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A Spider on a rotating disk conservation of momentum

  1. Jul 30, 2014 #1
    1. The problem statement, all variables and given/known data
    A 45 rpm record in the shape of a solid disk 25 cm in diameter and mass 0.1 kg rotates about a vertical axle through its center. A 15 g spider rides along the edge of the record. Calculate the final angular speed of the record if the spider drops off without exerting a torque on the record.


    2. Relevant equations
    Conservation of momentum



    3. The attempt at a solution
    So I attempted to use the conservation of momentum to solve this problem. So
    I first found the platforms moment of inertia
    I= 1/2Mr^2
    I = 1/2(.1)(.125)^2
    I = 7.8125 x 10^-4
    Then I found the spiders moment of inertia
    I = MR^2
    I = (.015)(.125)^2
    I = 2.34375 x 10 ^-4
    Then I found the momentum
    L = IW
    L = (7.8125 x 10^-4 + I = 2.34375 x 10 ^-4) (4.7129 rad/sec)
    L = .0047860211
    Then I compared that to the final momentum after the spider dropped off the disk
    .0047860211 = (7.8125 x 10^-4)w
    w = 6.126107 rad/sec or 58.50001265759 rpm. The correct answer is 45 rpm
    I = (.015)(
     
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  3. Jul 30, 2014 #2

    rcgldr

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    Angular momentum of the disk and spider is constant. What happens to the spider's component of angular momentum if it slides off without exerting a torque onto the disk?
     
  4. Jul 30, 2014 #3
    The spiders component becomes zero and is made up for by the increase in angular velocity.
     
  5. Jul 30, 2014 #4

    Orodruin

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    What is the angular momentum of a body in constant linear motion about a point not directly in its path?
     
  6. Jul 30, 2014 #5
    Its mass times its velocity? I don't get what you are saying.
     
  7. Jul 30, 2014 #6

    Orodruin

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    Angular momentum, not linear momentum. In particular, what is the angular momentum of the spider relative to the center of the disk?

    Also, if you make a free body diagram of the disk, what forces and torques are acting on it?
     
  8. Jul 30, 2014 #7
    Angular momentum is the angular velocity times the moment of inertia.
     
  9. Jul 30, 2014 #8

    Orodruin

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    So what is the angular velocity of the spider in said situation?

    Edit: Also consider ##{\bf L} = {\bf r}\times {\bf p}## for constant linear motion ...
     
  10. Jul 30, 2014 #9
    4.71239 rad/sec
     
  11. Jul 30, 2014 #10

    Orodruin

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    So what makes you think the angular momentum of the spider is zero?
     
  12. Jul 30, 2014 #11
    Because it is flung off the disk.
     
  13. Jul 30, 2014 #12

    Orodruin

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    This is not sufficient. Its angular momentum relative to any point on its trajectory is zero, but the angular momentum you have computed is not wrt a point on its trajectory.
     
  14. Jul 30, 2014 #13
    I'm sorry but I don't understand your point.
     
  15. Jul 30, 2014 #14

    rcgldr

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    The spider will move in a straight line that doesn't not intersect the center of the disk. It's angular momenum is equal to it's linear momentum (mv) times the shortest distance from the center of the disk to the linear path that the spider follows. In this case that distance is equal to the radius of the disk. You can also use the formula where the angular momentum = m v r sin(θ), where θ is the angle between the path line and a line that starts at the center of rotation, and crosses the path at some specific point. At the edge of the disk, θ would be 90°, so sin(θ) would be 1.
     
    Last edited: Jul 30, 2014
  16. Jul 30, 2014 #15
    I thought I included that in my calculation. I did the sum of all the moment of inertia's and then used that to find the total angular momentum. Then I removed the spiders moment of inertia and calculated what the speed had to be for the two momentum's to be equal.
     
  17. Jul 30, 2014 #16

    Orodruin

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    Angular momentum is always computed relative to a paricular point in space. In your case, the angular momentum (that is conserved) is computed around the center of the disk. You must therefore also express the angular momentum of the spider with respect to this. Hint: angular momentum for constant linear motion is the linear momentum multiplied by the shortest distance between the trajectory and the reference point.

    The alternative to using conservation of angular momentum is to draw the free body diagram for the disk. If the spider exerts no torque on the disk, what is the net torque on the disk?
     
  18. Jul 30, 2014 #17
    I'm very confused how the spider still be effecting the momentum of the disk when it is not on the disk?
     
  19. Jul 30, 2014 #18

    rcgldr

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    That's part of the issue. Since the spider slips off the disk without exerting any torque on the disk, then the spiders component of angular momentum doesn't change.
     
  20. Jul 30, 2014 #19

    rcgldr

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    In this case it never effects the angular momentum of the disk. It doesn't generate any torque on the disk when it slides off. It continues to move in a straight line that is tangent to the outer edge of the disk.
     
  21. Jul 30, 2014 #20

    Orodruin

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    It is not doing anyhing with the disk momentum by definition (no torque), but it is the total momentum that is conserved (that of the spider+disk system).
     
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