Angular Momentum for Man and Bullet

AI Thread Summary
The discussion focuses on a physics problem involving a man standing on a massless rod who shoots a bullet, analyzing the angular momentum of the system. It is established that there is no external torque acting on the man-bullet system, leading to the conclusion that angular momentum is conserved. The initial angular momentum is zero since the system starts at rest, and the challenge arises in calculating the changing radial vector of the bullet after it is fired. The provided solution indicates that the angular momentum of the bullet can be expressed as a function of its velocity and the angle θ, ultimately leading to a specific formula for the angular velocity of the man. The key takeaway is the importance of correctly accounting for the bullet's changing position in relation to the rod to solve the problem accurately.
GOPgabe
Messages
10
Reaction score
0

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?
 
Physics news on Phys.org
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.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top