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(D) Since the tension force acts along the radius,the torque about the center of rotation is zero. Therefore,the angular momentum of the mass is conse

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Increase by a factor of $4$

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(D) Since the tension force acts along the radius,the torque about the center of rotation is zero. Therefore,the angular momentum of the mass is conserved.Let $m$ be the mass,$v_1$ be the initial velocity,and $r_1 = r$ be the initial radius. Let $v_2$ be the final velocity and $r_2 = r/2$ be the final radius.By conservation of angular momentum,$L_1 = L_2$:$m v_1 r_1 = m v_2 r_2$$v_1 r = v_2 (r / 2)$$v_2 = 2 v_1$The initial kinetic energy is $KE_1 = \frac{1}{2} m v_1^2$.The final kinetic energy is $KE_2 = \frac{1}{2} m v_2^2 = \frac{1}{2} m (2 v_1)^2 = 4 (\frac{1}{2} m v_1^2) = 4 KE_1$.Thus,the kinetic energy increases by a factor of $4$.

$A$ small mass attached to a string rotates on a frictionless table top as shown. If the tension in the string is increased by pulling the string,causing the radius of the circular motion to decrease by a factor of $2$,the kinetic energy of the mass will

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