Step 1: Understanding the Concept:
When a body slides down a smooth incline, its initial potential energy converts entirely into translational kinetic energy.
When a body rolls down without slipping, its potential energy converts into both translational and rotational kinetic energy.
Therefore, a rolling body will reach the bottom with a slower linear velocity than a sliding body.
Step 2: Key Formula or Approach:
For sliding: Conservation of Energy gives $mgh = \frac{1}{2}mv^2$.
For rolling: Conservation of Energy gives $mgh = \frac{1}{2}mv'^2 + \frac{1}{2}I\omega^2$.
Rolling condition: $v' = r\omega$.
Moment of inertia for a ring: $I = mr^2$.
Step 3: Detailed Explanation:
Let the height of the incline be $h$.
Case 1: Body slides down smooth incline.
Potential energy at top = Translational kinetic energy at bottom.
\[ mgh = \frac{1}{2}mV^2 \]
From this, we find the square of the sliding velocity $V$:
\[ V^2 = 2gh \implies 2gh = V^2 \]
Case 2: Ring rolls down the incline.
Potential energy at top = Translational KE + Rotational KE at bottom. Let the new velocity be $v'$.
\[ mgh = \frac{1}{2}mv'^2 + \frac{1}{2}I\omega^2 \]
Substitute $I = mr^2$ for a ring and $\omega = v'/r$:
\[ mgh = \frac{1}{2}mv'^2 + \frac{1}{2}(mr^2)\left(\frac{v'}{r}\right)^2 \]
\[ mgh = \frac{1}{2}mv'^2 + \frac{1}{2}mv'^2 \]
\[ mgh = mv'^2 \]
Now, substitute $2gh = V^2$ or $gh = V^2/2$ from the sliding case into this equation:
\[ m\left(\frac{V^2}{2}\right) = mv'^2 \]
Cancel mass $m$ from both sides:
\[ \frac{V^2}{2} = v'^2 \]
Take the square root of both sides to find $v'$:
\[ v' = \frac{V}{\sqrt{2}} \]
Step 4: Final Answer:
The linear velocity of the rolling ring is $\frac{\text{V}}{\sqrt{2}}$.