Question

A particle is moving in a vertical circular motion such that tension in the string at the top most point B is zero the acceleration of the particle at point a is

Answer 1

Centripetal acceleration of the particle is given as

$a_c = \frac{v^2}{R}$

here v = speed at the given position

R = radius

Now for the speed we will use energy conservation

$\frac{1}{2}mv_i^2 + mgh' = \frac{1}{2}mv_f^2 + mgh$

$\frac{1}{2}m*(\sqrt{gR})^2 + mg*2R = \frac{1}{2}mv_f^2 + mg*R$

$\frac{5}{2}mgR - mgR = \frac{1}{2}mv_f^2$

$v_f = \sqrt{3gR}$

now for centripetal acceleration

$a_c = \frac{(\sqrt{3gR})^2}{R}$

$a_c = 3g$

now tangential acceleration is given by

$a_t = \frac{F_g}{m}$

$a_t = \frac{mg}{m}$

$a_t = g$

now the net acceleration is given as

$a_{net} = \sqrt{a_c^2 + a_t^2}$

$a_{net} = \sqrt{(3g)^2 + g^2}$

$a_{net} = 10\sqrt10 m/s^2$

so the correct option is 2nd option