Finding rod pendulum length from period of 2.24 s

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Homework Statement


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)

Homework Equations


T=2(pi) * (sqrt)(L/G)

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?
 
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Mr. Goosemahn said:
Anyone?

im getting the same as you :(
are you sure those are the correct numbers?
 
A_Munk3y said:
im getting the same as you :(
are you sure those are the correct numbers?

Yup, I copy pasted the problem right out of the assignment.
 
Here the length of the physical pendulum is the distance of center of mass from the point of suspension.
 
rl.bhat said:
Here the length of the physical pendulum is the distance of center of mass from the point of suspension.

Don't I need the mass (or masses) to calculate the center of mass for an object?
 
No. In this case center of mass is the center of gravity. It is the mid point of the rod.
 
rl.bhat said:
No. In this case center of mass is the center of gravity. It is the mid point of the rod.

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)

?
 
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?
 
In the physical pendulum, You have to take into 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.
 
rl.bhat said:
In the physical pendulum, You have to take into 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.

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