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Finding the length of a uniform steel bar given period?

A uniform steel bar swings from a pivot at one end with a period of 2.1 s. How long is the bar?The problem seems pretty simple, but when I do it, my online homework problem website keeps telling me the length is wrong.The equations I used:T = 2π * √(L/g) T/(2π) = √(L/g)T²/(4π²) = L/g[T²/(4π²)]*g = LSubstituting in the period in seconds for T and 9.8 for gravity (g).[2.1²/(4π²)]*9.8 = L ≃ 1.09 mI only need to be accurate to 2 significant figures, so shouldn’t this answer be right? Does anyone see problems with this?

2 Answers

T = 2πsqr(I/[mg(L_cm])I = (1/3)mL^2T^2/(2π)^2=(2/3)(L/g)Solve for L:L = (2.1^2/(2π)^2)9.8/(2/3)My buddy Steve’s answer is to bigby a factor of 2.... Show More
You are using the formula for a simple pendulum. A uniform bar is not a simple pendulum, but a physical pendulum with T = 2π√[I/mgL]; solving for L, L = 4π²I/(mgT²)I = ⅓mL² ; substituting,L = 3gT²/(4π²) = 39.82.1²/(4π²) = 3.284 m Edit: Al is right. I forgot that L in my first formula is the length to the CM of the rod, not the rod length itself. My bad, but I made a notation on my ‘cheat sheet’ to remind me to be more careful!Thanks, AlCorrect answer: L = 1.642 m... Show More

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