Animated Solution for Physics - Rotational Motion: A homogeneous solid cylindrical roller of radius $R$ and mass $m$ is pulled on a cricket pitch by a horizontal force. Assuming rolling without slipping, angular acceleration of the cylinder is

Imagine you are standing on a cricket pitch, and right in front of you is a massive, solid cylindrical roller. You grab the handle attached to its center and pull it forward with a horizontal force $F$. What happens next is a beautiful interplay of forces, torques, and the magic of pure rolling.

When you pull the roller, it doesn't just slide across the pitch like a block of ice. It rolls. But why does it roll? The secret lies at the very bottom of the cylinder, where it makes contact with the ground. As the force $F$ tries to drag the cylinder forward, the ground grips the bottom of the cylinder, exerting a backward force. This is static friction, denoted by $f$.

To solve this problem, we need to break the motion into two distinct parts: the translational motion (how the center of mass moves forward) and the rotational motion (how the cylinder spins around its center).

Let's start with the translational motion. According to Newton's Second Law, the net force acting on an object equals its mass times its acceleration ($F_{net}=ma$).

Looking at our cylinder, we have the applied force $F$ pulling it forward and the friction $f$ pulling it backward. Therefore, our first master equation is:

This equation tells us how the center of the cylinder accelerates linearly. But we have two unknowns here: the friction $f$ and the acceleration $a$. We need another equation to solve this puzzle.

This is where rotational dynamics comes into play. For an object to rotate, there must be a net torque acting on it. Torque is the rotational equivalent of force, calculated as the force multiplied by the perpendicular distance from the axis of rotation ($\tau =r\times F$).

Let's evaluate the torques about the center of mass $O$ of the cylinder. The applied force $F$ passes exactly through the center $O$. Because its perpendicular distance from the center is zero, it creates zero torque. It can only pull the cylinder forward; it cannot make it spin.

The friction $f$, however, acts at the bottom edge of the cylinder, at a distance $R$ (the radius) from the center. This force creates a torque that causes the cylinder to roll. According to Newton's Second Law for rotation, the net torque equals the moment of inertia $I$ times the angular acceleration $\alpha$ ($\tau =I\alpha$).

So, our second master equation is:

We are given a crucial piece of information: the cylinder rolls without slipping. This means the point of contact with the ground is instantaneously at rest. For this to happen, the linear acceleration $a$ of the center of mass must be perfectly synchronized with the angular acceleration $\alpha$. The mathematical condition for pure rolling is:

We also know that for a solid homogeneous cylinder, the moment of inertia $I$ about its central axis is:

Now, let's substitute these relationships into our torque equation to find the friction $f$.

Notice how beautifully the physics simplifies. One $R$ from the denominator cancels with one $R$ in the numerator, and the remaining $R$ cancels with the $R$ on the left side of the equation. We are left with a remarkably simple expression for friction:

This tells us that the friction required to maintain pure rolling is exactly half of the net force required to accelerate the mass!

Now, let's take this expression for friction and plug it back into our very first translational equation:

To solve for $a$, we add $\frac{1}{2}ma$ to both sides:

Rearranging this to isolate the linear acceleration $a$, we get:

We are almost there! The question asks for the angular acceleration $\alpha$. We simply use our pure rolling constraint one last time:

Substituting our value for $a$:

And there we have it! The angular acceleration of the solid cylindrical roller is $\frac{2F}{3mR}$. The beauty of this problem lies in how the linear and rotational equations seamlessly weave together, bound by the constraint of pure rolling, to reveal the final answer.