Homework Help: Oscillating sphere on a parabolic surface

1. Jul 11, 2010

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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. Jul 11, 2010

hikaru1221

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'.

3. Jul 11, 2010

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I thought w=v/r=x'/r

4. Jul 11, 2010

hikaru1221

Oops, sorry Yes, w=v/r=x'/r.

5. Jul 11, 2010

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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.

6. Jul 11, 2010

ehild

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?

ehild

Last edited: Jul 12, 2010
7. Jul 12, 2010

hikaru1221

Time "lies inside" x(t), x'(t), x"(t)

"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.

8. Jul 14, 2010

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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.