What Is the Angular Momentum of Two Masses on a Rotating Rod?

[MissPenguins]

Homework Statement

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.
Answer in units of kgm2/s.

Homework Equations

L_z=Iw
I = mr²
I = (1/12)ML²
w = v/r

The Attempt at a Solution

I used the above equations:
L_z=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
L_z=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.

 

[mitch987]

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

 

[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, $$x_{cm}= \frac{2.5\times -4.72 +9.1\times 4.72}{2.5+9.1}<br /> =2.68m$$

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

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

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

 

[MissPenguins]

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

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

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

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

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

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