An ice skater stands motionless with his arms outstretched on the (frictionless)

Question

An ice skater stands motionless with his arms outstretched on the (frictionless) ice. For his big finale, two of his friends throw objects of mass m_1 = 1.0 kg and = 0.50 kg at him from opposite sides, which he then catches simultaneously in his outstretched arms at distances of d_1 = 0.50 m and d_2 = 1.0 m from the center of his body, respectively. (See figure below, which shows a view of this scenario from above.) If the 1.0 kg object is thrown with a speed of v_1 = 10.0 m/s, with what speed must the 0.50 kg object be thrown such that, after the skater catches these, he only begins spinning and does not slide on the ice (i.e., there is no final translational/linear motion of the skater)? If the moment of inertia of just the skater about his center I_s = 0.75 kg middot m^2, what is the angular velocity of the skater just after he catches these two objects? (Don't forget to include the thrown objects' contributions to the final moment of inertia of the system.) What is the change in kinetic energy of the system (including both objects and the skater) before and just after he catches the objects, and is this change positive or negative? Note, rotational kinetic energy is given by 1/2 I omega^2, where I is the total moment of inertia of the system. Finally, if he pulls these objects to the center of his body, what is his new angular velocity? (to receive full credit. 25 points total)

Explanation / Answer

(A) applying momentum conservation,

0 = 1x 10 - (0.50 v2)

v2 = 20 m/s

(B) Applying angular momentum conservation,

m1 v1 r1 + m2 v2 r2 = I w

(1 x 0.50 x 10) + (0.50 x 1 x 20) = (0.75 + 1 x 0.5^2 + 0.50 x 1^2) w

w = 10 rad/s

(C) Ki = 1 x 10^2 /2 + 0.5 x 20^2 /2 = 150 J

Kf = I w^2 = (0.75 + 1 x 0.5^2 + 0.50 x 1^2) 10^2 / 2 = 75 J

deltaK = Kf - Ki = - 75 J

(D) 0

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