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A thin vertical rod sliding until it hits a lump of putty and rotates.

  1. May 10, 2013 #1
    1. The problem statement, all variables and given/known data
    A thin vertical rod of mass m and length d slides on a frictionless horizontal surface with velocity Vo to the right. It his and sticks to a small lump of putty and rotates about the putty at point O.
    (a) Find the angular momentum of the rod about O just before and just after it sticks to the putty.
    (b) Find the angular speed, w, of the rod just after it sticks to the putty.
    (c) find the speed of the free end of the rod just before and just after the rod sticks to the putty. Does the result surprise you?
    (d) What fraction of the initial kinetic energy is lost when the rod sticks to the putty?


    2. Relevant equations
    L=Iw=rxmv
    v=p/m
    KEi-KEf/KEi
    I=m(a^2)/3 for normal to rod at one end
    I=m(a^2)/12 for normal to rod at its center
    3. The attempt at a solution
    For part (a) I said L=Iw=rxmv=0 since r=0 and just after it sticks L=rxmv=mvd. I am not sure if the second part of this is right.

    For part (b) I said Iw=mvd so w=mvd/I=mvd/((md^2)/3)=3v/d
    I am confused on part c. I know that initially all the points on the rod move with a velocity Vo. After the rod sticks, the motion will be rotation about the axis through the putty; each part of the rod will have a different linear speed. This is speed, not velocity, so I do not need the direction of motion, but just the scalar speed. I just am not sure how to find this value. I assume I need to use energy conservation

    then for (d) KEi-KEf/KEi so (1/2)Irw^2 -(1/2)(Iw^2)/(1/2Irw^2) where Ir is moment of inertia for the rod and I is moment of the rod stuck to putty. I think I=((md^2)/3) +md^2 and Ir=((md^2)/3) since looking at normal to rod at one end
     
  2. jcsd
  3. May 10, 2013 #2

    Doc Al

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    Staff: Mentor

    Are you saying that the angular momentum before the rod sticks is zero? That isn't true. Hint: What's the definition of the angular momentum of a particle?
     
  4. May 10, 2013 #3
    The definition for a particle is L=mvrsinθ and θ=0 if it is moving in the horizontal direction and not rotating yet.
     
  5. May 10, 2013 #4

    Doc Al

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    In that definition, θ is the angle between the position vector r (which is vertical) and the velocity v (which is horizontal). So θ = 90°, not 0.
     
  6. May 10, 2013 #5
    oh right thanks!
     
  7. May 10, 2013 #6
    so L=mvd before and mvdsinθ after?
     
  8. May 10, 2013 #7

    Doc Al

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    The first thing you have to do is properly determine the angular momentum before the collision with the putty. Treat the rod as a stack of mass elements, each with their own distance from O. That will get you the answer to part (a).

    After the collision, the rod rotates about point O.
     
  9. May 10, 2013 #8
    Right, so L=Ʃri x (mi)(vi) and Ʃmi=m and v is the same for all the elements and Ʃr=d (length of the stick so why is L=mvd incorrect for before the rod hits the putty?
     
  10. May 10, 2013 #9
    Also, I do not understand conceptually how the rod can have angular momentum before it rotates. I thought to calculate angular momentum the object must be rotating.
     
  11. May 10, 2013 #10

    Doc Al

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    Each mass element of the stick has a different value of r. You must integrate.
     
  12. May 10, 2013 #11

    Doc Al

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    A particle isn't rotating, yet we saw before that it has an angular momentum about some point O. Same deal with the stick. (We are not saying that it has angular momentum about its center, just about point O.)
     
  13. May 10, 2013 #12
    Can I just use the center of mass so I do not have to do all those integrals? Our professor said the math wouldn't be very long on this problem.
     
  14. May 10, 2013 #13

    Doc Al

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    In general, you cannot do that. But in this case it will give you the correct answer.

    But try to set up the integral anyway. It's an easy one. (The hard part is setting it up.)
     
  15. May 10, 2013 #14
    So ∫ (r x mv dr) where the integral is from 0 to d? I am not sure how to integrate that then. I am just very confused.
     
  16. May 10, 2013 #15

    Doc Al

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    Almost. It should be like this: ∫V0 r dm. You have to express dm in terms of dr. Then you can integrate.
     
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