Chapter 11.1, Problem 14SSC

To determine

To Explain: The relative change in angular momentum and rotational kinetic energy if merry-go-round turns twice as fast.

Explanation of Solution

Given:

Let, initial angular velocity of merry-go-round be ${\omega }_o$ . Then final angular velocity would be $\omega =2{\omega }_o$

Formula Used:

Angular momentum:

  $L=I\omega$

Rotational kinetic energy:

  $K=\frac{1}{2}I{\omega }^2$

Here, Iis the moment of inertia and $\omega$ is the angular velocity.

Calculations:

The angular momentum is directly proportional to angular velocity.

  $\begin{array}{l}L\propto \omega \\ \frac{L}{L_o}=\frac{\omega }{{\omega }_o}\\ \frac{L}{L_o}=\frac{2{\omega }_o}{{\omega }_o}\\ L=2L_o\end{array}$

The angular momentum becomes twice when merry-go-round spins twice as fast.

Rotational kinetic energy is directly proportional to square of angular velocity.

  $\begin{array}{l}K\propto {\omega }^2\\ \frac{K}{K_o}=\frac{{\omega }^2}{{\omega }_o{}^2}\\ \frac{K}{K_o}=\frac{{\left(2{\omega }_o\right)}^2}{{\omega }_o{}^2}=4\end{array}$

The rotational kinetic energyis four times when merry-go-round spins twice as fast.