Finding Angular Velocity of Man with Gun in Physics Textbook

In summary: But what do you think about this situation: A man has a massless bullet which he fires horizontally at a massless target. Find the resultant velocity of the bullet.In summary, the man's resultant velocity is equal to the velocity of the bullet multiplied by the cosine of the angle between the bullet's velocity vector and the target's velocity vector.
  • #1
HSSN19
8
0

Homework Statement



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.

Homework Equations



L = r x mv = mrvsinθ
L = mr2ω

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.

Thanks in advance!
 

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  • #2
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
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
HSSN19 said:
What do guys think? Was I correct in using conservation of angular momentum? It is a very weird question compared to the rest.
I think your answer is fine, but your work needs a little correction. The total angular momentum is--and remains--zero.
 
  • #5
HSSN19 said:
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.

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
1994Bhaskar said:
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.
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
Doc Al said:
I think your answer is fine, but your work needs a little correction. The total angular momentum is--and remains--zero.

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
HSSN19 said:
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)?
Presumably, before the bullet is fired nothing is moving. So the total angular momentum is zero. And it remains zero, of course.
 
  • #9
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
Doc Al said:
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.

OK.I understood that.
 

1. What is angular velocity?

Angular velocity is a measurement of how fast an object is rotating or moving in a circular path. It is typically measured in radians per second (rad/s) or degrees per second (deg/s).

2. How is the angular velocity of a man with a gun calculated?

The angular velocity of a man with a gun can be calculated by dividing the change in angle (in radians or degrees) by the change in time. This is represented by the formula: ω = Δθ/Δt, where ω is the angular velocity, Δθ is the change in angle, and Δt is the change in time.

3. Why is angular velocity important in physics?

Angular velocity is important in physics because it helps us understand the motion of objects that are rotating or moving in circular paths. It is also a key component in the study of rotational motion and the laws of motion.

4. How does the angular velocity of a man with a gun affect the trajectory of the bullet?

The angular velocity of a man with a gun can affect the trajectory of the bullet by determining the direction and speed at which the gun is moving. This, in turn, can impact the initial velocity and angle of the bullet, ultimately affecting its path and trajectory.

5. Can the angular velocity of a man with a gun be changed?

Yes, the angular velocity of a man with a gun can be changed by altering the force or torque applied to the gun. This can be achieved by changing the direction or speed at which the man is moving, or by changing the weight or length of the gun, among other factors.

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