97;&#114;&#114;&#97;&#121;&#125;&#92;&#101;&#110;&#100;&#123;&#97;&#114;&#114;&#97;&#121;&#125;&#44;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"64\" width=\"95\" style=\"vertical-align: -26px;\" \/><\/div>\n<div class=\"conceptual-questions\" data-depth=\"1\" id=\"fs-id1977486\" data-element-type=\"conceptual-questions\">\n<h1 data-type=\"title\">Conceptual Questions<\/h1>\n<div data-type=\"exercise\" class=\"exercise\" data-element-type=\"conceptual-questions\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id2405338\">\n<p id=\"fs-id1903962\">If you wish to reduce the stress (which is related to centripetal force) on high-speed tires, would you use large- or small-diameter tires? Explain.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id3055510\" data-element-type=\"conceptual-questions\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id2449871\">\n<p id=\"fs-id1452850\">Define centripetal force. Can any type of force (for example, tension, gravitational force, friction, and so on) be a centripetal force? Can any combination of forces be a centripetal force?<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id3035023\" data-element-type=\"conceptual-questions\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id2447888\">\n<p id=\"fs-id1332111\">If centripetal force is directed toward the center, why do you feel that you are \u2018thrown\u2019 away from the center as a car goes around a curve? Explain.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id3286004\" data-element-type=\"conceptual-questions\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id2402563\">\n<p id=\"import-auto-id2655220\">Race car drivers routinely cut corners as shown in <a href=\"#import-auto-id2673863\" class=\"autogenerated-content\">(Figure)<\/a>. Explain how this allows the curve to be taken at the greatest speed.<\/p>\n<div class=\"bc-figure figure\" id=\"import-auto-id2673863\">\n<div class=\"bc-figcaption figcaption\">Two paths around a race track curve are shown. Race car drivers will take the inside path (called cutting the corner) whenever possible because it allows them to take the curve at the highest speed.<\/div>\n<p><span data-type=\"media\" id=\"import-auto-id2588311\" data-alt=\"In the figure, two paths are shown inside a race track through a steep curve, approximately equal to ninety degrees. Two cars are shown. One car is on the path one, which is the inside path along the track. The path of this car is shown with an arrow through the inside path. The second car is shown overtaking the first car, while taking a left turn, showing it to be crossing into the inside path from the second path. The path of this car is also shown with an arrow throughout.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/Figure_07_03_04a.jpg\" data-media-type=\"image\/jpg\" alt=\"In the figure, two paths are shown inside a race track through a steep curve, approximately equal to ninety degrees. Two cars are shown. One car is on the path one, which is the inside path along the track. The path of this car is shown with an arrow through the inside path. The second car is shown overtaking the first car, while taking a left turn, showing it to be crossing into the inside path from the second path. The path of this car is also shown with an arrow throughout.\" width=\"200\" \/><\/span><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id2405334\" data-element-type=\"conceptual-questions\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id2684442\">\n<p id=\"import-auto-id1824608\">A number of amusement parks have rides that make vertical loops like the one shown in <a href=\"#import-auto-id2452691\" class=\"autogenerated-content\">(Figure)<\/a>. For safety, the cars are attached to the rails in such a way that they cannot fall off. If the car goes over the top at just the right speed, gravity alone will supply the centripetal force. What other force acts and what is its direction if:<\/p>\n<p id=\"fs-id2382029\">(a) The car goes over the top at faster than this speed?<\/p>\n<p id=\"fs-id3064329\">(b)The car goes over the top at slower than this speed?<\/p>\n<div class=\"bc-figure figure\" id=\"import-auto-id2452691\">\n<div class=\"bc-figcaption figcaption\">Amusement rides with a vertical loop are an example of a form of curved motion.<\/div>\n<p><span data-type=\"media\" id=\"import-auto-id3026136\" data-alt=\"In the given line diagram, a circular amusement ride is shown from the front with a boat having four people seated in it going upward from the left to the right. The ride starts from the left in a horizontal direction, then goes upward, then turns sideways to the left, comes down from the right and moves horizontal to the right and then ends. It looks like a single knot of a thread, viewed from sideways. Some square iron blocks are also shown below the ride path.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/Figure_07_03_05a.jpg\" data-media-type=\"image\/jpg\" alt=\"In the given line diagram, a circular amusement ride is shown from the front with a boat having four people seated in it going upward from the left to the right. The ride starts from the left in a horizontal direction, then goes upward, then turns sideways to the left, comes down from the right and moves horizontal to the right and then ends. It looks like a single knot of a thread, viewed from sideways. Some square iron blocks are also shown below the ride path.\" width=\"300\" \/><\/span><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id1186294\" data-element-type=\"conceptual-questions\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id1845036\">\n<p id=\"fs-id2654826\">What is the direction of the force exerted by the car on the passenger as the car goes over the top of the amusement ride pictured in <a href=\"#import-auto-id2452691\" class=\"autogenerated-content\">(Figure)<\/a> under the following circumstances:<\/p>\n<p id=\"fs-id3016908\">(a) The car goes over the top at such a speed that the gravitational force is the only force acting?<\/p>\n<p id=\"fs-id2669590\">(b) The car goes over the top faster than this speed?<\/p>\n<p id=\"fs-id982124\">(c) The car goes over the top slower than this speed?<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id3011795\" data-element-type=\"conceptual-questions\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id1276085\">\n<p id=\"fs-id3037465\">As a skater forms a circle, what force is responsible for making her turn? Use a free body diagram in your answer.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" data-element-type=\"conceptual-questions\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id3064151\">\n<p id=\"fs-id2392046\">Suppose a child is riding on a merry-go-round at a distance about halfway between its center and edge. She has a lunch box resting on wax paper, so that there is very little friction between it and the merry-go-round. Which path shown in <a href=\"#import-auto-id2990767\" class=\"autogenerated-content\">(Figure)<\/a> will the lunch box take when she lets go? The lunch box leaves a trail in the dust on the merry-go-round. Is that trail straight, curved to the left, or curved to the right? Explain your answer.<\/p>\n<div class=\"bc-figure figure\" id=\"import-auto-id2990767\">\n<div class=\"bc-figcaption figcaption\">A child riding on a merry-go-round releases her lunch box at point P. This is a view from above the clockwise rotation. Assuming it slides with negligible friction, will it follow path A, B, or C, as viewed from Earth\u2019s frame of reference? What will be the shape of the path it leaves in the dust on the merry-go-round?<\/div>\n<p><span data-type=\"media\" id=\"import-auto-id2658360\" data-alt=\"The given figure shows the circular base of a merry-go-round, whose angular velocity is clockwise, shown here with an arrow. A single horse is shown on whom a child is sitting, with a vertical line shown passed through her, which goes from the bottom of the merry-go-round to the top of it. A point P is shown alongside the horse, through which three arrows in downward three directions are shown which depicts the three possible path of the fall of the lunch box.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/Figure_07_03_06a.jpg\" data-media-type=\"image\/jpg\" alt=\"The given figure shows the circular base of a merry-go-round, whose angular velocity is clockwise, shown here with an arrow. A single horse is shown on whom a child is sitting, with a vertical line shown passed through her, which goes from the bottom of the merry-go-round to the top of it. A point P is shown alongside the horse, through which three arrows in downward three directions are shown which depicts the three possible path of the fall of the lunch box.\" width=\"300\" \/><\/span><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id2404351\" data-element-type=\"conceptual-questions\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id2660428\">\n<p id=\"import-auto-id693773\">Do you feel yourself thrown to either side when you negotiate a curve that is ideally banked for your car\u2019s speed? What is the direction of the force exerted on you by the car seat?<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id3340905\" data-element-type=\"conceptual-questions\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id2530716\">\n<p id=\"import-auto-id3224639\">Suppose a mass is moving in a circular path on a frictionless table as shown in figure. In the Earth\u2019s frame of reference, there is no centrifugal force pulling the mass away from the centre of rotation, yet there is a very real force stretching the string attaching the mass to the nail. Using concepts related to centripetal force and Newton\u2019s third law, explain what force stretches the string, identifying its physical origin.<\/p>\n<div class=\"bc-figure figure\" id=\"import-auto-id2435986\">\n<div class=\"bc-figcaption figcaption\">A mass attached to a nail on a frictionless table moves in a circular path. The force stretching the string is real and not fictional. What is the physical origin of the force on the string?<\/div>\n<p><span data-type=\"media\" id=\"import-auto-id2671045\" data-alt=\"In the figure a table is shown. On the table a mass is attached to a nail at the center with the help of a string. The mass is moving on a circular path in counterclockwise direction.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/Figure_07_03_07a.jpg\" data-media-type=\"image\/jpg\" alt=\"In the figure a table is shown. On the table a mass is attached to a nail at the center with the help of a string. The mass is moving on a circular path in counterclockwise direction.\" width=\"250\" \/><\/span><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"problems-exercises\" data-depth=\"1\" id=\"fs-id2687518\">\n<h1 data-type=\"title\">Problems Exercise<\/h1>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id3245705\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id3035351\">\n<p id=\"import-auto-id2424366\">(a) A 22.0 kg child is riding a playground merry-go-round that is rotating at 40.0 rev\/min. What centripetal force must she exert to stay on if she is 1.25 m from its center? <\/p>\n<p id=\"import-auto-id789255\">(b) What centripetal force does she need to stay on an amusement park merry-go-round that rotates at 3.00 rev\/min if she is 8.00 m from its center? <\/p>\n<p id=\"import-auto-id2659195\">(c) Compare each force with her weight.<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"fs-id3025606\" data-element-type=\"problems-exercises\">\n<p id=\"import-auto-id2950375\">a) 483 N<\/p>\n<p id=\"import-auto-id2590108\">b) 17.4 N<\/p>\n<p id=\"import-auto-id1355232\">c) 2.24 times her weight, 0.0807 times her weight<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id1586925\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id3006657\">\n<p id=\"import-auto-id2437110\">Calculate the centripetal force on the end of a 100 m (radius) wind turbine blade that is rotating at 0.5 rev\/s. Assume the mass is 4 kg.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id1864125\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id3451983\">\n<p id=\"import-auto-id1900542\">What is the ideal banking angle for a gentle turn of 1.20 km radius on a highway with a 105 km\/h speed limit (about 65 mi\/h), assuming everyone travels at the limit?<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"fs-id3192173\">\n<p id=\"fs-id1920896\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-03d972c9aed1994af0dc822e027ee8d1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#52;&#92;&#116;&#101;&#120;&#116;&#123;&#46;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#49;&#52;&ordm;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"32\" style=\"vertical-align: -1px;\" \/><\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id2009808\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id2662619\">\n<p id=\"fs-id1849074\">What is the ideal speed to take a 100 m radius curve banked at a 20.0\u00b0 angle?<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"eip-id1167458029802\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"eip-id1167457015961\">\n<p id=\"fs-id2054959\">(a) What is the radius of a bobsled turn banked at 75.0\u00b0 and taken at 30.0 m\/s, assuming it is ideally banked?<\/p>\n<p id=\"fs-id2663295\">(b) Calculate the centripetal acceleration.<\/p>\n<p id=\"fs-id2682824\">(c) Does this acceleration seem large to you?<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"fs-id2053936\">\n<p id=\"import-auto-id1921665\">a) 24.6 m<\/p>\n<p id=\"import-auto-id3110349\">b) <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-43ed57742b0cfaf1f31768d77313121c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#51;&#54;&#46;&#54;&#32;&#109;&#47;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#115;&#125;&#125;&#94;&#123;&#50;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"70\" style=\"vertical-align: -5px;\" \/><\/p>\n<p id=\"import-auto-id1917809\">c) <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-f0e673ea15b449a9561829c54117a1d6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#97;&#125;&#95;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#99;&#125;&#125;&#61;&#51;&#46;&#55;&#51;&#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;&#109;&#97;&#116;&#104;&#114;&#109;&#123;&#103;&#46;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"89\" style=\"vertical-align: -3px;\" \/><br \/>\nThis does not seem too large, but it is clear that bobsledders feel a lot of force on them going through sharply banked turns.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id2992610\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id2953850\">\n<p id=\"import-auto-id1272129\">Part of riding a bicycle involves leaning at the correct angle when making a turn, as seen in <a href=\"#import-auto-id2682310\" class=\"autogenerated-content\">(Figure)<\/a>. To be stable, the force exerted by the ground must be on a line going through the center of gravity. The force on the bicycle wheel can be resolved into two perpendicular components\u2014friction parallel to the road (this must supply the centripetal force), and the vertical normal force (which must equal the system\u2019s weight).<\/p>\n<p id=\"fs-id2032385\">(a) Show that <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;\" \/> (as defined in the figure) is related to the speed <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 radius of curvature <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;\" \/> of the turn in the same way as for an ideally banked roadway\u2014that is,<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-702df8f8d6e7a209aa7847fafddb0963_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#116;&#104;&#101;&#116;&#97;&#32;&#61;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#116;&#97;&#110;&#125;&#125;&#94;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#45;&#49;&#125;&#125;&#123;&#118;&#125;&#94;&#123;&#50;&#125;&#47;&#92;&#109;&#97;&#116;&#104;&#114;&#109;&#123;&#114;&#103;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"112\" style=\"vertical-align: -5px;\" \/><\/p>\n<p id=\"import-auto-id2973190\">(b) Calculate <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;\" \/> for a 12.0 m\/s turn of radius 30.0 m (as in a race).<\/p>\n<div class=\"bc-figure figure\" id=\"import-auto-id2682310\">\n<div class=\"bc-figcaption figcaption\">A bicyclist negotiating a turn on level ground must lean at the correct angle\u2014the ability to do this becomes instinctive. The force of the ground on the wheel needs to be on a line through the center of gravity. The net external force on the system is the centripetal force. The vertical component of the force on the wheel cancels the weight of the system while its horizontal component must supply the centripetal force. This process produces a relationship among 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;\" \/>, the speed <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 the radius of curvature <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;\" \/> of the turn similar to that for the ideal banking of roadways.<\/div>\n<p><span data-type=\"media\" id=\"import-auto-id3259740\" data-alt=\"The given figure shows a boy riding a bicycle, from the front. The boy is sliding leftward to his left. Three vectors are shown. One is from the bottom the front cycle wheel to the right depicting centripetal force, another one is from the same point drawn vertically upward showing the force N, making an angle theta with the slope of the front cycle wheel. The third vector is drawn from the chest of the boy to vertically downward to the bottom showing his weight, w. An arrow from the bottom of the wheel to the chest point of the boy is also shown depicting the slope of the bicycle with force F exerting on it. A free-body diagram is also given alongside the figure showing the direction of weight and force vectors. And the values of net F equals to sum of N and centripetal force, and N equals to weight W also given alongside on the right.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/Figure_07_03_08a.jpg\" data-media-type=\"image\/jpg\" alt=\"The given figure shows a boy riding a bicycle, from the front. The boy is sliding leftward to his left. Three vectors are shown. One is from the bottom the front cycle wheel to the right depicting centripetal force, another one is from the same point drawn vertically upward showing the force N, making an angle theta with the slope of the front cycle wheel. The third vector is drawn from the chest of the boy to vertically downward to the bottom showing his weight, w. An arrow from the bottom of the wheel to the chest point of the boy is also shown depicting the slope of the bicycle with force F exerting on it. A free-body diagram is also given alongside the figure showing the direction of weight and force vectors. And the values of net F equals to sum of N and centripetal force, and N equals to weight W also given alongside on the right.\" width=\"300\" \/><\/span><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id3424499\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id1282324\">\n<p id=\"import-auto-id3253194\">A large centrifuge, like the one shown in <a href=\"#import-auto-id3099006\" class=\"autogenerated-content\">(Figure)<\/a>(a), is used to expose aspiring astronauts to accelerations similar to those experienced in rocket launches and atmospheric reentries.<\/p>\n<p id=\"fs-id2929873\">(a) At what angular velocity is the centripetal acceleration <\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-8ee2c1b46fdb25b1e605ca2991b06b27_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#109;&#97;&#116;&#104;&#114;&#109;&#123;&#49;&#48;&#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;&#103;\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"30\" style=\"vertical-align: -4px;\" \/><\/p>\n<p>if the rider is 15.0 m from the center of rotation?<\/p>\n<p id=\"fs-id1431970\">(b) The rider\u2019s cage hangs on a pivot at the end of the arm, allowing it to swing outward during rotation as shown in <a href=\"#import-auto-id3099006\" class=\"autogenerated-content\">(Figure)<\/a>(b). At what 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;\" \/> below the horizontal will the cage hang when the centripetal acceleration is <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-8ee2c1b46fdb25b1e605ca2991b06b27_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#109;&#97;&#116;&#104;&#114;&#109;&#123;&#49;&#48;&#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;&#103;\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"30\" style=\"vertical-align: -4px;\" \/>  ? (Hint: The arm supplies centripetal force and supports the weight of the cage. Draw a free body diagram of the forces to see what 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;\" \/> should be.)<\/p>\n<p id=\"import-auto-id3163736\">\n<div class=\"bc-figure figure\" id=\"import-auto-id3099006\">\n<div class=\"bc-figcaption figcaption\">(a) NASA centrifuge used to subject trainees to accelerations similar to those experienced in rocket launches and reentries. (credit: NASA) (b) Rider in cage showing how the cage pivots outward during rotation. This allows the total force exerted on the rider by the cage to be along its axis at all times.<\/div>\n<p><span data-type=\"media\" id=\"import-auto-id1932760\" data-alt=\"Figure a shows a NASA centrifuge n a big hall. In figure b, there is a girl sitting in the cage of the centrifuge. The centripetal force on the cage is directed toward left. The direction of the weight of the cage is downward and the force on the arm is directed in north-west direction.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/Figure_07_03_09a.jpg\" data-media-type=\"image\/jpg\" alt=\"Figure a shows a NASA centrifuge n a big hall. In figure b, there is a girl sitting in the cage of the centrifuge. The centripetal force on the cage is directed toward left. The direction of the weight of the cage is downward and the force on the arm is directed in north-west direction.\" width=\"300\" \/><\/span><\/p>\n<\/div>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"fs-id1344086\" data-element-type=\"problems-exercises\">\n<p id=\"import-auto-id2659571\">a) 2.56 rad\/s<\/p>\n<p id=\"import-auto-id2658565\">b) <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-8058433ce46180b988e4f7c6ead3d3bd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#53;&#46;&#55;&#49;&ordm;\" title=\"Rendered by QuickLaTeX.com\" height=\"14\" width=\"31\" style=\"vertical-align: -1px;\" \/><\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id2963096\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id2677993\">\n<p id=\"import-auto-id1572171\"><strong>Integrated Concepts<\/strong><\/p>\n<p id=\"import-auto-id3006132\">If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of the curve (a real problem on icy mountain roads). (a) Calculate the ideal speed to take a 100 m radius curve banked at 15.0\u00ba. (b) What is the minimum coefficient of friction needed for a frightened driver to take the same curve at 20.0 km\/h?<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"fs-id1870875\" data-element-type=\"problems-exercises\">\n<p id=\"import-auto-id1986122\">a) 16.2 m\/s<\/p>\n<p id=\"import-auto-id2669620\">b) 0.234<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id1933865\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id2673689\">\n<p id=\"import-auto-id1916232\">Modern roller coasters have vertical loops like the one shown in <a href=\"#import-auto-id2514816\" class=\"autogenerated-content\">(Figure)<\/a>. The radius of curvature is smaller at the top than on the sides so that the downward centripetal acceleration at the top will be greater than the acceleration due to gravity, keeping the passengers pressed firmly into their seats. What is the speed of the roller coaster at the top of the loop if the radius of curvature there is 15.0 m and the downward acceleration of the car is 1.50 g?<\/p>\n<div class=\"bc-figure figure\" id=\"import-auto-id2514816\">\n<div class=\"bc-figcaption figcaption\">Teardrop-shaped loops are used in the latest roller coasters so that the radius of curvature gradually decreases to a minimum at the top. This means that the centripetal acceleration builds from zero to a maximum at the top and gradually decreases again. A circular loop would cause a jolting change in acceleration at entry, a disadvantage discovered long ago in railroad curve design. With a small radius of curvature at the top, the centripetal acceleration can more easily be kept greater than <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-content\/ql-cache\/quicklatex.com-d208fd391fa57c168dc0f151de829fee_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#103;\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"9\" style=\"vertical-align: -4px;\" \/>  so that the passengers do not lose contact with their seats nor do they need seat belts to keep them in place.<\/div>\n<p><span data-type=\"media\" id=\"import-auto-id2681688\" data-alt=\"A teardrop shaped loop of a roller coaster is shown. The car of the roller coaster starts from the point A near the right of the base and covers the teardrop portion of the roller coaster and move to a point D at the left of base. Near the top of tear drop portion an upward arrow is shown labeled as r-minimum. Also at a point near the base toward A there is a label called r-maximum. The wire frame of the base is also shown.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/clalonde\/wp-content\/uploads\/sites\/280\/2017\/10\/Figure_07_03_10a.jpg\" data-media-type=\"image\/jpg\" alt=\"A teardrop shaped loop of a roller coaster is shown. The car of the roller coaster starts from the point A near the right of the base and covers the teardrop portion of the roller coaster and move to a point D at the left of base. Near the top of tear drop portion an upward arrow is shown labeled as r-minimum. Also at a point near the base toward A there is a label called r-maximum. The wire frame of the base is also shown.\" width=\"250\" \/><\/span><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"exercise\" id=\"fs-id2667652\" data-element-type=\"problems-exercises\">\n<div data-type=\"problem\" class=\"problem\" id=\"fs-id1863783\">\n<p id=\"import-auto-id3258600\"><strong>Unreasonable Results<\/strong><\/p>\n<p id=\"import-auto-id2403623\">(a) Calculate the minimum coefficient of friction needed for a car to negotiate an unbanked 50.0 m radius curve at 30.0 m\/s. <\/p>\n<p id=\"import-auto-id1771358\">(b) What is unreasonable about the result? <\/p>\n<p id=\"import-auto-id2663006\">(c) Which premises are unreasonable or inconsistent?<\/p>\n<\/div>\n<div data-type=\"solution\" class=\"solution\" id=\"fs-id1607940\">\n<p id=\"import-auto-id2677862\">a) 1.84<\/p>\n<p id=\"import-auto-id2661282\">b) A coefficient of friction this much greater than 1 is unreasonable .<\/p>\n<p id=\"import-auto-id3229848\">c) The assumed speed is too great for the tight curve.<\/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-id2658751\">\n<dt>centripetal force<\/dt>\n<dd id=\"fs-id2671027\">any net force causing uniform circular motion<\/dd>\n<\/dl>\n<dl class=\"definition\" id=\"import-auto-id969602\">\n<dt>ideal banking<\/dt>\n<dd id=\"fs-id2673973\">the sloping of a curve in a road, where the angle of the slope allows the vehicle to  negotiate the curve at a certain speed without the aid of friction between the tires and the road; the net external force on the vehicle equals the horizontal centripetal force in the absence of friction<\/dd>\n<\/dl>\n<dl class=\"definition\" id=\"import-auto-id2384610\">\n<dt>ideal speed<\/dt>\n<dd id=\"fs-id1594023\">the maximum safe speed at which a vehicle can turn on a curve without the aid of friction between the tire and the road<\/dd>\n<\/dl>\n<dl class=\"definition\" id=\"import-auto-id3149885\">\n<dt>ideal angle<\/dt>\n<dd id=\"fs-id3225958\">the angle at which a car can turn safely on a steep curve, which is in proportion to the ideal speed<\/dd>\n<\/dl>\n<dl class=\"definition\" id=\"import-auto-id3013678\">\n<dt>banked curve<\/dt>\n<dd id=\"fs-id3148861\"> the curve in a road that is sloping in a manner that helps a vehicle negotiate the curve<\/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-328","chapter","type-chapter","status-publish","hentry","license-all-rights-reserved"],"part":302,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-json\/pressbooks\/v2\/chapters\/328","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\/328\/revisions"}],"predecessor-version":[{"id":329,"href":"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-json\/pressbooks\/v2\/chapters\/328\/revisions\/329"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-json\/pressbooks\/v2\/parts\/302"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-json\/pressbooks\/v2\/chapters\/328\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-json\/wp\/v2\/media?parent=328"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-json\/pressbooks\/v2\/chapter-type?post=328"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-json\/wp\/v2\/contributor?post=328"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/ubcbatessandbox\/wp-json\/wp\/v2\/license?post=328"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}