Question

A dad pushes a small hand-driven merry-go-round tangentially and is able to accelerate it from rest to a frequency of 15 rpm in 10.0 s. Assume that the merry-go-round is a uniform disk of radius 2.5 m and has a mass of 560 kg, and two children (each with a mass of 25 kg) sit opposite each other on the edges. Calculate the torque required to produce the acceleration, neglecting the frictional torque. What force is required at the edge?

Short answer

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

Step-by-step solution

Concepts

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.

Given data

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.

Calculation of torque

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 \(320\;{\rm{N}} \cdot {\rm{m}}\).