Vertical Rod with two masses attached falling over table.

[DeLaHostia]

Homework Statement

This problem is very easy, but there is some conceptual point of view I am missing.
There is a vertical rod of length L without mass on its own that has two masses, m1 and m2, attached at each end standing on point P at a table. The rod is at unstable equilibrium at first. The only force is gravity and there is no friction between the m2 and the table. The systems is moved from it equilibrium, and the rod falls down.

Known data:
m1
m2
Lenght of rod: L
Rod has no mass
No friction with the table
P the point where the rod stands vertically.

Homework Equations

At what distance of P will each mass fall?

The Attempt at a Solution

I tried to work this out with conservation of linear momentum but I could not succeed.

 

[jackarms]

Having a bit of trouble seeing what the problem is asking. Do you mean at what distance from P will each mass fall?

Conserving momentum will be difficult to use here, since gravity will add momentum to the system, and the movement is going to involve rotation. Instead, try thinking about center of mass, or at least the components of it. Where will the center of mass of the rod be before and after it falls?

 

[haruspex]

Having a bit of trouble seeing what the problem is asking. Do you mean at what distance from P will each mass fall?

Conserving momentum will be difficult to use here, since gravity will add momentum to the system, and the movement is going to involve rotation. Instead, try thinking about center of mass, or at least the components of it. Where will the center of mass of the rod be before and after it falls?

That can be thought of as using conservation of linear momentum (horizontal component), but not merely from initial state to final state. Rather, one needs to consider how the horizontal velocity of the centre of mass might change at any point in the process.