At t = 0 s a flywheel is rotating at 30 rpm. A motor gives it a constant acceleration of 0.9 rad/s^2 until it reaches 70 rpm. The motor is then disconnected. How many revolutions are completed at t = 20 s?
Vo=30rev/min * 1min/60s * 6.28rad/1rev = 3.14rad/s.
d = Vo*t + 0.5a*t^2,
d = 3.14*20 + 0.45*(20)^2 = 242.8rad.
Rev = 242.8rad / 6.28rad/rev = 38.7.
To find the number of revolutions completed at t = 20 s, we need to first calculate the final angular velocity at t = 20 s.
Given:
Initial angular velocity, ω1 = 30 rpm
Final angular velocity, ω2 = 70 rpm
Acceleration, α = 0.9 rad/s²
Time, t = 20 s
Converting the angular velocities from rpm to rad/s:
ω1 = 30 rpm × (2π rad/1 min) × (1 min/60 s) = π rad/s
ω2 = 70 rpm × (2π rad/1 min) × (1 min/60 s) = (7π/3) rad/s
Using the equation of motion for rotational motion:
ω2 = ω1 + αt
Rearranging the equation to solve for the final angular velocity, ω2:
ω2 = ω1 + αt
(7π/3) = π + 0.9 × 20
(7π/3) = π + 18
Subtracting π from both sides and simplifying:
(7π/3) - π = 18
(4π/3) = 18
Simplifying further:
4π = 54
π = 13.5
Therefore, the final angular velocity at t = 20 s is 13.5 rad/s.
Now, to find the number of revolutions completed at t = 20 s, we need to calculate the change in angle during this time period.
Using the formula:
θ = ω1t + (1/2)αt²
Substituting the values:
θ = π × 20 + (1/2) × 0.9 × 20²
θ = 20π + 0.9 × 10 × 20
θ = 20π + 180
Converting the angle from radians to revolutions:
θ = (20π + 180) × (1 rev/2π rad)
θ = 10 + 180/π
Therefore, the number of revolutions completed at t = 20 s is approximately 10 + (180/π).