Animated Solution for Physics - Rotational Motion: A body rolls down an inclined plane without slipping. The kinetic energy of rotation is $50\%$ of its translational kinetic energy. The body is

Imagine you are standing at the top of an inclined plane, holding a mystery object in your hand. When you release it, it doesn't just slide down like a block of ice. It rolls.

This means the object is multitasking. It is moving forward (translating) and spinning around its own axis (rotating) simultaneously. Because it is doing two things at once, its total kinetic energy is split into two distinct parts: Translational Kinetic Energy and Rotational Kinetic Energy.

The problem gives us a beautiful, almost poetic clue to identify this mystery object. It tells us that the energy spent on spinning is exactly half of the energy spent on moving forward.

Let's translate this English sentence into the rigorous language of physics:

We already know the standard formulas for these energies. Rotational kinetic energy depends on the moment of inertia ($I$) and angular velocity ($\omega$), while translational kinetic energy depends on mass ($m$) and linear velocity ($v$).

Substituting these into our clue, we get our master equation:

Now, we hit a slight roadblock. We have $v$ on one side and $\omega$ on the other. How do we connect them?

But wait! The problem stated that the body rolls without slipping. This is the golden rule of pure rolling: the linear velocity $v$ of the center of mass is perfectly synchronized with the angular velocity $\omega$. They are locked together by the radius $R$:

Let's plug this golden rule into our energy equation to unify the variables:

Look at that! The ${\omega }^2$ term is present on both sides. It beautifully cancels out, leaving us with a pure, intrinsic property of the body that doesn't depend on how fast it's moving:

Now, we must search our memory banks. Which standard geometric body has a moment of inertia of $\frac{1}{2}m{R}^2$ about its central axis?

- A solid sphere? No, that's $\frac{2}{5}m{R}^2$. - A ring? No, that's $m{R}^2$. - A solid cylinder (or a uniform disc)? Yes! That is exactly $\frac{1}{2}m{R}^2$.

Therefore, our mystery body is definitively a solid cylinder.

What If?

As a physics student, you should always push the boundaries of a problem. What if the rotational energy was 100% of the translational energy?

In that scenario, the equation would yield $I=m{R}^2$. This means the body would be a ring or a hollow cylinder, where all the mass is concentrated at the maximum distance $R$ from the center, requiring much more energy to spin!