First, we can find the initial angular speed of the merry-go-round. Calling [tex]T=4.0 s[/tex] the period of revolution, the initial angular speed is
[tex]\omega _i = \frac{2 \pi}{T}= \frac{2 \pi}{4.0 s}=1.57 rad/s [/tex]
Then, the merry-go-round starts to decelerate, so it is an angular accelerated motion with angular acceleration [tex]\alpha[/tex], and the angular velocity follows the law
[tex]\omega(t) = \omega _i + \alpha t[/tex]
We know that after t=20 s, the merry go round stops, so [tex]\omega (20 s)=0[/tex]. So we can rewrite the previous equation and solve it to find [tex]\alpha[/tex]:
[tex]0= \omega _i + \alpha \cdot (20 s)[/tex]
[tex]\alpha = - \frac{\omega _i}{20.0 s}=-0.08 rad/s^2 [/tex]
where the negative sign means the merry-go-round is decelerating.
