Answer: a) 150 rev. b) 2105 rev.
Explanation:
a) Assuming a uniformly accelerated motion, we can use the equivalent kinematic equations, replacing linear variables by angular ones.
In order to get the number of revolutions executed, we can use this:
ωf² - ω₀² = 2 γ Δθ (1)
For the first part, we know that ω₀ = 0 (as it starts from rest).
We can find out the value of angular acceleration γ, just applying the definition of angular acceleration, as the change in angular velocity, regarding time, as follows:
γ = (ωf - ω₀) / Δt (2)
As we would want to use SI units, it is advisable to convert the value of ωf, from rpm to rad/sec.
3600 rev/min . (1min/60 sec) . (2π rad/rev) = 120π rad/sec
Replacing in (2), we get γ:
γ = 120 π / 5 rad/sec² = 24 π rad/sec²
Replacing in (1) and solving for Δθ:
Δθ = 120² π² / 2. 24 π = 300 π rad
As 1 rev = 2π rad, Δθ = 150 rev
b) For the second part, we can use exactly the same equations, taking into account that ω₀ = 120 π rad/sec, and that ωf = 0.
The new value for γ is as follows:
γ = -120π / 70 rad/sec² = -1.71 rad/sec²
Replacing in (1) and solving for Δθ, we get:
Δθ = -120² π² / 2. (-1.71) π = 4210 π rad
As 1 rev = 2π rad, Δθ = 2105 rev
