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A flywheel of mass 5000 kg and 2 m radius of gyration is rotating at a speed of 100 rpm. If the fluctuation of speed is limited to ±1% of mean speed. Find the maximum fluctuation of energy.
- a)21929 N-m
- b)10471 N-m
- c)43856 N-m
- d)10000 N-m
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A flywheel of mass 5000 kg and 2 m radius of gyration is rotating at a...
m = 5000 kg,(radius of gyration) = 2m
Maximum fluctuation of energy
∆E = 20000 × 10.4712 × 0.02 = 4385.67N - m × 10
E = 43856.73 N - m
Most Upvoted Answer
A flywheel of mass 5000 kg and 2 m radius of gyration is rotating at a...
Let's assume that the fluctuation of speed is limited to ΔN rpm.
The formula for the kinetic energy of a rotating object is:
KE = (1/2) * I * ω²
Where KE is the kinetic energy, I is the moment of inertia, and ω is the angular velocity.
The moment of inertia of a flywheel is given by the formula:
I = m * r²
Where m is the mass of the flywheel and r is the radius of gyration.
In this case, m = 5000 kg and r = 2 m.
So, the moment of inertia of the flywheel is:
I = 5000 kg * (2 m)² = 5000 kg * 4 m² = 20,000 kg * m²
The angular velocity of the flywheel is given by the formula:
ω = 2π * N / 60
Where N is the speed of rotation in rpm.
In this case, N = 100 rpm.
So, the angular velocity of the flywheel is:
ω = 2π * 100 / 60 = 20π / 3 rad/s
Now, let's calculate the kinetic energy of the flywheel at the given speed:
KE = (1/2) * I * ω²
= (1/2) * 20,000 kg * m² * (20π / 3 rad/s)²
= (1/2) * 20,000 kg * m² * (400π² / 9) m²/s²
= 400,000π² / 9 * 20,000 kg * m² m²/s²
= 800,000π² / 9 kg * m²/s²
Now, let's calculate the maximum fluctuation in kinetic energy due to the fluctuation in speed:
ΔKE = (1/2) * I * (Δω)²
Where ΔKE is the maximum fluctuation in kinetic energy and Δω is the maximum fluctuation in angular velocity.
Since the fluctuation in speed is limited to ΔN rpm, the maximum fluctuation in angular velocity is:
Δω = 2π * ΔN / 60
Substituting this into the formula for ΔKE, we get:
ΔKE = (1/2) * I * (2π * ΔN / 60)²
= (1/2) * 20,000 kg * m² * (2π * ΔN / 60)²
= 400,000π² / 9 * (2π * ΔN / 60)² kg * m²/s²
= 800,000π² / 9 * (π * ΔN / 30)² kg * m²/s²
= 800,000π⁴ / 9 * ΔN² / 900 kg * m²/s²
= 800,000π⁴ / 9 * ΔN² / 900 kg * m²/s²
So, the maximum fluctuation in kinetic energy is:
ΔKE = 800,000π⁴ / 9 * ΔN² / 900 kg * m²/s²
The formula for the kinetic energy of a rotating object is:
KE = (1/2) * I * ω²
Where KE is the kinetic energy, I is the moment of inertia, and ω is the angular velocity.
The moment of inertia of a flywheel is given by the formula:
I = m * r²
Where m is the mass of the flywheel and r is the radius of gyration.
In this case, m = 5000 kg and r = 2 m.
So, the moment of inertia of the flywheel is:
I = 5000 kg * (2 m)² = 5000 kg * 4 m² = 20,000 kg * m²
The angular velocity of the flywheel is given by the formula:
ω = 2π * N / 60
Where N is the speed of rotation in rpm.
In this case, N = 100 rpm.
So, the angular velocity of the flywheel is:
ω = 2π * 100 / 60 = 20π / 3 rad/s
Now, let's calculate the kinetic energy of the flywheel at the given speed:
KE = (1/2) * I * ω²
= (1/2) * 20,000 kg * m² * (20π / 3 rad/s)²
= (1/2) * 20,000 kg * m² * (400π² / 9) m²/s²
= 400,000π² / 9 * 20,000 kg * m² m²/s²
= 800,000π² / 9 kg * m²/s²
Now, let's calculate the maximum fluctuation in kinetic energy due to the fluctuation in speed:
ΔKE = (1/2) * I * (Δω)²
Where ΔKE is the maximum fluctuation in kinetic energy and Δω is the maximum fluctuation in angular velocity.
Since the fluctuation in speed is limited to ΔN rpm, the maximum fluctuation in angular velocity is:
Δω = 2π * ΔN / 60
Substituting this into the formula for ΔKE, we get:
ΔKE = (1/2) * I * (2π * ΔN / 60)²
= (1/2) * 20,000 kg * m² * (2π * ΔN / 60)²
= 400,000π² / 9 * (2π * ΔN / 60)² kg * m²/s²
= 800,000π² / 9 * (π * ΔN / 30)² kg * m²/s²
= 800,000π⁴ / 9 * ΔN² / 900 kg * m²/s²
= 800,000π⁴ / 9 * ΔN² / 900 kg * m²/s²
So, the maximum fluctuation in kinetic energy is:
ΔKE = 800,000π⁴ / 9 * ΔN² / 900 kg * m²/s²
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A flywheel of mass 5000 kg and 2 m radius of gyration is rotating at a speed of 100 rpm. If the fluctuation of speed is limited to ±1% of mean speed. Find the maximum fluctuation of energy.a)21929 N-mb)10471 N-mc)43856 N-md)10000 N-mCorrect answer is option 'C'. Can you explain this answer?
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