Question

A bucket of water of mass 15.6kg is suspended by a rope wrapped around a windlass, that is a solid cylinder with diameter 0.270m with mass 11.2kg . 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 10.2m to the water. You can ignore the weight of the rope.

A) What is the tension in the rope while the bucket is falling?

B) With what speed does the bucket strike the water?

C) What is the time of fall?

D) While the bucket is falling, what is the force exerted on the cylinder by the axle?

Answer 1

Let the tension in the rope is T ,

let the accleration of bucket is a ,

A)

Now, using the concept of effective force and mass ,

a = m*g/(m + Ipulley/r^2)

a = 15.6 * 9.8 /(15.6 + 0.5 * 11.2 * r^2/r^2)

a = 7.21 m/s^2

for the bucket ,

mg - T = m*a

15.6 * 9.8 - T = 15.6 * 7.21

T = 40.4 N

the tension in the rope is 40.4 N

B)

d = 10.2 m

using third equation of motion ,

v^2 = 2*10.2 *7.21

v = 12.13 m/s

the speed of bucket when hitting ground is 12.13 m/s

C)

let the time of fall is t ,

t = 12.13/7.21

t = 1.68 s

the time of fall is 1.68 s

D)

let the force is N

N = 11.2 * 9.8 + 40.4

N = 150.2 N

the force exerted on the cylinder by the axle is 150.2 N