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Conservation of Angular Momentum with SHM

  1. May 2, 2017 #1
    1. The problem statement, all variables and given/known data
    A 39.00 kg rod of length 2.8 m is hanging vertically by one of its ends that is free to swing in a complete circle about a frictionless axle/pivot. The rod has uniform mass density. Suddenly it is struck horizontally by a 5 kg putty that sticks to the center of mass of the rod upon impact. With what minimum speed must the putty strike the rod so that the combination (of rod and putty) is just barely able to make a complete circular loop?

    2. Relevant equations
    Lbefore=Lafter
    1/2mpv2=1/2Iw^2
    Where mp is the mass of the putty and I is for the system of the putty and the rod.

    3. The attempt at a solution
    I know that I have to use conservation of angular momentum and the kinetic energy conservation, but I have no clue what the condition for "barely able to make a complete circular loop" is in relation to the final velocity, so I'm completely stuck.
     
  2. jcsd
  3. May 2, 2017 #2
    Most impacts, and particularly impacts involving gobs of putty, do not conserve linear momentum. You need to consider this problem in two parts: (1) the impact of the putty on the rod, resulting in a velocity for the combination of rod and putty, and (2) the condition for which that rotating combination will barely make it to the top of the arc.

    PS: Work only in symbols, do not attempt to use any numbers until you have an algebraic solution in hand.
     
  4. May 2, 2017 #3
    I'm aware that I have to solve it in two parts and that it needs to be angular momentum, which is why I wrote the equation for angular momentum.

    I just need hint to figure out what barely making it to the top of the arc is. Do I have to solve for its amplitude? Is there a special condition for minimum velocity required for something to barely go in a complete circle?
     
  5. May 2, 2017 #4
    Barely making it to the top of the arc means arriving at the top with zero velocity. If it has positive velocity when it gets to the top, then that is more speed than the minimum.
     
  6. May 2, 2017 #5
    Thank you! That will help a lot. And I now understand what you meant by not conserving momentum.
     
  7. May 2, 2017 #6

    gneill

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    Staff: Mentor

    I think you need to be a bit more precise here. Impacts that don't involve external forces or torques will conserve both linear and angular momentum. It's when you have external forces or torques (such as those arising from having one end of the rod pinned in place) that you have to be wary. Linear momentum conservation is a lost cause in such cases, but a careful choice of the point about which you measure angular momentum can save its conservation law.
     
  8. May 2, 2017 #7
    What I said seemed to get the job done. Why is more needed?
     
  9. May 3, 2017 #8

    haruspex

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    Because what you wrote here is simply not true and very misleading:
    That would be true for conservation of work. As @gneill wrote, linear momentum is conserved unless there is an external impulse. In the present case, there will be an impulse from the axle. Even if the collision had been work conserving (perfectly elastic), momentum would not have been conserved.
    Likewise, angular momentum about a specified axis is conserved unless there is an external torsional impulse about that axis. In the present problem, no external force has a moment about the axle, so angular momentum is conserved about that.
     
  10. May 3, 2017 #9
    Lianne Evans, welcome to Physics Forums. Did you come up with a final answer? Sometimes I like to check my own work. Thank you.
     
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