POD30Prob#4) Rotational Inertia Two skaters, each of mass m 58kg are moving towa

Question

POD30Prob#4) Rotational Inertia Two skaters, each of mass m 58kg are moving toward each other along parallel paths separated by a distance of 4.0 m. They have opposite velocities of 1.6 m/s each. One skater carries on end of a long pole of negligible mass and the other skater grabs the other end as she passes. The skaters then rotate around the center of the pole. Assume that the friction on the ice is negligible 2 a. What is the radius of the circle formed by the skaters? b. What is the angular speed of the skaters? c. What is the total kinetic energy of the system? d. Now, imagine that the skaters reduce the distance between them (still hanging on to the pole) to a distance of 1.5 m. Compute the angular speed and kinetic energy of the system, and explain your answer...why did this happen?

Explanation / Answer

(A) r= L/2 = 4/2 = 2 m

(B) w= v /r = 1.6 / 2 = 0.8 rad/s

(C) KE = I w^2 /2 = 2(m r^2) (v / r)^2 / 2

Ke = m v^2 = 58x 1.6^2

KE = 148.5 J

(D) angular momentum conservation,

Li = Lf

2 ( 58 x 1.6 x 2) = 2(58 x v x 1.5/2)

vf = 4.3 m/s

wf = vf /r =5.7 rad/s

Kf = (58 x 4.3^2) = 1056 J  

kinetic energy increases. this energy comes from work done by skater.

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