## College Physics (4th Edition)

(a) The magnitude of the angular acceleration is $7.07~rad/s^2$ (b) It takes $~2.98~s~$ to rotate through $10\pi~radians$ (c) The tangential acceleration of a point located at a distance of 5.0 cm from the center of the disk is $0.35~m/s^2$
(a) We can find the angular acceleration $\alpha$: $\omega_f^2 = \omega_0^2+2\alpha~\Delta \theta$ $\alpha = \frac{\omega_f^2 - \omega_0^2}{2~\Delta \theta}$ $\alpha = \frac{(7\pi~rad/s)^2 - (2\pi~rad/s)^2}{(2)(10\pi~rad)}$ $\alpha = 7.07~rad/s^2$ The magnitude of the angular acceleration is $7.07~rad/s^2$ (b) We can find the time it takes to rotate through $10\pi~radians$: $\Delta \theta = \frac{1}{2}\alpha~t^2$ $t = \sqrt{\frac{2\Delta \theta}{\alpha}}$ $t = \sqrt{\frac{(2)(10\pi~rad)}{7.07~rad/s^2}}$ $t = 2.98~s$ It takes $~2.98~s~$ to rotate through $10\pi~radians$ (c) We can find the tangential acceleration of a point located at a distance of 5.0 cm from the center of the disk: $a = \alpha~r$ $a = (7.07~rad/s^2)(0.050~m)$ $a = 0.35~m/s^2$ The tangential acceleration of a point located at a distance of 5.0 cm from the center of the disk is $0.35~m/s^2$