# Confusion regarding application of Conservation of Momentum

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1. Mar 14, 2016

### Amartya Sen

1. The problem statement, all variables and given/known data
Actually, my confusion originated from solving two different problems...

1) A point object of mass 'm' moving horizontally hits the lower end of the uniform thin rod of length 'l' and mass 'm' and sticks to it. The rod is resting on a horizontal, frictionless surface and pivoted at the other end as shown in figure. Find out just after collision the angular velocity of the system.

2) A circular wooden loop of mass 'm' and radius 'R' rests flat on a horizontal frictionless surface. A bullet, also of mass 'm', and moving with a velocity 'V', strikes the loop tangentially and gets embedded in it. The thickness of the loop is much less than 'R'. The angular velocity with which the system rotates after the bullet strikes the loop is ...

2. Relevant equations

mrCMvCM +ICM ω2 = constant

3. The attempt at a solution

I solved both questions using the above formula exactly in the same way, but was able to solve only (1)....
What I am confused with is that whether to add the external particle moment of inertia in "ICMω2"

I applied the formula independently for both objects, that means the particle's MoI about its own CM is 0.

I get the answer in (2) if I add Consider the MoI of the ring to be 2mR2 instead of simply mR2...

If I follow this approach in (1), then I get another answer which is not correct....

What to do??
Any help would be greatly appreciated.

2. Mar 14, 2016

### BvU

Hello Amartya,

I hesitate to comment (see reason below), but I do notice two things:

That is correct and a good assumption: the size of the bullet isn't given and it can be thought of as a point mass. However, the axis of the rotational movement is executes after the collision is probably not through the COM of the bullet !
That may well be, but the ring has a moment of inertia for rotations about its axis through the center of mass (perpendicular to the plane) that really is no more than mR2.

Now my hesitation: After the collision you have an expression for the momentum of the center of mass. It's not in your list of relevant equations, but I do think you need it. The motion is split in two: the linear motion of the center of mass plus a rotation around it. So the axis of rotation is supposed to be the center of mass and you will need the moment of inertia of ring+bullet around this axis.

From the book answer it would almost seem the rotation could be around the center of the ring, which I think is not correct. @Doc Al ?

3. Mar 14, 2016

### haruspex

I get v/(3r), which I assume is correct, without such a cheat.
I took moments about the point of impact, as a point fixed in space.

4. Mar 14, 2016

### BvU

Thanks haru ! Smart approach!

5. Mar 14, 2016

### Amartya Sen

Yes.! That's right!! The correct answer is V/3R

Pardon, but I didn't understand your solution. You conserved angular momentum about point of contact ?

OK..., let me clear my basics first of all..,
• I suppose no matter about what point we are conserving angular momentum, we always use velocity of COM and MoI of COM of the object. (P.S.--if there are more that one objects, I write different mvr + Iω terms for the different COM of different objects in the system)
• I always conserve angular momentum about a point where external torque is 0.
Hence observing the points above, in question (2), I conserved momentum about centre of the ring.
That becomes :

mvr=mv1r+mr2ω : v1 is the velocity of bullet after it has embedded in the loop and ω is the common angular speed.

..OK, now I think I figured out my mistake..

I think I incorrectly built a relationship b/w v1 and ω as v1=ωr, which I now think is not correct.

But now I cant figure out the relationship b/w the above two variables.

Last edited: Mar 14, 2016