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physics question that i need help with. i need answer very soon.?
The 20 cm long wrench in the figure swings on its hook with a period of 0.85 s. when the wrench hangs from a spring of spring constant 370 N/m, it stretches the spring 2.6 cm. What is the wrench's moment of inertia?
Answer:
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 = 370 N/m
∆x = 0.026 m
T = 0.85 s
I = 0.032 kg-m²

























































































