91Ó°ÊÓ

The moment of inertia of the child can be calculated by multiplying the child's mass with the square of the child's distance from the center of a circle.

The moment of inertia of the merry-go-round is\({I_1} = 1260\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}\).

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

The initial angular velocity of the system is\({\omega _1} = 1.70\;{{{\rm{rad}}} \mathord{\left/

{\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.

\\{\rm{s}}}\).

The final angular velocity of the system is\({\omega _2} = 1.35\;{{{\rm{rad}}} \mathord{\left/

{\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.

\\{\rm{s}}}\).

Let\({I_2}\)be the moment of inertia of the child.

Apply the law of conservation of angular momentum to calculate the moment of inertia of the child.

\(\begin{aligned}{c}{L_{\rm{i}}} = {L_{\rm{f}}}\\{I_1}{\omega _1} = \left( {{I_1} + {I_2}} \right){\omega _2}\\{I_1}\frac{{{\omega _1}}}{{{\omega _2}}} = {l_1} + {I_2}\\{I_2} = {I_1}\left( {\frac{{{\omega _1}}}{{{\omega _2}}} - 1} \right)\end{aligned}\)

Substitute the values in the above expression.

\(\begin{aligned}{l}{I_2} = \left( {1260\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}} \right)\left[ {\frac{{\left( {1.70\;{{{\rm{rad}}} \mathord{\left/

{\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.

\\{\rm{s}}}} \right)}}{{\left( {1.35\;{{{\rm{rad}}} \mathord{\left/

{\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.

\\{\rm{s}}}} \right)}} - 1} \right]\\{I_2} = 326.6\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}\end{aligned}\)

The mass of the child can be calculated as:

\(\begin{aligned}{c}{I_2} = m{R^2}\\m = \frac{{{I_2}}}{{{R^2}}}\\m = \frac{{\left( {326.6\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}} \right)}}{{{{\left( {2.5\;{\rm{m}}} \right)}^2}}}\\m = 52.2\;{\rm{kg}} \approx {\rm{52}}\;{\rm{kg}}\end{aligned}\)

Thus, the mass of the child is \({\rm{52}}\;{\rm{kg}}\).