- #1

Cintdrix

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## 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.