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Homework Help: Oscillating sphere on a parabolic surface

  1. Jul 11, 2010 #1
    1. The problem statement, all variables and given/known data
    A sphere is making small oscillations on a parabolic surface . The equation of the parabola is
    y= 0.5kx^2 . Find the time period of this small oscillation.

    2. Relevant equations
    accln. (c.o.m)= 5/7 g sinѲ
    tanѲ= Ѳ ( for small angles)
    sinѲ=Ѳ (for small angles)

    3. The attempt at a solution
    I tried to find about the moment of the ball about the focus of the parabola but then in addition with theta it came up with an x variable and thus I am unable to apply SHM rules.
  2. jcsd
  3. Jul 11, 2010 #2
    I recommend you take time derivative of the mechanical energy instead. Force analysis here is somewhat complex I guess.
    Potential energy: U = mgy = 0.5mgkx^2
    Kinetic energy: K = 0.5mv^2 + 0.5Iw^2
    As for small oscillation around the lowest position of the parabola, we have v=x' and w=rv=rx'.
  4. Jul 11, 2010 #3
    I thought w=v/r=x'/r
  5. Jul 11, 2010 #4
    Oops, sorry :biggrin: Yes, w=v/r=x'/r.
  6. Jul 11, 2010 #5
    I still dont get it how to get time from the energy calculations. Can u explain some more.
    Moreover x'= something in terms of 1/U^(1/2) and dU/dt. And if we put U=0 in this equation then it is invalid so this also confuses me.
  7. Jul 11, 2010 #6


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    Homework Helper

    For small oscillation, you can consider the motion horizontal, along x. As the radius of the sphere was not mentioned, I think it can be taken negligible. In this case, the potential energy is mgy. Compare with the potential energy of SHM, U=1/2 Dx^2. How is D related to k? how the frequency of oscillation is related to D and m?

    Last edited: Jul 12, 2010
  8. Jul 12, 2010 #7
    Time "lies inside" x(t), x'(t), x"(t) :smile:

    "x'= something in terms of 1/U^(1/2) and dU/dt"
    For SHM, U=kx^2; dU/dt = 2kxx'. So: x' ~ dU/dt * 1/sqrt(U), is this what you mean? Look at U and dU/dt again. When U=0, x=0 and thus, dU/dt=0 too. So we cannot conclude anything about dU/dt * 1/sqrt(U) here.
  9. Jul 14, 2010 #8
    huh.... I realize just now that u dont need to take any derivative of Energy. The trick is converting the sinѲ in accln to tanѲ, which of course we can since Ѳ is small and then it is very easy to show that accln. is proportional to -x. from there we can just apply SHM rules to calculate time period.
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