Quick Question so I can finish this homework problem.

[moo5003]

I'm doing a physical pendulum problem involving a rigid rod of .720m in length attached to a string of 1.00 meters in length. I'm supposed to find its period. I can do the problem but I'm having trouble calculating the rods moment of inertia.

Parrallel Axis Theorem:
I = Icm + MD^2

Icm = I (1 + .720) ^2, Is md^2 the same thing...? I'm a little confused on this point.

 

[dicerandom]

The parallel axis theorem basically states that if you have the moment of inertia of an object about one axis, and you want to know its moment of inertia about another, parallel axis (one which also goes through the object's center of mass), then all you have to do is add on the moment of inertia of a point particle which is the same distance from the second axis as the center of mass of your object is. That's where the MD^2 comes from in your equation, the D is the distance to the CM of the rod.

Also note that, in this case, you could easily calculate the moment of inertia by simply doing the integral:

$$\int_{Mass}\rho r^2 dV$$

Which, for this case of a thin rod on the end of a massless string, reduces to:

$$\int_{1}^{1.720}\rho r^2 dr$$

 

[moo5003]

Thanks for the help.
The moment of inertia would then equal
1/12 M(.720m)^2 + M (1.00m + .720m)^2 Right?