# SHM question

1. Jan 9, 2013

### Saitama

1. The problem statement, all variables and given/known data
(see attachment for figure)
A small object is mounted to the perimeter of a hoop of radius r. The mass of the object and the hoop is same. The hoop is placed into a fixed semi-cylinder shaped rough trough of radius R, such that small mass is at top. Find the least r/R ratio such that the object performs simple harmonic motion.

2. Relevant equations

3. The attempt at a solution
I honestly have no idea on how should I begin to solve this problem, this is very different from the SHM problems I have done.

Any help is appreciated. Thanks!

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• ###### SHM.jpg
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Last edited: Jan 9, 2013
2. Jan 9, 2013

### Staff: Mentor

After a rotation by an angle of θ, can you calculate the position of the mass and the circle? Can you calculate the potential energy?

3. Jan 9, 2013

### Saitama

But then, how will I find the angle rotated by the small mass?

4. Jan 9, 2013

### Staff: Mentor

What do you mean with "the angle rotated by the small mass"?
Rotate the hoop by an angle θ in your trough, find the new positions of hoop and attached mass with geometry.

5. Jan 9, 2013

### Saitama

(see attachment)
I am having trouble finding the new position of the attached mass. Is angle α=θ?

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6. Jan 9, 2013

### Staff: Mentor

α=θ would imply that the mass always stays on top. In reality, the hoop rotates quicker.
Hmm, you use a different θ.

After rolling a distance of d on the track, the contact position (on the hoop) changed by α=d/r. As seen from the center of the track, the hoop moved by an angle θ=d/R. Putting this together, the point mass is an angle of α-θ = d/r - d/R away from the top of the hoop now.

7. Jan 9, 2013

### Saitama

Nice explanation mfb, thanks!

I still cannot find the potential energy of the system. (see attachment)
The blue dot represents the CM of hoop and the black dot represents the attached mass. How can I find h?

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8. Jan 9, 2013

### Staff: Mentor

The blue dot has a fixed distance from the center of the track (which?), and you can use θ...

9. Jan 9, 2013

### Saitama

R-r?

I worked on it to find h, is h=r+rcosθ?

10. Jan 9, 2013

### haruspex

No, not quite.
It's easier to think in terms of how far the blue dot is below the centre of the hoop. It starts R-r below.

11. Jan 10, 2013

### Saitama

I gave it a try again. This time, I got h=R(1-cosθ)+rcosθ. Is this right?

12. Jan 10, 2013

### haruspex

Yes.

13. Jan 10, 2013

### Saitama

What should be my next step?

14. Jan 10, 2013

### Saitama

Can I get some more help?

15. Jan 10, 2013

### Staff: Mentor

16. Jan 10, 2013

### Saitama

Potential energy or change in potential energy?

17. Jan 10, 2013

### Staff: Mentor

Potential energy (as function of some angle), as I wrote.

An expression for the kinetical energy can be interesting, too, but I think it is not required for the answer.

18. Jan 10, 2013

### Saitama

Potential energy assuming it is zero at the bottom of the trough is
$$U=mgh+mg(h+r\sin(\alpha-\theta))$$
$$U=2mgR(1-\cos \theta)+2mgr\cos \theta+mgr\cos(\alpha-\theta)$$
m is the mass of the hoop and the attached mass.

Last edited: Jan 10, 2013
19. Jan 10, 2013

### haruspex

Wouldn't r.sin(α-θ) represent a horizontal displacement?

20. Jan 10, 2013