# Angular Momentum

1. Dec 18, 2009

### yankees26an

1. The problem statement, all variables and given/known data

A 1.2 kg object travels with constant velocity (4i - 2j) m/s. When the object is at the point (0
m, 9 m, 0 m), what is the magnitude of its angular momentum about the origin?

2. Relevant equations

p = m*v
L = r*p

3. The attempt at a solution

v = sqrt(4^2 + 2^2)
p = m*v = 5.36

L = r*p = 9*5.36 = 48.2

Interestingly the answer I get this way is 2x the right answer. I wonder if this is a coincidence of not. Not sure how to do it the 'right' way.

2. Dec 18, 2009

### Staff: Mentor

These are not scalars, so this

is wrong.

3. Dec 18, 2009

### cepheid

Staff Emeritus
L is NOT simply given by r multiplied by p. The definition of L is

$$\textbf{L} = \textbf{r} \times \textbf{p}$$ ​

Where boldface quantities are vectors and the multiplication symbol signifies the vector cross product. Since L is the cross product of the r and p vectors, components of the momentum parallel to r do not contribute to the angular momentum. Are you sure that the right answer is 24.1, because I get a different answer both ways (i.e. by doing the full vector cross product or by using the fact that components of the velocity || to the position vector can be ignored)?

4. Dec 18, 2009

### yankees26an

Ok. On a diagram, 9 just turns out to be the height/radius. Not sure if that means anything.

x and z are 0 so they dont affect anything, so I just have the value of y left. If it's not scalar then do you have any suggestions how to continue the problem. :tongue2:

So a cross product would be r*Fsin(pheta). But how do I get to pheta?

Last edited: Dec 18, 2009
5. Dec 18, 2009

### cepheid

Staff Emeritus
First of all, to answer your question, the angle in the sine factor is the angle between the two vectors in the cross product.

Second of all, that is probably not the approach you want to take to calculate the cross product. It only tells you the magnitude, not the direction. You need to look up the definition of a cross product in Cartesian coordinates. Ie for input vectors of given x,y, and z components, what are the x y and z components of the result? This is closely related to the right hand rule.

Thirdly (minor point), there is no such Greek letter as 'pheta'. Perhaps you were thnking of theta?

6. Dec 18, 2009

### yankees26an

Ok you got me =P

How would I go about finding the angle between momentum and r?

7. Dec 18, 2009

### cepheid

Staff Emeritus
How would I go about finding the angle between momentum and r?[/QUOTE]

You would find it based upon their respective directions. But like I said, that is not the best approach. You should really read the second of my three points (the one you did not quote)

8. Dec 18, 2009

### yankees26an

You would find it based upon their respective directions. But like I said, that is not the best approach. You should really read the second of my three points (the one you did not quote)[/QUOTE]

Ok so the radius is in the Y direction? The momentum in the x direction?

I also remember some stuff about i,j,k, but seriously how do I apply it here