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.
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!