Animated Solution for Physics - Rotational Motion: A uniform disc of mass $m$ and radius $R$ is rolling up a rough inclined plane which makes an angle of $30^{\circ }$ with the horizontal. If the coefficients of static and kinetic friction are each equal to $\mu$ and the only forces acting are gravitational and frictional, then the magnitude of the frictional force acting on the disc is ……… and its direction is ……… (write up or down) the inclined plane.

Have you ever watched a wheel roll up a hill and wondered about the invisible forces orchestrating that perfect, smooth motion? It seems simple on the surface, but the interplay between gravity, inertia, and friction is a beautiful mechanical ballet. Let's dive into this classic JEE problem and unravel the mystery of rolling motion on an incline.

The Forces at Play

Imagine you are standing next to a rough inclined plane angled at $30^{\circ }$ to the horizontal. A uniform disc of mass $m$ and radius $R$ is rolling up this incline.

What forces are acting on it? First, we have gravity. The component of gravity acting parallel to the incline is $mg\sin 30^{\circ }$, and it's pulling the disc downwards, trying to slow its ascent.

Now, here is the tricky part: friction. Intuition might scream that since the disc is moving up, friction must act down to oppose the motion. But remember, static friction opposes relative slipping at the point of contact, not necessarily the overall motion of the center of mass.

As gravity slows down the linear velocity $v$ of the disc, the angular velocity $\omega$ must also decrease to maintain the pure rolling condition ($v=\omega R$). To decrease a clockwise angular velocity, we need a counter-clockwise torque. Since gravity acts through the center of mass, it provides zero torque. The only force capable of providing this necessary counter-clockwise torque is friction, and it must act up the incline to do so! Let's assume friction $f$ acts upwards and let the math confirm our physical intuition.

Let's set up our equations of motion. We will take the downward direction along the incline as positive for our deceleration $a$.

For translation, Newton's Second Law gives us:

For rotation about the center of mass, the torque is provided entirely by friction. The lever arm is the radius $R$:

Because the disc is rolling without slipping, its linear and angular motions are perfectly synchronized. The linear deceleration $a$ is directly tied to the angular deceleration $\alpha$ by the constraint equation:

This is the bridge that connects our translational and rotational worlds. Let's use it to simplify our torque equation. We know the moment of inertia of a uniform disc is $I=\frac{1}{2}m{R}^2$. Substituting this and $\alpha =\frac{a}{R}$ into the torque equation yields:

Notice how beautifully the $R$'s cancel out! This leaves us with a direct relationship between friction and linear acceleration:

Now, we bring it all together. We substitute our expression for $a$ back into the translational equation:

Since $\sin 30^{\circ }=\frac{1}{2}$, the equation simplifies to:

Moving $f$ to the right side, we get:

Finally, solving for $f$ gives us our answer:

Because our calculated value for $f$ is positive, it confirms that our initial assumption was absolutely correct. The frictional force has a magnitude of $\frac{mg}{6}$ and it acts up the inclined plane. This elegant result shows how friction acts as the unsung hero, perfectly balancing the linear and rotational deceleration to keep the disc rolling smoothly.