Oscillating sphere on a parabolic surface

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Homework Statement

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.

Homework Equations

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

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.

 

[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'.

 

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

 

[hikaru1221]

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

 

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I still don't 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.

 

[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

 

[hikaru1221]

I still don't 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.

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.

 

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huh... I realize just now that u don't 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.