A point mass m connected to one end of inextensible string of length l and other end of string is fixed at peg. String is free to rotate in vertical plane. Find the minimum velocity given to the mass in horizontal direction so that it hits the peg in its subsequent motion.

A point mass m connected to one end of inextensible string of length l

1.	A point mass m connected to one end of inextensible string of length l and other end of string is fixed at peg. String is free to rotate in vertical plane. Find the minimum velocity given to the mass in horizontal direction so that it hits the peg in its subsequent motion.
Answer» Solution :Tension in string is zero at point P in its subsequent motion, after this point its motion is projectile. Velocity at point P, `T=0`. `impliesmgcostheta=(mv^2)/(L) v=sqrt(glcostheta)` ASSUME its projectile motion start at point P and it passes through point C. So that EQUATION of TRAJECTORY satisfy the co-ordinate of `C(lsintheta, -lcostheta)` Equation of trajectory, `y=xtantheta-(gx^2)/(2v^2cos^2theta)` `=(lsintheta)tantheta-(g(lsintheta)^2)/(2(glcostheta)cos^2theta)` `=-costheta=(sin^2theta)/(costheta)-1/2(sin^2theta)/(2cos^3theta)` `implies-2cos^4theta=2sin^2thetacos^2theta-sin^2theta` `impliessin^2theta=2sin^2thetacos^2theta+2cos^4theta` `=2cos^2theta(sin^2theta+cos^2theta)` `impliestan^2theta=2` `impliestantheta=sqrt2` hence, `costheta=1/sqrt3`, `sintheta=sqrt(2/3)` From energy CONSERVATION between point P and A, `1/2m u^2=1/2mv^2+mg(1+costheta)` `impliesu^2=v^2+2gl(1+costheta)` `=2gl+3glcostheta=2gl+3gl(1)/(sqrt3)` `impliesu=[(2+sqrt3)gl]^(1//2)`