# Pivot Putty Question

1. Nov 20, 2011

### sarahmlipinsk

pivot putty question? Any help would be greatly appreciated!

Last edited by a moderator: May 5, 2017
2. Nov 20, 2011

### grzz

Try to use conservation of angular momentum of the whole system about the pivot of the rod.

3. Nov 20, 2011

### sarahmlipinsk

I tried that. I compared the energy lost to the initial kinetic energy with the equation: KEi-KEf/KEi and then substituted my equations and got:
(1/2mv^2-1/2*1/3ML^2+mL^2*mvL)/1/2mv^2
but that didn't work...

4. Nov 20, 2011

### nasu

Your formula is unclear and seems to have wrong dimensions in some terms.
What is the final kinetic energy?

5. Nov 20, 2011

### sarahmlipinsk

I had the final kinetic energy as:
(1/2)(1/3)ML^2+mL^2(MvL)

6. Nov 20, 2011

### sarahmlipinsk

And my initial kinetic energy was:
(1/2)mv^2

7. Nov 20, 2011

### nasu

This is not a kinetic energy.
Must be something wrong in your calculations.
If you mean
$$\frac{1}{2}\frac{1}{3}ML^2+mL^2 MvL$$
then the two terms have diferent dimensions.
If you mean
$$(\frac{1}{2}\frac{1}{3}ML^2+mL^2 ) MvL$$
then it is a valid formula but it's not kinetic energy.
Try to use enough brackets to make the formulas unambiguous.

8. Nov 20, 2011

### sarahmlipinsk

The beginning part was supposed to be the moment of inertia and the second part is the omega.

9. Nov 20, 2011

### nasu

You mean MvL is the angular speed?
You realize also that the final velocity is not the same as initial velocity, right?

10. Nov 20, 2011

### sarahmlipinsk

Yes and yes.

11. Nov 20, 2011

### nasu

Then you have some conceptual confusions.
The angular speed is omega=v/R or in this case, v/L.
MvL is not angular speed.

12. Nov 20, 2011

### nasu

How did you write conservation of angular momentum? You need this to find final velocity.

13. Nov 20, 2011

### sarahmlipinsk

Oh, I see what you mean. I tried it that way, with no luck. I think my formula is just wrong.

14. Nov 20, 2011

### sarahmlipinsk

I'm not quite sure. I said the moment of inertia for the rod is (1/12)ML^2
And the moment of inertia for the putty is mL^2

15. Nov 20, 2011

### nasu

Try to find the velocity after collision.
The equality of angular momentum before (just the putty ball) and after collision (putty ball + rod) will provide an equation to calculate this velocity. Let's call it vf. Do you know how to do this?
Then the KE after collision will be $$\frac{1}{2} I \omega_{f}^2= \frac{1}{2} I (\frac{v_f}{L})^2$$ where I is the total moment of inertia. I see that you got the total moment of inertia right, by the way.

16. Nov 20, 2011

### sarahmlipinsk

No, I did not know how to do that... so then the final velocity would be equal to the final angular velocity times L?

17. Nov 20, 2011

### nasu

Yes, this is the relationship between angular and linear velocities.
Angular momentum relative to the pivot is
initial:Li= mvL
final: $$L_f=I \omega_f = I \frac{v_f}{L}$$
where I is the total angular momentum.
I hope now you can solve it.

18. Nov 20, 2011

### sarahmlipinsk

Okay, I think I got it. The only thing is,how do I find the final angular velocity?

19. Nov 21, 2011

### nasu

By equating the angular monetum before and after collision and solving the equation for final velocity (angular or linear).

20. Nov 21, 2011

### sarahmlipinsk

Okay, I got it! Thank you so much for all of your help!