# Homework Help: Moment of Inertia and the parallel axis theorem

1. Oct 19, 2007

### mdk31

1. The problem statement, all variables and given/known data
I have a problem that involves a vertical axis that can rotate through a fixed axis located at the very top of the rod, at point A (the axis is coming out towards us, the rod's rotation would look like a pendulum). The rod is nonuniform and there is a block attached to the rod. A bullet is fired at the block, causing the block-bullet-rod system to rotate. I must find the rotational inertia of the block-rod-bullet system. We are told to treat the block as a particle

The mass of the bullet is .01 kg
The mass of the block .50 kg.
The rotational inertia of the rod alone is .060 kg-m^2
The length of the rod is .60 m.
The mass of the rod is .50 kg.

2. Relevant equations
I=I[center of mass] + Mh^2

3. The attempt at a solution

I was thinking about using the parallel axis theorem, where the .060 was the I[cm] and Mh^2 was the total mass of the bullet and block times the length of the rod squared. But I then I realized that the rotational axis was not at the rod's center of mass. I'm really not sure how to proceed...I'm stuck.

2. Oct 19, 2007

### learningphysics

"The rotational inertia of the rod alone is .060 kg-m^2"

is this rotational inertia about the rod's center of mass, or about the end of the rod?

3. Oct 19, 2007

### mdk31

It is the rotational inertia of the rod about the fixed axis at A, at the top end of the rod.

4. Oct 19, 2007

### learningphysics

Ok. So add to that number... the moment of inertias of the box and the bullet... treat them as point particles... the moment of inertia of a point particle is mr^2.

5. Oct 19, 2007

### mdk31

So this problem really has nothing to do with the parallel axis theorem does it?

6. Oct 19, 2007

### learningphysics

I don't think so...

7. Oct 19, 2007

### mdk31

No, it doesn't seem to. I was being stupid about it.

Thanks for you help though!

8. Oct 31, 2007

### mdk31

I just looked at this problem again and realized that it also wants us to find the speed of the bullet just before impact. The angular speed of the system immediately after impact is given as 4.5 rad/s.

I know I'm supposed to use conservation of angular momentum rm[0]v[0]sin(x)=Iw

(w is omega)

and I understand that. But why can't I use rm[0]v[0]sin(x)=rm[1]v[1]sin(x) where m[0] is the mass of just the bullet, m[1] is the mass of the bullet, block, and rod, and v[1] is wr. When I try to find the velocity this way, I get the wrong answer.

9. Oct 31, 2007

### learningphysics

The angular momentum after the collision is not: rm[1]v[1]sin(x)... this presumes the entire mass m[1] is located at a distance r from the pivot. it's not. rmv only holds for point masses. you need to use the moment of inertia of the bullet/block/rod system.

10. Oct 31, 2007

### saket

Yes, you ought to get wrong answer.. because, when you do so, you are assuming that rod can also be treated as a point mass located at the impact-point.. which is NOT true... as rod is having mass (non-uniform).

11. Oct 31, 2007

### mdk31

I see, that makes sense.

Thanks!