# Homework Help: Moment of inertia for a thin rod

1. Apr 3, 2013

### dinospamoni

1. The problem statement, all variables and given/known data
The thin rod shown has a length L = 100 cm, and a density that varies from 26 g/cm at the origin to 3 g/cm at the far end. Determine a) the moment of inertia about axis 1 (passing through the center of mass of the rod), and b) the moment of inertia about axis 2 (passing through the heavy end of the rod).

picture attached

2. Relevant equations

Mass of rod = 1.45*10^3 g
Center of mass of rod = 36.8 cm from the more dense end

The density as a function of position I found to be: ρ(x)=.23x + 3

3. The attempt at a solution
For the first part:

I used I = ∫ r^2 * m/L dr
where -36.8 < r < 63.2

and got 1.46*10^6 g*cm^2

For the second part I know it has to do with the parallel axis theorem:

I = I_cm+mh^2

where h = 36.78 cm

and got 3.4*10^6 but that's not right.

Any ideas?

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Last edited: Apr 3, 2013
2. Apr 3, 2013

### voko

Since the density is not uniform, you can't just assume it is m/L. Use the density function you found above. Pay attention to the relation between r in the integral and x in the density function.

3. Apr 3, 2013

### SammyS

Staff Emeritus

Your density function gives a linear density of 3 g/cm at the origin rather than at x = 100 cm ,

and a linear density of 26 g/cm at x = 100 cm rather than at the origin.