InterviewSolution
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InterviewSolution
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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 |
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Answer» 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)` |
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