# Angular momentum problem

## Homework Statement

Four uniform rods, each of mass m and length 2L are joined rigidly to form a square frame ABCD of side 2L. The frame is placed with all four sides at rest on a smooth horizontal table. An inextensible string has one end attached to the corner A. A particle of mass 4m is tied to the other end of the string. The particle, initially at A, is projected with speed u in the direction DA. Given that the speed of the particle immediately after the string becomes taut is V, show that the initial angular speed of the square frame about an axis through its centre of gravity perpendicular to the plane of the frame is w where w=(2V-u)/L. Show that V=(7u)/11, and that immediately after the string becomes taut the kinetic energy of the particle and the frame is (14mu^2)/11.

## Homework Equations

Momentum conservation.
Impulsive moment = change in angular momentum.
Moment of inertia of a rod of mass m and length 2L about an axis through its centre = (1/3)mL^2.
Parallel axis theorem.

## The Attempt at a Solution

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Doc Al
Mentor
Show what you've done so far. You'll need both conservation of linear and angular momentum. Hint: What's the speed of the frame's center of mass immediately after the string becomes taut?

Well first I calculated the moment of inertia of the 4 rods about the centre of the frame, and found that to be (16/3)mL^2.

Then the way I thought about it was considering the linear momentum before and after.

Before = 4mu
After = 4mV + 4mV - (Iw)/L

For the linear momentum afterwards I know the particle is moving with speed V so it definitely has momentum 4mV. Then I thought the frame would start moving translationally with speed V also, since it's an inextensible string and there's no friction. That would give it linear momentum 4mV also. Then I thought there would be some linear movement backwards by virtue of its rotation (not sure about this, but that's what I thought at the time). I know it's angular momentum is Iw, so I just divided by L to get the linear momentum.

Then I put these equal to each other due to conservation of momentum, but that gives me something similar to the answer but not quite right?

Doc Al
Mentor
Well first I calculated the moment of inertia of the 4 rods about the centre of the frame, and found that to be (16/3)mL^2.
Good!
Then the way I thought about it was considering the linear momentum before and after.

Before = 4mu
After = 4mV + 4mV - (Iw)/L
When considering linear momentum, just worry about mass, not moment of inertia. (You'll factor in w soon enough.)
For the linear momentum afterwards I know the particle is moving with speed V so it definitely has momentum 4mV.
Good.
Then I thought the frame would start moving translationally with speed V also, since it's an inextensible string and there's no friction.
Good thinking, but not quite right. The edge of the frame (side DA) would be moving at speed V, but that's not the speed of the center of mass.
That would give it linear momentum 4mV also.
No. Use conservation of linear momentum to figure out the speed of the center of mass.
Then I thought there would be some linear movement backwards by virtue of its rotation (not sure about this, but that's what I thought at the time). I know it's angular momentum is Iw, so I just divided by L to get the linear momentum.
You'll find w by comparing the instantaneous speeds of side DA and the center of mass.

I'm really stumped as to how to find the speed of the centre of mass. I understand that side DA is moving with speed V and has mass m, so it's momentum is mV, therefore the momentem of the remaining part of the frame must be 4mu-mV. But where do I go from there? :s

I was thinking maybe that side BC wouldn't be moving at all yet at that point, which would give the speed of the centre of mass to be (4u-V)/2 but i'm not sure.

Doc Al
Mentor
You're thinking way too much! To find the speed of the center of mass, just set up a simple conservation of linear momentum equation. Hint: As far as linear momentum goes, the frame can be treated as a point mass. (Its speed will automatically be that of the center of mass.)