College Physics (4th Edition)

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

Chapter 8 - Problems - Page 320: 107


(a) $I = 1.35\times 10^{-5}~kg~m^2$ (b) The required muscular force is $526.5~N$

Work Step by Step

(a) We can find the rotational inertia: $I = \frac{1}{3}MR^2$ $I = \frac{1}{3}(0.028~kg)(0.038~m)^2$ $I = 1.35\times 10^{-5}~kg~m^2$ (b) We can find the angular acceleration: $\omega_f = \omega_0+\alpha~t$ $\alpha = \frac{\omega_f - \omega_0}{t}$ $\alpha = \frac{175~rad/s - 0}{1.50\times 10^{-3}~s}$ $\alpha = 1.17\times 10^5~rad/s^2$ We can find the required force: $\tau = I~\alpha$ $r~F = I~\alpha$ $F = \frac{I~\alpha}{r}$ $F = \frac{(1.35\times 10^{-5}~kg~m^2)(1.17\times 10^5~rad/s^2)}{3.00\times 10^{-3}~m}$ $F = 526.5~N$ The required muscular force is $526.5~N$.
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