Respuesta :
Answer:
Moment of inertia, [tex]I = 0.0987\ kg-m^2[/tex]
Explanation:
It is given that,
Mass of the monkey, m = 1.8 kg
Distance from the center of mass, d = 0.25 m
The time period of oscillation, t = 0.94 s
It is assumed to find the moment of inertia of the wrench about an axis through the pivot. the time period of the oscillation is given by :
[tex]t=2\pi \sqrt{\dfrac{I}{mgd}}[/tex]
[tex]I=\dfrac{mgdt^2}{4\pi^2}[/tex]
[tex]I=\dfrac{1.8\times 9.8\times 0.25\times (0.94)^2}{4\pi^2}[/tex]
[tex]I = 0.0987\ kg-m^2[/tex]
So, the moment of inertia of the wrench about an axis through the pivot is [tex]0.0987\ kg-m^2[/tex]. Hence, this is the required solution.
