easy question

the answer given is 2√gl but i think it should be √5gl. pls help which is right?(hcverma-1 pg-135 q.49)

It is √5gl for a string and not rod

In the qs a rod is given

So ans is 2√gl

thnks got it.

n.b.....you are getting that answer because you are supposing that a string and a light rod are the same !! there is a subtle difference between the two !

when the mass is at any pt in the 4th quadrant
T₁ = mg cosθ +mv₁ ²/l
when the mass is at any pt in the 1st quadrant
T₂ = mv₂ ²/l - mg cosθ......(suppose the two angles are equal,,,can be different also)

T₁ cannot be zero as it is an addition of 2 positive quantities, but v₁ can be zero. when v₁ becomes zero, the body stops and comes back again ! thus the 4th quadrant cn become a zone of oscillation !

on the other hand v₂ cannot be zero as that would make T₂= 0 at some point, which is impossible because a string cannot take compression ! thus velocity cannot be zero but tension can be zero in the 1st quadrant ! when T₂ becomes 0 , c.p. force ceases to act and circular motion stops,,the body now follows projectile motion !
but when we use a massless rod, the velocity can become zero in the 1st quadrant.

applying conservation of energy theorem ,
1/2 mv₁² = mg(2l) + 1/2 mv₂²
....putting v₂ = 0 for light massless rod, we get v₁=2√gl

there is no tension at rod. so P.E at highest point is equal to K.E at lowest point .

since length of rod is l.therefore distance betn lowest & highest points is 2l

mg(2l)=1/2mv² ,v=velocity at lowest point

v=2√(gl)