- #1
vetgirl1990
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- 3
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
Homework Equations
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
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