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Another circular motion problem

  1. May 5, 2003 #1
    A pendulum comprising a string of length L and a sphere swings in the vertical plane. The string hits a peg located a distance d below the point of suspension (see attached)

    A) show that if the sphere is released from a height below that of the peg, it will return to this height after striking the peg.

    B) Show that if the pendulum is released from the horizontal position (theta = 90 degrees) and is to swing in a complete circle centered on the peg, then theminimum value of d must be 3L/5.

    I've come up with the following equations that I'm trying to relate to show the height after striking the peg is the same as before.

    The initial height is L - Lcos[the], the height after striking the peg is L - d - (L - d)cos[psi].

    mgh(initial) = (1/2)mv^2(bottom of the arc)
    mgh(final) = (1/2)mv^2(bottom of the arc).

    I'm not sure where to go from here.

    Any suggestions would be greatly appreciated.

    Thanks.
     

    Attached Files:

  2. jcsd
  3. May 5, 2003 #2

    Tom Mattson

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    The way you started the problem is not going to get you anywhere, because you used the exact same state for the initial and final states, namely the state at the bottom of the arc.

    Choose instead the state at release as the initial state, and the maximum height above the peg reached by the sphere. You should be able to show that the sphere does not clear the peg, and so instead of wrapping the rope around it, it comes back to its original height.

    Here you have to note what the critical condition is for the ball going "over the top". That critical condition is that the speed has to be zero at the top of the swing. Use that as your (known) final state to deduce the (unknown) initial state.
     
  4. May 6, 2003 #3
    Re: Re: Another circular motion problem

    Thanks for replying.

    The initial state at release of the sphere is mgh. There's only gravitational potential energy.

    The maximum height reached by the sphere is:
    ((radius x angular velocity)^2)/2g or (velocity tangential^2)/2g.

    I can't relate these using the conservation of energy, so I'm still stuck.
     
  5. May 6, 2003 #4

    Tom Mattson

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    Take a closer look at what I said:

    Choose instead the state at release as the initial state, and the maximum height above the peg reached by the sphere. You should be able to show that the sphere does not clear the peg, and so instead of wrapping the rope around it, it comes back to its original height.


    You've got the initial state right, so use the stopping point at the other end of the swing as the final state, and show that it cannot make a loop around the peg. Then you will know that it has to come back to its initial position.
     
  6. May 6, 2003 #5
    Thanks again for taking the time to reply.

    OK, the stopping point at the other end of the swing is mgh'

    so if I equate the two states (does the peg steal away any energy from the system or is energy conserved?)...

    mgh = mgh'

    mg(L - Lcos[the]) = mg(L - d - (L - d)cos [psi])

    Lcos[the] >= d

    If Lcos[the] = d then (L - d)cos[psi] = 0 so either L = d or
    [psi] = 90 degrees.

    So if the release point is at the peg the angle the sphere makes with the vertical after striking the peg is 90 degrees.

    If I assume Lcos[the] to be greater than d then I get
    (L - d)cos[psi] = some positive number meaning the angle is less than 90 degrees or greater than 270 degrees.

    Why do I think there's a cleaner easier way to do this?

    Am I at least on the right track?
     
  7. May 7, 2003 #6

    Tom Mattson

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    The two heights will be the same if friction and the diameter of the peg are negligble.

    I think you're thinking too hard about this. All you have to do is note that the ball cannot get "over the hump" of swinging around the peg. Since the ball can't get to the point directly over the peg, it can't go over.
     
  8. May 8, 2003 #7
    Thanks Tom,

    That makes part b) simple

    At the top the sum of the centripetal forces = mv^2/R
    so T + mg = m^2/R, but at the top T = 0
    so vtop = (g(L-d))^1/2

    By conservation of mechanical energy
    mgL = 2mg(L-d) + 1/2mg(L-d)
    2L = 4L - 4d + L - d
    d = 3L/5
     
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