A figure skater on ice with arms extended,spins at a rate of 2.5 rev/s.After he draws his arms in,he spins at 6.0 rev/s.By what factor does the skater's kinetic energy change when he draws his arms in?
A)2.4
B)1.0
C)0.42
D)0.12

Question

Quizplus · Accepted Answer

The kinetic energy of a rotating object is given by:$$K = \frac{1}{2}I\omega^2$$where I is the moment of inertia and ω is the angular velocity.When the skater draws his arms in, his moment of inertia decreases. This is because the mass of his arms is now closer to the axis of rotation. As a result, his angular velocity must increase in order to conserve angular momentum.The factor by which the skater's kinetic energy changes is:$$\frac{K_f}{K_i} = \frac{\frac{1}{2}I_f\omega_f^2}{\frac{1}{2}I_i\omega_i^2} = \frac{I_f\omega_f^2}{I_i\omega_i^2}$$Since the skater's angular momentum is conserved, we have:$$I_i\omega_i = I_f\omega_f$$Substituting this into the equation above, we get:$$\frac{K_f}{K_i} = \frac{\omega_f^2}{\omega_i^2}$$Plugging in the given values, we get:$$\frac{K_f}{K_i} = \frac{(6.0 	ext{ rev/s})^2}{(2.5 	ext{ rev/s})^2} = 2.4$$Therefore, the skater's kinetic energy increases by a factor of 2.4 when he draws his arms in.

A Figure Skater on Ice with Arms Extended,spins at a Rate