Chapter 12, Problem 48A

If a trapeze artist rotates twice each second while sailing through the air, and contracts to reduce her rotational inertia to one-third, how many rotations per second will result?

To determine

The number of rotations per seconds for the artist.

Answer to Problem 48A

The number of rotations per seconds for the artist is $6\text{Rotations}/\text{s}$ .

Explanation of Solution

Given:

The rotation is ${\omega }_1=2\text{Rotations}/\text{s}$ .

The reduced rotational inertia is $I_2=\frac{I_1}{3}$ .

Formula used:

The expression for the law of conservation of momentum is,

  $I_1{\omega }_1=I_2{\omega }_2$

Here, $I_2$ is the rotational inertia and ${\omega }_2$ is the angular velocity.

Calculation:

The number of rotations per seconds for the artistis

  $\begin{array}{l}{I}_1{\omega }_1=I_2{\omega }_2\\ I_1\left(2\text{Rotations}/\text{s}\right)=\frac{I_1}{3}{\omega }_2\\ {\omega }_2=6\text{Rotations}/\text{s}\end{array}$

Conclusion:

Thus, the number of rotations per seconds for the artist is $6\text{rotations}/\text{s}$ .