A merry-go-round rotates at the rate of 0.25 rev/s with an 80-kg man standing at

Question

A merry-go-round rotates at the rate of 0.25 rev/s with an 80-kg man standing at a point 2.0 m from the axis of rotation.

(a) What is the new angular speed when the man walks to a point 0.5 m from the center? Assume that the merry-go-round is a solid 25-kg cylinder of radius 2.0 m.
________ rad/sec

(b) Calculate the change in kinetic energy due to this movement.
_______ J

How do you account for this change in kinetic energy?

A. Work is done by the merry-go-round on the man as he moves towards the center.

B. The moment arm of the man about the axis of rotation decreases.

C. Work is done by the man on the system as he moves towards the center.

D. Wind resistance varies as the square of the speed.

Explanation / Answer

I_disk = M R^2 /2 = 25 x 2^2 /2 = 50 kg m^2

initially, Ii = 50 + (80 x 2^2) = 370 kg m^2

wi =0.25 rev/s = 0.25 x 2pi rad/s = 1.57 rad/s

If = 50 + (80 x 0.5^2) = 70 kg m^2

(A) Applying angular momentum conseravtion,

Ii wi = If wf

370 x 1.57 = 70 wf

wf = 8.30 rad/s ........Ans

(B) Ki = Ii wi^2 / 2 = 456 J  

Kf = 70 x 8.30^2 /2 = 2411 J  

change in kE Kf - Ki = 1955 J ........Ans

Ans: C. Work is done by the man on the system as he moves towards the center.

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