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 3.0 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 1

here,

the diameter of cyclinder , d = 1.2 m

radius , r = d/2 = 0.6 m

the initial velocity of elevator , u = 3 m/s

the initial angular velocity , w0 = u /r

w0 = 3 /0.6 rad/s = 5 rad/s

the final angular velocity , w = 0 rad/s

the angle covered , theta = 1 rev = 2pi rad

let the angular acceleration be alpha

using third equation of motion

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

0^2 - 5^2 = 2 * alpha * 2pi

solving for alpha

alpha = - 2 rad/s^2

using first equation of motion

w = w0 + alpha * t

0 = 5 - 2 * t

t = 2.5 s

the time taken is 2.5 s