Question

A spinning ice skater can increase his rate of rotation by bringing his arms and free leg closer to his body. How does this procedure affect the skater's angular momentum and kinetic energy.

• A. angular momentum remains the same while kinetic enery increases
• B. angular momentum remains the same while kinetic enegry decreases
• C. both angular momentum acid kinetic energy remain the same
• D. angular momentum increases while energy remains the same

Answer 1

Answer: (A)

Option 1

Answer 2

• Effect on Moment of Inertia (I): When the ice skater brings his arms and free leg closer to his body, he is redistributing his mass closer to the axis of rotation. This action decreases his moment of inertia ($I$).

• Effect on Angular Momentum (L): In the absence of external torques (assuming negligible air resistance and friction from the ice), the angular momentum of the skater remains conserved. The skater is an isolated system in terms of rotational motion.

  • Angular momentum, $L = I\omega$, where $I$ is the moment of inertia and $\omega$ is the angular velocity.
  • Since $L$ is constant and $I$ decreases, the angular velocity $\omega$ must increase, which is why the skater spins faster.

• Effect on Kinetic Energy (K): The rotational kinetic energy is given by: $K = \frac{1}{2}I\omega^2$ We can express kinetic energy in terms of angular momentum $L$. Since $L = I\omega$, we have $\omega = \frac{L}{I}$. Substituting this into the kinetic energy equation: $K = \frac{1}{2}I\left(\frac{L}{I}\right)^2 = \frac{1}{2}I\frac{L^2}{I^2} = \frac{L^2}{2I}$ Since the angular momentum $L$ remains constant and the moment of inertia $I$ decreases, the rotational kinetic energy $K$ must increase. The skater does work by pulling his arms and legs in, converting chemical energy into rotational kinetic energy.

Therefore, the skater's angular momentum remains the same, while his kinetic energy increases.