A merry-go-round rotates at the rate of 0.20 rev/s with an 85-kg man standing at

Question

A merry-go-round rotates at the rate of 0.20 rev/s with an 85-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 0.9 m from the center? Assume that the merry-go-round is a solid 25-kg cylinder of radius 2.0 m.
4.12

rad/sec

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



Your response differs from the correct answer by more than 100%. J
How do you account for this change in kinetic energy?
Wind resistance varies as the square of the speed.
The moment arm of the man about the axis of rotation decreases.
Work is done by the merry-go-round on the man as he moves towards the center.
Work is done by the man on the system as he moves towards the center.

Explanation / Answer

initial K.E of the person and man

.5(Im+Ip)(.2*2*)^2=0.5*(.5*25*2^2+ 85*2^2)(.2*2*)^2=307.93 J

 

final K.E of the person and man

=.5(Im+Ip')(4.12) ^2=1008.7 J

so change in K.E= 1008.7-307.93=700.8 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.