Question

A small heavy block is attached to the lower end of a light rod of length l which can be rotated about its clamped upper end. What minimum horizontal velocity should the block be given so that it moves in a complete vertical circle.
Concept of Physics - 1 , HC VERMA , Chapter " Work and Energy"

Answer 1

Let the velocity at the point be v. Assume that the lower end the Potential energy of the block is zero. 

Now Total energy,
= K.E + P.E 
= 1/2 mv² +0
=1/2mv²

Now, At the maximum point ,  Potential energy  . P.E 
= (mg) 2.l
So by law of conservation of energy . (K.E = P.E)
1/2mv² = mg.2l
1/2 v² = 2gl
v² = 4gl 
v  =√4gl
$v= 2 \sqrt{gl}$ .



Hope it Helps. :-)

Answer 2

Answer:

Explanation:

1/2mv^2 = mg 2l (i) (law of conservation of energy )

Initially PE =0 And KE at the top = 0

PE at top is at height 2×l

this from 1

V=2 root gl