Question

A(n) 0.95-kg bucket is tied to a rope of negligible mass that is wrapped around a pole mounted horizontally on frictionless bearings. The cylindrical pole has a diameter of 0.340 m and a mass of 2.60 kg. When the bucket is released from rest, how long will it take to fall to the bottom of the 20.0 m well?

Answer 1

The difficulty is to find the acceleration with which the bucket will fall. It will not fall with "g" because of the tension in the rope.

On pole torque ,τ = Tr = Iα where I is moment of inertia of pole =Mr^2/2 and r is the radius of the pole

α = Tr/I ..........................................(1)

Applying newtons law on bucket,

ma = mg-T a=g-T/m .................................................(2)
 
If the rope does not slip

a=αr

g-T/m = Tr^2/I

Putting I = Mr^2/2 ,we get

g-T/m = 2T/M T = g/(1/m+2/M)
 
Putting this value of T in eqn (2), we will get the value of "a" the downward accelration of bucket.

(Please find a by putting the values of m and M yourself).

Now that we know the value of "a", find time taken to travel 20m is very easy.
 
Apply the equation of motion s=ut+1/2at^2
 
u=0

s=20m

I hope my answer has come to your help. Have a nice day ahead and may God bless you always!