A particle of mass 100 g is suspended from the end of a weightless string of length 100 cm and is rotated in a vertical plane. The speed of the particle is `200 cm*s^(-1)` when the string makes an angle of `theta =60^@` with the verctcal. Determine
(ii) the speed of the particle at the lowest position . Acceleration due to gravity = `980 cm *s^(-2)`.

Explanation

From the principle of conservation of energy we get, kinetic energy of the body at the lowest point A = sum of the potential and kinetic energies at the point B
`or, 1/2 mv_A^2=1/2mv^2+mg*AC`
[`v_A` =speed of the particle at the point A ]
`or, v_A^2=v^2+2g*AC`
Here, AC=OA-OC=OA-OB `cos theta`
`=l-lcos theta =100 -100 cos 60^@=50cm`
`therefore v_A^2=(200)^2+2xx980xx50=13.8xx10^4`
`or, v_A= 371.5 cm *s^(-1)`.