# Angular Momentum of two masses

1. Apr 8, 2010

### MissPenguins

1. The problem statement, all variables and given/known data
A light, rigid rod l = 9.44 m in length rotates
in the xy plane about a pivot through the
rod’s center. Two particles of masses m1 =
9.1 kg and m2 = 2.5 kg are connected to its
ends.

Determine the angular momentum of the
system about the origin at the instant the
speed of each particle is v = 2.1 m/s.

2. Relevant equations
Lz=Iw
I = mr2
I = (1/12)ML2
w = v/r

3. The attempt at a solution
I used the above equations:
Lz=Iw
I = mr2
I = (1/12)ML2
w = v/r
I = 1/12(9.1kg)((9.44)2)=67.578
I = 1/12(2.5kg)((9.44)2)=18.57
sum of I = 86.1431
Used w = v/r = ((2.1m/s))/(9.44/2)=0.444915
Lz=Iw
(86.1431)(0.444915)=38.32639
I got it wrong, please help, what did I do wrong? Did I even approach the right way? Thanks.

2. Apr 8, 2010

### mitch987

To find the correct value of I, you would need the moment of inertia about the centre of mass?

3. Apr 9, 2010

### mitch987

For the MOI about the centre of mass, take take the radius as 4.72m and the centre x=0.
Hence,

centre of mass, $$x_{cm}= \frac{2.5\times -4.72 +9.1\times 4.72}{2.5+9.1} =2.68m$$

MOI, $$I_{cm}=\frac{1}{12}(M_{system})(x_{cm})^{2}=\frac{1}{12}(11.6)(2.68)^{2}=6.94$$

$$\omega=\frac{v}{r} =\frac{2.1}{4.72} = 0.44 rad/s$$

Therefore, $$L=I\omega=6.94\times 0.44=3.05 kgm^{2}/s$$

4. Apr 9, 2010

### MissPenguins

I tried that, but it is not right either. ;(