# When is Conservation of angular momentum valid?

## Homework Statement

[/B]
A thin uniform bar 2.00 m long and weighing 90.0 N is hanging from the ceiling by a frictionless pivot. It is suddenly struck 1.50 m below the pivot by a small 3.00 kg ball initially travellimg horizontally at 10.0 m/s. The ball rebounds and moves in oppossite direction with a velocity of 6 m/s. Find the angular speed of the bar after the collision.

## Homework Equations

I decided to use the principle of conservation of angular momentum.

## The Attempt at a Solution

And I got the answer pretty easily too. But, I did not understand why I could use it. I have read its valid when external torques sum up to 0. So can you explain how it is 0 in this case? And is the linear momentum conserved?

Last edited by a moderator:

Related Introductory Physics Homework Help News on Phys.org
scottdave
Homework Helper
It sounds like you took the ball as an instantaneous velocity, tangent to a circle centered at the pivot point? There is not a torque on the bar when it is vertical.

It sounds like you took the ball as an instantaneous velocity, tangent to a circle centered at the pivot point? There is not a torque on the bar when it is vertical.
I'm sorry I didn't get you

kuruman
Homework Helper
Gold Member
Your system is hanging bar + ball. Angular momentum of the system about the pivot is conserved instantaneously under the assumption that the collision is over so quickly that the bar does not rotate appreciably over the time interval Δt that it is in contact with the ball. Linear momentum is not conserved, not even instantaneously, because of the pivot that constrains the top end of the bar from moving. If the ball were to collide with the bar right at the pivot, there would be no motion of any kind.

• Nabin kalauni and scottdave
scottdave
Homework Helper
I'm sorry I didn't get you
So when you used angular momentum, what did you use to figure the ball's angular momentum? Was it assumed to be a point mass, orbiting the pivot point?

haruspex
Homework Helper
Gold Member
its valid when external torques sum up to 0
Except in special circumstances, torques and angular momenta should always be considered relative to some stated axis.
(In the kinetic cases, that had better be an axis fixed in an inertial frame, or the mass centre of the rigid body being considered, or its instantaneous centre of rotation.)
Conservation of angular momentum about an axis is valid when there are no external torques about that axis. If your system here is bar and ball then you can eliminate all external torques by suitable choice of axis.

Conservation of angular momentum about an axis is valid when there are no external torques about that axis. If your system here is bar and ball then you can eliminate all external torques by suitable choice of axis.
Can you explain it in this example?

So when you used angular momentum, what did you use to figure the ball's angular momentum? Was it assumed to be a point mass, orbiting the pivot point?
Yes thats what I did but I'm not sure

Your system is hanging bar + ball. Angular momentum of the system about the pivot is conserved instantaneously under the assumption that the collision is over so quickly that the bar does not rotate appreciably over the time interval Δt that it is in contact with the ball. Linear momentum is not conserved, not even instantaneously, because of the pivot that constrains the top end of the bar from moving. If the ball were to collide with the bar right at the pivot, there would be no motion of any kind.
Thanks a lot

haruspex
• 