## College Physics (4th Edition)

(a) The angular acceleration of the merry-go-round is $~0.111~rad/s^2$ (b) After 4.0 seconds, the angular speed is $~0.444~rad/s$
(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 $~0.444~rad/s$.