Conservation Laws in Elastic Collision of Particle with Rotating Square

[Fibo112]

Homework Statement

A particle of mass m and and velocity v collides with a square of mass M (at rest)whose movement is confined to rotation about its centet. I must now solve for the angular velocity and the velocity of the particle after the collision (elastic collision)

Homework Equations

Kinetic energy and angular momentum
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The Attempt at a Solution

Since the collision if elastic the sum of the kinetic energies is certainly maintained. Linear momentum seems to not be maintained. Angular momentum about the point of rotation must be maintained since the only exfernal force seems to be the object the square if mounted on which cannot have a torque at that sams point. So convervation of energy gives me one equation but I can't seem to formulate an equation with conservation of angular momentum.

 

[NFuller]

I can't seem to formulate an equation with conservation of angular momentum.

What is the formula for the angular momentum of a point particle?
What is the distance from the particle to the center of the cube at the point of collision?
Once you know these, it should be straight forward to calculate the angular momentum of the particle. To find the angular momentum of the cube, you need the moment of inertia for the cube which you can either derive or look up.

 

[Fibo112]

What is the formula for the angular momentum of a point particle?
What is the distance from the particle to the center of the cube at the point of collision?
Once you know these, it should be straight forward to calculate the angular momentum of the particle. To find the angular momentum of the cube, you need the moment of inertia for the cube which you can either derive or look up.

I know how to calculate the angular momentum before the collision and I can calculate the angular momentum of the square based on its angular velocity. What i am having trouble with is calculating the angular momentum of the particle after the collision according to its velocity. For this I need to know the direction its traveling in.

 

[NFuller]

I know how to calculate the angular momentum before the collision and I can calculate the angular momentum of the square based on its angular velocity.

You should show this in your "attempt at a solution" so we know where exactly you are stuck.

What i am having trouble with is calculating the angular momentum of the particle after the collision according to its velocity. For this I need to know the direction its traveling in.

Are you treating the angular momentum and the velocity as a scalar or a vector? If you are doing this correctly, the velocity of the particle will be a vector and contain the direction.

 

[haruspex]

For this I need to know the direction its traveling in.

What is the direction of the impulse on the particle? (Treat the impulse as taking negligible time.)

 

[Fibo112]

What is the direction of the impulse on the particle? (Treat the impulse as taking negligible time.)

Perpendicular to the square?

 

[haruspex]

Perpendicular to the square?

Right. So what is the post-collision direction of the particle (ignoring the sign)?

 

[Fibo112]

Right. So what is the post-collision direction of the particle (ignoring the sign)?

Also perpendicular to the initial position of the square. In this case I can solve the problem. Is this a general assumption when solving problems involving collisions that the collision occurs in negligible time?

 

[haruspex]

Also perpendicular to the initial position of the square. In this case I can solve the problem. Is this a general assumption when solving problems involving collisions that the collision occurs in negligible time?

I can imagine a problem where you should not assume that, but you would need to be given information about the elasticities of the bodies, and it would be at a very advanced level.

 

[Fibo112]

I can imagine a problem where you should not assume that, but you would need to be given information about the elasticities of the bodies, and it would be at a very advanced level.

Ok. Thanks for your help.