Answer
(a) $v = 0.868~m/s$
(b) $a = 4.06~m/s^2$
Work Step by Step
(a) We can find the angular acceleration.
$\tau = I\alpha$
$\alpha = \frac{\tau}{I}$
$\alpha = \frac{R~F}{\frac{1}{2}MR^2}$
$\alpha = \frac{2~F}{MR}$
$\alpha = \frac{(2)(30.0~N)}{(40.0~kg)(0.200~m)}$
$\alpha = 7.50~rad/s^2~$
We can find the angular velocity after the disk turns through 0.200 revolution.
$\omega^2 = \omega_0^2+2\alpha \theta$
$\omega = \sqrt{2\alpha \theta}$
$\omega = \sqrt{(2)(7.50~rad/s^2)(2\pi~rad/rev)(0.200~rev)}$
$\omega = 4.34~rad/s$
We can find the tangential velocity.
$v = \omega ~R$
$v = (4.34~rad/s)(0.200~m)$
$v = 0.868~m/s$
(b) We can find the tangential acceleration.
$a_t = \alpha ~R$
$a_t = (7.50~rad/s^2)(0.200~m)$
$a_t = 1.50~m/s^2$
We can find the centripetal acceleration.
$a_c = \omega^2~R$
$a_c = (4.34~rad/s)^2(0.200~m)$
$a_c = 3.77~m/s^2$
We can find the magnitude of the resultant acceleration.
$a = \sqrt{(a_t)^2+(a_c)^2}$
$a = \sqrt{(1.50~m/s^2)^2+(3.77~m/s^2)^2}$
$a = 4.06~m/s^2$
