Question

A 1.80-kg monkey wrench is pivoted 0.250 m from its center of mass and allowed to swing as a physical pendulum. The period for small-angle oscillations is 0.940 s. (a) What is the moment of inertia of the wrench about an axis through the pivot

Answer 1

`I=0.0987kg.m^2`

Step-by-step explanation:

From the question we are told that:

Mass
`M=1.80kg`

Deviation
`d=0.250`

Time
`t=0.940s`

Generally the equation for moment of inertia is mathematically given by

`I=(T)/(2\pi)^2(mgd)`

`I=(0.94)/(2.3.142)^2(1.80*9.8*0.250)`

`I=0.0987kgm^2`