Calculating Angular Momentum: Particle+Meter Stick

[jmwachtel]

Homework Statement

A particle of mass 0.500 kg is attached to the 100 cm mark of a meter stick of mass 0.175 kg. The meter stick rotates on a horizontal, frictionless table with an angular speed of 5.00 rad/s.

(a) Calculate the angular momentum of the system when the stick is pivoted about an axis perpendicular to the table through the 30.0 cm mark.

(b) What is the angular momentum when the stick is pivoted about an axis perpendicular to the table through the 0 cm mark?

Homework Equations

1/12(mD^2) + mD^2

The Attempt at a Solution

I am close to the correct answer, but only within 10%. I know it has to do with the fact the mass is not in the center of mass. So in the second part of the equation stated above I have substituded D=.7D^2. Any ideas? I am having trouble with the concept on this problem too.

 

[CompuChip]

At what position did you find the center of mass to be?

 

[jmwachtel]

At what position did you find the center of mass to be?

I was just using the 1/12 formula? Does that calculate it at 50cm?

 

[Hootenanny]

I was just using the 1/12 formula? Does that calculate it at 50cm?

Does it make sense that the centre of mass is at 50cm? How in general does one calculate the centre of mass of a system?

 

[jmwachtel]

Does it make sense that the centre of mass is at 50cm? How in general does one calculate the centre of mass of a system?

No it would be between 30 and 100 I would assume. Would it just be in the middle of that? I'm not following...

 

[Hootenanny]

In general one can define the centre of mass ($\underline{x}_c$) of a system of N particles of masses $m_i$ at positions $\underline{x}_i$ thus,

$$\underline{x}_c = \frac{\sum_{i=1}^{N}{m_i\underline{x}_i}}{\sum_{i=1}^{N}{m_i}}$$

Usings this definition, it is possible to calculate the centre of mass of any body. However, it is extreamly cumbersome.

Perhaps an easier method to use would be two the fact that the net torque about the centre of mass of anybody must be zero. That is,

$$\sum_{i=1}^{N}F_i\cdot d_i = 0$$

Where $d_i$ is the perpendicular distance from the centre of mass to the line of application of the force. I assume that you know that the weight of a body acts through it's centre of mass, therefore one may write,

$$m_\text{ruler}\left(x_c\right-0.5) + m_\text{particle}\left(x_c-1\right) = 0$$

(where we define an anti-clockwise rotation as positive).