# Minimum velocity for particle to lose contact

1. May 30, 2015

### Tanya Sharma

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

2. Relevant equations

3. The attempt at a solution

Honestly I have very little idea about this problem . The contact force has to vanish for particle B to loose contact .Not sure how to relate the velocity of upper particle with the normal contact force of the lower particle .

I don't think any conservation laws would work.

I would be grateful if somebody could help me with the problem.

#### Attached Files:

• ###### rod.GIF
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2. May 30, 2015

### theodoros.mihos

Assuming that the rope have constant length and the B particle can move only horizontally, this problem must ask for the limit velocity of A just it is over B particle. If A move to right, the distance go to be larger that impossible, so B start to move with some acceleration. Think about what can be this acceleration.

Or, change frame, take origin to A particle so the problem is the same that to give horizontal velocity $-v$ to B particle. The rope is constant (again).

3. May 30, 2015

### Staff: Mentor

If B loses contact, there is no force between B and the ground any more. What happens to the center of mass? Where is B relative to the center of mass, where is it relative to the ground?

4. May 30, 2015

### theodoros.mihos

It is a limit process $t\to0$.

5. May 30, 2015

### Tanya Sharma

The CM moves with speed v/2 towards right . The only external force acting on the system would be gravitational force .

B moves with speed v/2 towards left relative to CM .

Since it just looses contact , it is effectively at rest relative to the ground .

6. May 30, 2015

### ehild

I haven't solved the problem, but would start as follows.

Remember, motion of a rigid body can be considered as motion of the CM with acceleration determined by the external forces and rotation about the CM according to the external torques.
At t=0, the system has some momentum and angular momentum.
You can write the motion of ball B in terms of the motion of the CM and the rotation about it. B should stay on the ground or lift from it: its vertical acceleration can not be negative even without a normal force.

7. May 30, 2015

### Tanya Sharma

Sorry , I can't see how to implement the above strategy to deal with this problem.

8. May 30, 2015

### TSny

If B were to stay in contact with the floor, what would be the height, $y$, of the CM of the system as a function of angle of rotation of the system?

Can you use this relation to find an expression for $\ddot{y}$?

9. May 31, 2015

### Tanya Sharma

Taking upward positive and $\theta$ to be angle made by rod with the vertical .

$N - 2mg =2m\ddot{y}$

$\ddot{y} = -\frac{l}{2}[cos\theta \dot{\theta}^2+sin\theta\ddot{\theta}]$

10. May 31, 2015

### TSny

What doe the second equation reduce at the time of interest?

11. May 31, 2015

### Tanya Sharma

Are you getting v = 2m/s ?

12. May 31, 2015

### TSny

My answer is in terms of the length of the rod and g. I haven't yet put in numbers. Lets see....

Yes, I get 2 m/s.

13. May 31, 2015

### Tanya Sharma

$v=\sqrt{2gl}$ . Right ?

14. May 31, 2015

### TSny

Yep.

15. May 31, 2015

### Tanya Sharma

I have been thinking about this problem for last two days and here you come and make this problem look so ordinary .

How do you think so clearly ?

16. May 31, 2015

### TSny

I don't know. It always starts out kind of foggy and then the fog lifts (sometimes ).

You got the answer very quickly after I gave a hint. So, I think I might have given too much away. It's always difficult to find the right nudge.

17. May 31, 2015

### Tanya Sharma

Thank you very very much .

Good night .