Animated Solution for Physics - Rotational Motion: A round uniform body of radius $R$, mass $M$ and moment of inertia $I$, rolls down (without slipping) an inclined plane making an angle $\theta$ with the horizontal. Then, its acceleration is

Imagine a round body—perhaps a solid sphere, a hollow cylinder, or a simple ring—placed at the top of an inclined plane. When released, it doesn't just slide down; it rolls. This beautiful combination of translational and rotational motion is a classic scenario in mechanics. Let's dive deep into the physics to find out exactly how fast it accelerates.

To understand the motion, we first need to identify the forces acting on the body.

1. Gravity ($Mg$): This force acts straight down. We can resolve it into two components: $Mg\cos \theta$ perpendicular to the incline, and $Mg\sin \theta$ parallel to the incline. The parallel component is the driving force that pulls the body down. 2. Normal Force ($N$): This acts perpendicular to the surface, perfectly balancing $Mg\cos \theta$ since the body doesn't jump off the plane. 3. Static Friction ($f$): This is the crucial force. It acts up the incline at the point of contact. Without friction, the body would simply slide. Friction provides the necessary torque to make the body rotate.

We analyze the motion by writing two separate equations: one for linear motion and one for rotational motion.

For the translational motion down the incline, Newton's second law gives us:

Here, $a$ is the linear acceleration of the center of mass.

For the rotational motion, the only force that exerts a torque about the center of mass is friction. The torque $\tau$ is the force times the perpendicular distance (the radius $R$). Using Newton's second law for rotation:

Here, $I$ is the moment of inertia and $\alpha$ is the angular acceleration.

Because the body rolls without slipping, the linear and angular accelerations are strictly locked together by the kinematic constraint:

Substituting this constraint into our torque equation allows us to express the friction force entirely in terms of the body's acceleration and inertia:

Now, we substitute this expression for friction back into our translational equation:

To solve for $a$, we group the acceleration terms on one side:

Finally, isolating $a$ and dividing the numerator and denominator by $M$, we arrive at our elegant final result:

Notice the profound physical meaning here: If the body were just sliding without friction, its acceleration would be simply $g\sin \theta$. The term $\frac{I}{M{R}^2}$ in the denominator acts as a 'rotational inertia penalty'. A body with a larger moment of inertia (like a ring) will have a larger penalty and thus accelerate slower than a body with a smaller moment of inertia (like a solid sphere)!