A round uniform body of radius $$R$$, mass $$M$$ and moment of inertia $$I$$, rolls down (without slipping) an inclined plane making an angle $$\theta$$ with the horizontal. Then its acceleration is

When a round uniform body rolls down an inclined plane without slipping, it experiences both translational and rotational motion.

1. Equations of Motion:

   Let a be the linear acceleration of the body down the incline, and alpha be its angular acceleration.

The force acting down the inclined plane is the component of gravity:

$$Mg \sin \theta$$

The static friction force f acts up the incline to cause rolling without slipping. Applying Newton's second law for translational motion along the incline:

$$Mg \sin \theta - f = Ma$$

Applying Newton's second law for rotational motion about the center of mass:

$$\tau = I\alpha$$

$$fR = I\alpha$$

1. Condition for rolling without slipping:

   The relationship between linear acceleration and angular acceleration is:

   $$a = \alpha R$$
2. $$\alpha = \frac{a}{R}$$
3. Substituting alpha into the torque equation:

   $$fR = I\left(\frac{a}{R}\right)$$

4. $$f = \frac{Ia}{R^2}$$
5. Substituting the friction force f into the translational equation:$$Mg \sin \theta - \frac{Ia}{R^2} = Ma$$

Move the acceleration terms to one side:

$$Mg \sin \theta = Ma + \frac{Ia}{R^2}$$

Factor out the acceleration a:

$$Mg \sin \theta = a\left(M + \frac{I}{R^2}\right)$$

Divide both sides by M to isolate a:

$$g \sin \theta = a\left(1 + \frac{I}{MR^2}\right)$$

$$a = \frac{g \sin \theta}{1 + \frac{I}{MR^2}}$$