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The flywheel of a machine having the weight of 4500 N and a radius of gyration of 2 m has a cyclic fluctuation of speed from 125 r.p.m. to 120 r.p.m. Assuming g = 10 m/s², the maximum fluctuation of energy is
  • a)
    12822 N-m
  • b)
    24200 N-m
  • c)
    14822 N-m
  • d)
    12100 N-m
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
The flywheel of a machine having the weight of 4500 N and a radius of...
Given data:
Weight of the flywheel (W) = 4500 N
Radius of gyration (r) = 2 m
Cyclic fluctuation of speed (ΔN) = 125 rpm - 120 rpm = 5 rpm
Acceleration due to gravity (g) = 10 m/s²

To find: Maximum fluctuation of energy

The maximum fluctuation of energy can be calculated using the formula:

ΔE = W * ΔV

Where:
ΔE = Maximum fluctuation of energy
W = Weight of the flywheel
ΔV = Change in angular velocity

Let's calculate ΔV first.

- Angular velocity (ω) can be calculated using the formula:

ω = 2πN / 60

Where:
ω = Angular velocity in rad/s
N = Speed in rpm

- ΔV can be calculated by finding the difference between the angular velocities at maximum and minimum speeds:

ΔV = ωmax - ωmin

Let's calculate ωmax and ωmin.

ωmax = 2πNmax / 60
= 2π * 125 / 60
≈ 13.09 rad/s

ωmin = 2πNmin / 60
= 2π * 120 / 60
≈ 12.57 rad/s

ΔV = ωmax - ωmin
≈ 13.09 - 12.57
≈ 0.52 rad/s

Now, let's calculate ΔE using the formula:

ΔE = W * ΔV
= 4500 * 0.52
≈ 2340 N-m

Therefore, the maximum fluctuation of energy is approximately 2340 N-m.

The correct answer is not among the given options.
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The flywheel of a machine having the weight of 4500 N and a radius of gyration of 2 m has a cyclic fluctuation of speed from 125 r.p.m. to 120 r.p.m. Assuming g = 10 m/s², the maximum fluctuation of energy isa)12822 N-mb)24200 N-mc)14822 N-md)12100 N-mCorrect answer is option 'D'. Can you explain this answer?
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