Question

A merry-go-round with a moment of inertia equal to $\mathbf{1260}\mathbf{k}\mathbf{g}\cdot \mathbf{m}\mathbf{2}$ and a radius of 2.5 m rotates with negligible friction at \({\bf{1}}{\bf{.70}}\;{{{\bf{rad}}} \mathord{\left/ {\vphantom {{{\bf{rad}}} {\bf{s}}}} \right. \{\bf{s}}}\). A child initially standing still next to the merry-go-round jumps onto the edge of the platform straight toward the axis of rotation, causing the platform to slow to \({\bf{1}}{\bf{.35}}\;{{{\bf{rad}}} \mathord{\left/{\vphantom {{{\bf{rad}}} {\bf{s}}}} \right.

\{\bf{s}}}\). What is her mass?

Short answer

The mass of the child is $52\mathrm{k}\mathrm{g}$.

Step-by-step solution

Determination of moment of inertia of the child

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.

Given information

The moment of inertia of the merry-go-round is$I_1=1260\mathrm{k}\mathrm{g}\cdot {\mathrm{m}}^2$.

The radius of the merry-go-round is$R=2.5\mathrm{m}$.

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}}}\).

Evaluation of the moment of inertia of the child

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.

$\begin{array}{r}c{L}_{\mathrm{i}}=L_{\mathrm{f}}\\ I_1{\omega }_1=\left(I_1+I_2\right){\omega }_2\\ I_1\frac{{\omega }_1}{{\omega }_2}=l_1+I_2\\ I_2=I_1\left(\frac{{\omega }_1}{{\omega }_2}-1\right)\end{array}$

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}\)

Evaluation of the mass of the child

The mass of the child can be calculated as:

$\begin{array}{r}c{I}_2=m{R}^2\\ m=\frac{I_2}{R^2}\\ m=\frac{\left(326.6\mathrm{k}\mathrm{g}\cdot {\mathrm{m}}^2\right)}{{\left(2.5\mathrm{m}\right)}^2}\\ m=52.2\mathrm{k}\mathrm{g}\approx 52\mathrm{k}\mathrm{g}\end{array}$

Thus, the mass of the child is $52\mathrm{k}\mathrm{g}$.