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Homework Help: Question about conservation of angular momentum

  1. Feb 11, 2010 #1
    First of all: this is not a homework; I am posting it here because I extracted and adapted this problem from a book; it is a simple problem that involves conservation of angular momentum.
    I know the solution to this problem, and I put it below to follow the forum's rules, but my problem is not with the solution. I ask the actual doubt at the bottom of this message.

    1. The problem statement, all variables and given/known data

    A carousel has a radius of 2 m and a moment of inertia of 500 kg·m², and can rotate around a frictionless axis. A person of 25 kg runs tangentially towards the edge of this carousel, which is at rest, with an initial velocity v = 2.5 m/s, and jumps into the toy. What is the final angular velocity of the person and the carousel together?

    2. Relevant equations

    [tex]L=mvr[/tex]
    [tex]I=mr^2[/tex]

    3. The attempt at a solution

    Here is the solution, but it is actually not the doubt I have:

    Since there is no external torque acting on the system, angular momentum is conserved. The initial angular momentum of the system is the person's angular momentum, since the carousel is at rest. Since the person runs tangentially towards the carousel, the distance from the axis of rotation perpendicular to the direction of motion is the radius of the carousel (R).
    Li = mvR = (25)·(2.5)·(2) = 125 kg·m²/s
    After the person jumps into the edge of the carousel, his moment of inertia is mr² = 25·2² = 100 kg·m²; the moment of inertia of the carousel is 500 kg·m²; so, the final angular momentum of the system is: Lf = (100 + 500)ωf; equalling Li and Lf, I find that ωf = 0.208 rad/s.

    As I said, this is actually not the doubt yet. The doubt is: the book says that the linear momentum is not conserved in the above situation. Why?
     
  2. jcsd
  3. Feb 11, 2010 #2

    tiny-tim

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    Hi pc2-brazil! :smile:
    Because there is an external reaction force from the axle acting on the carousel, opposite in direction to the way the person was running.

    It has no torque about the centre (because it's at the centre!), so there is no external torque, but you don't have that excuse for the external force! :wink:
     
  4. Feb 11, 2010 #3
    Thank you for the response.
    I think I understood. But what exactly does this external force do to the system? How does it change the linear momentum?
     
    Last edited: Feb 11, 2010
  5. Feb 11, 2010 #4

    tiny-tim

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    The change in total momentum of the person-and-carousel equals the (impulsive) external force on the system.

    That external force is the reaction I mentioned.

    It reduces the total momentum from mpersonv to 0. :smile:
     
  6. Feb 12, 2010 #5
    Thank you.
    Let me see if I understood: during the collision of the person with the carousel, friction between them generates decrease of the angular momentum of the person, and increase of the angular momentum of the carousel, which applies a force in the axle; and this makes the axle apply a reaction force in the carousel. Is this reasoning correct?
    What do you mean by "impulsive" external force?
    Thank you in advance.
     
  7. Feb 12, 2010 #6

    tiny-tim

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    Hi pc2-brazil! :smile:
    (I'd call it the reaction rather than friction … if he just jumps onto a horse and grabs the neck, or into a backward-facing seat, friction doesn't come into it)

    Yes, the two equal and opposite reaction forces between them have equal and opposite torques about the axle, and so the angular momentum lost by the person equals the angular momentum gained by the carousel.
    The carousel is stuck on the axle, so there has to be a pair of reaction forces there.
    Impulse is force times time.

    A force applied suddenly is too difficult to measure as a rate-of-change of momentum, so instead we measure the total change of momentum, which is an impulsive force or impulse.
     
  8. Feb 12, 2010 #7
    I think I understood what you said in the previous post.
    But I still don't understand how the external force exerted by the axle acts to decrease the linear momentum of the person. It is a reaction force that acts on the carousel and exerts no torque, but how does it affect the person?
    Couldn't I consider it as an inellastic collision, and that, instantaneously, the particles of the carousel and those of the person have linear momentum during the rotational movement?
     
  9. Feb 12, 2010 #8

    tiny-tim

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    Yes, it is an inelastic collision … in fact, it's what's called a perfectly inelastic collision, since the two bodies move as one body after the collision.

    The linear momentum of the person isn't changed by the external force exerted by the axle, it's changed by the force from the horse (or the seat or whatever the person is in contact with).

    The force at the axle is a result of the force by the person on the carousel, and has nothing to do with any force on the person.
     
  10. Feb 12, 2010 #9
    Thank you for the help.
    It is clearer to me now. The fact that the person looses all his linear momentum is just because he jumps into the carousel... it has nothing to do with the rotation or the angular momentum; it would also happen if there was no rotation of the carousel involved.
     
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