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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)
- a)1 kgm2
- b)2 kgm2
- c)1.688 kgm2
- d)1.5kgm2
Correct answer is option 'B'. Can you explain this answer?
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When a 12000 joule of work is done on a flywheel, its frequency of ro...
Calculation of moment of inertia of flywheel
Given:
Work done on flywheel = 12000 J
Initial frequency of rotation = 10 Hz
Final frequency of rotation = 20 Hz
π2 = 10 (given)
Formula used:
Work done = Change in kinetic energy
Change in kinetic energy = 1/2 I (ωf2 - ωi2)
Where,
I = Moment of inertia
ωi = Initial angular velocity
ωf = Final angular velocity
Calculation:
Change in kinetic energy = 12000 J
ωi = 2πfi = 2π x 10 = 20π rad/s
ωf = 2πff = 2π x 20 = 40π rad/s
12000 = 1/2 I (402π2 - 202π2)
I = 2 x 12000 / (3π2)
I = 2 kgm2
Therefore, the moment of inertia of the flywheel about its axis of rotation is 2 kgm2.
Given:
Work done on flywheel = 12000 J
Initial frequency of rotation = 10 Hz
Final frequency of rotation = 20 Hz
π2 = 10 (given)
Formula used:
Work done = Change in kinetic energy
Change in kinetic energy = 1/2 I (ωf2 - ωi2)
Where,
I = Moment of inertia
ωi = Initial angular velocity
ωf = Final angular velocity
Calculation:
Change in kinetic energy = 12000 J
ωi = 2πfi = 2π x 10 = 20π rad/s
ωf = 2πff = 2π x 20 = 40π rad/s
12000 = 1/2 I (402π2 - 202π2)
I = 2 x 12000 / (3π2)
I = 2 kgm2
Therefore, the moment of inertia of the flywheel about its axis of rotation is 2 kgm2.
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Community Answer
When a 12000 joule of work is done on a flywheel, its frequency of ro...
Given, work done. W = 12000J,
.
Initial frequency, f1 = 10Hz and final frequency, f2 = 20Hz
Angular velocity for rotational motion is given by ω = 2πf
∴ ω1 = 2πf1 = 2π x 10 = 20π rad/s and ω2 = 2πf2 = 2π x 20 = 40π rad/s
According to work-energy theorem, work done in rotation = change in rotational kinetic energy
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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)a)1 kgm2b)2 kgm2c)1.688 kgm2d)1.5kgm2Correct answer is option 'B'. Can you explain this answer?
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