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Perfectly Inelastic Disk and Stick Collision- Angular Speed

  1. Jul 15, 2010 #1
    I think I'm not understanding something about this problem. Can someone help me out?

    1. The problem statement, all variables and given/known data
    A 2.0 kg disk traveling a 3.0 m/s strikes a 1.0 kg stick of length 4.0 m that is lying flat on nearly frictionless ice. The disk strikes the endpoint of the stick, at a distance r = 2.0 m from the stick's center. Suppose the collision is perfectly inelastic so that the disk adheres to the stick at the endpoint at which it strikes. The moment of inertia of the stick about its center of mass is 1.33 kg*m2. Find the angular speed of the system after the collision.

    2. Relevant equations
    (I think) Parallel axis theorem: I= Icm + mr2
    L= r cross p
    L= I*omega
    m= mass disk

    3. The attempt at a solution
    L = L'
    r cross p = I(system)*omega
    mvr = (I(stick) + I(disk) + mvr2)*omega
    mvr = (I(stick) + (1/2)mr2 + mr2)*omega
    omega = mvr/(I(stick) + 2/3(mr2))
    omega = 2.0kg(3.0m/s)(2.0m)/(1.33kg*m2+1.5(2.0kg)(2.0m)2)
    omega= .90 rad/s

    The answer is 1.0 rad/s.
     
  2. jcsd
  3. Jul 15, 2010 #2

    tiny-tim

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    Science Advisor
    Homework Helper

    hi bcca! :smile:

    (have an omega: ω :wink:)

    nope, start again …

    you need to decide which point you're taking moments about

    the easiest point is the combined centre of mass (if you take it about anywhere else, you'll need to add mrc.o.m. x vc.o.m. to get the final angular momentum) …

    what do you get? :smile:
    (and you can assume the disk is a point)
     
  4. Jul 15, 2010 #3
    I got it! Thanks :)
     
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