Question

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

Answer 1

Hint: In order to solve this question, we will first find the tangential acceleration on the particle due to increasing velocity as given in question and then we will find radial acceleration on the particle due to moving in circular path and then will compute the net acceleration acting on the particle.

Formula used: If ${a_T},{a_R}$ denotes for tangential and radial acceleration acting on a body then net acceleration is found as $a = \sqrt {{a^2}_T + {a^2}_R} $

Complete step by step answer:
According to the question, we have given that velocity is changing at the rate of g per second and rate of change of velocity is acceleration in the direction of velocity so,
tangential acceleration on the particle can be written as ${a_T} = g$
Now, it’s given that radius of circular path is r and let u be the initial velocity of the particle and v denoted for velocity at any instant of time t with an acceleration of g (as given in question) then from newton’s equation of motion we have,
$v = u + gt$ and with this velocity the radial acceleration is calculated as
${a_R} = \dfrac{{{v^2}}}{r}$ on putting the values we get,
${a_R} = \dfrac{{{{(u + gt)}^2}}}{r}$
Now, the net acceleration can be calculated using formula $a = \sqrt {{a^2}_T + {a^2}_R} $ on putting the values we get,
$a = \sqrt {{g^2} + \dfrac{{{{({{(u + gt)}^2})}^2}}}{{{r^2}}}} $
$ \Rightarrow a = \sqrt {{g^2} + \dfrac{{{{(u + gt)}^4}}}{{{r^2}}}} $
Hence, the net acceleration acting on the particle is $a = \sqrt {{g^2} + \dfrac{{{{(u + gt)}^4}}}{{{r^2}}}} $.

Note: It should be remembered that, the tangential component of net acceleration is in the direction of increasing velocity and the radial component of net acceleration is directed towards the centre of the circular path, both components are perpendicular to each other.