# Change in rotational kinetic energy

1. Oct 26, 2016

### Minestra

1. The problem statement, all variables and given/known data
A disk of mass m1 is rotating freely with constant angular speed ω. Another disk of mass m2 that has the same radius is gently placed on the first disk. If the surfaces in contact are rough so that there is no slipping between the disks, what is the fractional decrease in the kinetic energy of the system? (Use any variable or symbol stated above as necessary. Enter the magnitude.)

It wants ΔK/Ki

3. The attempt at a solution
I really don't think I understand what is being asked of me here. I have attempted to find ΔK by using the moment of inertia of a disk, and then adding a second disk of mass m2. But that got me nowhere. I wish I could offer up a better attempt but I'm at a loss.

2. Oct 26, 2016

### Staff: Mentor

Think collision. What type of collision is it? What's conserved?

3. Oct 26, 2016

### Minestra

Well I do believe it's conservation of momentum, but I'm having trouble figuring out what they're asking me to solve for once I write out the equation.

4. Oct 26, 2016

### Staff: Mentor

You wrote:
So it looks like they want you to find the fraction of the rotational KE that was lost in the collision.

5. Oct 26, 2016

### Minestra

So I would solve for the new velocity and use that to find the kinetic energy after the collision?

6. Oct 26, 2016

### Minestra

Then I would basically set the difference over the original and that would be the answer?

7. Oct 26, 2016

### Staff: Mentor

That would be my thinking, yes.

8. Oct 26, 2016

### Minestra

Thanks for talking it out with me, gotta work an overnight now, I'll do the algebra tomorrow.

9. Oct 26, 2016

### Minestra

So I think I solved the problem do you mind looking over my work?

#### Attached Files:

• ###### IMG_20161026_234556255.jpg
File size:
20.8 KB
Views:
40
10. Oct 26, 2016

### haruspex

Posting algebra as images, even if readable, makes it hard to make comments referencing specific lines. At the least, please number the equations, but better still, please take the trouble to type the algebra into the post.
Your third equation looks like conservation of energy, which is clearly not valid.
There are two errors in the next equation.

11. Oct 26, 2016

### Staff: Mentor

You should express $I_1$ and $I_2$ in terms of m and r. The r's are the same for both disks so you can be pretty sure that the r's will disappear somewhere along the way. But the important thing is to incorporate the given information that the disks differ in their mass.

12. Oct 26, 2016

### Minestra

Sorry for the photo reply I'm on my moble. I'll do my best to type it out once I'm back on my desktop.

If I rework using I = mr2 and ω = v/r, would the rest of my thought process be correct in using the angular velocity at moment two, to find a new kinetic energy, which could then be subtracted from the original to determine its change?

13. Oct 26, 2016

### haruspex

Not the right formula for a disk.
As I mentioned, there are errors in some of your previous equations.

14. Oct 27, 2016

### Staff: Mentor

Leave the ω's alone. There's no need to introduce linear velocity. The idea is to incorporate the different masses, hence the differing moments of inertia for the disks. Just use the (correct) form for the MOI of a disk and analyze the collision.

15. Oct 27, 2016

### Minestra

Sorry for the long delay I have been busy with work.

Here is my attempt at this problem:

Angular Momentum:

$$\frac{1}{2}m_{1}r^{2}\omega_{1} = \frac{1}{2}(m_{1}+m_{2})r^{2}\omega _{2}$$

solving for ω2 and canceling out what you can, you get:

$$\omega_{2} = \frac{m_{1}}{m_{1}+m_{2}}\omega _{1}$$

Next we take this new ω which should be our speed after collision and substitute it into the kinetic energy equation:

$$\frac{1}{4}m_{1}r^{2}\omega _{1}^{2} = \frac{1}{4}(m_{1}+m_{2})r^{2}\frac{m_{1}}{m_{1}+m_{2}}$$

Setting this equal to zero would have you subtract kinetic energy, thus giving you the difference as follows:

$$\frac{1}{4}m_{1}r^{2}\omega _{1}^{2} - \frac{1}{4}(m_{1}+m_{2})r^{2}\frac{m_{1}}{m_{1}+m_{2}} = \Delta K$$

Putting this over Ki as the question requests would get you:

$$-\frac{1}{4}r^{2}m_{1}$$

Which of course results in an incorrect answer from web-assign. I understand I made some errors somewhere, I would greatly appreciate being told exactly what I did wrong. I know it's not the best for my education but I need to submit the assignment by tomorrow night and my professor is grossly unavailable when it comes to help (numerous emails throughout the semester with not a single response, I have confirmed with other students that he does this) and I fear I will not be able to solve it.

16. Oct 27, 2016

### haruspex

Not sure what you are doing there. I think what you mean is that the expression on the left is initial KE and the expression on the right is the final KE. They are clearly not equal.
If so, you have made a mistake or two in substituting for ω2 on the right. Try that step again.