The cable lifting an elevator is wrapped around a 1.0-m-diameter cylinder that is turned by the elevator's motor. The elevator is moving upward at a speed of 1.9 m/s. It then slows to a stop, while the cylinder turns one complete revolution. Part A How long does it take for the elevator to stop

Answer: 3.31s

Explanation:

Given that,

Radius = diameter/2 = 1/2 = 0.5m

U = 1.9m/s

V=0m/s

The elevator stopped as the cylinder made one complete oscillation.

Vertical distance travelled by the elevator is the same as the circumference of the cylinder.

Hence, s = 2πr

S = 2*3.142*0.5

S = 3.142m

V2=u2+2as

0 = 1.9square + 2*ax3.142

0 = 3.61 + 6.284a

-3.61 = 6.284a

a = 3.61/6.284

a = - 0.574m/s2

The time taken for the elevator to stop

V = u+at

0 = 1.9 + (-0.574) t

0 = 1.9-0.574t

-1.9 = - 0.574t

t = 1.9/0.574

t = 3.31 s

In the formulary used above,

V is the final velocity,

U is the initial velocity

A is the acceleration

T is the time

Given:

theta = 1 rev

= 2pi rads

S = r * theta

= 0.5 * 2pi

= 3.142 rad

Vo = 1.9 m/s

D = 1 m

R = 0.5 m

w * r = V

wo = 1.9/0.5

= 3.8 rad/s

wf = 0 rad/s

Using angular equations of motion,

wf^2 = wo^2 + 2aoS

ao = 3.8^2/2 * 6.284

= 1.15 rad/s^3

wf = wo + (ao * t)

0 = 3.8 + 1.15t

t = 3.3 s.