Question

A bucket of water of mass 15.1 kg is suspended by a rope wrapped around a windlass, that is a solid cylinder with diameter 0.330 m with mass 13.0 kg . The cylinder pivots on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls a distance 11.0 m to the water. You can ignore the weight of the rope.

a-What is the tension in the rope while the bucket is falling?

Take the free fall acceleration to be g = 9.80 m/s2 .

b-With what speed does the bucket strike the water?

Take the free fall acceleration to be g = 9.80 m/s2 .

c-What is the time of fall?

Take the free fall acceleration to be g = 9.80 m/s2 .

d-While the bucket is falling, what is the force exerted on the cylinder by the axle?

Take the free fall acceleration to be g = 9.80 m/s2 .

Answer 1

MI of the cylinder   I = MR²/2    M= 13.0 kg and R= 0.33 m

let a be the acceleration of the bucket then motion of the bucket is governed by

mg-T = ma   m = 15.1 kg

T = 15.1(g-a) -------------(1)

if is angualr acceleration of the cylinder then

TR is the torque on the cylinder

T*R = I        = a/R

   = MRa/2

T = Ma/2 = 13.0*a/2 = 6.5a   -----------(2)

from 1 and   2

15.1(9.8 -a) = 6.5a

a = 15.1*9.8/21.6

     = 6.85 m/s²

Tension in the rope

a) T = 6.5*6.85 = 44.53 N

b) distance of fall s = 11m

v² = 2as , a = 6.85

speed v = sqrt(2*6.85*11)

               = 12.28 m/s

c) time of fall t is given by

t = sqrt(2s/a) = sqrt(2*11/6.85)

= 1.79 s

d) force acting on the cylinder

force on the cylinder are

1) Tension of the rope - downward

2) weight of teh cylinder - downward

3) normal reaction of the axle N

      T = 44.53 N

normal reaction of the axle N = 13.0*9.8 +44.53

                                               =171.93 N