Consider a perdulum consisting of a mssless string with a mass at one end. The mass is held with the string horizontal and then released. The mass swings down, and on its way back up, the string is cut at point P when it makes an angle of with theta the vertical. Find the angle for theta which horizontal range .R. is maximum.

Consider a perdulum consisting of a mssless string with a mass at one

1.	Consider a perdulum consisting of a mssless string with a mass at one end. The mass is held with the string horizontal and then released. The mass swings down, and on its way back up, the string is cut at point P when it makes an angle of with theta the vertical. Find the angle for theta which horizontal range .R. is maximum.
Answer» SOLUTION :The speed at ANGLE is given by conservation of energy. `(1)/(2)mv^(2)=mghrArr(1)/(2)mv^(2)=mglcostheta` `v=sqrt(2glcostheta)` When string is cut particle MOVES as a PROJECTILE with velocity component. `v_(x)=vcosthetav_(y)=vsintheta` The time of flight is `t=2((v_(y))/(g))` Range is, R = `v_(x)t = v_(x) ((2y_(y))/(g))` `=(2v_(x)v_(y))/(g)=(2(vcostheta)(vsintheta))/(g)=(2v^(2)sin thetacostheta)/(g)` `=(2(2glcostheta)sinthetacostheta)/(g)=4lcos^(2)thetasintheta` Differentiating .R. with respect to `.theta.` and equatingit to zero we get `0=4l(-2costhetasintheta)sintheta+cos^(2)thetacostheta` or `,sin^(2) theta cos theta = cos^(3) theta or, TAN theta = 1 // sqrt(2)`