Question 20.14: A bucket of mass m for drawing water from a well is attached...

i) Consider the energy of the cylinder and the bucket and let the zero level of potential energy be the initial position of the bucket.
The tension T in the rope does no work because it is an internal force.

Suppose the speed of the bucket is v when the angular speed of the cylinder is ω.
While the cylinder rotates through an angle θ, the bucket falls a height x where

x = rθ.

Differentiating    ⇒     v = rω.

After falling a height h from rest, the total energy of the bucket is \frac{1}{2}mv^{2}  –  mgh.

The cylinder also has a kinetic energy due to its rotation of

\frac{1}{2}Iω² = \frac{1}{2}\left(\frac{1}{2}Mr^{2}\right)ω²

= \frac{1}{4}Mr^{2}ω².

So the total energy of the bucket and cylinder when the bucket has fallen a height h is

\frac{1}{4}Mv^{2}  +  \frac{1}{2}mv^{2}  –  mgh.

The initial energy is zero so, by the principle of conservation of energy,

\frac{1}{4}Mv^{2}  +  \frac{1}{2}mv^{2}  –  mgh = 0

⇒    \frac{1}{4}(M  +  2m)v^{2} = mgh

⇒    v^{2} = \sqrt{\frac{4mgh}{2m  +  M}}

⇒    v = \sqrt{\frac{4mgh}{2m  +  M}}.

ii) When there is a resistive couple C, the work done against the couple is

∫ C dθ = Cθ

= \frac{Ch}{r}       (x = h = rθ).

Then the gain in kinetic energy is the difference between the work done by gravity and the work done against the resistance

⇒      \frac{1}{2}mv^{2}  +  \frac{1}{4}Mv^{2} = mgh  –  \frac{Ch}{r}

⇒      \frac{1}{4}(2m  +  M)v^{2} = \frac{(mgrh  –  Ch)}{r}

⇒      v^{2} = \frac{4(mgr  –  C)h}{(2m  +  M)r}

⇒      v = \frac{4(mgr  –  C)h}{(2m  +  M)r}