# Moment of Inertia of a Rod with Two Uniform Masses Attached

1. Apr 12, 2013

### jcd2012

1. The problem statement, all variables and given/known data

A uniform thin rod with mass 4.59 kilograms pivots about an axis through its center and perpendicular to its length. Two small bodies with a mass of 0.543 kilograms, are attached to the ends of the rod. What is the length of the rod such that the moment of inertia of the three-body system with respect to the described axis is 0.941 kg·m^2?

2. Relevant equations

Length of Rod = L (unknown)

For the rod, MoI = 1/12ML

For the two small masses, MoI of each is: MoI = MR^2

MoI-rod + MoI-mass + MoI-mass = ƩI

3. The attempt at a solution

Moment of Inertia of Rod: 1/12(4.59 kg )L

Moment of Inertia of Mass: (0.543 kg) * (L/2)^2

For moment of inertia of small mass, I picked the radius from the axis of rotation to be L/2 since the axis of rotation is at the rod's center and perpendicular to its length. I thought the radius from that point to the small mass would be L/2.

So the setup is: 1/12(4.59)L + (0.543)(L/2)^2 + (0.543)(L/2)^2 = 0.941

When I solve for L, I get L = 1.28609 m, which is apparently incorrect. I do not know where I could have screwed up on this problem.

2. Apr 12, 2013

### TSny

Check this formula. All moments of inertia should have dimensions of mass*length2

3. Apr 12, 2013

### cepheid

Staff Emeritus
Welcome to PF,

Your equation for the moment of inertia of the thin rod is obviously wrong just by inspection, because it has the wrong dimensions. It should have dimensions of mass*length^2, but it has dimensions of mass*length.