Question

The kinetic energy of a rotating body is generally written as K=12Iω2, where I is the moment of inertia (also known as rotational inertia) of the body. Find the moment of inertia of the particle described in the problem introduction with respect to the axis about which it is rotating. Assume ω = 34.0 rad/s .

Answer 1

Answer

given,

expression of Kinetic energy of rotating body

$K = \dfrac{1}{2}I\omega^2$

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

$I= \dfrac{2K}{\omega^2}$..............(1)

we know

$K = \dfrac{1}{2}mv^2$

v = r ω

$K = \dfrac{1}{2}mr^2\omega^2$

putting value in equation (1)

$I= \dfrac{2\dfrac{1}{2}mr^2\omega^2}{\omega^2}$

$I =mr^2$

$I =13 \times 1.25^2$

I = 20.3125 kg.m²