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 = L

Substituting in the period in seconds for T and 9.8 for gravity (g).

[2.1²/(4π²)]*9.8 = L ≃ 1.09 m

I only need to be accurate to 2 significant figures, so shouldn’t this answer be right? Does anyone see problems with this?

That makes sense, however the answer still comes out saying it is wrong.

### 2 Answers

T = 2

*π*sqr(I/[mg(L_cm])I = (1/3)mL^2

T^2/(2

*π)^2=(2/3)*(L/g)Solve for L:

L = (2.1^2/(2

*π)^2)*9.8/(2/3)My buddy Steve’s answer is to big

by a factor of 2.

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) = 3

*9.8*2.1²/(4π²) = 3.284 mEdit: 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

Source(s): My ‘cheat sheet’ of physics formulae