# Kinetic Energy Ratio/Rotational Equilibrium

1. Feb 21, 2012

### Victorzaroni

1. The problem statement, all variables and given/known data

Two equal masses, each m, are resting at the ends of a uniform rod of length 2a and negligible mass. The system is in equilibrium about the center C of the rod. A piece of clay of mass m is dropped down on the mass at the right end, hits it with velocity v as shown below and sticks to it.

The ratio of the kinetic energy Ef just after the collision to the kinetic energy Ei just before the collision, Ef/Ei, would be:

(A) 1
(B) 3/4
(C) 2/3
(D) 1/2
(E) 1/3

2. Relevant equations

Not really sure.

3. The attempt at a solution

If the system is initially in equilibrium, then it's not moving, and I don't understand how it could have kinetic energy if it's not moving. I want to write: Ef/0, but that's obviously wrong.

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2. Feb 21, 2012

### tiny-tim

Hi Victorzaroni!
"The system" includes the clay, and that is moving!

3. Feb 21, 2012

### Victorzaroni

ooohhh. wow i feel dumb. that makes sense. so its the initial kinetic energy of the system is the kinetic energy of the clay.

4. Feb 21, 2012

### Victorzaroni

So Ei=(1/2)mv2. Now what? I know torque is involved somehow.

5. Feb 21, 2012

### tiny-tim

torque has nothing to do with it

now find Ef (as a function of vf)

6. Feb 21, 2012

### genericusrnme

Energy is a function of velocities (kinetic) and positions (potential), energy isn't a function of accelerations and torques

7. Feb 22, 2012

### Victorzaroni

Alright... So the new kinetic energy after the clay has stuck to the mass would be KE=(1/2)(M+M)r2? How do I account for the mass on the other end of the board?

8. Feb 22, 2012

### tiny-tim

(it'll have the same speed, won't it? )​

9. Feb 22, 2012

### Victorzaroni

Alright so I have: KEi=(1/2)Mv^2, and KEf=(1/2)(M+M+M)v^2. Dividing KEf by KEi yields (3/2)/(1/2), which is 3, which is not a choice? Dividing them the other way yields 1/3, choice E...

10. Feb 22, 2012

### tiny-tim

i'm lost

what conservation equation are you using?

11. Feb 22, 2012

### Victorzaroni

I'm not using conservation, KEf/KEi is what the question is asking for and the other stuff is what you've been saying. lol

12. Feb 23, 2012

### tiny-tim

how are you going to find the relation between before and after without a conservation law?

13. Feb 23, 2012

### Victorzaroni

So how do i use a conservation law?

14. Feb 23, 2012

### tiny-tim

what do you think will be conserved?

(and i'll repeat my previous question, since i'm still wondering …

how were you intending to find the relation between before and after without a conservation law?)​

15. Feb 23, 2012

### Victorzaroni

The kinetic energy of the clay equals the rotational kinetic energy of the system?

16. Feb 23, 2012

### tiny-tim

no, this is a perfectly inelastic collision
… so energy is not conserved