# Boy and Merry-Go-Round

1. Apr 19, 2004

### masamune

A boy of mass m = 50 kg running with speed v = 4 m/s jumps onto the outer edge of a merry-go-round of mass M = 150 kg and radius R = 2 m, as shown in the picture above. The merry-go-round is initially at rest, and can rotate about a frictionless pivot at its center. You may assume that the inital velocity of the boy is tangent to the edge of the merry-go round.

Treat the boy as a point particle and the merry-go-round as a uniform solid disk. What is the angular velocity of the merry-go-round after the boy has jumped onto it?

I don't know if I can do this, but I set the linear momentum of the boy equal to the angular momentum of the merry-go-round with the boy.
Basically, mv = Iw
For my moment of inertia, I used the sum of both masses and plugged my given information into ((M+m)R^2)/2. This was how I calculated moment of inertia. Then I plugged the boy's mass and his initial speed divided by my moment of inertia and tried to get omega (w). I got 0.5 exactly, but it's not correct. Any help would be appreciated.

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2. Apr 20, 2004

### cair0

linear momentum and angular momentum are not the same thing.

What you want to do is set the boys rotational inertia to the rotational intertia of the entire system.

you should get something like

mvr = Iw

I is the I of the system, im sure you can figure that out..

3. Apr 20, 2004

### masamune

Is my method of calculating the I of the system correct? I took the sum of the boy and the merry-go-round, multiplied by the square of the radius and all that divided by 2.This should give me the the I of the ystem right?

4. Apr 20, 2004

### cair0

you need to add the I of the boy and the I of the disk

I for a uniform disk rotating about the center of mass is $$\frac{1}{2}MR^2$$

I for a point mass is $$MR^2$$

add them together you get $$(\frac{1}{2}M_{merry-go-round} + M_{boy})R^2$$