1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Conservation of angular momentum

  1. Jun 16, 2012 #1
    1. The problem statement, all variables and given/known data
    There's a system of 4 masses, all connected to a cross which has a negligible mass, and which is positioned on a smooth surface. The distance of each mass from the center of the cross is L and the cross spins around its center in a constant radial velocity of ω0 rad/sec:
    Now mass m4 disconnects from the cross.

    What is the the radial velocity of the system after m4 disconnected, considering m1=m3 and m2=m4=M?

    2. Relevant equations
    Conservation of momentum:

    Conservation of angular momentum:

    3. The attempt at a solution
    I calculated using conservation of momentum that the linear velocity of the system after m4 disconnected was v2=Mω0L/(M+2m)

    Now I think I should use the law of conservation of angular momentum but I'm not sure how. I think that the center of mass is L/2 to the right from the center of the cross so the distance of m1 and m3 from the center of mass is √((0.5L)2+L2). What should I do next?
  2. jcsd
  3. Jun 16, 2012 #2
    Relevant equations

    Moment of Inertia =?
    Angular Momentum=?
  4. Jun 16, 2012 #3
    I know that the moment of inertia is I=Ʃmiri2 and the angular momentum L can be expressed as ωI, so I tried:

    L=Ʃmiviri=ωI = ω(Ʃmiri2) and I can get the value of ω this way, but I'm not sure what the the distance from each mass to the center of mass is. I mean, what are the values of ri in this sum: Ʃmiri2 ?
  5. Jun 17, 2012 #4
    You have to start with conservation of energy.
    All masses have equal tangential velocity.
    As mass m4 detached from the cross(it follows a tangential path), the total energy of the system remains.

    Using consevation of momentum requires the momentum of detached mass m4, which follows a straight line.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook