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A flywheel connected to a punching machine has to supply energy of 400 Nm while running at a mean angular speed of 20 radians/s. If the total fluctuation of speed is not to exceed ±2%, the mass moment of inertia of the flywheel in kg − m2 is
  • a)
    25
  • b)
    50
  • c)
    100
  • d)
    125
Correct answer is option 'A'. Can you explain this answer?
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2 radians/s, what should be the moment of inertia of the flywheel?

We can use the formula for the kinetic energy of a rotating object:

KE = (1/2) Iω^2

where KE is the kinetic energy, I is the moment of inertia, and ω is the angular speed.

We know that the energy to be supplied by the flywheel is 400 Nm, and the mean angular speed is 20 radians/s. Therefore, the kinetic energy of the flywheel is:

KE = 400 Nm

We also know that the total fluctuation of speed should not exceed 2 radians/s. This means that the minimum and maximum angular speeds are:

ω_min = 20 - 1 = 19 radians/s
ω_max = 20 + 1 = 21 radians/s

Using the formula for the kinetic energy, we can solve for the moment of inertia:

KE = (1/2) Iω^2

400 Nm = (1/2) I (19 radians/s)^2
400 Nm = (1/2) I (21 radians/s)^2

Solving for I, we get:

I = 105.26 kg m^2

Therefore, the moment of inertia of the flywheel should be 105.26 kg m^2 to ensure that the total fluctuation of speed does not exceed 2 radians/s.
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A flywheel connected to a punching machine has to supply energy of 400 Nm while running at a mean angular speed of 20 radians/s. If the total fluctuation ofspeed is not to exceed ±2%, the mass moment of inertia of the flywheel inkg − m2 isa)25b)50c)100d)125Correct answer is option 'A'. Can you explain this answer?
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