Relating moment of Inertia and pendulum oscillation

[heatherro92]

I took a picture of the question to help.

I know that the moment of inertia for a rod (at one end, not the center) is:
1/3ML^2

And I know that the period of oscillation is:
T=2∏√(L/g)

But I don't know how to relate them... I tried doing Torque=Iα=Fdcosθ and solve in terms of Time... but it wasn't becoming the equation and I had a random ω I couldn't get rid of. And I don't think I'm allowed to use the oscillation equation at all since I'm supposed to be deriving it, I'm just not sure what to do.

 

[ehild]

T=2∏√(L/g) is valid for a mathematical pendulum (hanging point mass).
Use the formula valid for physical pendulum.
http://hyperphysics.phy-astr.gsu.edu/hbase/pendp.html

ehild

 

[heatherro92]

so am I solving for α in that equation and that's identical to the period?

 

[ehild]

No, α is the angular acceleration. The period is T.

ehild

 

[Nickie]

The moment of inertia and the period of oscillation of a pendulum are related through the equation T=2∏√(I/mgd), where I is the moment of inertia, m is the mass of the pendulum, g is the acceleration due to gravity, and d is the distance from the pivot point to the center of mass of the pendulum. This equation can be derived from the equation for torque, as you have attempted. However, it is important to note that the moment of inertia for a pendulum is not the same as the moment of inertia for a rod. The moment of inertia for a pendulum is given by I=ml^2, where l is the length of the pendulum. Therefore, the correct equation to use would be T=2∏√(ml^2/mgd), which simplifies to T=2∏√(l/g). This is the equation for the period of oscillation of a simple pendulum, and it shows that the period is directly proportional to the length of the pendulum and inversely proportional to the acceleration due to gravity.