Animated Solution for Physics - Rotational Motion: As shown in the figure, a bob of mass $m$ is tied by a massless string whose other end portion is wound on a flywheel (disc) of radius $r$ and mass $m$. When released from rest the bob starts falling vertically. When it has covered a distance of $h$, the angular speed of the wheel will be

This problem is a classic and beautiful demonstration of the Conservation of Mechanical Energy applied to a coupled translational and rotational system. Let's embark on a journey to understand exactly how the falling bob transfers its energy to the spinning disc.

Imagine the physical reality of the setup. We have a solid disc (flywheel) of mass $m$ and radius $r$ mounted on a frictionless horizontal axis. A massless string is wound around its rim, and a bob of the exact same mass $m$ hangs from the free end of the string.

When the system is released from rest, gravity pulls the bob downwards. As the bob falls through a vertical distance $h$, it loses gravitational potential energy. But where does this energy go? Because there are no non-conservative forces (like friction or air resistance) doing work on the system, the total mechanical energy must remain strictly conserved.

The lost potential energy is entirely converted into two forms of kinetic energy: 1. The translational kinetic energy of the falling bob. 2. The rotational kinetic energy of the spinning disc.

We can express this energy transformation mathematically. The decrease in potential energy of the bob is $\Delta PE=mgh$. This must equal the sum of the kinetic energies gained by the system:

Here, $v$ is the linear velocity of the bob, $I$ is the moment of inertia of the disc, and $\omega$ is the angular velocity of the disc.

To solve this, we need to link the linear world of the bob to the rotational world of the disc. Because the string is tightly wound and does not slip over the disc, the linear speed of the string (and thus the bob) must exactly match the tangential speed of the disc's rim. This gives us our crucial no-slip condition:

Furthermore, we know the standard formula for the moment of inertia of a uniform solid disc rotating about its central axis:

Now, we substitute our constraints back into the master energy equation. We replace $v$ with $r\omega$ and $I$ with $\frac{1}{2}m{r}^2$:

Notice how elegant this becomes! Everything is now expressed in terms of our target variable, $\omega$. Let's expand the terms:

We can now combine the kinetic energy terms. Adding one-half and one-fourth gives us three-fourths:

Interestingly, the mass $m$ appears on both sides of the equation. This means the specific mass of the bob and disc (as long as they are equal) doesn't actually affect the final kinematic result! Let's cancel $m$ and isolate ${\omega }^2$:

Taking the square root of both sides yields our final, pristine answer:

What if the flywheel was a hollow ring instead of a solid disc? A ring has its mass distributed further from the axis of rotation, giving it a larger moment of inertia ($I=m{r}^2$). If you run through the exact same derivation, you'll find that the ring absorbs a larger fraction of the energy, resulting in a slower final angular speed. Physics is beautifully consistent!