Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 8 - Rotational Kinematics - Problems - Page 215: 63

Answer

The difference between the angular speeds of the wheels is $1.47rad/s$

Work Step by Step

Since 2 wheels have different $\omega$, the rotation circle of each wheel is different from each other. The distance between the left and right tires is $1.6m$, so if the radius of the rotation circle of the wheels on one side is $R$, that of the wheels on the other side is $R+1.6m$ The perimeter of each circle is $2\pi R$ and $2\pi (R+1.6)$ $t$ is the time finished to finish 1 lap. The linear velocity of each wheel, then, is $$v_1=\frac{2\pi R}{t}$$ $$v_2=\frac{2\pi(R+1.6)}{t}$$ Therefore, $$\omega_1-\omega_2=\frac{v_1}{r_{wheel}}-\frac{v_2}{r_{wheel}}=\frac{2\pi R}{rt}-\frac{2\pi(R+1.6)}{rt}$$ $$\omega_1-\omega_2=\frac{2\pi R}{rt}-\frac{2\pi R}{rt}-\frac{2\pi\times1.6}{rt}=-\frac{3.2\pi}{rt}$$ We know $t=19.5s$ and $r_{wheel}=0.35m$. Therefore, $$\omega_1-\omega_2=-1.47rad/s$$ So the difference between the angular speeds of the wheels is $1.47rad/s$
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