A heavy particle is suspended by a stirng of length l. The particle is given a horizontal velocity `v_0`. The stirng becomes slack at some angle and the particle proceeds on a parabola.find the value of `v_0` if the prticle psses through the ponit of suspension

Explanation

Suppose the string becomes slack when the particle reaches the point P ure. Suppose the string OP makes an angle `theta` with the upward vertical. The only force actig on the particle at P is its weight mg.The radia component of the force is `mg costheta`. As the particle moves on the circle upto P,
`mg costheta=m(v^2/l)`
or, `v^2=glcostheta` ...........i
where v is its speed at P. Using conservation of energy,
`1/2 mv_0^2=1/2mv^2+mgl(1+costheta)`
or `v^2=v^2_0-2gl(1+costheta)`......ii
From i and ii `v_0^2-2gl(1+costheta)=glcostheta`
or, `v_0^2=gl(2+3costheta)` .........iii
Now onwards the particle goes in a parabola under the action of gravity. As it passes through the point of suspension O, the equations for horizontal and verticla motions give,
`lsintheta=(v costheta)t`
and `-lcostheta=(vcostheta)t-1/2gt^2`
or, `-lcostheta=(vsintheta((lsintheta)/(vcostheta))-1/2g((lsintheta)/(vcostheta))^2`
or `-cos^2theta=sin^2theta-1/2g(lsin^2theta)/(v^2costheta)`
or `-cos^2thet=1-cos^2theata-1/2 (glsin^2theta)/(2glcos^2theta) [from i]`
or, `=1/2tan^2theta`
`or tantheta=sqrt2`
From iii `v_0=[gl(2+sqrt3)]^(1/2)`