Correct Answer - Option 1 : 395

__Concept:__

Flywheels

A flywheel is used to control the variation in speed during each cycle of the engine.

The kinetic energy of the flywheel can be calculated by:

\(KE= \frac{1}{2}I{\omega ^2}\)

where

I = moment of inertia of thin disc = mR2

ω = angular velocity of the flywheel = \(\frac{{2\pi N}}{{60}}\)

__Calculation:__

__Given:__

Mass of flywheel (m) = 20 kg, radius (R) = 100 mm = 0.1 m, N = 600 rpm

I = mR^{2}

I = 20 × 0.1^{2}

I = 0.2 kg/m^{2}

\(w = \frac{{2\pi N}}{{60}} = \frac{{2~ ×~ \pi ~×~ 600}}{{60}} = 62.84\;rad/s\)

∴ \({\rm{K}}.{\rm{E}} = \frac{1}{2}{\rm{I}}{{\rm{\omega }}^2} = \frac{1}{2} × 0.2 × {\left( {62.84} \right)^2} = 395{\rm{\;J}} \)

Moment of inertia of a thin disc is mR2

Moment of inertia of a thick disc or flywheel is \(\frac{{m{R^2}}}{2}\)

Since here in the question thickness is very very less so it is a thin disc