## College Physics (4th Edition)

The motor must deliver a torque of $1.24\times 10^{-3}~N \cdot m$
We can find the angular acceleration: $\omega_f^2 = \omega_0^2+2\alpha~\Delta \theta$ $\alpha = \frac{\omega_f^2 - \omega_0^2}{2~\Delta \theta}$ $\alpha = \frac{(3.49~rad/s)^2 - 0}{(2)(4\pi~rad)}$ $\alpha = 0.485~rad/s^2$ We can find the required torque: $\tau = I~\alpha$ $\tau = \frac{1}{2}MR^2~\alpha$ $\tau = \frac{1}{2}(0.22~kg)(0.1525~m)^2~(0.485~rad/s^2)$ $\tau = 1.24\times 10^{-3}~N \cdot m$ The motor must deliver a torque of $1.24\times 10^{-3}~N \cdot m$