Answer
given,
expression of Kinetic energy of rotating body
[tex]K = \dfrac{1}{2}I\omega^2[/tex]
ω = 34.0 rad/s
Assuming mass of the particle equal to 13 Kg
and perpendicular distance from the particle to the axis is r = 1.25 m
now,
moment of inertia of particle = ?
from the given expression
[tex]I= \dfrac{2K}{\omega^2}[/tex]..............(1)
we know
[tex]K = \dfrac{1}{2}mv^2[/tex]
v = r ω
[tex]K = \dfrac{1}{2}mr^2\omega^2[/tex]
putting value in equation (1)
[tex]I= \dfrac{2\dfrac{1}{2}mr^2\omega^2}{\omega^2}[/tex]
[tex]I =mr^2 [/tex]
[tex]I =13 \times 1.25^2 [/tex]
I = 20.3125 kg.m²
