Question

The cable lifting an elevator is wrapped around a 1.2-m-diameter cylinder that is turned by the elevator's motor. The elevator is moving upward at a speed of 2.7m/s. It then slows to a stop, while the cylinder turns one complete revolution.
How long does it take for the elevator to stop?

Answer 1

Answer: 2.8s

Explanation:

1) Since the cable is tense all the time, and it is neither sliping nor stretching, the cable will run a linear distance equal to the distance run by the elevator.

2) Since the cylinder turns one complete revolution, the distance run by the cable equals the circumference of cylinder.

3) Hence, you need to calculate the circumference of the cylinder, which is calcualted with the formula C = 2π × radius.

The radius is half the diameter: 1.2 m / 2 = 0.6 m

Hence, C = 2π × 0.6m = 3.8m

⇒ Distance run by the elevator = 3.8m

4) Now, assuming that the elevator went from the initial speed of 2.7 m/s to the complete stop (Vf = 0), with uniform acceleration (deceleration), you use the corresponding formulas:

i) Vf² = Vo² - 2ad ⇒ 2ad = Vo² - Vf² ⇒ a = [Vo² - Vf²] / (2d)

a = [2.7 m/s]² / (2×3.8m) = 0.96 m/s²

ii) Vf = Vo - at ⇒ t = [Vo - Vf] / t = [2.7 m/s] / (0.96 m/s²) = 2.8 s

Answer 2

C = PI*d = 1.2*PI meters 
2C = Vin*t 
t = 2C/Vin = 2.4*PI/2.7= 0.88*PI in sec.(2.79 approx)