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Converting tangential momentum to angular momentum

  1. Jul 7, 2011 #1
    "converting" tangential momentum to angular momentum

    If someone is running and jumps onto a merry-go-round, momentum is still conserved, correct? (ignoring friction). So, would the momentum of the person while they were running be the same as the angular momentum of the merry-go-round after they jumped onto it? If not, how would I "convert" (for lack of a better term) their momentum into angular momentum? I know about

    L = IW
    and all, but I am really confused.


  2. jcsd
  3. Jul 7, 2011 #2
    Re: "converting" tangential momentum to angular momentum

    For simplicity's sake assume that the merry-go-round has no mass and therefore possesses no momentum of its own and the person's speed is the same before he or she jumps on (when they are running) and after (when they are sitting on the merry-go-round). Also assume no friction and that the person is a "point" mass (again just for simplicity).

    Linear momentum is conserved, but in this case the person's momentum is transferred to the Earth through the merry-go-round as the merry-go-round exerts a force on the person towards its center. In other words, linear momentum is covered overall (if you include the Earth) but if you only include the person and the merry-go-round in your system then linear momentum is NOT conserved. Note the magnitude of the person's linear momentum stays the same since the person continues to travel at the same speed, but the momentum changes since momentum is a vector quantity and the person's direction is continually changing as they go around the merry-go-round.

    To convert, just get the person's speed before the jump and combine that with the distance from the center to come up with the omega (the W, angular speed) value. So for example if the person is going 15 ft per second and is 5 feet from the center of the merry-go-round, they are doing speed (ft/sec) / circumference (ft/revolution) = 15/(2*5*pi) revolutions/sec = .477 rev/sec = 3 radians/sec (I'm not sure if my math is right, but you get the idea).

    To get I:http://en.wikipedia.org/wiki/List_of_moments_of_inertia" [Broken]

    use the one for point mass (m*r*r) so you get mass of person * 5 * 5

    multiply I and W together to get angular momentum! Note that this problem might be doable if you are not given the mass of the person, set it up and see if the m's cancel. Good luck!
    Last edited by a moderator: May 5, 2017
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