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Conservation of Angular Momentum

  1. Nov 24, 2008 #1
    1. The problem statement, all variables and given/known data
    A 500 gram piece of putty falls straight down and sticks to a 4 kg rotating solid disk below. The uniform disk has a radius of 3 meters, and the putty hits the disk 2.5 meters from its center. You may neglect all forces other than that of the collision.
    a- If the disk was initially rotating at 10.0 rad/s, find its final angular velocity.
    b- The putty takes 15 ms to collide with the disk. Calculate the avergae force on the putty.


    2. Relevant equations
    I = mr^2 and .5 mr^2, L = I omega


    3. The attempt at a solution

    For part a,
    I found moment of inertia of the disk (I_d) = 1/2 * 4kg* (3m) ^2 =
    18 kg m^2
    moment of inertia of putty = .5kg*(2.5m)^2 = 3.13 kgm^2
    Since the putty falls straight down, i assumed the initial angular momentum of the putty was 0. So the initial angular momentum of the system is L_d = 18kgm^2*10 rad/s) = 180 kgm^2/s
    Lf = Li, so (I_p + I_d) * angular velocity = 180, angular velocity = 8.52 m/s.

    I'm really unsure about this, so I just wanted someone to doublecheck it. Also, I dont really know how to find the average force for part b.
     
  2. jcsd
  3. Nov 24, 2008 #2
    Part (a) is right, except angular velocity isn't m/s.

    Hint for part (b): Look up the impulse-momentum theorem
     
  4. Nov 24, 2008 #3
    I looked up the impulse-momentum theorem, and I see that force = change in velocity over time. How can I apply this if I don't know the initial velocity of the putty? Can I use angular velocity and say vf - vi = 8.52 rad/s- 0? So force on putty would just be 8.52/.015 seconds?
     
  5. Nov 24, 2008 #4
    1. That is not an accurate statement of the theorem. Re-look at it.
    https://www.physicsforums.com/library.php?do=view_item&itemid=53

    2. No, you cannot simply use the angular velocity. You have to convert it into a linear velocity. How?

    You do not know the vertical velocity, so you are right, you cannot say anything about the vertical force that acts on the putty.
     
  6. Nov 24, 2008 #5
    I converted the final angular to linear using omega = v/r
    8.52 = v/ 2.5
    v= 3.41

    Since the putty is dropped straight down, can I say the initial velocity in the horizontal was 0 and the final is 3.41 m/s?
     
  7. Nov 25, 2008 #6
    Answer to your question: Yes.

    Whether 8.52 = 3.41 / 2.5: No.
     
  8. Nov 25, 2008 #7
    OK, so now I have vf = 8.52rad/s * 2.5 m = 21.3 m/s

    p initial = 0
    p final = mvfinal = 21.3 m/s(.5kg) = 10.7 kg m/s

    pf - pi = 10.7 kg m/s = J

    Force avg = J/ time = 107. kg m/s / .015 s = 713 N

    Please let me know if I have done this correctly.
     
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