# Simple harmonic oscillation: uniform rod

## Homework Statement

A uniform rod of mass m and length L is freely pivoted at one end. What is the period of its oscillations? Icm for a uniform rod rotating about its centre of mass is 1/12mL2

(a) √3g/2L
(b) 2π √3L/2g
(c) 2π √2L/3g
(d) 2π √L/g
(e) none of the above

ω2 = mgL/I

Icm = 1/12mL2
I = Icm +1/12mL2

Period: T = 2π/ω

## The Attempt at a Solution

I'm fairly certain that the answer is "none of the above", but I'd just like to make sure I'm not neglecting anything in what seems to be a simple plug-and-chug question.

I = Icm +1/12mL2 = 1/12mL2 + mL2 = 13/12 mL2

ω = √mgL/I= √(mgL)/(13/12)mL2 = √12g/13L

T = 2π/ω = 2π √13L/12g

Last edited:

What is the distance between the Centre of mass of rod and the end? Answer this and you'll realize what you did wrong.

• vetgirl1990
What is the distance between the Centre of mass of rod and the end? Answer this and you'll realize what you did wrong.
I just realized what I did wrong. The distance is L/2, so the moment of inertia equation turns into: I = 1/3mL2

So then ω2 = mg(L/2) / (1/3)mL2

Therefore T = 2π √2L/3g

Thanks!

Thanks!
Welcome :)