# Ring on a smooth hoop

1. Oct 18, 2015

### omegasquared

1. The problem statement, all variables and given/known data

A ring of mass m slides on a smooth circular hoop with radius r in the vertical plane. The ring is connected to the top of the hoop by a spring with natural length r and spring constant k.

By resolving in one direction only show that in static equilibrium the angle the spring makes to the vertical is θ where:

$$cos\theta = \frac{1}{2}\cdot \frac{1}{1-\frac{mg}{kr}}$$

2. Relevant equations

$$T=kx$$

3. The attempt at a solution

To find the equation for tension:

$$L=2rcos\theta\\ T = kr(2cos\theta - 1)$$

Resolving vertically I got:

$$Tcos\theta = Rcos\theta + mg$$

and horizontally:

$$Rsin\theta = Tsin\theta$$

The above can't be correct and also I've resolved in two directions where the question has only asked for one. The forces I have drawn on my diagram are tension and reaction parallel but in opposite directions and weight straight down.

2. Oct 18, 2015

### TSny

Hello. Welcome to PF!

A diagram of the setup would be helpful. Based on some of your work, I think I understand the system.

Are you sure the reaction force (normal force), R, makes the angle θ to the vertical?

Since you are asked to resolve forces in one direction only, try to resolve the forces in an appropriate direction that is neither horizontal nor vertical.

3. Oct 18, 2015

### omegasquared

I've attached the provided diagram of the system.

I assume the action force would make the same angle θ as it's been pulled in that direction by the tension.

Resolving parallel to the reaction & tension would yield a result still involving R though.

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4. Oct 18, 2015

### TSny

Thanks for the diagram. That makes the setup clear.

I don't think your assumption about the direction of the reaction force is correct. You are apparently supposed to assume no friction between the ring and the hoop. So, the reaction force is a "normal force".

You won't be able to make one resolution parallel to the reaction and tension since those two forces are not in the same direction.

Last edited: Oct 18, 2015
5. Oct 18, 2015

### omegasquared

If it's a normal force, what is it normal to? The surface/edge of the hoop?

6. Oct 18, 2015

### TSny

Yes, it's normal to the surface of the hoop.

7. Oct 18, 2015

### omegasquared

Like in the modified diagram attached?

#### Attached Files:

• ###### FullSizeRender.jpg
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8. Oct 18, 2015

### TSny

No. In your diagram, R is not perpendicular to the surface of the ring.

9. Oct 18, 2015

### omegasquared

Sorry, noticed that almost straight away. Thanks for your help there.

I'm still unsure how it is possible to resolve to get a value for cosθ which doesn't involve R though.

10. Oct 18, 2015

### TSny

To get a relation that does not involve R, resolve along a direction for which R has zero component.

11. Oct 18, 2015

### omegasquared

Thanks, that pointed me in the right direction! Much appreciated!