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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

Respuesta :

okpalawalter8

Answer:

[tex]I=0.0987kg.m^2[/tex]

Explanation:

From the question we are told that:

Mass [tex]M=1.80kg[/tex]

Deviation [tex]d=0.250[/tex]

Time [tex]t=0.940s[/tex]

Generally the equation for moment of inertia is mathematically given by

 [tex]I=\frac{T}{2\pi}^2(mgd)[/tex]

 [tex]I=\frac{0.94}{2.3.142}^2(1.80*9.8*0.250)[/tex]

 [tex]I=0.0987kgm^2[/tex]