A merry-go-round has a mass of 1440 kg and a radius of 7.50 m. How much net work is required to accelerate it from rest to a rotation rate of 1.00 revolution per 7.00 s? Assume it is a solid cylinder.

The given data can be listed below as:

• The mass of the merry-go-round is\(M = 1440{\bf{ }}{\rm{kg}}\).
• The radius of the merry-go-round is \(R = 7.50{\rm{ m}}\).
• The initial angular velocity ofthe merry-go-roundis\({\omega _i} = 0{\rm{ rad/s}}\).
• The final angular velocity of the merry-go-round is \({\omega _f} = \left( {\frac{{1{\rm{ rev}}}}{{7{\rm{ s}}}}} \right)\left( {\frac{{2\pi {\rm{ rad}}}}{{1{\rm{ rev}}}}} \right) = 0.897{\rm{ rad/s}}\).

A merry-go-round starts from rest and moves with angular velocity. During its motion, the energy possessed by it is the rotational kinetic energy.From the work-energy theorem, the net work done for its acceleration equals the change in its rotational kinetic energy.

Assume that the merry-go-round is of solid cylindrical shape. Then, the moment of inertia can be expressed as:

\(I = \frac{1}{2}M{R^2}\)

Substitute the values in the above equation.

\(\begin{align}I &= \frac{1}{2} \times 1440{\bf{ }}{\rm{kg}} \times {\left( {7.50{\rm{ m}}} \right)^2}\\ &= 40500{\rm{ kg}} \cdot {{\rm{m}}^2}\end{align}\)

From the work-energy theorem, the change in the rotational kinetic energy is equal to the work done. This can be expressed as:

\(\begin{align}W &= {K_f} - {K_i}\\W &= \frac{1}{2}I\omega _f^2 - \frac{1}{2}I\omega _i^2\end{align}\)

Here,\({K_f}\)is the final rotational kinetic energy and\({K_i}\)is the initial rotational kinetic energy,\({\omega _f}\)is the final angular speed, and\({\omega _i}\)is the initial angular speed.

Substitute the values in the above equation.

\(\begin{align}W &= \frac{1}{2} \times 40500{\rm{ kg}} \cdot {{\rm{m}}^2} \times {\left( {0.897{\rm{ rad/s}}} \right)^2} - \frac{1}{2} \times 40500{\rm{ kg}} \cdot {{\rm{m}}^2} \times {\left( {0{\rm{ rad/s}}} \right)^2}\\ &= 16293.33{\rm{ kg}} \cdot {{\rm{m}}^2}{\rm{/}}{{\rm{s}}^2}\left( {\frac{{1{\rm{ J}}}}{{1{\bf{ }}{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{s}}^2}}}} \right)\\ &= 16293.33{\rm{ J}}\end{align}\)

Thus, the net work required to accelerate the merry-go-round is \(16293.33{\rm{ J}}\).