Chapter 8, Problem 62P

To determine

The angular velocity of the merry-go-round when people are standing on it and when people jumped off it.

Answer to Problem 62P

The angular velocity of merry-go-round when people are standing is $0.48\text{rad/s}$

The angular velocity of merry-go-round when people jump off it is $1.32\text{rad/s}$

Explanation of Solution

Given:

Diameter of merry-go-round, $d=4.2m$

Initial angular velocity, ${\omega }_1=0.8\text{rad/s}$

Mass of each person, $m=65kg$

Moment of inertia of merry-go-round, $I_1=1760kg-m^2$

Formula used:

Moment of inertia of people, $I=M{r}^2$

Radius, $r=\frac{d}{2}$

Angular momentum, $L=I\omega$

Law of conservation of angular momentum, $L_1=L_2$

Calculation:

Substituting values for radius,

  $r=\frac{4.2}{2}=2.1m$

When people jump on the platform:

Substituting values for moment of inertia of persons,

  $\begin{array}{l}$ I_2=(4)(65){(2.1)}^2$I_2=1146.6kg-m^2\end{array}$

Substituting values for initial angular momentum,

  $\begin{array}{l}{L}_1=(1760)(0.8)\\ L_1=1408{\text{kg-m}}^{\text{2}}\text{/s}\end{array}$

Apply law of conservation of angular momentum,

  $\begin{array}{l}{L}_1=(I_1+I_2){\omega }_2\\ 1408=(1760+1146.6){\omega }_2\\ {\omega }_2=0.48\text{rad/s}\end{array}$

When people jump off the platform:

Substituting values for initial angular momentum,

  $\begin{array}{l}{L}_1=(I_1+I_2){\omega }_1\\ L_1=(1760+1146.6)(0.8)\\ L_1=2325.28{\text{kg-m}}^{\text{2}}\text{/s}\end{array}$

Apply law of conservation of angular momentum,

  $\begin{array}{l}{L}_1=I_1{\omega }_2\\ 2325.28=(1760){\omega }_2\\ {\omega }_2=1.32\text{rad/s}\end{array}$

Conclusion:

The angular velocity of merry-go-round when people are standing is $0.48\text{rad/s}$

The angular velocity of merry-go-round when people jump off it is $1.32\text{rad/s}$