QJEEM24Phcf47: JEE Main 2024 Physics - Class 11 JEEM24Phcf47 Solution: A cylinder is rolling down on an inclined plane of inclination…

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

Now using $a = r \alpha$ :

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:

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

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

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