# Conservation of Angular Momentum - Man firing a bullet on a metal rod at an angle

1. Apr 15, 2012

### HSSN19

1. The problem statement, all variables and given/known data

This is a question from the physics textbook Don't Panic Volume I, chapter XIV.

A man, mass M, stands on a massless rod which is free to rotate about its center in the horizontal plane. The man has a gun (massless) with one bullet, mass m. He shoots the bullet with velocity vb, horizontally. Find the anglular velocity of the man as a function of the angle θ which the bullet's velocity vector makes with the rod.

2. Relevant equations

L = r x mv = mrvsinθ
L = mr2ω

3. The attempt at a solution

Initial L = m(l/2)vbsinθ
Final L = M(l/2)2ω

Therefore, ω = (2mvbsinθ)/Ml

What do guys think? Was I correct in using conservation of angular momentum? It is a very weird question compared to the rest.

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2. Apr 15, 2012

Yes it's correct.But should we neglect mass of man in comparison to mass of bullet???
If not then conserve linear momentum along cos component of velocity, i.e.
M*v=-m*vB*cos(θ).
'-' sign indicates man gets velocity in opposite direction of motion of bullet.Then the problem bcms complex bcs. now man's distance from centre of table keeps decreasing.
Check the answer.If it's simple which mostly should be then what you have done is seems correct.
I think you must have missed '2' in (l/2)2 in initial angular momentum term.

3. Apr 15, 2012

### HSSN19

Nah I don't think linear momentum should be used in problems like this. About the 2, if you're talking about the square, then no there shouldn't be a square. The definition of angular momentum is the cross product between r and linear momentum. The other equation has a square because it's derived for the special case of circular motion.

4. Apr 15, 2012

### Staff: Mentor

I think your answer is fine, but your work needs a little correction. The total angular momentum is--and remains--zero.

5. Apr 15, 2012

Oh!Sorry.I didn't think about that.Well then all terms written by you are correct.But linear momentum should be conserved in such problems if mass of man is not much greater than mass of bullet.However in your problem it seems that book has assumed the mass of man to be much greater than mass of bullet.So what you have done seems fine.

6. Apr 15, 2012

### Staff: Mentor

Conservation of linear momentum would not apply here--the rod is fixed on an axis, so linear momentum is not conserved.

No need for any assumption about the relative masses of bullet and man.

7. Apr 15, 2012

### HSSN19

Wait, what? I'm confused now. Why is the total L zero? Don't you mean the change in L is zero (L is conserved)?

8. Apr 15, 2012

### Staff: Mentor

Presumably, before the bullet is fired nothing is moving. So the total angular momentum is zero. And it remains zero, of course.

9. Apr 15, 2012

### HSSN19

Oh man, I already submitted the homework. I knew my "initial" and "final" L's didn't make any sense since the events are simultaneous. So the answer should be the same but negative now, I guess. Thanks!

10. Apr 15, 2012