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. ;(