# 2 sphere system, Conservation of Energy and Momentum

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1. Sep 13, 2015

### Better WOrld

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

A sphere of radius $1m$ and mass $25 Kg$ is put on another sphere of radius $5 m$ and $7 Kg$ which is placed on a smooth ground. Now the upper sphere is pushed very slightly from it's equilibrium position and it begins to fall.

Now when the line joining the centre of the 2 spheres makes an angle $\theta$ with the vertical the upper sphere loses contact with lower sphere. Then find $\theta$ in degrees.

Note that there is no Friction between the 2 spheres.

2. Relevant equations

$$g=9.8 ms^{-2}$$
$$m_1u_1+m_2u_2=m_1v_1+m_2v_2$$
$$U_{initial} + K_{initial}=U_{final}+K_{final}$$

3. The attempt at a solution

Unfortunately, I haven't been able to get anywhere with the solution.

I took the small sphere (of radius r) as $m$ and the large sphere (of radius R) as $M$. The velocity of the small sphere is $v_m$ and the velocity of the large sphere is $v_M$.

When the small sphere will fall off, the Normal Reaction would be 0. Using this relation,
at the time of falling off (taking radial components),

$$\dfrac{mv_m^2}{R+r}=mg\cos\theta$$

I was told that the next step would be to conserve linear momentum in the horizontal direction since there was no external force in that direction. Lastly I was to apply the Conservation of Mechanical Energy for the system sphere+sphere+Earth.

Also, I took the 0 Potential Energy reference level as the common horizontal tangent to both spheres.

Unfortunately I'm unable to form the Equations for Conservation of Linear Momentum and Mechanical Energy.

I would be grateful for any help with this question. Many thanks in advance!

PS. My (incorrect?) equation for Conservation of Energy is as follows: Taking 0 Potential Energy at the top of the sphere, $$-MgR+mgr=\dfrac{1}{2}MV_M^2+\dfrac{1}{2}mv_m^2-MgR-mgR(1-\cos\theta)$$

Last edited: Sep 13, 2015
2. Sep 13, 2015

### Staff: Mentor

The big sphere will move as well.
For the initial positions?
I guess the floor would be more intuitive.

If there is friction, the spheres will rotate as well, but the problem statement is a bit unclear in several aspects.

3. Sep 13, 2015

### Better WOrld

Sir, I'm really sorry. I had meant to write that there is no friction. Is the problem more clear now?
Sir, why would the floor be more intuitive, please?

Also, could you please show how you arrived at $$\vec{F} = a^2 \cos(\theta) \cdot \{ \sin(\theta) \cos(\varphi), \sin(\theta) \sin(\varphi), \cos(\theta)^2 \}?$$

4. Sep 13, 2015

### Staff: Mentor

It is the usual way to define "zero height".

That equation in my post was a copy&paste error, sorry. Still had a formula from an unrelated thread in the clipboard. I wanted to copy this formula:
The big sphere will move as well.

5. Sep 13, 2015

### Better WOrld

Sir, I realised that it would move. This would be because of the Normal Force between the 2 spheres. However, I still cannot understand how to apply the law of Conservation of Momentum.

6. Sep 13, 2015

### Staff: Mentor

Then you should take this motion into account.

How can you describe the horizontal momentum in terms of the velocities of the spheres?
What do you know about the horizontal position of the center of mass?

7. Sep 13, 2015

### Better WOrld

I'm sorry Sir, I don't know the answer to the first question. Had there been 2 separate spheres with velocities v1 and v2, the momentum (initial) of them would be $m1v1$ and $m2v2$. However, the confusion is arising since these 2 spheres are in physical contact and it thus seems that the motion of the bottom sphere will influence the motion of the top sphere. Thus, the velocity of the top sphere would not simply be $v_m$. This is what is bothering me, Sir.

Regarding the second question, I didn't understand. Wouldn't the center of mass of each sphere always be at the center of each sphere respectively?

8. Sep 13, 2015

### Staff: Mentor

The contact does not matter.
That is true, but it does not change for formulas for their momentum.
You can define the velocity to be v_m. By the way, for LaTeX you can use two \$ or two # here: $v_m$.
The center of mass of the spheres yes, but I meant the center of mass of the whole system.

9. Sep 13, 2015

### Better WOrld

Sir, regarding the center of mass, in the horizontal direction, it would be at the same level (ie the $Y$ coordinate would be the same) since there is no external force acting in the horizontal direction.