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Conservation of Angular Momentum of Ball

  1. Dec 8, 2007 #1
    1. The problem statement, all variables and given/known data
    A teatherball is attached to a pole with a 2.0m rope. It is circling at 0.20 rev/s. As the rope wraps around the pole it shortens. How long is the rope when the ball is moving at 5.0 m/s.

    Note: I can't use the html formating with this site!!!

    Givens:
    f(initial)=20 rev/s
    length(initial)=2.0m
    v(final)=5.0 m/s

    2. Relevant equations
    L(initial)=L(final)
    mR^2omega(intial)=mR^2omega(final)
    f=2PiOmega
    omega=v/r

    3. The attempt at a solution

    Now I set angular momentums equal and solved and got an answer of 1 by getting the masses to cancel I ended up with
    (2.0 meters)^2*2Pi*(.20rev/s)/[5.0m/s]=R
    Now this works and I get the desired answer but I do not think this is physicaly possible because the teather ball would hand at some angle theta below the horizon and would not circle straight out without some sort of upwards normal force.

    To start with I drew an FBD. I measured theta below the horizon so that Ftx=Ft cos theta &Fty=Ft sin theta
    I also set x=r= 2 cos theta for the intial conditions.
    Then I balanced out the two component forces. Gravity and Centripial
    I set Fty=mg & Ftx=mv^2/r

    I did a lot of algebra and after I got some trig functions to cancel I ended up with
    Fty=[mv^2/cos theta] * sin theta =mg

    I end up with sin theta= g/(2 *omega^2) = g/[2 (2Pi f)] but g/[2 (2Pi f)] is greater than 1 so there is no angle possible.

    Now if the angle then plugs back into the conservation of momentum and eventaully cancels then I would be wrong but my math breaks down at that point so I don't know that the angle would be constant in such a problem.

    thanks a ton,
    dave
     
    Last edited: Dec 8, 2007
  2. jcsd
  3. Dec 8, 2007 #2

    Dick

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    Not this one again. I hope it's being presented as a trick question. Either that or the author is being careless. Ask yourself, "Self, if the ball is being accelerated, where does the energy come from?". Energy is conserved, not angular momentum. The ball will never reach 5m/sec.
     
  4. Dec 8, 2007 #3
    Are you sure?
    I think that momentum is always conserved even if energy is not. The system is changed by the shortening of the rope and therefore responds with an increase in velocity. If you can make the rope infinitely short, it could go infinitely fast.

    Dave
     
  5. Dec 8, 2007 #4

    Dick

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    That's yet another problem with the exercise. But you are making it to hard (be glad your name is not Dick). If the radius of the pole is zero so the the force is central, and angular momentum is conserved, then the rope will never shorten. If it is bigger than zero then the force on the ball is not directed straight at the pole. It's directed at where the rope contacts the pole. So angular momentum is not conserved. The whole exercise is damaged. Don't work too hard at it.
     
  6. Dec 8, 2007 #5

    Dick

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    I'm really sure. Screamed about this problem to people in grad school. Doesn't 'infinitely fast' scream 'implausible'???
     
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