Answer: a) 1.3 rev. b) 1,573 rev.
Explanation:
Let's split the process in 3 phases: start-up, playing, and slow-down.
As the turntable starts from rest, ω₀ = 0.
If we choose θ₀ = 0º, assuming a constant angular acceleration, we can write the following expression for θ (in rads):
θ = 1/2 γ t² (1)
The angular acceleration γ, by definition is the rate of change of the angular velocity, so:
γ = ωf - ω₀ / t . (2)
As ω₀ = 0, we can put the following:
γ = ωf / t
As we have ωf in rpm, it is advisable to convert it to rad/sec, as follows:
33 1/3 rpm (1 min/60 sec) . (2π rad/1 rev) = 111/100π rad/sec
Replacing in (2), we have:
γ = 37/150 π rad/sec²
Replacing in (1):
θ₁ = 999/400 π rad = 1.3 rev
During this phase, the turntable rotates at constant angular speed, so we can apply the definition of angular velocity to get the angle rotated, as follows:
θ₂ = ωf . t = (111/100)π . 2,827 sec = 313797/100 π rad = 1,569 rev.
Assuming a constant deceleration, as ωf = 0, we can find the value of the angular acceleration during this phase, as follows:
γ = -111/100 π / 8 rad/sec² = -0.44 rad/sec²
Now, we can apply the kinematic equation for the angle rotated under constant angular acceleration condition, as follows:
θ₃ = ω₀ t -1/2γ t²
Replacing by the values, we get:
θ₃ = 222/25 π - 111/25 π = 111/25 π rad = 2.2 rev
Adding the the three angles, we have:
θ = θ₁ + θ₂ + θ₃ = 1.3 + 1,569 + 2.2 = 1,573 rev
