Question: (II) A 4.2-m-diameter merry-go-round is rotating freely with an angular velocity of \({\bf{0}}{\bf{.80}}\;{{{\bf{rad}}} \mathord{\left/{\vphantom {{{\bf{rad}}} {\bf{s}}}} \right.} {\bf{s}}}\). Its total moment of inertia is \({\bf{1360}}\;{\bf{kg}} \cdot {{\bf{m}}^{\bf{2}}}\). Four people standing on the ground, each of mass 65 kg, suddenly step onto the edge of the merry-go-round. (a) What is the angular velocity of the merry-go-round now? (b) What if the people were on it initially and then jumped off in a radial direction (relative to the merry-go-round)?

The angular momentum of a rotating object may be obtained by examining the value of the angular velocity of the object and the object’s moment of inertia.

Given data:

The diameter of the merry-go-round is \(D = 4.2\;{\rm{m}}\).

The initial angular velocity of the system is \({\omega _{\rm{i}}} = 0.80\;{{{\rm{rad}}} \mathord{\left/{\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.} {\rm{s}}}\).

The moment of inertia of the merry-go-round is \({I_{\rm{m}}} = 1360\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}\).

The number of people is \(n = 4\).

The mass of each person is \(m = 65\;{\rm{kg}}\).

(a)

Let\({I_{\rm{P}}}\)be the moment of inertia of each person.

The expression for the final moment of inertia of the system can be written as:

\(\begin{aligned}{c}{I_{\rm{f}}} &= {I_{\rm{m}}} + n{I_{\rm{P}}}\\{I_{\rm{f}}} &= {I_{\rm{m}}} + 4\left( {m{R^2}} \right)\\{I_{\rm{f}}} &= {I_{\rm{m}}} + 4m{\left( {\frac{D}{2}} \right)^2}\\{I_{\rm{f}}} &= {I_m} + m{D^2}\end{aligned}\)

Now, apply the conservation of angular momentum to find the final angular velocity of the system.

\(\begin{aligned}{c}{I_{\rm{f}}}{\omega _{\rm{f}}} &= {I_{\rm{m}}}{\omega _{\rm{i}}}\\{\omega _{\rm{f}}} &= \frac{{{I_{\rm{m}}}{\omega _{\rm{i}}}}}{{{I_{\rm{f}}}}}\\{\omega _{\rm{f}}} &= \frac{{{I_{\rm{m}}}{\omega _{\rm{i}}}}}{{{I_{\rm{m}}} + m{D^2}}}\end{aligned}\)

Substitute the values in the above expression.

\(\begin{aligned}{l}{\omega _{\rm{f}}} &= \frac{{\left( {1360\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}} \right)\left( {0.80\;{{{\rm{rad}}} \mathord{\left/{\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.} {\rm{s}}}} \right)}}{{\left( {1360\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}} \right) + \left( {65\;{\rm{kg}}} \right){{\left( {4.2\;{\rm{m}}} \right)}^2}}}\\{\omega _{\rm{f}}} &= 0.43\;{{{\rm{rad}}} \mathord{\left/{\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.} {\rm{s}}}\end{aligned}\)

Thus, the final angular velocity of the system is \(0.43\;{{{\rm{rad}}} \mathord{\left/{\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.} {\rm{s}}}\).

(b)

The torque exerted by the people jumping off the merry-go-round is zero. Therefore, there will be no variation in the angular momentum of the merry-go-round. Thus, the merry-go-round will continue to revolve with the initial angular momentum.

Thus, the final angular velocity when people jump off it will be \(0.80\;{{{\rm{rad}}} \mathord{\left/{\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.} {\rm{s}}}\).