Answer :
The number of revolutions through which the wheel turns during this interval is 50 + 3π / r.
To solve this problem, we can use the formula for uniform acceleration:
ω₂ - ω₁ = αt
where ω₁ and ω₂ are the initial and final angular velocities, α is the angular acceleration, and t is the time interval.
We know that ω₁ = 300 rev/min, ω₂ = 900 rev/min, and t = 6 seconds. We need to find α to use the above formula.
We can use the formula for angular acceleration:
α = τ / I
where τ is the torque applied and I is the moment of inertia of the flywheel.
We don't have the moment of inertia, but we can use the fact that the acceleration is uniform to relate α to the linear acceleration a and the radius r of the flywheel:
a = αr
We can also use the formula for linear acceleration:
a = F / m
where F is the force applied and m is the mass of the flywheel.
We don't have the mass, but we can use the fact that the torque is related to the force by:
τ = Fr
where r is the radius of the flywheel.
Putting all of these equations together, we can solve for α:
α = τ / I = Fr / Ir = (a / r) / (ω₂ - ω₁) = (2π / 60) (900 - 300) / (6 / 60) / r
= 2π (600 / 360) / r
= π / (3r)
Now we can use the formula for angular displacement:
θ = ω₁t + (1/2)αt²
to find the number of revolutions N:
N = θ / (2π) = [(300)(6) + (1/2)(π/3r)(6)²] / (2π)
= 50 + 3π / r
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