Simple Harmonic Motion of a Rigid Body

[Cintdrix]

Homework Statement

A very light, rigid rod with a length of 0.570 m extends straight out from one end of a meter stick. The stick is suspended from a pivot at the far end of the rod and is set into oscillation. Determine the period of oscillation.

Homework Equations

omega = (Mgd/I)^1/2
omega = 2pi/period.

The Attempt at a Solution

I solved this question, but I was just looking for clarification as to why it was right. The moment of inertia of the rod would be 1/3 ML^2 and using the parallel axis theorem I would have to add Md^2 on top of that.

The masses cancel in the first equation, but I was wondering about the "d" in the numerator. What does it represent? I originally thought it was the distance from the centre of mass to the pivot but it wasn't the case for this problem. Any insight is appreciated.

I when i subbed d=0.57 into the equation i got the correct answer of 2.15s, but I don't understand why.

 

[kuruman]

The Attempt at a Solution

I solved this question, but I was just looking for clarification as to why it was right. The moment of inertia of the rod would be 1/3 ML^2 and using the parallel axis theorem I would have to add Md^2 on top of that.

The parallel axes theorem says that the moment of inertia about distance d from the center of mass is given by

I = I_CM + md², where I_CM = (1/12)mL² for a uniform rod not (1/3)mL². If you got the right answer, you must have made two errors that canceled each other.

 

[Cintdrix]

True. I was using the moment of inertia about one end of the stick, which is indeed (1/3)mL². In my case i guess the D was still the distance from the old axis to the new axis. In using the moment of inertia about the CM, and making d the distance between the pivot and the CM, I was able to again get the correct answer.

Thanks for your help