Question

A particle moving along a circular path with speed increased by g per second. If the radius of angular path be r. Then the net acceleration of the particle is..???

Answer 1

We have a case of accelerated circular motion. Given rate of change of velocity

`(dv)/dt=g=veca_t`

Assuming `u` be initial velocity, velocity `v` after time `t` of the particle is given by the kinematic expression

`v=u+g t`

Tangential acceleration `veca_t` is parallel to velocity of the particle.

Radial acceleration is given by

`veca_r=romega^2=(v^2)/r=(u+g t)^2/r`
here we see that `omega=v/r` is also changing with time due to accelerated motion.

As both accelerations are orthogonal to each other, net acceleration of the particle can be found from the expression

`a_n=sqrt(|veca_t|^2+|veca_r|^2)`
`a_n=sqrt(g^2+((u+g t)^2/r)^2)`
`a_n=sqrt(g^2+(u+g t)^4/r^2)`