# Pendulum in Simple Harmonic Motion

## Homework Statement

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).

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

## 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!

## Answers and Replies

LowlyPion
Homework Helper
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

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