# SHM r/R ratio question

## Homework Statement

(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.

## 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!

#### Attachments

• SHM.jpg
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mfb
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?

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?

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

mfb
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.

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.

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

#### Attachments

• SHM1.jpg
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mfb
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.

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.

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?

#### Attachments

• SHM2.jpg
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mfb
Mentor
The blue dot has a fixed distance from the center of the track (which?), and you can use θ...

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

R-r?

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

haruspex
Homework Helper
Gold Member
2020 Award
R-r?

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

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.

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.

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

haruspex
Homework Helper
Gold Member
2020 Award
I gave it a try again. This time, I got h=R(1-cosθ)+rcosθ. Is this right?
Yes.

Yes.

What should be my next step?

Can I get some more help?

mfb
Mentor
can you calculate the position of the mass and the circle? Can you calculate the potential energy?

Potential energy or change in potential energy?

mfb
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.

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.

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:
haruspex
Homework Helper
Gold Member
2020 Award
Potential energy assuming it is zero at the bottom of the trough is
$$U=mgh+mg(h+r\sin(\alpha-\theta))$$
$$U=2mgR(1-\cos \theta)+2mgr\cos \theta+mgr\sin(\alpha-\theta)$$
m is the mass of the hoop and the attached mass.
Wouldn't r.sin(α-θ) represent a horizontal displacement?

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

What should I do next?

ehild
Homework Helper
First express alpha with theta.

The potential function should have minimum at equilibrium. What is the condition that your U(θ) function has minimum at θ=0?

ehild

First express alpha with theta.

How?

Do I need to use the relation posted by mfb: α-θ = d/r - d/R ?

ehild
Homework Helper
What is d?

Think what rolling means. The point where the ball and cylinder touch each other covers equal lengths of arc both of the cylinder and ball. That gives a relation between alpha and theta.

Draw a big circle and cut out a small one, and try.

ehild

What is d?
Rθ?

ehild said:
Think what rolling means. The point where the ball and cylinder touch each other covers equal lengths of arc both of the cylinder and ball. That gives a relation between alpha and theta.

Draw a big circle and cut out a small one, and try.

ehild
I am still clueless.

haruspex