A cylinder is rolling down on an inclined plane of inclination $60^{\circ}$. It's acceleration during rolling down will be $\frac{x}{\sqrt{3}} m / s^2$, where $x=$ ________ (use $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$).

<p>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.</p> <p>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., $a = r \alpha$.</p> <p>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</p> $$ \alpha = \frac{\tau}{I} = \frac{mg \sin \theta \cdot r}{\frac{1}{2} m r^2} = \frac{2g \sin \theta}{r} $$ <p>Now using $a = r \alpha$:</p> $a = r \left( \frac{2g \sin \theta}{r} \right) = 2g \sin \theta$ <p>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.</p> <p>The acceleration of the center of mass for the cylinder is therefore:</p> $a = \frac{2}{3} g \sin \theta$ <p>Now we plug in the values of $\theta = 60^{\circ}$ (which has $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$) and $g = 10 \; m/s^2$:</p> $a = \frac{2}{3} \cdot 10 \cdot \frac{\sqrt{3}}{2} = \frac{10 \sqrt{3}}{3}$ <p>To match the given expression $\frac{x}{\sqrt{3}} m / s^2$, let's manipulate our expression for $a$:</p> $$ a = \frac{10 \sqrt{3}}{3} = \frac{10 \sqrt{3}}{3} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{10 \cdot 3}{3 \cdot \sqrt{3}} = \frac{10}{\sqrt{3}} m/s^2 $$ <p>Hence, the value of $x$ is $10$.</p>