Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 10 - Rotation - Questions - Page 286: 8

Answer

We can rank the values of $\phi$ according to the magnitude of the angular acceleration of the bar: $90^{\circ} \gt 70^{\circ} = 110^{\circ}$

Work Step by Step

The torque due to $F_1$ tends to make the bar rotate counterclockwise. The torque due to $F_2$ also tends to make the bar rotate counterclockwise. To maximize the angular acceleration, we should maximize the torque. The expression for the torque due to $F_2$ is: $~~\tau_2 = r_2~F_2~sin~\phi$ The torque is a maximum when $\phi = 90^{\circ}$ Since $~~sin~70^{\circ} = sin~110^{\circ},~~$ the torque is equal at both of these angles of $\phi$ We can rank the values of $\phi$ according to the magnitude of the angular acceleration of the bar: $90^{\circ} \gt 70^{\circ} = 110^{\circ}$
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