College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Review & Synthesis: Chapters 6-8 - Review Exercises - Page 323: 11


(a) $I = 0.502~kg~m^2$ (b) The motor must provide a torque of $16.8~N \cdot m$

Work Step by Step

(a) We can find the rotational inertia: $I = \frac{1}{2}MR^2$ $I = \frac{1}{2}(20.0~kg)(0.224~m)^2$ $I = 0.502~kg~m^2$ (b) We can express the angular speed in units of rad/s: $1200~rpm \times \frac{2\pi~rad}{1~rev} \times \frac{1~min}{60~s} = 40\pi~rad/s$ We can find the angular deceleration as the wheel slows to a stop: $\omega_f = \omega_0+\alpha~t$ $\alpha = \frac{\omega_f - \omega_0}{t}$ $\alpha = \frac{0-40\pi~rad/s}{60.0~s}$ $\alpha = -2.09~rad/s^2$ We can find the frictional torque: $\tau_f = I~\alpha$ $\tau_f = (0.502~kg~m^2)(-2.09~rad/s^2)$ $\tau_f = -1.05~N \cdot m$ We can find the angular acceleration as the wheel speeds up: $\omega_f = \omega_0+\alpha~t$ $\alpha = \frac{\omega_f - \omega_0}{t}$ $\alpha = \frac{40\pi~rad/s-0}{4.00~s}$ $\alpha = 31.4~rad/s^2$ We can find the required torque that the motor must provide: $\sum \tau = I~\alpha$ $\tau_m+\tau_f = I~\alpha$ $\tau_m = I~\alpha-\tau_f$ $\tau_m = (0.502~kg~m^2)(31.4~rad/s^2)-(-1.05~N \cdot m)$ $\tau_m = 16.8~N \cdot m$ The motor must provide a torque of $16.8~N \cdot m$.
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