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 2.5 m/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

here,

the diameter of cyclinder , d = 1.2 m

radius , r = d/2 = 0.6 m

the initial speed , u = 2.5 m/s

the initial angular speed , w0 = u/r

w0 = 2.5 /0.6 rad/s = 4.17 rad/s

the angle covered , theta = 1 * 2pi = 6.28 rad

the final velocity , w = 0 rad/s

let the angle acceleration be alpha

using third equation of motion

w^2 - w0^2 = 2 * alpha * theta

0 - 4.17^2 = 2 * alpha * 6.28

solving for alpha

alpha = - 1.38 rad/s^2

let the time taken be t

using first equation of motion

w = w0 + alpha * t

0 = 4.17 - 1.38 * t

solving for t

t = 3.02 s

the time taken is 3.02 s