Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 8 - Rotational Motion - Problems - Page 223: 35

Answer

(a) $\tau = 7.8~m \cdot N$ (b) The force required by the muscle is 310 N.

Work Step by Step

(a) We can think of the ball as a point mass a distance of 0.31 m from the axis of rotation. We can find the moment of inertia of the ball relative to the axis of rotation. $I = MR^2 =(3.6~kg)(0.31~m)^2 = 0.346~kg\cdot m^2$ We can find the torque required to accelerate the ball. $\tau = I \alpha = I ~\frac{a}{r}= ( 0.346~kg\cdot m^2)(\frac{7.0~m/s^2}{0.31~m})$ $\tau = 7.8~m \cdot N$ (b) The distance $r$ from the muscle to the axis of rotation is 0.025 m. $r \cdot F = \tau$ $F = \frac{\tau}{r} = \frac{7.8~m \cdot N}{0.025~m} = 310~N$ The force required by the muscle is 310 N.
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