Animated Solution for Physics - Rotational Motion: A small object of uniform density rolls up a curved surface with an initial velocity $v$. It reaches up to a maximum height of $\frac{3v^2}{4g}$ with respect to the initial position. The object is

Imagine standing at the base of a curved hill, holding a small, mysterious object. You give it a push, and it starts rolling up the incline with an initial velocity $v$. As it climbs, it fights against gravity, slowing down until it reaches a momentary standstill at a specific height $h=\frac{3v^2}{4g}$. Our mission? To uncover the true identity of this object just by observing how high it climbs!

Before we dive into the equations, let's visualize the physical reality of the situation. At the very bottom of the track, the object is in a state of pure rolling. This means it's not just sliding forward like a block of ice; it's spinning around its own center of mass. It possesses a linear velocity $v$ and an angular velocity $\omega$.

As the object ascends the curved surface, gravity pulls it downward, doing negative work. The object's speed decreases, and its rotation slows down. Eventually, it reaches its maximum height, where it momentarily stops. At this exact peak, both its linear velocity and angular velocity are zero. All the kinetic energy it had at the bottom has been stored as gravitational potential energy.

You might be wondering, 'What about friction? Doesn't it steal energy?' Here is the beautiful catch about pure rolling: the point of the object in contact with the surface is instantaneously at rest. Because there is no relative sliding at the contact point, the static friction does absolutely zero work!

This is a massive conceptual breakthrough. It means the total mechanical energy of the system is perfectly conserved. We can confidently write our master equation:

Let's set our reference level for potential energy at the bottom of the track, meaning $U_i=0$. At the maximum height, the kinetic energy $K_f=0$. Our equation elegantly simplifies to:

Now, we need to be careful. The initial kinetic energy isn't just $\frac{1}{2}m{v}^2$. Because the object is rolling, it has two distinct energy bank accounts: translational kinetic energy and rotational kinetic energy.

The translational kinetic energy depends on the motion of the center of mass: $\frac{1}{2}m{v}^2$. The rotational kinetic energy depends on the spinning motion: $\frac{1}{2}I{\omega }^2$, where $I$ is the moment of inertia.

So, the total initial kinetic energy is:

At the maximum height, the gravitational potential energy is simply $mgh$. Equating the two, we get:

We are given the maximum height $h=\frac{3v^2}{4g}$. We also know the golden rule of pure rolling: the angular velocity $\omega$ is tied to the linear velocity $v$ by the relation $\omega =\frac{v}{R}$, where $R$ is the radius of the object.

Let's substitute these known values into our energy equation:

Now, let's simplify the right side. The acceleration due to gravity, $g$, cancels out perfectly:

Take a close look at this equation. Every single term contains $v^2$. This means the maximum height doesn't depend on how fast we push it, the relationship holds universally! We can divide the entire equation by $v^2$:

To isolate the moment of inertia, let's move the $\frac{1}{2}m$ to the right side:

Finally, multiply both sides by $2R^2$:

We have found our answer! The moment of inertia of the object is $\frac{1}{2}m{R}^2$. If we look up our standard moments of inertia, we immediately recognize this signature. A uniform solid sphere has $I=\frac{2}{5}m{R}^2$, a ring has $I=m{R}^2$, but a uniform solid disc has exactly $I=\frac{1}{2}m{R}^2$.

The mysterious object rolling up the hill is a disc!

Think about the profound nature of this result. Just by measuring how high an object rolls, we can determine its internal mass distribution without ever breaking it open. If it were a ring, it would have stored more energy in its rotation and climbed even higher. Physics allows us to see the invisible!

A Thought Experiment

Let's take this concept a step further. Imagine we have three objects: a solid sphere, a solid disc, and a thin ring. They all have the exact same mass $m$ and the exact same radius $R$. We line them up at the bottom of our curved track and launch them all simultaneously with the exact same initial velocity $v$.

Which one wins the race to the top? Which one climbs the highest?

To answer this, we have to look at their moments of inertia. The moment of inertia is a measure of how much an object resists rotational acceleration. It depends on how the mass is distributed relative to the center. - The solid sphere has its mass packed closely to the center, giving it the lowest moment of inertia: $I=0.4m{R}^2$. - The solid disc has its mass spread out evenly, giving it a medium moment of inertia: $I=0.5m{R}^2$. - The thin ring has all its mass concentrated at the very edge, giving it the highest moment of inertia: $I=1.0m{R}^2$.

Because the ring has the highest moment of inertia, it requires the most energy to get it spinning at the angular velocity $\omega =v/R$. Therefore, for the same linear velocity $v$, the ring stores the maximum amount of rotational kinetic energy.

When they roll up the hill, this massive reservoir of rotational energy is converted into gravitational potential energy. The object with the most total initial energy will climb the highest.

- The sphere will reach a height of $h=\frac{7v^2}{10g}$. - The disc, as we calculated, reaches $h=\frac{3v^2}{4g}$ (or $\frac{7.5v^2}{10g}$). - The ring will reach a towering height of $h=\frac{v^2}{g}$ (or $\frac{10v^2}{10g}$).

The ring is the undisputed champion of the hill climb! This beautiful thought experiment showcases how the abstract concept of moment of inertia dictates the physical reality of motion. It's not just about how heavy an object is, but where that heaviness is located.