A stone tied to an inextensible string of length l = 1 m is kept horizontal. If it is released, find the angular speed of the stone when the string makes an angle theta = 30^@ with horizontal.

A stone tied to an inextensible string of length l = 1 m is kept horiz

1.	A stone tied to an inextensible string of length l = 1 m is kept horizontal. If it is released, find the angular speed of the stone when the string makes an angle theta = 30^@ with horizontal.
Answer» Solution :When the stone is released, it moves down executing a circular motion. At any INSTANT, the stone accelerates towards (w.r.t.) the center of its revolution `0`, that is, centripetal ACCELERATION `a_r` and simultaneously accelerates down with an acceleration, that is, gravitational acceleration `g , vec a_r = omega^2 r`. Resolving `vec a_r` and along the tangent, we obtain the tangential acceleration of the stone as `a_t = g COS theta`. Angular acceleration, `prop = (a_t)/( r)` `rArr prop = (g cos theta)/(l)` `(since r = 1)` PUTTING `prop = (omega d omega)/(d theta),` we obtain `(omega d omega)/(d theta) = (g cos theta)/( r)` `rArr omega d omega = (g//l) cos theta d theta`. Integrating both the sides, we obtain `int_0^(omega) omega d theta = (g)/(l) int_0^(theta)cos theta d theta` `rArr (omega^2)/(2) = (g)/(l) sin theta or omega = sqrt((2 g)/(l)) sin theta` Putting `l = 1 m and theta = 30^@`, we obtain `omega = sqrt((2(10) sin 30^@)/((1))) = sqrt(10) rArr omega ~= 3.1 rad s^-1`. .