# Homework Help: Real Question This Time about a Merry go Round

1. Jun 6, 2007

### MJC8719

A disk-shaped merry-go-round of radius 2.74 m and mass 168 kg rotates freely with an angular speed of 0.737 rev/s. A 51.7 kg person running tangential to the rim of the merry-go-round at 2.86 m/s jumps onto its rim and holds on. Before jumping on the merry-go-round, the person was moving in the same direction as the merry-go-round's rim. What is the final angular speed of the merry-go-round?

Last edited: Jun 6, 2007
2. Jun 6, 2007

### MJC8719

Actually, ignore all of this...the kinetic energy is not relevant...i need to focus on angular momentum

therefore the equation should be:
rmv(child) + Iw = (i + mr^2)(w)

is this correct?

3. Jun 6, 2007

### MJC8719

So, when I tried to solve this i got:

(2.74)(51.7)(2.86) + (1/2(168(2.74^2)(0.737) = [(.5(168)(2.74^2) + (51.7 x 2.74^2)] w

then i solved for w and got 0.8539 which is not correct

what am i missing?

4. Jun 7, 2007

### Staff: Mentor

In order for Iw to equal angular momentum, w must be given in standard units of radians/sec (not rev/sec).