# Another Angular Momentum

1. Dec 2, 2004

### Jacob87411

A light, rigid rod 1 meter in length roatets about an axis perpendicular to its length and its center. Two particles of masses 4 KG and 3 KG are connected to the ends of the rod. What is the angular momentum of the system if the speeds of earch particle is 5 m /s

Curious if this is right

L=Iw
Both are point masses so I=MR^2, so I
=(4KG)(.5M^2) + (3KG)(.5^2)

w=r/v
w=.5/5 = 1/10

So plug this all back into L=IW

[(4KG)(.5M^2)+(3KG)(.5M^2)]x.1

2. Dec 2, 2004

### BobG

One slight change.

$$v=r\omega$$

so:

$$\frac{v}{r}=\omega$$

3. Dec 2, 2004

### Jacob87411

Ah so its 5/.5 so 10, thanks!