{"id":314,"date":"2020-04-19T15:58:59","date_gmt":"2020-04-19T19:58:59","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/humanbiomechanics\/chapter\/6-2-centripetal-acceleration-2\/"},"modified":"2020-08-22T14:23:55","modified_gmt":"2020-08-22T18:23:55","slug":"6-2-centripetal-acceleration-2","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/humanbiomechanics\/chapter\/6-2-centripetal-acceleration-2\/","title":{"raw":"5.4 Linear Accelerations","rendered":"5.4 Linear Accelerations"},"content":{"raw":"
<\/div>\r\nWhen a limb is rotating at an increasing angular velocity, the tangential (linear) velocity of a point on the limb is increasing as well. The angular and linear accelerations of a point are related but unlike velocity, there are two<\/strong> linear accelerations to consider on rotation objects: tangential<\/strong> and centripetal acceleration<\/strong>.\r\n

Tangential Acceleration<\/h1>\r\nThe linear acceleration tangent to the circular path of a point on a rotating segment is called: tangential acceleration. It represents the chance in tangential (linear) velocity discussed in the previous section. \u00a0Tangential acceleration is related to angular acceleration in the following way:\r\n

at<\/sub> = \u03b1r<\/strong><\/p>\r\nKeep in mind that in order for this equation to be true, angular acceleration must be expressed in rads\/s2<\/sup>.\r\n\r\nTangent:<\/strong> a line is tangent to a circle if the line intersects the circle at just one point. A line from this point to the centre of the circle (radius) is perpendicular to the tangent line.\r\n

Centripetal Acceleration<\/h1>\r\n

We know from kinematics that acceleration is a change in velocity, either in its magnitude or in its direction, or both. In uniform circular motion, the direction of the velocity changes constantly, so there is always an associated acceleration, even though the magnitude of the velocity might be constant. You experience this acceleration yourself when you turn a corner in your car. (If you hold the wheel steady during a turn and move at constant speed, you are in uniform circular motion.) What you notice is a sideways acceleration because you and the car are changing direction. The sharper the curve and the greater your speed, the more noticeable this acceleration will become. In this section we examine the direction and magnitude of that acceleration.<\/p>\r\n

Figure 1<\/a> shows an object moving in a circular path at constant speed. The direction of the instantaneous velocity is shown at two points along the path. Acceleration is in the direction of the change in velocity, which points directly toward the center of rotation (the center of the circular path). This pointing is shown with the vector diagram in the figure. We call the acceleration of an object moving in uniform circular motion (resulting from a net external force) the centripetal acceleration <\/span><\/strong>(a<\/em>c<\/sub><\/strong>); centripetal means \u201ctoward the center\u201d or \u201ccenter seeking.\u201d<\/p>\r\n\r\n

<\/figcaption>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"250\"]\"The Figure 1.<\/strong> The directions of the velocity of an object at two different points are shown, and the change in velocity \u0394v<\/strong> is seen to point directly toward the center of curvature. (See small inset.) Because ac = \u0394v<\/em>\/\u0394t<\/em><\/strong>, the acceleration is also toward the center; ac is called centripetal acceleration. (Because \u0394\u03b8<\/em> <\/strong>is very small, the arc length \u0394s<\/em><\/strong> is equal to the chord length \u0394r<\/em><\/strong> for small time differences.)[\/caption]<\/figure>\r\n

The direction of centripetal acceleration is toward the center of curvature, but what is its magnitude? Note that the triangle formed by the velocity vectors and the one formed by the radii r<\/strong><\/em> and \u0394s<\/em><\/strong> are similar. Both the triangles ABC and PQR are isosceles triangles (two equal sides). The two equal sides of the velocity vector triangle are the speeds v<\/em>1<\/sub> = v<\/em>2<\/sub> = v<\/em><\/strong>. Using the properties of two similar triangles, we obtain<\/p>\r\n\r\n

[latex]\\boldsymbol{\\frac{\\Delta{v}}{v}}\\boldsymbol{=}\\boldsymbol{\\frac{\\Delta{s}}{r}}.[\/latex]<\/div>\r\n

Acceleration is \u0394v<\/em>\/\u0394t<\/em><\/strong>, and so we first solve this expression for \u0394v<\/em><\/strong>:<\/p>\r\n\r\n

[latex]\\boldsymbol{\\Delta{v}\\:=}\\boldsymbol{\\frac{v}{r}}\\boldsymbol{\\Delta{s}}.[\/latex]<\/div>\r\n

Then we divide this by \u0394t<\/em><\/strong>, yielding<\/p>\r\n\r\n

[latex]\\boldsymbol{\\frac{\\Delta{v}}{\\Delta{t}}}\\boldsymbol{=}\\boldsymbol{\\frac{v}{r}\\times\\frac{\\Delta{s}}{\\Delta{t}}}.[\/latex]<\/div>\r\n

Finally, noting that \u0394v<\/em>\/\u0394t<\/em> = a<\/em>c<\/sub><\/strong> and that \u0394s<\/em>\/\u0394t<\/em> = v<\/em><\/strong>, the linear or tangential speed, we see that the magnitude of the centripetal acceleration is<\/p>\r\n\r\n

[latex]\\boldsymbol{a_{\\textbf{c}}\\:=}\\boldsymbol{\\frac{v^2}{r}},[\/latex]<\/div>\r\n

which is the acceleration of an object in a circle of radius r<\/strong><\/em> at a speed v<\/strong><\/em>. So, centripetal acceleration is greater at high speeds and in sharp curves (smaller radius), as you have noticed when driving a car. But it is a bit surprising that a<\/em>c<\/sub><\/strong> is proportional to speed squared, implying, for example, that it is four times as hard to take a curve at 100 km\/h than at 50 km\/h. A sharp corner has a small radius, so that a<\/em>c<\/sub><\/strong> is greater for tighter turns, as you have probably noticed.<\/p>\r\n

It is also useful to express a<\/em>c<\/sub><\/strong> in terms of angular velocity. Substituting v<\/em> = r<\/em>\u03c9<\/strong> into the above expression, we find a<\/em>c <\/sub>= (r<\/em>\u03c9)2<\/sup>\/r<\/em> = r<\/em>\u03c92<\/sup><\/strong>. We can express the magnitude of centripetal acceleration using either of two equations:<\/p>\r\n\r\n

[latex]\\boldsymbol{a_{\\textbf{c}}\\:=}\\boldsymbol{\\frac{v^2}{r}}\\boldsymbol{;\\:a_{\\textbf{c}}=r\\omega^2}.[\/latex]<\/div>\r\n

Recall that the direction of a<\/em>c<\/sub><\/strong> is toward the center. You may use whichever expression is more convenient, as illustrated in examples below.<\/p>\r\n\r\n

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Example 1: How Does the Centripetal Acceleration of a Car Around a Curve Compare with That Due to Gravity?<\/h3>\r\nWhat is the magnitude of the centripetal acceleration of a car following a curve of radius 500 m at a speed of 25.0 m\/s (about 90 km\/h)? Compare the acceleration with that due to gravity for this fairly gentle curve taken at highway speed. See Figure 2<\/a>(a).\r\n

Strategy<\/strong><\/p>\r\n

Because v<\/strong><\/em> and r<\/strong><\/em> are given, the first expression in [latex]\\boldsymbol{a_{\\textbf{c}}=\\frac{v^2}{r};\\:a_{\\textbf{c}}=r\\omega^2}[\/latex] is the most convenient to use.<\/p>\r\n

Solution<\/strong><\/p>\r\n

Entering the given values of v<\/em> = 25.0 m\/s<\/strong> and r<\/em> = 500 m<\/strong> into the first expression for a<\/em>c<\/sub><\/strong> gives<\/p>\r\n\r\n

[latex]\\boldsymbol{a_{\\textbf{c}}\\:=}\\boldsymbol{\\frac{v^2}{r}}\\boldsymbol{=}\\boldsymbol{\\frac{(25.0\\textbf{ m\/s})^2}{500\\textbf{ m}}}\\boldsymbol{=\\:1.25\\textbf{ m\/s}^2}.[\/latex]<\/div>\r\n

Discussion<\/strong><\/p>\r\n

To compare this with the acceleration due to gravity (g<\/em> = 9.80 m\/s2<\/sup><\/strong>), we take the ratio of a<\/em>c<\/sub> \/g<\/em> = (1.25 m\/s2<\/sup>)\/(9.80m\/s2<\/sup>) = 0.128<\/strong>. Thus, a<\/em>c <\/sub>= 0.128 g<\/em><\/strong> and is noticeable especially if you were not wearing a seat belt.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"198\"]\"In Figure 2.<\/strong> (a) The car following a circular path at constant speed is accelerated perpendicular to its velocity, as shown. The magnitude of this centripetal acceleration is found in Example 1<\/a>. (b) A particle of mass in a centrifuge is rotating at constant angular velocity . It must be accelerated perpendicular to its velocity or it would continue in a straight line. The magnitude of the necessary acceleration is found in Example 2<\/a>.[\/caption]<\/figure>\r\n

Of course, a net external force is needed to cause any acceleration, just as Newton proposed in his second law of motion. So a net external force is needed to cause a centripetal acceleration.<\/p>\r\n\r\n

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Section Summary<\/h1>\r\n