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
View all questions of this test
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'.
The speed of an engine varies from 210 rad/s to 190 rad/s. During the ...
Yes answer should be 'A'