Find the angle .theta. with the lower vertical at which the resultant acceleration of the bob is along the horizontal, when the perpendicular is released from horizontal position.

Find the angle .theta. with the lower vertical at which the resultant

1.	Find the angle .theta. with the lower vertical at which the resultant acceleration of the bob is along the horizontal, when the perpendicular is released from horizontal position.
Answer» <P> SOLUTION :Since the resultant acceleration is along the horizontal, the VERTICAL acceeleration at .P. should be zero. ie., `a_(c) sin(90 - theta) = a_(t) sin theta` `(a_(c))/(a_(1)) = tan theta ..... (1)` But `a_(c) = (v^(2))/(l)(2glcostheta)/(l)=2gcostheta` and `a_(1) = g sin theta` Then `(a_(c))/(a_(t)) = 2 cot theta` SUBSTITUTING in EQUATION (1) `2 cot theta = Tan theta` Hence `theta = tan^(-1) (sqrt(2))`