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

Answer:

$I=0.0987kg.m^2$

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

 $I=\frac{T}{2\pi}^2(mgd)$

 $I=\frac{0.94}{2.3.142}^2(1.80*9.8*0.250)$

 $I=0.0987kgm^2$