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.
T = 2πsqr(I/[mg(L_cm])
I = (1/3)mL^2
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.
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!
Correct answer: L = 1.642 mSource(s): My ‘cheat sheet’ of physics formulae