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Q18CQ

Expert-verifiedFound in: Page 354

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**Describe how work is done by a skater pulling in her arms during a spin. In particular, identify the force she exerts on each arm to pull it in and the distance each moves, noting that a component of the force is in the direction moved. Why is angular momentum not increased by this action?**

The work is done by applying a force perpendicular to the length of the arm. The angular momentum won’t increase because the moment of inertia will decrease and the angular velocity will increase.

**The mass, rotation, and speed of an object, as well as the radius of the object, determine its angular momentum.**

The work is done by the net torque applied by the skater for a rotation and is equal to the product of net torque and the angle rotated by the skater.

\(\begin{array}{c}W = \tau \theta \\ = rF\theta \end{array}\)

Here, \(W\) is the work done, \(\tau \) is the torque, \(\theta \) is the angle of rotation and is the angular displacement, \(r\) is the length of the arm, and \(F\) is the applied force.

The work done in rotational motion is equal to the change in rotational kinetic energy.

The net work done during this process is expressed as follows:

\(W = \frac{1}{2}I{\omega ^2} - \frac{1}{2}{I_0}\omega _0^2\)

Here, \(I\) is the final moment of inertia, \(\omega \) is the final angular velocity, \({I_0}\) is the initial moment of inertia, \({\omega _0}\) is the initial angular velocity.

Angular momentum is conserved. Because angular momentum is the product of the moment of inertia and angular velocity.

\(\begin{array}{c}L = I\omega \\ = m{r^2}\omega \end{array}\)

Here, \(L\) is the angular momentum, \(I\) is the moment of inertia, \(\omega \) is the angular velocity, \(m\) is the mass of the skater and \(r\) is the length of the arm.

When the skater pulls in her arms, the length of the arm decreases and angular velocity increases. So, the angular momentum remains unchanged.

Thus, the work is done by applying a force perpendicular to the length of the arm, and the angular momentum remains unchanged.

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