JEE Main 2024 — Rotational Motion Question with Solution

To determine the acceleration of a cylinder rolling down an inclined plane without slipping, we can use Newton's second law and the concept of rolling motion. For an inclined plane at an angle $$\theta$$, the component of gravitational acceleration along the plane is $$g\sin \theta$$. However, because the cylinder is rolling and not sliding, not all of this component accelerates the center of mass; some of it goes into causing rotational acceleration about the center of mass.

For a rolling cylinder, the moment of inertia $$I$$ is $$I=\frac{1}{2}m{r}^2$$, where $$m$$ is the mass of the cylinder and $$r$$ is the radius. The condition for rolling without slipping is that the linear acceleration $$a$$ of the center of mass is equal to the radius $$r$$ times the angular acceleration $$\alpha$$, i.e., $$a=r\alpha$$.

To find the linear acceleration $$a$$, we use the torque $$\tau$$ about the center of mass caused by the gravitational force down the incline. The torque due to gravity is $$\tau =mg\sin \theta \cdot r$$, and from Newton's second law for rotation, the angular acceleration is given by

$$\alpha =\frac{\tau }{I}=\frac{mg\sin \theta \cdot r}{\frac{1}{2}m{r}^2}=\frac{2g\sin \theta }{r}$$

Now using $$a=r\alpha$$:

$$a=r\left(\frac{2g\sin \theta }{r}\right)=2g\sin \theta$$

But we must account for the fact that only a portion of the gravitational acceleration goes into translating the cylinder down the plane due to the rolling condition. This is where we apply the concept of the ``rolling factor'' for a cylinder, which is $$\frac{2}{3}$$ for a solid cylinder, meaning $$\frac{2}{3}$$ of the gravitational component is used for translation.

The acceleration of the center of mass for the cylinder is therefore:

$$a=\frac{2}{3}g\sin \theta$$

Now we plug in the values of $$\theta =60^{\circ }$$ (which has $$\sin 60^{\circ }=\frac{\sqrt{3}}{2}$$) and $$g=10m/s^2$$:

$$a=\frac{2}{3}\cdot 10\cdot \frac{\sqrt{3}}{2}=\frac{10\sqrt{3}}{3}$$

To match the given expression $$\frac{x}{\sqrt{3}}m/s^2$$, let's manipulate our expression for $$a$$:

$$a=\frac{10\sqrt{3}}{3}=\frac{10\sqrt{3}}{3}\cdot \frac{\sqrt{3}}{\sqrt{3}}=\frac{10\cdot 3}{3\cdot \sqrt{3}}=\frac{10}{\sqrt{3}}m/s^2$$

Hence, the value of $$x$$ is $$10$$.