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 π 2 = 10

Step-by-step solution

Given, work done, W = 12000 J, Initial frequency, f 1 = 10 H z Angular velocity for rotational motion is given by ω = 2 π f ∴ ω 1 = 2 π f 1 = 2 π × 10 = 20 π r a d s a n d ω 2 = 2 π f 2 = 2 π × 20 = 40 π r a d s According to work-energy theorem, Work done in rotation = change in rotational kinetic energy ⇒ W = 1 2 ω 2 2 - 1 2 ω 1 2 ∵ K E r o t a t i o n a l 1 2 ω 2 = 1 2 / ω 2 2 - ω 1 2 …(i) Where, l = moment of inertia of the flywheel. Substituting given values in Eq. (i), we get 12000 1 2 1600 π 2 - 400 π 2 ⇒ = 1