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: 

$$For\ an\ object\ rolling\ without\ slipping\ down\ an\ inclined\ plane\ with\ inclination\ \theta,\ $$

$$the\ linear\ acceleration\ (a)\ of\ the\ center\ of\ mass\ is:\ a=\frac{g\sin\theta}{1+\frac{I}{mr^2}}$$

$$For\ a\ uniform\ solid\ cylinder,\ the\ moment\ of\ inertia\ is\ I=\frac{1}{2}mr^2$$

$$Substituting\ this\ and\ \theta=45^{\circ}\ into\ the\ formula:$$

$$a = \frac{g \sin 45^\circ}{1 + \frac{1}{2}mr^2 / mr^2}$$

$$a = \frac{g \cdot \frac{1}{\sqrt{2}}}{1 + \frac{1}{2}}$$

$$a = \frac{2g}{3\sqrt{2}}$$

$$a = \frac{\sqrt{2}}{3} g$$