Animated Solution for Physics - Rotational Motion: A solid disc of radius $a$ and mass $m$ rolls down without slipping on an inclined plane making an angle $\theta$ with the horizontal. The acceleration of the disc will be $\frac{2}{b}g\sin \theta$, where $b$ is ……… . (Round off to the nearest integer) ($g=\text{acceleration due to gravity}$) ($\theta =\text{angle as shown in figure}$)

Imagine a solid disc of mass $m$ and radius $a$ placed at the top of an inclined plane. The plane makes an angle $\theta$ with the horizontal. As the disc is released, it begins to roll down the incline without slipping.

When a rigid body rolls without slipping, it undergoes both translational motion (its center of mass moves down the incline) and rotational motion (it spins about its center of mass). The static friction acting up the incline provides the necessary torque for rotation, while the component of gravity along the incline, $mg\sin \theta$, pulls it downwards.

For any round object rolling down an incline without slipping, the linear acceleration $a_{cm}$ of its center of mass can be derived using Newton's second law for translation and rotation. The standard formula is:

where $I$ is the moment of inertia of the object about its central axis, $m$ is its mass, and $a$ is its radius. This beautiful equation tells us that the acceleration depends not on the absolute mass or radius, but purely on the shape of the object, dictated by the factor $\frac{I}{m{a}^2}$.

We are given that the object is a solid disc. The moment of inertia of a solid disc about its central axis is:

Let's substitute this into our master equation:

Notice how the $m{a}^2$ terms elegantly cancel out! This leaves us with:

Simplifying the denominator, $1+\frac{1}{2}=\frac{3}{2}$. Flipping this to the numerator gives:

The problem states that the acceleration is $\frac{2}{b}g\sin \theta$. By directly comparing our result with the given expression, we can clearly see that: