A playground marry-go-round has radius $2.40\mathrm{m}$ and moment of inertia $2100kg{m}^2$ about a vertical axle through its center, and it turns with negligible friction. (a) A child applies an $18.0-N$ force tangentially to the edge of the merry-go-round for $15.0s$. If the merry-go-round is initially at rest, what is its angular speed after this $15.0-s$ interval? (b) How much work did the child do on the merry-go-round? (c) What is the average power supplied by the child?

Moment of inertia of the merry-go-round $\left(I\right)=2100kg{m}^2$.

Radius of the marry-go-round$\left(R\right)=2.40m$.

Force exerted by the child $\left(F\right)=18.0N$.

Time interval $\left(t\right)=15.0s$.

The torque is given by,

Here, F is the force, R is the radius, $\alpha$is angular acceleration and I is the moment of inertia.

The angular acceleration is:

$\alpha =\frac{FR}{I}$ ……. (2)

Here, F is the force, R is the radius, and I is the moment of inertia.

Consider the formula for the angular speed:

$\omega =\alpha \Delta t$ …… (3)

Here, $\alpha$is angular acceleration and $\Delta t$is the time interval.

The initial time is $t_i=0$s and final time is $t_f=15s$.

Therefore, the time interval is $\Delta t=15s$.

Then using equation$\left(2\right)$in $\left(3\right)$, angular speed is given by,

Hence, the angular speed is, $\omega =0.309\frac{\text{rad}}{\text{s}}$.

The work done on the merry-go-round by the child is equal to the rotational kinetic energy.

Therefore,

Hence, the work done on the merry-go-round by the child is$W=100 J$.

The average power supplied by the child is given by,

$P=\frac{W}{\Delta t}$.

Substituting values, we get,

Hence, the average power supplied by the child is, $P=6.67 W$.