Answer
(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 $0.35~m/s^2$
Work Step by Step
(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 $0.35~m/s^2$
