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The speed of an engine varies from 210 rad/s to 190 rad/s. During the cycle the change in kinetic energy is found to be 400 Nm. The inertia of the flywheel in kg/m2 is
  • a)
    0.10
  • b)
    0.20
  • c)
    0.30
  • d)
    0.40
Correct answer is 'A'. Can you explain this answer?
Verified Answer
The speed of an engine varies from 210 rad/s to 190 rad/s. During the ...
Given ω1 = 210 rad/ sec, ω2 = 190 rad/ sec, ΔE= 400 NmAs the speed of flywheel changes from ω1 to ω2, the maximum fluctuation of energy,ΔE = 1/2I [(ω1)^2 (ω2)^2] I = 0.10 kgm^2
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The speed of an engine varies from 210 rad/s to 190 rad/s. During the ...
Given:
The speed of the engine varies from 210 rad/s to 190 rad/s.
The change in kinetic energy during the cycle is 400 Nm.

To find:
The inertia of the flywheel in kg/m2.

Solution:
The kinetic energy of a rotating object is given by the equation:
KE = (1/2) * I * ω2
where KE is the kinetic energy, I is the moment of inertia, and ω is the angular velocity.

We are given that the change in kinetic energy during the cycle is 400 Nm. This can be expressed as:
ΔKE = (1/2) * I * (ω2 - ω1)
where ω2 is the final angular velocity and ω1 is the initial angular velocity.

We are also given that the speed of the engine varies from 210 rad/s to 190 rad/s. This means that ω2 = 190 rad/s and ω1 = 210 rad/s.

Substituting these values in the equation, we have:
400 = (1/2) * I * (1902 - 2102)

Simplifying the equation further:
400 = (1/2) * I * (36100 - 44100)
400 = (1/2) * I * (-8000)
400 = -4000I

Dividing both sides of the equation by -4000, we get:
I = -400/4000
I = -0.1 kg/m2

Since moment of inertia cannot be negative, we take the absolute value of I:
I = 0.1 kg/m2

Therefore, the inertia of the flywheel is 0.1 kg/m2.

Hence, the correct answer is option 'A'.
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The speed of an engine varies from 210 rad/s to 190 rad/s. During the ...
Yes answer should be 'A'
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The speed of an engine varies from 210 rad/s to 190 rad/s. During the cycle the change in kinetic energy is found to be 400 Nm. The inertia of the flywheel in kg/m2 isa)0.10b)0.20c)0.30d)0.40Correct answer is 'A'. Can you explain this answer? for Mechanical Engineering 2025 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about The speed of an engine varies from 210 rad/s to 190 rad/s. During the cycle the change in kinetic energy is found to be 400 Nm. The inertia of the flywheel in kg/m2 isa)0.10b)0.20c)0.30d)0.40Correct answer is 'A'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The speed of an engine varies from 210 rad/s to 190 rad/s. During the cycle the change in kinetic energy is found to be 400 Nm. The inertia of the flywheel in kg/m2 isa)0.10b)0.20c)0.30d)0.40Correct answer is 'A'. Can you explain this answer?.
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