Recent content by Bobby345

I think all my answers are sorted except part 3 could be buffed up a little bit. I was thinking to maybe discuss how there is 0 torque from the force applied by the axle at point 0 and therefore assert that the net torque on the system remains unchanged so angular momentum is conserved.

Perhaps this has something to do with net torques?

Even though the question asks why the angular momentum of the system is conserved?

So back to the original questions. 1. What is the linear momentum of the system before the collision in terms of L, m1, and ω? (m1Lω)/2 2. What is the linear momentum of the system after the collision in terms of L, m1, and ω? (m1Lω)/3 3. Why is the angular momentum about point X conserved...

Center of mass should be at L/2, right? Thus the rod's initial linear velocity is (Lω)/2 at the center of mass, making it's momentum (m1Lω)/2?

Ohhh. So you can just negate m2 by means of multiplying it by v. I was just confused because I hadn't tried the multiplication and automatically assumed that m2 would be in the answer. Thus, final linear momentum = (m1Lω)/3

I don't believe it produces a torque because the radius at point X is 0, correct? That must be the reason why the angular momentum is conserved, but not the linear momentum. Now I just need to establish the initial and final linear momentum of the system.

Yep. Given in a diagram that X is on one end. Correct. The tip of the rod strikes the ball.

So the linear momentum of the system is just m1Lω, considering v = Lω?The velocity of the rod is now 0, but the ball has a velocity of v now. I am not exactly sure how to express this momentum in terms of m1, L, and ω, considering that m1 doesn't have any momentum anymore, but m2 does. Any...

Homework Statement A system has a ball and a uniform rod. The rod is rotating about point X on a frictionless table until it strikes the ball. The rod stops and the ball moves away. Variables: Rod's mass: m1 Ball's mass: m2 Rod's original angular velocity: ω Ball's final velocity: v Rod's...