Pendulum Motion Equation and Period Calculation

[Icetray]

Hi guys,

I was doing my lab report and stumbled onto this question and I would really appreciate it if you guys could assist me on this. (:

Homework Statement

For a simple pendulum consisting of a mass M attached to a very thin light string of length L, in the absence of air resistance, derive the equation of motion for the simple pendulum in terms of the angular displacement θ relative to its equilibrium position? For “small” oscillation, namely θ is less than 5˚, what is the period T of oscillation? Compare with the derived result in Exercise 4 above.

Homework Equations

From Exercise 4:
T = 2π√(I/(MgL_g))

The Attempt at a Solution

I'm guessing that I will just be M for a simple pendulum and I'll end up with:
T = 2π√(1/(gL_g ))
which doesn't make much sense.

Looking forward to your replies! (:

 

[Spinnor]

What are you using for I?

 

[Icetray]

What are you using for I?

It's the moment of Inertia of the pendulum (taken to be a I = ML_g².

We assume that the pendulum of mass M is attached to a very light thin string of length L_g.

I googled and found out that the answer should be:

T = 2π√(L_g/g) but I have no clue how I am to get to this. ):

 

[Spinnor]

Hi guys,

...

From Exercise 4:
T = 2π√(I/(MgL_g))

The Attempt at a Solution

I'm guessing that I will just be M for a simple pendulum and I'll end up with:
T = 2π√(1/(gL_g ))
which doesn't make much sense...

(:

I goes as L^2. In your first equation if you substitute for I you should get L in the numerator and not the denominator as you have.

 

[Icetray]

I goes as L^2. In your first equation if you substitute for I you should get L in the numerator and not the denominator as you have.

Why does I become L^2? ):

 

[Icetray]

Can anyone assit me with this? ):