Question

The 20-cm-long wrench in the figure swings on its hook with a period of 0.94 s . When the wrench hangs from a spring of spring constant 340 N/m , it stretches the spring 3.2 cm . What is the wrench's moment of inertia about the hook?

Answer 1

The period of a pendulum is

T = 2*π*√[Rg/g], where Rg is the radius of gyration.

Rg = T²/(2*π)² * g

The radius of gyration is Rg = √[I/m], where m = mass, so I = m*Rg²

When the wrench hangs from the spring, it force is m*g and

m*g = k*∆x

m = k*∆x/g

I = m*Rg² = k*∆x/g * [T²/(2*π)² * g]²

I = k*∆x*g*T^4/(2*π)^4

k = 340 N/m
∆x = 0.032 m
T = 0.94 s

I = 0.053 kg-m²