Sir, my name is Cyril Johnson and I am from Jabalpur.I am greatly thankful to elessar_iitkgp  and biki(303) for helping me in the previous kinematics problem.Sir, I am new to this site and I dont know how to rate the Genius like you all.So, kindly excuse me.I will learn it soon.

Sir,now i have a different question for you.

a 2Kg block is fixed at one end of a 3 mt long string(radius of the exibited motion) and released from horizontal position.find the minimum velocity required to keep the circular motion going.

Sir, is this practically proven?Please kindly explain me the physics behind it.

thanks .... frnd.... for the appraisal....

let me try this.
taking the lowest position of the body as the reference level of potential energy, the body is thus at a height of (radius) m from the ref. level.
So using energy conservation .... the total energy at this point must be equal to the K.E. at the lowest position (as P.E. at lowest position =0 ).
Let v₀ = velocity at lowest position for which the body completes circular motion.

let T = tension in the string at highest position.

The limiting condition for completing circular motion in case the body is attached to a string is that the string does not become slack at the highest point.

in case of rods... the centrifugal force is not reqd. at highest position as the rod will never become slack. So in case of rods , we take velocity at top position as zero and just equate energy conservation principle between initial and final state (without considering any other things) 

Now, coming back to the given problem, the minimum condition for this to happen is that all forces in opposite direction become equal at highest point.

let v = velocity at top position.

So mg + T = mv²/r ________(1)

Now using conservation of mech. energy at top and bottom positions....

(1/2)mv₀² = (1/2)mv² + mg(2r)

or mv² = mv₀² - 4mgr _____(2)

using (2) in (1)...

mg+ T = (mv₀² - 4mgr) /r
or mv₀² = 5mgr + Tr.

Now the v₀ will be minimum if T is minimum.

Now T_min =0

So mv₀² = 5mgr

or v₀ =  5gr.

So v₀ =  (5 x 9.8 x 3) = 73 m/s

Now let v ^/ = velocity at initial horizontal position (at height = r from ref. level)

So again by conservation of mech. energy...

(1/2) m.(v ^/ )² + mgr = (1/2) m.v₀²

Putting values of m,g,r and v₀ in (3)..

v ^/ = 9.39 m/s