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Simple harmonic oscillation: uniform rod
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[QUOTE="vetgirl1990, post: 5451665, member: 573977"] [h2]Homework Statement [/h2] A uniform rod of mass m and length L is freely pivoted at one end. What is the period of its oscillations? I[SUB]cm[/SUB] for a uniform rod rotating about its centre of mass is 1/12mL[SUP]2[/SUP] (a) √3g/2L (b) 2π √3L/2g (c) 2π √2L/3g (d) 2π √L/g (e) none of the above [h2]Homework Equations[/h2] ω[SUP]2[/SUP] = mgL/I I[SUB]cm[/SUB] = 1/12mL[SUP]2[/SUP] I = I[SUB]cm[/SUB] +1/12mL[SUP]2[/SUP] Period: T = 2π/ω [h2]The Attempt at a Solution[/h2] 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[SUB]cm[/SUB] +1/12mL[SUP]2[/SUP] = 1/12mL[SUP]2[/SUP] + mL[SUP]2[/SUP] = 13/12 mL[SUP]2[/SUP] ω = √mgL/I= √(mgL)/(13/12)mL[SUP]2[/SUP] = √12g/13L T = 2π/ω = 2π √13L/12g [/QUOTE]
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Simple harmonic oscillation: uniform rod
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