# Rotation of a turntable

1. Jun 1, 2009

### talaroue

1. The problem statement, all variables and given/known data
A 2.05 kg, 33.64 cm diameter turntable rotates at 365 rpm on frictionless bearings. Two 280 g blocks fall from above, hit the turntable simultaneously at opposite ends of a diagonal, and stick. What is the turntable's angular velocity, in rpm, just after this event?

2. Relevant equations

I= m(r^2)/2
Ki=I(w^2)/2

3. The attempt at a solution

I changed the 365 rpm to 38.22 rad/s
Then i found that the Ki=mt(r^2)(wi^2)/2
Then i found that the Kf=(mt+mb1+mb2)(r^2)(wi^2)/2)

Then i solved for wf and got the equation mt*wi^2/(the mass of the turntable and blocks). Then the square root of that and got 33.87 rad/s

Then turned 33.87 rad/s to rpm and got 323.46.....Did i use the wrong I maybe? I used the I from above. and combined it with the Kinetic energy.

2. Jun 1, 2009

### Staff: Mentor

You cannot assume that kinetic energy is conserved. But something else is.

3. Jun 5, 2009

### talaroue

momentuem.....ahhh i see. stupid mistake on my part.