# Moment of Inertia and pivot joint

1. Mar 30, 2008

### glasshut137

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

Consider a rod length L and mass m which is pivoted at one end. An object with mass m attached to the free end of the rod. g=9.8 m/s^2. Note: Contrary to the diagram shown, consider the mass at the end of the rod to be a point particle.

Basically it looks like a rod pivoted at a point on the origin, and 23 degrees below the x-axis.

1) determine the moment of Inertia, I, of the system with respect to the pivot joint.
I tried doing the sum of the moments about the pivot joint, I= m(L)^2/12 for the rod plus I= m (L)^2 of the point mass and got (13/12)*m*(L)^2 but it was wrong.

2) Determine the position of the center of mass from the pivot point, i.e., find C.
I used xcm= (mL+mL)/(m+m) and got C=L. Can someone tell me if i did that correctly?

3. The attempt at a solution

2. Mar 30, 2008

### Dr. Jekyll

There is something called http://en.wikipedia.org/wiki/Parallel_axis_theorem" [Broken]. It's not that simple.

Unfortunately, no. You get that xcm=L (assuming that the pivoted end has x coordinate equal to 0), which can't be true. Center of mass will be somewhere between the middle of the rod and mass m. Mass (m) will "move" the center of mass towards itself.

Last edited by a moderator: May 3, 2017
3. Mar 30, 2008

### glasshut137

ok I got the first part but I'm still not sure about the second one. So since the center of mass of the rod alone is L/2 and L for the point mass then the center of mass of the system would be at (3/4)L ?

4. Mar 30, 2008

### Dr. Jekyll

Yes, but only if mass of the rod is equal to mass m:

$$x_{cm}=\frac{m\frac{L}{2}+mL}{m+m}=\frac{3}{4}L$$.