Question

A disc of mass m and radius r is free to rotate about its centre. a string is wrapped over its rim and a block of mass $m$ is attached to the free end of the string. The system is released from rest. what will be the speed of the block as it descends through a height $h$?

Answer 1

Hint: As according to the law of conservation of energy describes that the entire energy of an isolated system remains constant or it is said to be conserved over time. So, apply the law of conservation of energy and then calculate the velocity of the block.

Formula used:
 Loss in the kinetic energy of block = gain in kinetic energy of block + pulley
$mgh=\dfrac{1}{2}m{{v}^{2}}+\dfrac{1}{2}I{{\omega }^{2}}$
$m$= mass of the block
$h$= height
$I$= moment of inertia
$\omega $= angular velocity

Complete step-by-step answer:
As given in question string is wound over the pulley so angular speed and linear speed is related to each other
$v=R\omega $
now plug in all values
$\begin{align}
  & mgh=\dfrac{1}{2}m{{v}^{2}}+\dfrac{1}{2}I{{\omega }^{2}} \\
 & \Rightarrow \dfrac{1}{2}m{{v}^{2}}+\dfrac{1}{2}I{{\left( \dfrac{v}{r} \right)}^{2}} \\
\end{align}$
here moment of inertia is given as
$I=\dfrac{1}{2}m{{R}^{2}}$
 

Now we will have
$mgh=\dfrac{1}{2}m{{v}^{2}}+\dfrac{1}{2}\left( \dfrac{1}{2}m{{R}^{2}} \right){{\left( \dfrac{v}{R} \right)}^{2}}$
$mgh=\dfrac{3}{4}m{{v}^{2}}$
$v=\sqrt{\dfrac{4gh}{3}}$
So, the velocity of the block as it descends $v=\sqrt{\dfrac{4gh}{3}}$.

Additional Information: A consequence of the law of conservation of energy is that a motion machine of the primary kind cannot exist, that's to mention, no system without an external energy supply can deliver a vast amount of energy to its surroundings. For systems which don't have time translation symmetry, it's going to not be possible to define conservation of energy.

Note: In problems involving the utilization of conservation of energy, the trail taken by the thing is often ignored. The sole important quantities are the object's velocity (which gives its kinetic energy) and height above the point of reference (which gives its gravitational potential energy).