# Vertically oscillating rod

## Homework Statement

Assume all oscillators are frictionless.
e) A uniform thin rod (Mass M, length Lo) oscillates vertically on a frictionless axis perpendicular to the rod and passing through one end with period To. Now, a second uniform rod is made of the same material and mounted the same way, so it oscillates vertically about an axis through one end with period T. The length of the second pendulum is L = 9.47Lo. You may assume the rod is essentially one-dimensional.
Find the ratio T/To.

## Homework Equations

I = 1/3 ML2, T = 2∏√(I/Mgd)

## The Attempt at a Solution

T/To = [2∏√(1/3 M(9.47Lo)2/Mgd)] / [2∏√(1/3 MLo2/Mgd)]
T/To = [√(1/3 M(9.47Lo)2/Mgd)] / [√(1/3 MLo2/Mgd)]
T2/To2 = [(1/3(9.47Lo)2/gd)] / [(1/3Lo2/gd)]
T2/To2 = (3gd(9.47Lo)2)/(3gd(Lo)2)
T2/To2 = (9.47Lo)2/(Lo)2
T/To = (9.47Lo)/(Lo)
T/To = 9.47

The ratio I calculated is wrong. I believe my calculation is in order. Any help in pointing out any possible mistakes is appreciated!

Last edited:

Doc Al
Mentor
I = 1/3 ML2, T = 2∏√(I/Mgd)
Write d in terms of L. What's T in terms of L and g?

Thank you for your response Doc Al. I'm thinking d in terms of L would be L/2. Also T = 2∏√(2L/3g). Is this correct?

Doc Al
Mentor
I'm thinking d in terms of L would be L/2. Also T = 2∏√(2L/3g). Is this correct?
Looks good to me.

T/To = 2∏√(2*9.47Lo/3g) / 2∏√(2Lo/3g)
T2/To2 = (2*9.47Lo/3g) / (2Lo/3g)
T2/To2 = (2*3g*9.47Lo) / (2*3g*Lo)
T2/To2 = 9.47
T/To = 3.077336511

I redid the problem with the substitution of the terms, Is it right that I ended up canceling out most of the terms?

Doc Al
Mentor
I redid the problem with the substitution of the terms, Is it right that I ended up canceling out most of the terms?
Definitely. You can write the period as T = constants*√(L). Those constants don't matter, since all you want is the ratio.

Thank you for your help, I don't think I would've ever thought to make those substitutions on my own.