The cable lifting an elevator is wrapped around a 1.0-mdiameter cylinder that is turned by the elevator’s motor. The elevator is moving upward at a speed of 1.6 m/s. It then slows to a stop as the cylinder makes one complete turn at constant angular acceleration. How long does it take for the elevator to stop?

Solution To solve this problem we have require knowledge about one dimensional kinematics.the elevator will advance an amount equal to circumference of the cylinder allow y = 2r For one dimensional kinematics v v = 2a02 v = v0+at Combine first two equation solve for a ,and insert the result into third equation that is t = 4R/v 0 t = 4(0.50 m)/(1.6 m/s) t = 3.9 s