# Angular momentum of a particle

1. Apr 15, 2009

### joemama69

1. The problem statement, all variables and given/known data
Determine the Angular Momentum of each particle about the origin

Point A Located @ (-8,12), 6kg,4m/s, headed 60degrees below the horizon to the right
Point B (2,1.5),4kg, 6m/s, headed 30 degrees above the horizon to the right
Point C (6,-2), 2kg, 2.6m/s, headed on a 5,12,13 angled to the left and down

2. Relevant equations

L=r x p
p=mv

3. The attempt at a solution

A. -8i+12j X 24(-i-8.66j)=
B. 2i+1.5j X 24(i+.5j)=
c. 6i-2j X 5.2(-12i-5j)=

If this is right i dont know how to calculate it

2. Apr 15, 2009

### LowlyPion

Not quite right. To the right is +i isn't it?
A. -8i+12j X 24(-i-8.66j)=
should be
A. (-8i + 12j) X 24*(.5i - .866j)=

B. 2i+1.5j X 24(i+.5j)=
should be
B. (2i + 1.5j) X 24*(.866i + .5j)=

c. 6i-2j X 5.2(-12i-5j)=
should be
c. (6i - 2j) X 5.2*(-12i - 5j -13k) =

To take the cross product ...
http://en.wikipedia.org/wiki/Cross_product#Computing_the_cross_product

Last edited: Apr 15, 2009
3. Apr 15, 2009

### rl.bhat

(-8i + 12j) X 24*(.5i - 8.66j)

It should be (-8i + 12j) X 24*(.5i - .866j)

4. Apr 15, 2009

### LowlyPion

Indeed it should. Thanks for the catch.

5. Apr 16, 2009

### joemama69

ok i see my mistakes.

just to clarify @ point C

13 is not the k value, it is the third side of the triangle/hipotinuse

6. Apr 16, 2009

### joemama69

Ok so here what i got

A. (-8i+12j)X(12i-20.784j)=22.72k
B. (2i+1.5j)X(2.784i+12j)=-7.176k
C. (6i-2j)X(62.4i-26j)=280k

7. Apr 16, 2009

### LowlyPion

A. (-8 i + 12 j) X (12 i - 20.784 j)= 22.272 k

C. (6i-2j)X(62.4i-26j)=280k
Looking at the original direction again I think it should be -26 i -62.4 j. This yields a different result.

8. Apr 16, 2009

### joemama69

Nope the triangle is not oriented like that

the bottom is 12 or -12i, the side is on the right and up -5j.

9. Apr 16, 2009

### LowlyPion

OK. I can't see the picture, so going with that ... so using -62.4 i -26 j
shouldn't it be -156 - 124.8 = -280.8 k