Question

What is the relationship between the translational and rotational momentum in an elastic collision based on the distance from an axis, based on problem below.

Answer 1

I'm looking for a general approach also possibly an explanation of the following. I know that if the collision occurs at the CoM of the rod. The kinetic energy of the rod is only translational, and i know that if it hits the farthest distance of the rod the energy is completely rotational. How do we consider a distance in between?

Answer 2

L = r x p

You can find the angular momentum of the sphere just before it strikes the bar. Then use conservation of angular momentum to find the angular momentum imparted to the bar (easy in this case, since the sphere goes to rest).

Answer 3

Isn't that just the angular momentum? Is there no translational?

Answer 4

I would assume that is the correct approach if the rod was fixed.

Answer 5

Oh, you're right. I made an assumption that it says nothing about. It isn't going to just rotate about it's CoM.

Answer 6

Quote: "and i know that if it hits the farthest distance of the rod the energy is completely rotational. How do we consider a distance in between?"

This is not the case, that would imply that the centre of mass of the rod remains fixed.

Answer 7

Both translational and angular momentum are conserved.

Answer 8

I could really use a deeper explanation because I just feel like I don't understand the distinction between translational and angular momentum. I.e How can both translational and angular momentum be conserved, because neither is initially rotating yet one will begin to rotate.

Answer 9

I'm still working the problem (and probably making a circus of it), but Vincent is correct.

In a collision, both linear and rotational momentum are conserved. This has always been the case, but usually you didn't need to worry about it. Further, rotational momentum is conserved around any point.

The sphere isn't rotating initially as you think of it, but it has angular momentum with respect to some point in space. The point in question is the center of the rod. Just before the collision, it has angular momentum of L = r x p = dmv, where d is the distance it strikes the rod from the center of mass.

Answer 10

Are you able to check answers somehow? If so, please see if 4/7 is correct. If it is, I'll show you how I got to it.

Answer 11

that is correct

Answer 12

4/7 = 0.57142857... (if needed in decimal form)

Answer 13

You can speak conceptually, I dont need the calculation.

Answer 14

Woo. Terrible that it took me this long, but hey. Ok, so to do this problem, we simply just use the three applicable conservation laws: Kinetic Energy, translational momentum, and rotational momentum.

Kinetic Energy: \[\frac{1}{2}mv^2 = \frac{1}{2}Mu^2+\frac{1}{2}I\omega^2\]where u is the translational speed of the bar after the collision.

Linear momentum: \[mv = Mu\]

Rotational momentum: \[mvd = I\omega\]

We know that the moment of inertia I of the bar is:
\[I = \frac{1}{12}ML^2\]

And all we have to do is use these three equations together to solve for m/M.

First, I solved the kinetic energy equation to find an expression for m/M. Then, I used the rotational momentum equation to find an equation for omega, and plugged that in. Simplify. I then used the momentum equation to find an expression for u/v, and plugged that in.

At this point, you end up with m/M in a lot of places, but it's pretty easy algebra from there. If you get confused easily, you may want to make a substitution for m/M, such as m/M = a, and then just solve for a.

Answer 15

Thank you

Answer 16

Very sorry that it took that long.

Conceptually, the take-away is that you have to conserve both linear and angular momentum, and the kinetic energy also includes rotational kinetic energy. These things are actually always true, but in earlier, easier problems, we didn't have to worry about the angular components.

Answer 17

I had everything but \[mv=Mu\] I guess I should have recalled that linear momentum is always true about the CoM. Again thanks.