1. Apr 19, 2007

### Wurz

A horizontal disk of diameter 12.0cm is spinning freely about a vertical axis through its centre at an angular speed of 72 revolutions per minute. A piece of putty of mass 5.0g drops on to and sticks to the disk a distance of 4.0cm from the centre. The angular speed reduces to 60 revolutions per minute.

1. I need to calculate the moment of inertia of the disc. I can assume that no external torques are applied to the system during this process.

2. A constant tangential force is now applied to the rim of the disk which brings it to rest in 6.0s. Calculate the magnitude of this force.

3. Calculate the rotational energy of the system before and after the putty is added to the disc.

2. Apr 19, 2007

### denverdoc

wurz, we need to have you show some work or thought about the problem before helping out. I'll add a suggestion that maybe the first question is like a collision which sticks, where Mvi=(M+m)vf where vi and vf are initial and final velocities.

3. Apr 19, 2007

### Atomicbomb22

I think you must use these equations for the questions: L=lw and you must know that I=mr^2. You must also use the analogous equations--similar to those in linear mechanics-- of rotational mechanics equations to solve the problem.

Just plug in and solve.

4. Apr 22, 2007

### Wurz

I understand Denverdoc, thanks for the lead... I think I've got a good answer using the conservation laws: m1v1 = m2v2, and I=mr^2
the answer is very small however, with I = 2.00x10-5kg m^2 does this sound anything like what you would expect??

5. Apr 22, 2007

### Wurz

6. Apr 22, 2007

### denverdoc

Wurx,

uh, might have taken what I suggersted too literally, Its a rotational problem using angular quantities and conservation of these. Without doing the work looks ballpark where 72X=(x+2)60

7. Apr 22, 2007

### hage567

I think you are missing a factor of 2 in your answer. Use conservation of angular momentum, I'm sure that's what the question intended.

8. Apr 22, 2007

### hage567

How did you get a 2 in there? I'm assuming X and x are representing the same thing.

9. Apr 22, 2007

### denverdoc

Yes, the moment of inertia of the disk. For that I recall being 1/2mr^2, while the point is mr^2

Last edited: Apr 22, 2007