# Conservation of angular momentum

1. Jul 27, 2012

### Northbysouth

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

A 0.5-kg particle is located at the point r = 3m i + 4m k and is moving with a velocity v = 5m/s i -2m/s k . What is the angular momentum of this particle about the origin?
A) 13 kgm^2/s k
B) 13 kgm^2/s j
C) 26 kgm^2/s k
D) 26m^2/s j
E) 13 kgm^2/s i

2. Relevant equations
L = RxP
where L is angular momentum, R is the distance from the origin and P is the momentum and the x stands for the cross product
P = mv

3. The attempt at a solution

P = 0.5(5m/s i - 2m/s k)
= 5/2 kgm^2/s i - 1kgm^2/s

L = (3i + 4i) x (5/2 kgm^2/s i - 1kgm^2/s)
= -3j - 10j
= -13j

I'm fairly certain that the answer is B because I believe that the cross product should give me a value that is perpendicular to the two vectors, which in this case are i and k, thus the cross product should, I think, include j. Also, I believe that I can eliminate D because the units are incorrect. What I don't understand is why my answer is negative.

2. Jul 27, 2012

### szynkasz

$L=m\cdot r\times v=\frac{1}{2}\cdot \det\begin{bmatrix}i&j&k\\3&0&4\\5&0&-2\end{bmatrix}=13j$