Rolling, Torque, and Angular Momentum

[ranjaxt2012]

A girl of mass 28.1 kg stands on the rim of a frictionless merry-go-round of radius 1.58 m and rotational inertia 433 kg·m2 that is not moving. She throws a rock of mass 670 g horizontally in a direction that is tangent to the outer edge of the merry-go-round. The speed of the rock, relative to the ground, is 7.78 m/s. Afterward, what are (a) the angular speed of the merry-go-round and (b) the linear speed of the girl?

please help me solve this question.

Mass M = 28.1 kg
radius r = 1.58 m
R-Inertia = 433 Kg.m2
mass of rock = 670g = .67 kg (tangent to merry-go-round)
speed of rock = 7.78 m.s
now is the angular speed ω of merry-go-round = ?
we know ω = L/Inertia or (m.r.v)/inertia.
Now what is velocity of the system? is it 7.78?

if yes then

ω = [(28.1+.67)*1.58*7.78)]/433 ≈ 0.81674 rad/sec

is linear speed of the girl = 0.816748*r

 

[NascentOxygen]

Hi ranjaxt2012! http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

This sounds like a homework question, so should really be posted in the homework sub-forum.