When the power is turned on, the unit attains the rated speed in 5 seconds & when the power is turned off, the unit comes to rest in 90 seconds. Assuming uniformly accelerated motion, determine the number of revolutions the unit turns: i) to attain the rated speed & ii) to come to rest.

i) acceleration

$N2 = 1740 rpm$ (final speed)

$T = 5 s$ (time of acceleration)

$N1 = 0$ (initial speed)

ii) retardation

$T = 90 s$ (time of retardation)

$$N1 = 1740 rpm$$

$$N2 = 0$$

Solution:

i)

$w2 = 2πN_2/60 = 182.2 rad/s \\ w1 = 0$

Using first kinematical equation

$w2 = w1 + αt \\ 182.2 = 0 + α \times 5 \\ . α = 36.44 rad/sec^2$

Using second equation, $θ = w1t + 0.5 \times α \times t^2 \\ Θ = 0 + 0.5 \times 36.44 \times 52 = 455.5 rad $

No. of turns $= θ/2π = 72.5 $

ii) Retardation

$w1 = 2πN1/60 = 182.2 rad/s \\ w2 = 0$

Using first kinematical equation

$w2 = w1 + αt \\ 0 = 182.2 + α \times 90 \\ . α = -2 rad/sec^2 $

Using second equation, $θ = w1t + 0.5 \times α \times t^2 \\ Θ = 182.2 \times 90 - 0.5 \times 2 \times 90^2 = 8298 rad $

No. of turns $= θ/2π = 1320.7$