Animated Solution for Physics - Rotational Motion: Two identical uniform discs roll without slipping on two different surfaces $AB$ and $CD$ (see figure) starting at $A$ and $C$ with linear speeds $v_1$ and $v_2$, respectively, and always remain in contact with the surfaces. If they reach $B$ and $D$ with the same linear speed and $v_1=3\text{ m/s}$, then $v_2$ in m/s is (Take $g=10{\text{ m/s}}^2$)

The journey of a rolling object is one of the most elegant phenomena in classical mechanics. When an object rolls without slipping, it perfectly synchronizes its translational motion (moving forward) with its rotational motion (spinning). In this problem, we are tasked with comparing the journeys of two identical uniform discs rolling down two different tracks.

Let's dive into the physics and unravel the mystery of their final speeds!

Imagine two identical uniform discs, each poised at the top of a different track.

The first disc starts on surface $AB$ at a height of $h_1=30\text{ m}$ with an initial push, giving it a speed of $v_1=3\text{ m/s}$. The second disc starts on surface $CD$ at a slightly lower height of $h_2=27\text{ m}$ with an unknown initial speed $v_2$.

The crucial piece of information given to us is that both discs reach the bottom of their respective tracks with the exact same final linear speed. This is our golden key to unlocking the value of $v_2$.

When an object rolls without slipping, the point of contact with the surface is instantaneously at rest. Because there is no relative sliding at this contact point, the static friction does absolutely zero work.

This is a profound realization! It means that no mechanical energy is dissipated as heat. The total mechanical energy of the disc—the sum of its gravitational potential energy and its total kinetic energy—remains perfectly conserved throughout its descent.

To apply energy conservation, we first need to understand the kinetic energy of a rolling disc. A rolling object possesses two forms of kinetic energy: 1. Translational Kinetic Energy ($K_T=\frac{1}{2}m{v}^2$) due to the center of mass moving forward. 2. Rotational Kinetic Energy ($K_R=\frac{1}{2}I{\omega }^2$) due to the object spinning about its center.

For a uniform solid disc, the moment of inertia is $I=\frac{1}{2}m{R}^2$. In pure rolling, the angular velocity is locked to the linear velocity by the relation $\omega =\frac{v}{R}$.

Substituting these into our kinetic energy equation: $K=\frac{1}{2}m{v}^2+\frac{1}{2}\left(\frac{1}{2}m{R}^2\right){\left(\frac{v}{R}\right)}^2$

Notice how beautifully the $R^2$ terms cancel out! We are left with: $K=\frac{1}{2}m{v}^2+\frac{1}{4}m{v}^2=\frac{3}{4}m{v}^2$

Now, applying the conservation of mechanical energy from the top of the track to the bottom, the loss in potential energy equals the gain in kinetic energy: $mgh=K_f-K_i$

The mass $m$ cancels out from every term, showing that the motion is independent of the disc's mass. Rearranging for the final velocity squared, we get our master equation: $v_f^2=v_i^2+\frac{4}{3}gh$

The problem states that both discs reach the bottom with the same final speed. Therefore, we can equate the final velocity squared for both tracks: $v_{f1}^2=v_{f2}^2$

Substituting our master equation for both discs: $v_1^2+\frac{4}{3}g{h}_1=v_2^2+\frac{4}{3}g{h}_2$

Now, it's just a matter of plugging in the given values. We know $v_1=3\text{ m/s}$, $h_1=30\text{ m}$, $h_2=27\text{ m}$, and $g=10{\text{ m/s}}^2$.

Let's simplify the terms carefully: $9+\frac{4}{3}(300)=v_2^2+\frac{4}{3}(270)$

Subtracting 360 from both sides: $v_2^2=409-360=49$

Taking the square root gives us a perfect integer: $v_2=7\text{ m/s}$

This problem beautifully illustrates the power of energy conservation in rotational dynamics. But what if the objects were solid spheres instead of discs?

The moment of inertia would change to $I=\frac{2}{5}m{R}^2$, making the total kinetic energy $\frac{7}{10}m{v}^2$. This would alter the multiplier for the $gh$ term, leading to a completely different final speed. Always pay close attention to the geometric shape of the rolling object, as it dictates how energy is distributed between translation and rotation!