JEE Main 2025 — Rotational Motion Question with Solution

The acceleration of the center of mass for a rolling object down an incline is given by:

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

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

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

Substituting this into the expression for acceleration:

$$a=\frac{g\sin \theta }{1+\frac{\frac{1}{2}m{r}^2}{m{r}^2}}=\frac{g\sin \theta }{1+\frac{1}{2}}=\frac{g\sin \theta }{\frac{3}{2}}=\frac{2}{3}g\sin \theta$$

For an incline of $$45^{\circ }$$, we have:

$$\sin 45^{\circ }=\frac{\sqrt{2}}{2}$$

Thus, the acceleration becomes:

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

Therefore, the linear acceleration of the cylinder's axis is:

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