Pivot Putty Question

  • #1

Main Question or Discussion Point

Can someone please help me solve this: http://wug.physics.uiuc.edu/cgi/courses/shell/per/phys111/ie.pl?11/IE_putty_plus_rod [Broken]
pivot putty question? Any help would be greatly appreciated!
 
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Answers and Replies

  • #2
993
13
Try to use conservation of angular momentum of the whole system about the pivot of the rod.
 
  • #3
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
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417
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...
Your formula is unclear and seems to have wrong dimensions in some terms.
What is the final kinetic energy?
 
  • #5
I had the final kinetic energy as:
(1/2)(1/3)ML^2+mL^2(MvL)
 
  • #6
And my initial kinetic energy was:
(1/2)mv^2

Sorry about the clarity issues!
 
  • #7
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I had the final kinetic energy as:
(1/2)(1/3)ML^2+mL^2(MvL)
This is not a kinetic energy.
Must be something wrong in your calculations.
If you mean
[tex] \frac{1}{2}\frac{1}{3}ML^2+mL^2 MvL [/tex]
then the two terms have diferent dimensions.
If you mean
[tex] (\frac{1}{2}\frac{1}{3}ML^2+mL^2 ) MvL [/tex]
then it is a valid formula but it's not kinetic energy.
Try to use enough brackets to make the formulas unambiguous.
 
  • #8
The beginning part was supposed to be the moment of inertia and the second part is the omega.
 
  • #9
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417
You mean MvL is the angular speed?
You realize also that the final velocity is not the same as initial velocity, right?
 
  • #10
Yes and yes.
 
  • #11
3,740
417
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
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How did you write conservation of angular momentum? You need this to find final velocity.
 
  • #13
Oh, I see what you mean. I tried it that way, with no luck. I think my formula is just wrong.
 
  • #14
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
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417
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 [tex]\frac{1}{2} I \omega_{f}^2= \frac{1}{2} I (\frac{v_f}{L})^2[/tex] where I is the total moment of inertia. I see that you got the total moment of inertia right, by the way.
 
  • #16
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
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No, I did not know how to do that... so then the final velocity would be equal to the final angular velocity times L?
Yes, this is the relationship between angular and linear velocities.
Angular momentum relative to the pivot is
initial:Li= mvL
final: [tex]L_f=I \omega_f = I \frac{v_f}{L}[/tex]
where I is the total angular momentum.
I hope now you can solve it.
 
  • #18
Okay, I think I got it. The only thing is,how do I find the final angular velocity?
 
  • #19
3,740
417
Okay, I think I got it. The only thing is,how do I find the final angular velocity?
By equating the angular monetum before and after collision and solving the equation for final velocity (angular or linear).
 
  • #20
Okay, I got it! Thank you so much for all of your help!
 

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