Answer
(a) The angular acceleration of the merry-go-round is $~0.111~rad/s^2$
(b) After 4.0 seconds, the angular speed is $~0.444~rad/s$
Work Step by Step
(a) We can find the angular acceleration of the merry-go-round:
$\tau = I~\alpha$
$2~(R\times F) = I~\alpha$
$\alpha = \frac{2~(R\times F)}{I}$
$\alpha = \frac{2~(R\times F)}{\frac{1}{2}MR^2}$
$\alpha = \frac{4~F}{MR}$
$\alpha = \frac{(4)(10.0~N)}{(180~kg)(2.0~m)}$
$\alpha = 0.111~rad/s^2$
The angular acceleration of the merry-go-round is $~0.111~rad/s^2$
(b) We can find the angular speed after 4.0 seconds:
$\omega_f = \omega_0+\alpha~t$
$\omega_f = 0+(0.111~rad/s^2)(4.0~s)$
$\omega_f = 0.444~rad/s$
After 4.0 seconds, the angular speed is $~0.444~rad/s$.
