A merry-go-around rotates at the rate of 0.20 rev/s with an 80kg man standing at

Question

A merry-go-around rotates at the rate of 0.20 rev/s with an 80kg man standing at a point 2.0m from the axis of rotation.
a.What is the new angular speed what the man walks to a point 1.0m from the center?
p.s: I already have the answer for paer 'a', dont need it!
Assume that the merry go round is a solid 25kg cylinder of radius 2.0m.

b. Calculate the moment of inertia of the merry-go-round alone?
c. Calculate the moment of inertia of the man standing 2.0m from the center. Assume he is a point mass? (I=mr^2)
d. Find the initial moment of inertia of the man plus merry-go-round.
e. Write down the formula for conservation of angular momentum.
f. Solve for the final angular speed and plug in numbers to calculate it.

Explanation / Answer

b) moment inertia for merry go round as cylinder as is:
I merry= 1/2mr2 = 1/2(25)(2)2=50kgm2

c) I man = I = 1/2mr2 = 1/2 (80)(2)2 = 160kgm2

d) I merry + I man = 210kgm2

e) L = I

f)

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