# Homework Help: Rotational Collision

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1. Mar 2, 2015

### Soniteflash

1. The problem statement, all variables and given/known data
A 2kg , 0.2 m diameter turntable rotates at 100 rpm on frictionless bearings. Two 0.5 kg block fall from above, hit the turntable simultaneously at opposite ends of the diameter, and stick. What is the turntable's angular velocity (in rpm) just after?

2. Relevant equations
L= I x ω
Li = Lf
Moment of inertia for disk I = (1/2) MR2
Moment of inertia for point mass = MR2

3. The attempt at a solution
I converted 100 rpm to rotations per sec by dividing by 60 which gives me (5/3).
I thought about the problem and came up with the idea of conservation of angular momentum so
Li = Lf.

Li = I x ω= (1/2)MR2 x ω. I am confused. The answer has to be in rpm but ω has the SI units of rad/s. Does unit consistency play a role here since I have ω on both sides of the equation.

I thought that before doing it with numbers, using only variables to make sure I understand what I am doing.

Lf= Isys x ωf So my question here is, does this reflect what is happening when the blocks stick to the plate? I reasoned that the blocks added moment of inertia to the system.

I continued and
Isys = [((1/2)MR2 x ω) + ( MBlock R2 + MBlock R2)].

Is that correct?

Then I set them equal with Li =Lf
and solved for angular velocity.

I continued and plugged in the given values and got a completely different answers. I feel like my approach is missing out on something.

2. Mar 2, 2015

### SammyS

Staff Emeritus
There should be no ω in this expression for moment of inertia.

Isys = [((1/2)MR2 × ω) + ( MBlock R2 + MBlock R2)].

Is that correct?

3. Mar 2, 2015

### Staff: Mentor

Careful, rotations per second is not the same as radians per second. How many radians in a rotation?
Since rpm and radians per second are related by a proportionality constant you can get away with using "100 rpm" as the angular velocity in your equations. Note that the value you get for angular momentum will have "funny units" as a result, but that's resolved when you solve for the new angular velocity where the results will be in rpm.
Yes, that's right.
There's no "ω" in the moment of inertia calculation. ω gets involved when you are looking for the angular momentum.
Your problem may be due to the extraneous "ω" in your moment of inertia calculation, but it's hard to tell without seeing more of your intermediate steps.

4. Mar 2, 2015

### Soniteflash

I see the mistake with omega!
For the conversion from rotations to radians. So i got the proportionality constant which gives me the number of rotations per second (5/3 rotations per sec). Then I can can multiply the PC by 2 PI and that gives me the radians?

5. Mar 3, 2015