# Angular Momentum of a Jogger

1. Apr 13, 2012

### eagles12

1. The problem statement, all variables and given/known data
Jogger 1 in the figure has a mass of 62.3 and runs in a straight line with a speed of 3.35 .

2. Relevant equations

L=r*p*sinθ

3. The attempt at a solution
The first part of the problem was to find the magnitude of the jogger's linear momentum, which i found to be 908 kgm/s

The second part says to find the magnitude of the jogger's angular momentum with respect to the origin.

There is a picture that shows the coordinates and says the jogger is at (8,5)
using this i found r=√(8^2+5^2)=9.433
i also found θ=arctan(5/8)=32.0053
when i multiplied these by 908(the p i found in part 1) it is saying my answer is incorrect

2. Apr 13, 2012

### Staff: Mentor

How did you get this?
What direction is the jogger running? You need the angle between r and the jogger's momentum vector.

3. Apr 13, 2012

### eagles12

I got the linear momentum using p=mv
i plugged in the mass and velocity given and got p=(62.3)(3.35)=908

The runner is running at a northeast direction toward the point (8,5) from the origin

4. Apr 13, 2012

### Staff: Mentor

I thought the runner was located at the point (8,5)?

If he's running northeast, what angle does his momentum make with the x-axis (east)?

5. Apr 13, 2012

### eagles12

I meant 209 I was already told that my linear momentum is correct

That's right the runner is located at (8,5) and is running east. So the angle is the angle made with a vector pointed to (8,5) from the origin.

6. Apr 13, 2012

### Staff: Mentor

OK, so he's running east (not northeast). So his momentum vector points east.
That is the position vector r.

OK, looking back at your first post, show your calculation for the angular momentum. Looks like you have the correct values for r and p (after you correct your typo) and the angle.