Consider an elevator which works by winding a cable around a 1.2m diameter cylin

Question

Consider an elevator which works by winding a cable around a 1.2m diameter cylinder which is turned by a motor at a rate of 100 rpm. If the motor reaches this rate from rest in just 3 seconds what acceleration does this imply to the elevator car attached to the bottom of the cable ? If the combined weight of the car and passengers is 13600 kg, what tension does the cable exert on the car ? If in a test of the elevator using only dummies for the passengers to achieve full weight, the cable is cut and it takes about ten seconds for the elevator to hit the bottom of the shaft, how fast is the elevator falling ? If the previous collision with the ground takes place in just 0.3 seconds and the force is assumed constant during the collision, what is the force that acts on the passengers ?

Explanation / Answer

a)

it reached 100 rpm in 3 seconds

100 rpm = 100*2pi /60 = 10.472 rad/s

wf = wi + at

10.472 = 0 + angular acceleartion * 3

angular acceleration= 3.491 rad/s^2

so acceleration a = r * angular accelertion = 3.491*0.6 = 2.0944 m/s^2

so acceleration of elevator = 2.0944 m/s^2

b)

T - mg = ma

T = mg + ma

T = 13600*9.8 + 13600*2.0944

T = 161763.84 N

c) it took 10 seconds, and it is in free fall..

so V = u + gt = 0+9.8*10 = 98 m/s

velocity when it hits ground is 98 m/s

d)

force * time of impact = change in momentum

force * 0.3 = 13600 * (98 -0)

force = 442666.67 N

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