Finding the length of a uniform steel bar given period?

Question

A uniform steel bar swings from a pivot at one end with a period of 2.1 seconds. How long is the bar? The problem seems straightforward, but my online homework platform keeps indicating that my answer is incorrect. 
Here are the equations I used:
T = 2π * √(L/g)
Rearranging gives:
T/(2π) = √(L/g)
Squaring both sides results in:
T²/(4π²) = L/g
Thus,
L = g * [T²/(4π²)]  
I substituted the period (T) in seconds and used 9.8 m/s² for gravity (g):
L = [2.1²/(4π²)] * 9.8 ≈ 1.09 m  
I only need to be accurate to two significant figures, so shouldn’t this answer be correct? Does anyone see any issues with my calculations?

Anonymous · Accepted Answer

T = 2πsqr(I/[mg(L_cm])
I = (1/3)mL^2
T^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 big
by a factor of 2.

Anonymous · Answer

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, Al

Correct answer: L = 1.642 m

Anonymous

Finding the length of a uniform steel bar given period?

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