# Pendulum in Simple Harmonic Motion

1. Dec 7, 2008

### smile05

1. The problem statement, all variables and given/known data

A 20 cm long (L) wrench swings on its hook with a period of 0.90s. When the wrench hangs from a spring of spring constant 360 N/m (k), it stretches the spring 3.0 cm (x). What is the wrench's moment of inertia about the hook.
A diagram also shows that the centre of mass for the wrench is 14cm (d) from the hook (i.e pivot point).

2. Relevant equations

Fsp = -ks
Fg = mg
I = Icm + md^2

3. The attempt at a solution

Fsp = -10.8 N
Fsp = Fg gives m = 1.1 kg
Icm for a rod with a pivot in one of the edges is 1/3mL^2
Therefore, I = 1/3mL^2 + md^2 = 0.036 kg m^2

However, that answer is not correct, the correct answer according to the textbook is 0.031 kg m^2. Any help is appreciated. Thank you!

2. Dec 7, 2008

### LowlyPion

I think you have assumed a moment of inertia initially as being a rod and this is not what they are suggesting.

What they give you is information about a physical pendulum.

Such a pendulum can be determined to have as a period of oscillation

T = 2π(I/m*g*L)½

To find I then:

I = m*g*L*T2/2π

3. Dec 7, 2008

4. Dec 7, 2008

### smile05

Oh now I understand! Thank you so much! :)