Question

A particle is given an initial speed u inside smooth spherical shell of radius r = 1 m that it is just able to complete the circle. acceleration of the particle when its velocity is vertical is

Answer 1

The minimum speed required to just complete the circular motion in vertical plane is given by

$v = \sqrt{5Rg}$

here we have

R = 1 m

g = 9.8 m/s^2

$v = \sqrt{5*1*9.8}$

$v = 7 m/s$

now we will use energy conservation to find the speed where its velocity is vertical

its velocity is vertical at the height same as that of level of center.

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

$\frac{1}{2}*7^2 = 9.8*1 + \frac{1}{2}v_f^2$

$v_f^2 = 29.4$

now acceleration of the particle towards the center is given by

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

$a_c = 29.4 m/s^2$

also tangential acceleration is given by

$a_t = g = 9.8 m/s^2$

now net acceleration is given by

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

$a_{net} = \sqrt{29.4^2 + 9.8^2}$

$a_{net} = 31 m/s^2$

so above is the net acceleration of the object

Answer 2

Speed to complete a circle u=7gR

v^2-u^2=2gR

v^2=3gR

We have tangential acceleration=gm/s^2

Centripetal acceleration=v^2/r

=3g

Total a=√g^2+9g^2

=g√10