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Homework Help: Missing Linear Momentum

  1. Nov 19, 2011 #1
    I'm not exactly sure what is wrong with my analysis for this problem concerning the conservation of angular and linear momentum.

    Problem Statement:
    Suppose you have a uniform disk of mass M and radius R that can rotate about its central axis. A particle with mass M and velocity V strikes the rim of the disk (along a path tangent to the disk), gets lodged into it, and causes the disk to spin. Show that the linear momentum of the system is conserved.

    Attempted Solution:
    I start by using the conservation of angular momentum:

    MVR = (0.5MR^2 + MR^2)ω

    The left side of the equation is the total angular momentum before the collision (it is just the angular momentum of the particle since the disk is stationary) and the right side is the angular momentum after the collision. The expression in parenthesis is the moment of inertia of the disk-particle system. With a bit of algebra you can conclude that:

    ω = (2V)/(3R)

    To analyze the linear momentum after the collision I look at the linear velocity of the center of mass of the disk-particle system. The center of mass is in between the center of the disk and the lodged particle, so the linear momentum is given by:

    (0.5Rω)(2M) = (V/3)(2M) = (2/3)MV

    However, the linear momentum before the collision is MV. Does anyone know what I am doing wrong?
  2. jcsd
  3. Nov 20, 2011 #2


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    The problem is if the disk rotates around a fixed axis or it is a free disk the particle is colliding with. If the disk rotates about a fix axis the linear momentum is not conserved as the axis exerts force.
    Try to solve the problem with free disk. Then the whole system will rotate about the common centre of mass.

  4. Nov 20, 2011 #3
    The axis is fixed for this problem. Are you saying that the missing momentum is going into whatever is holding the axis fixed? If that is the case, problem solved! :p

    Perhaps it is important to say that the disk is spinning with respect to the axis through the center of the disk (like a merry go round, not a coin spinning on its side) and not through the center of mass of the combined disk-particle system.
    Last edited: Nov 20, 2011
  5. Nov 20, 2011 #4


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    You know that the change of linear momentum is equal to the impulse of the force exerted. FΔt=Δ(Ʃmv). The axis exerts force, but you can imagine that it "takes on" momentum when the particle collides with the disk (it bends a bit) and then starts to vibrate and dissipates the momentum and energy at the end.

  6. Nov 20, 2011 #5
    Ah, I see now! Thanks, Ehild! :biggrin:
  7. Nov 20, 2011 #6


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    Welcome! (the green grin is very nice!):wink:

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