Animated Solution for Physics - Rotational Motion: The following bodies, 1. a ring 2. a disc 3. a solid cylinder 4. a solid sphere of same mass $m$ and radius $R$ are allowed to roll down without slipping simultaneously from the top of the inclined plane. The body which will reach first at the bottom of the inclined plane is ……… . (Mark the body as per their respective numbering given in the question)

The classic "rolling race" down an inclined plane is one of the most beautiful demonstrations of rotational dynamics. It perfectly illustrates how the distribution of mass within an object dictates its motion. Let's dive deep into the physics behind this phenomenon and mathematically prove which shape takes the crown.

Forces on a Rolling Body

Imagine a round body—be it a ring, a disc, a solid cylinder, or a solid sphere—placed at the top of an inclined plane of angle $\theta$. As it begins its descent, three primary forces dictate its fate:

1. Gravity ($mg$): Acting straight down, its component parallel to the incline, $mg\sin \theta$, is the driving force pulling the object down. 2. Normal Force ($N$): Acting perpendicular to the surface, balancing the perpendicular component of gravity, $mg\cos \theta$. 3. Static Friction ($f$): This is the unsung hero of rolling. It acts up the incline. Why? Because without it, the body would simply slide. Friction grabs the bottom edge of the object, providing the necessary torque to make it spin.

Newton's Laws in Action

To determine the winner, we need to find out which body has the highest linear acceleration $a$. A higher acceleration means it will cover the distance $s$ in less time, according to the kinematic equation $t=\sqrt{\frac{2s}{a}}$.

Let's apply Newton's Second Law for translational motion along the incline:

Next, we apply Newton's Second Law for rotational motion about the center of mass. The only force causing torque is friction:

Since the body is rolling without slipping, the linear acceleration $a$ and angular acceleration $\alpha$ are perfectly synchronized by the relation:

Radius of Gyration

To make our equations universal for any round shape, we express the moment of inertia $I$ using the radius of gyration $K$:

Now, let's substitute $I$ and $\alpha =\frac{a}{R}$ into our torque equation to solve for friction $f$:

This tells us exactly how much friction is required to keep the body rolling. We can now plug this back into our translational equation:

Notice something beautiful? The mass $m$ appears in every term and cancels out completely! This means a heavy sphere and a light sphere will roll down at the exact same rate. Rearranging to solve for acceleration $a$:

Our generalized acceleration formula reveals the secret: the acceleration depends entirely on the geometric factor $\frac{K^2}{R^2}$. To maximize acceleration $a$, we need to minimize the denominator, which means we need the smallest possible value for $\frac{K^2}{R^2}$.

Let's look at the contenders: - Ring: All mass is at the edge. $I=m{R}^2⟹\frac{K^2}{R^2}=1$ - Disc / Solid Cylinder: Mass is spread evenly. $I=\frac{1}{2}m{R}^2⟹\frac{K^2}{R^2}=0.5$ - Solid Sphere: Mass is concentrated closest to the center. $I=\frac{2}{5}m{R}^2⟹\frac{K^2}{R^2}=0.4$

The solid sphere has the smallest $\frac{K^2}{R^2}$ value. Because its mass is packed tightly around its center, it requires the least amount of energy to get spinning. Consequently, more of the gravitational potential energy is converted into linear kinetic energy, giving it the highest linear acceleration.