Question

When a 12000 joule of work is done on a flywheel, its frequency of rotation increases from 10 Hz to 20 Hz. The moment of inertia of flywheel about its axis of rotation is ______. (π² = 10)

Options

• 1 kgm²

• 2 kgm²

• 1.688 kgm²

• 1.5 kgm²

Answer 1

When a 12000 joule of work is done on a flywheel, its frequency of rotation increases from 10 Hz to 20 Hz. The moment of inertia of flywheel about its axis of rotation is 2 kgm².

Explanation:

Given, work done, W = 12000J,

Initial frequency, f₁ = 10 Hz

and final frequency, f₂ = 20 Hz

Angular velocity for rotational motion is given by ω = 2πf

∴ ω₁ = 2πf₁ = 2π × 10 = 20 π rad/s

and ω₂ = 2πf₂ = 2π × 20 = 40 π rad/s

According to work-energy theorem,

work done in rotation = change in rotational kinetic energy

`=> "W" = 1/2"k" omega_2^2 - 1/2 "l" omega_1^2    ....[because "KE"_"rotational" = 1/2 "l" omega^2]`

`= 1/2 "l" (omega_2^2 - omega_1^2)`    ....(i)

where, I = moment of inertia of the flywheel.

Substituting given values in Eq. (i), we get

`12000 = 1/2 "l"(1600 pi^2 - 400 pi^2)`

`=> = 1/2 "l"(1200 xx 10)`   ....(π² = 10)

`=> "l" = (12000 xx 2)/12000` = 2 kg m²