A Hula-hoop problem with a paradox?

Okay, sorry Anish, I was being too hasty and thought too much about the problem.

This is a simple energy conservation problem. $mg\frac{R}{2}=\frac{1}{2}mv^2 \Rightarrow v=\sqrt{gR}\approx 1.714\,m/s$

I haven't figured out the reason for the second part yet.

I won't present a solution for the first part, as it is identical the rest here.

I would like to point out that the particle WILL NEVER reach the bottommost point, while satisfying the constraint that the hoop remains in contact with the ground and does not slip.

Why?

Consider the situation when $\theta >90\deg$ , in the frame of reference of the hoop(non-rotating). In order to force the particle into the circular motion, a radial centripetal force is required. The vertical component of this centripetal force points upwards, which cannot be exerted by the hoop, since a) the hoop is massless and has no weight, and b) normal force between ground and hoop must be positive.

Hence, once the particle reach the lower quadrant, the hoop will break contact with the ground, and the paradox is resolved.

Let $M$ be the mass of the hula loop. As there is no slipping, so $u=\omega R$ , where $u$ is the velocity of the centre of mass of the hula loop and $\omega$ is its angular velocity.

Also, by simple vector addition, ${ \upsilon }^{ 2 }={ u }^{ 2 }+{ (\omega { R }) }^{ 2 }+2.u.\omega R.cos(\theta )=2{ (\omega R) }^{ 2 }(1+cos(\theta ))$

The energy conservation equation will actually contain 2 other terms for the rotational and translational kinetic energy of the hula loop, and hence will be: $mgR(1-cos(\theta ))=\frac { 1 }{ 2 } m{ \upsilon }^{ 2 }+\frac { 1 }{ 2 } M{ u }^{ 2 }+\frac { 1 }{ 2 } M{ R }^{ 2 }{ \omega }^{ 2 }=m{ (\omega R) }^{ 2 }(1+cos(\theta ))+M{ R }^{ 2 }{ \omega }^{ 2 }$

Thus, given a ratio between the mass of the loop and the mass of the particle, it is possible to find the angular velocity of the loop and hence the velocity of the particle at any angle $\theta$ from the vertical.

But in this case, as it is given that the loop is 'light', so $M\rightarrow 0$ and hence we can neglect the two additional terms (of the kinetic energy of the hula loop) in all cases except one where $1+cos(\theta )\rightarrow 0$ i.e. $\theta \rightarrow { 180 }^{ 0 }$ . In that case, $\omega ,u\rightarrow \infty$ although $\upsilon$ will still be zero, and hence, it can be said that the potential energy will be converted into the kinetic energy of the 'light' hula loop.

In a practical situation, the mass of the loop will never be infinitely small and hence the loop will gain only a finite, but high angular velocity (which will get even higher as the loop is made lighter), hence not violating the energy conservation principle. (This also means that $\upsilon =1.714m/s$ is not an entirely correct answer)