# Rotational Dynamics problem

1. Jun 25, 2014

### Vibhor

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

A small body A is fixed to the inside of a thin rigid hoop of radius R and mass equal to that of the body A. The hoop rolls without slipping over a horizontal plane; at the moments when the body A gets into the lower position, the center of the hoop moves with velocity v0. At what values of v0 will the hoop move without bouncing?

Ans : √(8gR)

2. Relevant equations

3. The attempt at a solution

The center of mass is located at a distance R/2 from the center of the hoop.CM rotates in a circle of radius R/2 about the center of hoop.

The forces acting on the system(hoop+mass A) are normal from the surface and weight 2mg.

Applying ∑F = ma , N-2mg=2may ,where ay is the acceleration of the CM in vertical direction.

The hoop will move without bouncing if normal force due to the surface is not zero.

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Last edited: Jun 25, 2014
2. Jun 25, 2014

### dauto

Use the centripetal acceleration formula to calculate the vertical acceleration.

3. Jun 25, 2014

### Vibhor

The only forces acting are Normal and the weight ,both vertical .Which forces provide centripetal acceleration to the center of mass ?

Please elaborate the general idea to solve the problem.

4. Jun 26, 2014

### Vibhor

Could somebody help me with the problem ?

5. Jun 27, 2014

### Zondrina

Does this mean the car is locked into place? As in it isn't accelerating in the vertical direction at the bottom?

If so, think along the lines of:

$\sum F_y = 0 \Rightarrow F_N = m_Ag$

There is still an acceleration towards the center though when the car reaches the top and $v_0$ changes direction.

$\sum F_n = m_A(\frac{v_0^2}{R})$

I seem to get $v_0 = \sqrt{2gR}$, not sure where the $8$ in the answer came from though.

Last edited: Jun 27, 2014
6. Jun 27, 2014

### Vibhor

Thanks for trying to help me :) .

This is incorrect .

If A is experiencing centripetal acceleration at the bottom ,then it means it surely has acceleration in upwards direction.

Again this is wrong . $v_0$ is the speed of the CM of the hoop ,not that of mass A .

The answer I have given in OP is correct .

Last edited: Jun 27, 2014
7. Jun 27, 2014

### Zondrina

Oh wow, I'm sorry. That will teach me to not relax after an exam, pretty tired.

On topic though... I'm thinking that conservation of energy will be of use here as the point mass moves from the bottom to the top.

8. Jun 27, 2014

### Zondrina

I'm not entirely sure about this, but you mentioned that the mass of A was $2m$. Put a datum through the bottom at $A$. So at the top the car posses potential. Then:

$\frac{1}{2} m v_0^2 = (2m)g(2R)$
$v_0 = \sqrt{8gR}$

9. Jun 27, 2014

### Vibhor

No. Mass of A is m .

This is incorrect .

On the left you have considered only the translational KE of the hoop whereas the hoop also has rotational KE . On the right you are considering only the PE of the mass A ?

Do you believe the hoop comes to rest when the mass A is at the top ? If not ,where is the KE of the hoop ?

10. Jun 27, 2014

### TSny

For the minimum initial speed vo for which bouncing first occurs, it was not obvious to me where the mass A will be located at the instant the normal force goes to zero. It is tempting to just assume that A will be at the top of the hoop at that instant, but in principle you would need to justify that assumption. I found it messy to prove. But maybe there is a nice, simple argument for it.

Can you find an expression for the y coordinate of the CM of the system in terms of the angle of rotation θ of the hoop (with θ = 0 when mass A is at the bottom)?

If so, try to use it to determine ay of the CM of the system in terms of the speed v of the center of the hoop at the instant mass A is at the top of the hoop.