A ring is at the top of the incline of angle θ and height ‘h’. If it is left from rest and goes down rolling purely, find its speed at the bottom.

√ (gh) m/s

Friction will be acting on the ring, which is the reason it is able to roll purely.

But this friction will not do any work on the ring since point of contact remains at rest. Hence work done by frictional force is zero.

The potential energy at the top will be converted into kinetic energy at the bottom. Kinetic energy is given by: 

$ K.E = \frac{1}{2} M V^2 + \frac{1}{2} I \omega^2$

I for a ring = MR² & ω = V/R.

$\Rightarrow \text{kinetic energy}= \frac{1}{2} M V^2 + \frac{1}{2} MR^2 \frac{V^2}{R^2} = MV^2$

Potential energy at the top = Mgh.

Therefore, from Energy conservation 

Mgh = MV² OR V = √(gh) m/s.