Consider a pendulum consisting of a nuscless string with a mass on the 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 theta with the vertical. Find theta if that mass covers maximum horizontal range.

Consider a pendulum consisting of a nuscless string with a mass on the

1.	Consider a pendulum consisting of a nuscless string with a mass on the 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 theta with the vertical. Find theta if that mass covers maximum horizontal range.
Answer» Solution :The speed at ANGLE `theta` is given by conservation of energy. `(1)/(2) mv^(2) = mgh rArr (1)/(2) mv^(2) = mg R COS theta v = sqrt(2 g R cos theta)` When string is cut PARTICLE MOVES as a projectile with velocity component. `v_(x) = v cos theta v_(y) = v sin theta` The time of flight is `t = 2 ((v_(y))/(g))` Range is, `x = v_(x) t = v_(x) ((2v_(y))/(g))` `= (2v_(x)v_(y))/(g) = (2(v cos theta) (v sin theta))/(g)` `= (2v^(2) sin theta cos theta)/(g) = (2 (2 g R cos theta)sin theta cos theta)/(g) = 4 R cos^(2) theta sin theta` Take the derivative of this FUNCTION and equate it to zero Thus we get `0 = 4R [(-2 cos theta sin theta) sin theta + cos^(2) theta cos theta]` (or) `2 sin^(2) theta cos theta = cos^(3) theta (or) tan theta = 1//sqrt(2)`