A small ball is suspended from point O by a thread of length l. A nail is driven into the wall at a distance of `l//2` below O, at A. The ball is drawn aside so that the thread takes up a horizontal position at the level of point O and then released. Find
a. At what angle from the vertical of the ball's trajectory, will the tension in the thread disappear?
b. What will be the highest point from the lowermost point of circular track, to which it will rise?

Answer

a. `cos^-1(2//3)`; b. `(50l)/(54)`

Explanation

If at point P, tension is zero. Then
`mgcos theta=(mv^2)/(r)`
Using conservation of energy, `v^2=gl(1-costheta)`
`:.mg costheta=(mgl)/(l//2)(1-costheta)`
`impliestheta=cos^-1(2//3)`
Therefore, height of point,
`P=l/2+l/2cos theta=(5l)/(6)`,
from lowest point
`v^2=gl(1-2/3)=(gl)/(3)impliesv=sqrt((gl)/(3))`
Now the particle describes parabolic path.
The height attained by the particle, from point P.
`h=((vsintheta)^2)/(2g)=(5l)/(54)`
Therefore, highest point from loweset point will be
`((5l)/(6)+(50l)/(54))=(50l)/(54)`