91Ó°ÊÓ

Torque is the product of the moment of inertia and the square of the distance. For this problem, first, find out the angular acceleration of the system and multiply it with the total moment of inertia of the system.

The initial angular velocity of the merry-go-round is \({\omega _1} = 0\).

The final angular velocity of the merry-go-round is \({\omega _2} = 15\;{\rm{rpm}} = \left( {15 \times 2\pi } \right)\;{\rm{rad/s}}\).

The time taken by the merry-go-round is \(\Delta t = 10.0\;{\rm{s}}\).

The radius of the merry-go-round is \(r = 2.5\;{\rm{m}}\).

The mass of the disk is \(M = 560\;{\rm{kg}}\).

The mass of each child is \(m = 25\;{\rm{kg}}\).

The children are at the edge of the merry-go-round.

You can assume the children as point masses at the edge of the disc.

Let \(\tau \) be the torque required to produce such acceleration.

The total moment of inertia of the merry-go-round and the child system is

\(I = \frac{1}{2}M{r^2} + 2\left( {m{r^2}} \right)\).

Now, the angular acceleration of the system is \(\alpha = \frac{{{\omega _2} - {\omega _1}}}{{\Delta t}}\).

So, the required torque is

\(\begin{align}\tau &= I\alpha \\ &= \left\{ {\frac{1}{2}M{r^2} + 2\left( {m{r^2}} \right)} \right\}\frac{{{\omega _2} - {\omega _1}}}{{\Delta t}}\\ &= \left\{ {\left( {\frac{1}{2} \times \left( {560\;{\rm{kg}}} \right) \times {{\left( {2.5\;{\rm{m}}} \right)}^2}} \right) + 2\left( {\left( {25\;{\rm{kg}}} \right) \times \times {{\left( {2.5\;{\rm{m}}} \right)}^2}} \right)} \right\}\frac{{\left( {\frac{{15 \times 2\pi }}{{60}}} \right)\;{\rm{rad/s}} - 0}}{{10.0\;{\rm{s}}}}\\ &= 323.81\;{\rm{N}} \cdot {\rm{m}} \approx 320\;{\rm{N}} \cdot {\rm{m}}\end{align}\).

Hence, the required torque at the edge is \(320\;{\rm{N}} \cdot {\rm{m}}\).