91Ó°ÊÓ

The mass of the merry-go-round is \(m = 31,000\;{\rm{kg}}\).

The radius of the merry-go-round is\(r = 7\;{\rm{m}}\).

The angular velocity of the merry-go-round is\(\omega = 0.68\;{\rm{rad/s}}\).

The time taken by the merry-go-round to accelerate is \(t = 34\;{\rm{s}}\).

Moment of inertia is a quantity that expresses a body’s tendency to resist angular acceleration about the axis of rotation.

It is given as the product of the mass of a rigid body and the square of the distance from the axis of rotation.

\(I = m{r^2}\)

The initial angular velocity of the merry-go-round is \({\omega _0} = 0\). Using the kinematic equation of rotational motion \(\omega = {\omega _0} + \alpha t,\)the angular acceleration is obtained as:

\(\begin{align}\alpha &= \frac{{\omega - {\omega _0}}}{t}\\ &= \frac{{0.68\;{\rm{rad/s}} - 0}}{{34\;{\rm{s}}}}\\ &= 0.020\;{\rm{rad/}}{{\rm{s}}^2}\end{align}\)

The merry-go-round is considered a uniform disk. Therefore, the moment of inertia of the disk is given as:

\(\begin{align}I &= \frac{1}{2}m{r^2}\\ &= \frac{1}{2}\left( {31000\;{\rm{kg}}} \right){\left( {7\;{\rm{m}}} \right)^2}\\ &= 759,500\;{\rm{kg}}\;{{\rm{m}}^2}\end{align}\)

The torque \(\tau \) and angular acceleration \(\alpha \) are related as:

\(\tau = I\alpha \)

Here, I is the moment of inertia.

The net torque required to accelerate the merry-go-round is:

\(\begin{align}\tau &= I\alpha \\ &= \left( {759,500\;{\rm{kg}}\;{{\rm{m}}^2}} \right)\left( {0.020\;{\rm{rad/}}{{\rm{s}}^2}} \right)\\ &= 1.5 \times {10^4}\;{\rm{m}}\;{\rm{N}}\end{align}\)