Simple harmonic oscillation: uniform rod

[vetgirl1990]

Homework Statement

A uniform rod of mass m and length L is freely pivoted at one end. What is the period of its oscillations? I_cm 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

Homework Equations

ω² = mgL/I

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

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

T = 2π/ω = 2π √13L/12g

 

[AbhinavJ]

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!

 

[AbhinavJ]

Thanks!

Welcome :)