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!