Question

A merry-go-round rotates at the rate of.2rev/s with an 80kg man standing at a point 2m from the axis ofrotation. (A)What is the new angular speed when the man walksto a point 1m from the center? Assume that the merry-go-round is a solid 25kg cylinder of radius of 2m.
B. calculate the change in kinetic energy due to man moving
C. how do you account for this change in kinetic energy?

Answer 1

We will gladly critique your work

Answer 2

angular momentum (L) = Inertia (I) * angular velocity (w)

Using conservation of angular momentum (L):
Inertia initial total (Ii) * angular velocity initial (wi) = Inertia final total (If) * angular velocity final (wf)

Inertia of disks (such as the merry-go-round) = .5 * mass (m) * radius squared (r)

Inertia of a point-mass (such as the person) = m * r^2

with this you should get the following:

Li=Lf
Ii*wi=lf*wf
(I + Ii)*wi=(I + If)*wf
(.5MR^2 + mri^2)wi=(.5MR^2 + mrf^2)wf

where
M=mass of the merry-go-round
R=radius of the merry-go-round
m=mass of the person
ri=initial radius of the person

re-arrange and solve for wf for part a

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change in KE(rotational) is KEf-KEi

KE(rot) = .5*I*w^2

you should get a positive change. This makes sense because if you think of radius between the person and the pivot-point as the distance associated with Potential Energy, as the distance decreases, the speed along with Kinetic energy should increase due to conservation of energy.

PEi + KE(rot)i = PEf + KE(rot)f
big + small = small + big

change in KE = big - small = positive change

I hope that makes sense :)