# Conservation of Angular Momentum

## Homework Statement

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?

I1ω1=I2ω2

## 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?

ehild
Homework Helper
The radii are different for the children.

ehild

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

ehild
Homework Helper

ehild

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

ehild
Homework Helper

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

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

ehild

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?

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

ehild

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

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

ehild