Understanding Momentum in Rotational Motion

[Whitebread1]

I have a conceptual question to ask of you.

Suppose you have a point mass moving in a straight line at any arbitary speed with no net external force, so the particles speed does not change. At a later time, the point mass has a completely inelastic collision with another point mass, which causes both points to rotate around an axis with a constant radius.

Now my question, will the momentum of the first point mass be equal to the product of the two point masses times the tangental velocity of the two masses?

mathmatically:
m1*v(i)=(m1+m2)*v(tangental) ?

 

[OlderDan]

I have a conceptual question to ask of you.

Suppose you have a point mass moving in a straight line at any arbitary speed with no net external force, so the particles speed does not change. At a later time, the point mass has a completely inelastic collision with another point mass, which causes both points to rotate around an axis with a constant radius.

Now my question, will the momentum of the first point mass be equal to the product (you mean the sum, as per your equation) of the two point masses times the tangental velocity of the two masses?

mathmatically:
m1*v(i)=(m1+m2)*v(tangental) ?

Yes it will for point (or nearly so for small) masses. If the moving point mass collides with an extended object like a pivoting rod (with mass), it will not be true. In that case you would need to use conservation of angular momentum rather than linear momentum. For two point masses, either linear or angular momentum gives you the same result.

 

[Whitebread1]

Yes, I meant sum, my mistake.
Now regarding an extended object. Would this condition hold true if the extended object was struck at its center of mass?

 

[americanforest]

Yes, I meant sum, my mistake.
Now regarding an extended object. Would this condition hold true if the extended object was struck at its center of mass?

It would have no rotational momentum because \omega=0 so I would assume all of the momentum would remain linear. So yes, I think so.

 

[Whitebread1]

Alright, thank you

 

[Whitebread1]

It would have no rotational momentum because \omega=0 so I would assume all of the momentum would remain linear. So yes, I think so.

Wait, I was not clear. I mean an extended object where the axis of rotation is not at the center of mass. Thing of a flag hanging off of a flag pole. I'm wondering because ideally you can treat any object as a point located at the center of mass when doing calculations.

 

[OlderDan]

Yes, I meant sum, my mistake.
Now regarding an extended object. Would this condition hold true if the extended object was struck at its center of mass?

No it would not. The problem you first stated said the two masses would rotate around an axis. If the extended object has a pivot point, linear momentum will not be conserved. The pivot would have to exert a force on the extended object with a component opposite the initial velocity of the ball. Angular momentuym about the pivot point would be conserved, but linear momentum would not. Linear momentum will be conserved if the extended object is free from any other forces, even if it is not struck at its center of mass. It could move away with all the momentum of the incident ball while spinning about the CM of the system of the extended object plus the stuck ball.

 

[OlderDan]

Wait, I was not clear. I mean an extended object where the axis of rotation is not at the center of mass. Thing of a flag hanging off of a flag pole. I'm wondering because ideally you can treat any object as a point located at the center of mass when doing calculations.

Not when it is rotating. Even something as simple as a rolling ball is very different from a sliding block. We often ignore the rotation in early problems like billiard balls colliding, but some aspects of the motion have to be handled differently for rolling than they do for sliding objects.