Angular Momentum for Man and Bullet

  • Thread starter GOPgabe
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Homework Statement


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 angular velocity of the man as a function of the angle θ which the bullet's velocity vector makes with the rod.


Homework Equations


Torque = r x F
L = r x mv
T = dL/dt

The Attempt at a Solution


There's no external torque acting on the system of the man and the bullet. Torque = 0 = dL/dt means that L is conserved. The system was at rest meaning Lint was 0. The angular momentum for the man is Mmanω(l/2)^2 post firing. Here's my problem, the bullet's radial vector is changing. r = < l/2 + Vbcos(θ)t, vbsin(theta)t>x<mbVbcosθ, mbVbsinθ>
The book gives the answer as mbvbsinθ/(mman(l/2))
Where did I go wrong?
 

Answers and Replies

  • #2
gneill
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The angular momentum of something that has no external torques applied is conserved, so it is sufficient to consider the angular momentum of all parties concerned at a particular representative instant. The instant after the bullet is fired (t = 0+) seems like a reasonable choice...

You could, if you were so inclined, write the bullet's radius in the form of a vector function of time, r(t), and express its velocity as a vector, too. Then its angular momentum would be given by mb r(t) x v, which you'll probably find comes out as a constant value after slogging through the math.
 

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