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Cut a string

  1. Oct 6, 2014 #1
    1. The problem statement, all variables and given/known data
    Use energy conservation to answer the following question.
    A mass is attached to one end of a massless string, the other end of which is attached to a fixed support. The
    mass swings around in a vertical circle as shown. Assuming that the mass has the minimum speed
    necessary at the top of the circle to keep the string from going slack, at what location should you cut the string
    so that the resulting projectile motion of the mass has its maximum height located directly above the center of
    the circle?

    2. Relevant equations
    kinetics, energy conservation


    3. The attempt at a solution
    The speed at the top of the circle is sqrt(rg). The energy at the top of the circle is E=1/2mrg+2rmg=5/2mrg, given we take 0 as the bottom of the circle. e=(Ke+Pe). The energy is conserved around the system, so choose another point a; Ea=Etop.... 5/2mrg=1/2mv^2+mg(r+rcosx).... then v= sqrt(rg(3-2cosx).

    If the max height did not need to be at the top of the circle, the angle would be 90. I tried to deal with this condition by, x-dir: d=vt, then t=rsinx/(sqrt(rg(3-2cosx))cosx. This yields an unwieldy derivative when I try to maximize distance in the y-direction.

    I've done a lot of useless math on this one. Please help!
     
    Last edited: Oct 6, 2014
  2. jcsd
  3. Oct 6, 2014 #2

    gneill

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    Staff: Mentor

    Actually, the question statement says only that the maximum height needs to be directly over the center of the circle. I take that to mean that the apex of the trajectory lies on the y-axis (assuming that the circle's center is at the origin. So I'm thinking things like range equation, launch angle, speed at launch...

    Also, you might save yourself a bit of math agony if you assume your PE zero is at the top of the circle. You've already got your baseline KE from what the speed has to be there. While the projectile is lower than the top it has "extra" KE associated with the PE difference between its current location and the top.
     
  4. Oct 6, 2014 #3

    OldEngr63

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    Gold Member

    Might help if you would show some of that "useless math" so we can see where you are.

    (This is an interesting problem.)
     
  5. Oct 6, 2014 #4
    gneill: good point, I suppose as long as you are consistent the zero can be placed anywhere. Actually, I got the same velocity wherever the zero was placed.

    here is some of the math:

    min speed at top so string is not slack:
    (using a for angle)
    mv^2/r=mg+T, T=0, so v=root(rg)

    b is an arbitrary point, Ea=Eb, then 5/2mrg=1/2mv(b)^2 +mg(r+rcosa)

    from which, v(b)=sqrt(rg(3-2cosa)), which has components vx and vy.

    dy=do+v(y)t-1/2gt^2+do(y)

    the time here is the time to get to the vertical axis passing through the circle origin, by d(x)=v(x)t=r(sina) then t=rsina/v(x)

    So I substituted that t into the equation for dy, tried to take the derivative to maximize it and solve for the angle. But it is not so easy...

    tell me if I can make anything more clear
     
  6. Oct 6, 2014 #5

    gneill

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    Staff: Mentor

    If you have the velocity and launch angle at the release point, why not apply the range equation to find the x-distance of the apex of the trajectory? The trajectory is parabolic so its apex will occur at half the range. You know what the x-distance must be because you know the point b.

    I've just done some tinkering with the math and I'm not convinced that there's a real solution (other than cutting the string at the top of the circle so it's already at its apex and over the center)... more (or better) tinkering required perhaps :)
     
  7. Oct 6, 2014 #6

    BvU

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    Isn't the time derivative of dy known as vy(t) = vy, 0 - gt ? Whence the need to take a derivative ?
     
  8. Oct 6, 2014 #7
    gneill: I can give that a shot. I think the way I set it up could be the cause of the lack of real solutions.

    BvU: I think you're on the money. I'm seeing what I can get now if I apply that.
     
  9. Oct 6, 2014 #8
    Okay so I took the time derivative of d(y) and let it equal zero. Then after some algebra I obtained a quadratic expression,

    2cos^2a-3cosa-1=0, which corresponds to the roots a=pi/3 and a=0. Of these, the second does not make sense. So in conclusion, if the math is right, the angle at which the string is cut such that the mass attains a maximum displacement above the center of the circle corresponds to an angle of 60 degrees to the vertical. I would like to thank you both very very much for your help.

    If this answer appears incorrect, please let me know!
    regards,
    Matt W
     
  10. Oct 6, 2014 #9

    gneill

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    Looks fine. I just found the same result using the range equation approach and better attention pushing the symbols around :smile:

    I think that the ##\theta = 0## result is also acceptable, since the projectile is at the peak of its trajectory when launched horizontally.
     
  11. Oct 6, 2014 #10

    BvU

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    I got the same result
    Well done ! :-)
     
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