A uniform solid cylinder of mass ' m ' and radius ' r ' rolls along an inclined rough plane of inclination $45^{\circ}$. If it starts to roll from rest from the top of the plane then the linear acceleration of the cylinder's axis will be

<p>The acceleration of the center of mass for a rolling object down an incline is given by:</p> <p>$a = \frac{g \sin\theta}{1 + \frac{I}{m r^2}}$</p> <p>For a uniform solid cylinder, the moment of inertia about its central axis is:</p> <p>$I = \frac{1}{2} m r^2$</p> <p>Substituting this into the expression for acceleration:</p> <p>$$ a = \frac{g \sin\theta}{1 + \frac{\frac{1}{2} m r^2}{m r^2}} = \frac{g \sin\theta}{1 + \frac{1}{2}} = \frac{g \sin\theta}{\frac{3}{2}} = \frac{2}{3} g \sin\theta $$</p> <p>For an incline of $45^\circ$, we have:</p> <p>$\sin 45^\circ = \frac{\sqrt{2}}{2}$</p> <p>Thus, the acceleration becomes:</p> <p>$a = \frac{2}{3} g \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2} g}{3}$</p> <p>Therefore, the linear acceleration of the cylinder's axis is:</p> <p>$\frac{\sqrt{2} g}{3}$</p>