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

Step 1: Forces Acting on the Cylinder

For a rolling object on an inclined plane, the forces acting are:

Gravitational force ($mg$):

• Component along the incline: $mg \sin\theta$
• Component perpendicular to the incline: $mg \cos\theta$

Normal reaction force ($N$): Acts perpendicular to the inclined surface.

Friction force ($f$): Provides the necessary torque for rolling motion (without slipping). It acts up the incline.

Step 2: Equations of Motion

1. Translational Motion

From Newton's Second Law along the incline:

$$mg \sin\theta - f = ma$$

where $a$ is the linear acceleration of the cylinder.

2. Rotational Motion

For rotation about the center, using Torque Equation:

$$f r = I \alpha$$

For a solid cylinder, the moment of inertia about its axis is:

$$I = \frac{1}{2} m r^2$$

Using the pure rolling condition: $\alpha = \frac{a}{r}$,

$$f r = \left(\frac{1}{2} m r^2\right) \frac{a}{r}$$

 $$f = \frac{1}{2} m a$$ 

Step 3: Solve for Acceleration

Substituting $f = \frac{1}{2} m a$ into the translational equation:

$$mg \sin\theta - \frac{1}{2} m a = ma$$

 $$mg \sin\theta = ma + \frac{1}{2} ma$$ $$mg \sin\theta = \frac{3}{2} ma$$ $$a = \frac{2}{3} g \sin\theta$$

Substituting $\theta = 45^\circ$, we get:

$$a = \frac{2}{3} g \sin 45^\circ$$

 $$a = \frac{2}{3} g \times \frac{1}{\sqrt{2}}$$ $$a = \frac{2g}{3\sqrt{2}}$$ $$a = \frac{\sqrt{2} g}{3}$$ 

Final Answer:

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