# A uniform steel bar swings from a pivot

1. Apr 23, 2013

### chicagobears34

1. The problem statement, all variables and given/known data
A uniform steel bar swings from a pivot at one end with a period of 1.3 sec.
Solve for Length of bar

2. Relevant equations
T=2pi*sqrt(L/g)

3. The attempt at a solution
since the period is 1.3 seconds, I just plug in 1.3 for T and 9.8m/s^2 for g and solve for L
i get L= .42m, which is wrong.
What am I doing wrong?

Last edited: Apr 23, 2013
2. Apr 23, 2013

### Yosty22

So are you solving for the length of the bar? It is unclear in the information above what you are solving for.

3. Apr 23, 2013

### chicagobears34

yes I am supposed to solve for the length of the bar, sorry about that

4. Apr 23, 2013

### ehild

Your formula is valid for a simple (mathematical) pendulum. It is a "physical pendulum" now.

ehild

5. Apr 23, 2013

### chicagobears34

I used T=2pi * sqrt(2L/3G) and got the correct answer.
is this the formula for period of a pendulum with a bar or something?

6. Apr 23, 2013

### ehild

7. Dec 10, 2013

### k0ss

I read your source and can't figure out how they came to that equation. When I do the math out of $$\frac{I}{Mgh}$$ and substitute $$\frac {ML^2}{3}$$for I, I get $$2\pi\sqrt{\frac {L}{3g}}$$instead of $$2\pi\sqrt{\frac{2L}{3g}}$$ Where did the 2 come from in the numerator in the correct equation?

EDIT: Was missing the equation higher up in your source article where it stated that $$h=\frac{L}{2}$$ where I was assuming h was the same as L.

Last edited: Dec 10, 2013
8. Dec 10, 2013

### ehild

The time period of a physical pendulum is

$$T=2\pi \sqrt{\frac{I}{mgh}}$$

where I is the moment of inertia of the swinging body with respect to the pivot,
m is the mass,
h is the distance of the centre of mass from the pivot.

In case of a homogeneous thin bar of length L, I=mL2/3, and the CM is at the middle, so h=L/2.

ehild

Last edited: Dec 10, 2013