# Conservation of Angular Momentum

1. Oct 31, 2012

### icf927

1. The problem statement, all variables and given/known data
A solid cylinder merry-go-round of mass 250kg and radius of 1.9m spinning at 1 revolution every 5 seconds has a 40kg child sitting at 1.1m from the axis. A 50kg child, running tangentially at 3m/s, jumps on the merry go round at the outer edge. What is the new rotation rate in rev/s?

2. Relevant equations
I1ω1=I2ω2

3. The attempt at a solution
(1/2MR^2 +m1r^2)ω1 + (m2r^2)(v/r)=(1/2MR^2 +m1r^2 +m2r^2)ω2
I plugged in all the variables and solved for ω2 and got 0.214 rev/s. Did I set up the equation correctly and come about the right answer?

2. Oct 31, 2012

### ehild

The radii are different for the children.

ehild

3. Oct 31, 2012

### icf927

yes I plugged in the different radii. I just forgot to put a 1 and 2 behind them in the equations.

4. Oct 31, 2012

### ehild

ehild

5. Oct 31, 2012

### icf927

solved for angular velocity(w2) and got 0.214rev/s. is this the right set up?

6. Oct 31, 2012

### ehild

You left out the first child on the left hand side.

ehild

7. Oct 31, 2012

### icf927

I did it on paper just forgot to type it on here. When I include the first child on the left hand side, is that the right setup?

8. Oct 31, 2012

### ehild

It is, but I got different result. Show calculations in detail.

ehild

9. Oct 31, 2012

### icf927

Did you change the initial angular velocity from revolution/sec to rad/sec and then change the final answer back to revolutions/sec

10. Oct 31, 2012

### ehild

Ops! You are right. I calculated w. So your result is correct.

ehild