# Angular Momentum of a bar

1. Nov 15, 2009

### Dark Visitor

http://session.masteringphysics.com/problemAsset/1035643/5/jfk.Figure.P09.32.jpg

Use the link above to get the picture for this problem. I could use some help with both parts of this problem, and it is due tonight.

Part 1) What is the magnitude of the angular momentum of the 670 g rotating bar in the figure?

Part 2) What is the direction of the angular momentum of the bar? (2 options below)
-into the page
-out of the page

2. Nov 15, 2009

### SeanGillespie

For part 2 there is a "right hand rule", curl your right hand with your fingers in the direction of the rotation, your thumb will be in the direction of your momentum.

3. Nov 15, 2009

### cepheid

Staff Emeritus
Well, I think that the easiest way of going about doing part 1 would be to calculate the moment of inertia of the bar. You already know the angular velocity of the bar, and you should know what the relationship between these two quantities and angular momentum is.

For part 2, it's just a convention (i.e. a rule that everybody agrees upon for the sake of consistency). For a quantity like this, we define the direction of a vector as being related to the direction of rotation by the right hand rule.

4. Nov 15, 2009

### Dark Visitor

Does that mean I would use the equation L = Iw? And if so, how do I find I (moment of inertia)?

5. Nov 15, 2009

### cepheid

Staff Emeritus
Yeah, that's right. Aren't moments of inertia of objects with simple geometry given in your book or your notes?

6. Nov 15, 2009

### Dark Visitor

The only thing I have is a sheet my teacher gave me, which has Inertia equations for a few objects. The closest one is a thin rod, but the problem calls it a bar, so I don't know if I should try it anyways. Otherwise, I have nothing to go on.

7. Nov 15, 2009

### Dark Visitor

Nevermind. I used it, and it was right. Thanks dude. I appreciate the help. I will be posting 3 more problems in a little bit which are harder for me. So if you feel like helping, I would be very grateful. Otherwise, thanks anyways.

8. Nov 15, 2009

### cepheid

Staff Emeritus
Hey Dark Visitor,

I'll see what I can do.

Whether you call it a bar or a rod, that formula is basically for an object that we can idealize as being 1-dimensional, i.e. all of the mass can be considered to be concentrated along a line of length L, and the dimensions perpendicular to that don't contribute in a significant way to the rotational inertia.

9. Nov 16, 2009

### ice2morrow

So what was the I in this problem? I've been given the same problem, but don't know how to calculate the answer.

10. Nov 16, 2009

### cepheid

Staff Emeritus
ice2morrow,

You have two options for determining the moment of inertia of an object fulfilling the criteria I described in post #8 (i.e. a long, thin bar/rod)

1. You could look it up:

2. If you understand how (and why this is true), you could compute it as:

$$I = \frac{M}{L}\int_{-L/2}^{L/2} x^2\, dx$$ ​

where x is the dimension along the length of the rod with mass M and total length L (with x=0 being at the rod centre).