# Moment of Inertia question

1. Oct 5, 2008

### pchalla90

1. The problem statement, all variables and given/known data
Show that the rotational inertia of the rod about an axis through its center and perpendicular to its length is ML^2/12, where M is Mass and L is length from axis of rotation.

2. Relevant equations
∫r^2dm=Mr^2

3. The attempt at a solution

Well, you see the integral that I did in the relevant equations? I got most of the work done there, assuming that R=L. My problem is understanding where the /12 comes from in the question. There were no numbers given in the initial question, just M and L.

2. Oct 5, 2008

### Dick

Great. dm should have a dr in it. What's R? What are the limits on your integral? Take rho to be the density of the rod per unit length. What's M in terms of rho and L? What's the moment of inertia in terms of rho and L? Now eliminate rho.

3. Oct 6, 2008

### pchalla90

well, dm=pdL is what im assuming...

so ∫dm=∫pdL
M=pL

but shouldnt the right side of the above equation be a definite integral while the left is an indefinite integral?

i think the right side of the above equation should go from -L/2 to L/2, but that raises the question of the limits of the left side of the equation...

either way, assuming both sides are indefinite integrals, substituting pL for M in ML^2 gives me

pL^3=I

but thats as much as i think i understood of your post.

4. Oct 6, 2008

### Dick

Now you want to integrate r^2*rho*dr for r from -L/2 to +L/2 to get the moment of inertia. Then use M=rho*L to get rid of the rho.

5. Oct 6, 2008

### pchalla90

Okay, so....

∫(-L/2,L/2) pR^2dr= (1/3)R^3p|(-L/2,L/2)
=(1/3)(L/2)^3p-(1/3)(-L/2)^3p
=(1/3)(L/2)^3p+(1/3)(L/2)^3p
=(2/3)(L^3/8)p
=L^3/12p

p=M/L

=(L^3/12)(M/L)
=ML^2/12

Thank you so much for your help.