Q.Discuss rolling on an inclined plane and arrive at the expression for the acceleration.

Step 1. Setting up the forces.

Consider a round object of mass $m$, radius $R$ and radius of gyration $K$, rolling WITHOUT SLIPPING down an incline of angle $\theta$. Along the incline, two forces act: the driving component of gravity, $mg\sin\theta$, and an opposing static frictional force $f$ at the point of contact (the perpendicular component of gravity, $mg\cos\theta$, is simply balanced by the normal reaction $N$ and plays no role in the motion along the incline).

Step 2. The translational equation of motion.

Applying Newton's second law along the incline, taking the direction of motion as positive:

Here $a$ is the linear acceleration of the object's center of mass down the incline, and $f$ is still an unknown at this stage.

Step 3. The rotational equation of motion.

Now take torques about the object's own center. Gravity acts through the center itself, so its line of action passes exactly through the axis and it contributes NO torque here. Only the friction force $f$, acting tangentially at the rim (distance $R$ from the center), contributes a torque:

where $I$ is the object's moment of inertia about its own central axis and $\alpha$ its angular acceleration.

Step 4. Applying the rolling (no-slip) condition.

Because the object rolls without slipping, its linear and angular accelerations are locked together by $a=R\alpha$, i.e. $\alpha=a/R$. Also write the moment of inertia in terms of the radius of gyration, $I=mK^2$. Substituting both into the rotational equation:

Step 5. Combining the two equations.

Substitute this expression for $f$ back into the translational equation (i):

Factor $a$ out of the two terms on the right:

Step 6. Solving for the acceleration.

Dividing through by $m\left(1+K^2/R^2\right)$:

Step 7. Extensions — final speed and time (brief). …

Simultaneous translational (Newton's second law along the incline) and rotational (torque about the center) equation …

• Forgetting that gravity, acting through the object's own center of mass, produces ZERO torque about that same center — only the friction force, acting at the rim, contributes to the rotational equation.
• Dropping the friction force from the translational equation altogether (treating it as frictionless sliding) — without $f$ there is nothing to produce rotation at all, and the (wrong) result collapses to the frictionless value $a=g\sin\theta$ regardless of shape. …