A small disk is attached to the end of a ligh | vedclass

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(B) Since the tension force acts towards the center (the hole),the torque about the center is zero. Therefore,the angular momentum of the disk is cons

Vedclass · Accepted Answer

$W = (\eta^2 - 1) K_0$

Vedclass · Answer

(B) Since the tension force acts towards the center (the hole),the torque about the center is zero. Therefore,the angular momentum of the disk is conserved.Let $m$ be the mass of the disk. The initial angular momentum is $L_i = m v_1 R = m (\omega_0 R) R = m R^2 \omega_0$.The final angular momentum is $L_f = m v_2 (R/\eta) = m (\omega_f R/\eta) (R/\eta) = m (R^2/\eta^2) \omega_f$.By conservation of angular momentum,$L_i = L_f \Rightarrow m R^2 \omega_0 = m (R^2/\eta^2) \omega_f \Rightarrow \omega_f = \omega_0 \eta^2$.The initial kinetic energy is $K_0 = \frac{1}{2} m v_1^2 = \frac{1}{2} m (\omega_0 R)^2 = \frac{1}{2} m R^2 \omega_0^2$.The final kinetic energy is $K_f = \frac{1}{2} m v_2^2 = \frac{1}{2} m (\omega_f \frac{R}{\eta})^2 = \frac{1}{2} m (\omega_0 \eta^2 \frac{R}{\eta})^2 = \frac{1}{2} m R^2 \omega_0^2 \eta^2 = K_0 \eta^2$.By the work-energy theorem,the work done $W$ by the pulling force is equal to the change in kinetic energy:$W = K_f - K_0 = K_0 \eta^2 - K_0 = K_0 (\eta^2 - 1)$.

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