# Rod Pendulum Problem

1. Nov 18, 2009

### Mr. Goosemahn

1. The problem statement, all variables and given/known data
A rod suspended on its end and acting as a physical pendulum swings with a period of 2.24 s. What is its length? (g = 9.80 m/s2)

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

3. The attempt at a solution
Okay, so I'm given the period, which is 2.24 seconds.

I'm also given G, as it's 9.8.

I plug in the values and the resulting equation is what I get:

2.24=2pi * (sqrt)(L/9.8)

I divide 2.24 by 2pi, which leaves me with this:

.356507=(sqrt)(L/9.8)

I square both sides and get:

.127097=L/9.8

Multiply 9.8 * .127097 should give me L.

I get L=1.24555. I put it into my online homework assignment, and it's wrong.

What am I doing incorrectly?

2. Nov 18, 2009

### Mr. Goosemahn

Anyone?

3. Nov 18, 2009

### A_Munk3y

im getting the same as you :(
are you sure those are the correct numbers?

4. Nov 18, 2009

### Mr. Goosemahn

Yup, I copy pasted the problem right out of the assignment.

5. Nov 18, 2009

### rl.bhat

Here the length of the physical pendulum is the distance of center of mass from the point of suspension.

6. Nov 18, 2009

### Mr. Goosemahn

Don't I need the mass (or masses) to calculate the center of mass for an object?

7. Nov 18, 2009

### rl.bhat

No. In this case center of mass is the center of gravity. It is the mid point of the rod.

8. Nov 18, 2009

### Mr. Goosemahn

I don't know if I understood this right, but the L in the equation must then be doubled?

So...

2.24=2pi * (sqrt)(2L/9.8)

?

9. Nov 18, 2009

### rl.bhat

No. Length should be L/2.

10. Nov 18, 2009

### Mr. Goosemahn

It still doesn't work.

I plugged in (L/2) for the L, so the equation looks like this:

2.24 = 2pi * (sqrt)((L/2)/9.8)

Solving for L:

2.24/(2pi) = (sqrt)((L/2)/9.8) ----> 0.356507 = (sqrt)((L/2)/9.8)

(0.356507)^2 = ((sqrt)((L/2)/9.8))^2 -----> 0.127097 = (L/2)/9.8

L/2 = (0.127097)*9.8 -----> L/2 = 1.24555

L = 1.24555 * 2 -----> L = 2.49111

This value is also incorrect. Where am I making a mistake?

11. Nov 18, 2009

### rl.bhat

In the physical pendulum, You have to take in to account the moment of inertia and torque.
The formula for period of oscillation is given by
T = 2π*sqrt(I/τ), where I = 1/3*M*L^2 and τ = M*g*L/2
So T = 2π*sqrt(2L/3g)
Now solve for L.

12. Nov 18, 2009

### Mr. Goosemahn

It worked, finally! Thanks for the help, it was really useful!