A uniform meter stick is hung at its center from a thin wire. It is twisted and oscillates with a period of 5 s. The meter stick is then sawed off to a length of 0.76 m, rebalanced at its center, and set into oscillation.
With what period does it now oscillate?
L = 1m
L(final) = 0.76m
T = 5s
Moment of Inertia: I = (1/12)*M*L^2
period: T = 2pi*sqrt(I/K) Where K is the torsional constant
The Attempt at a Solution
I first found the torsional constant:
K = [(1/12)*M*(1^2)]/[(T/(2pi))^2] = 0.13459M
So no I have K = 0.13459M
T(final) = 2pi*[sqrt((1/12)*M*(0.76^2)/0.13159M)] = 3.8s
The answer looks correct but I'm not sure, any ideas? Thanks!
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