A particle moves on a circle of radius `r` with centripetal acceleration as function of time as `a_c = k^2 r t^2`, where `k` is a positive constant. Find the following quantities as function of time at an instant :
(a) The speed of the particle
(b) The tangential acceleration of the particle
( c) The resultant acceleration, and
(d) Angle made by the resultant with tangential direction.

Explanation

(a) In the problem, it is given that `a_c = k^2 rt^2`
Since `a_c = (v^2)/( r) rArr (v^2)/( r) = k^2 r t^2`
`rArr v^2 = k^2 r^2 t^2`
Taking square root on both sides, we get `v = krt`
(b) Tangential acceleration,
`a_t = (d v)/(d t) = (d)/( dt) (k r t) = kr`
( c) Resultant acceleration,
`a_(res) = sqrt(a_c^2 + a_t^2)`
=`sqrt((k^2 r t^2)^2 + (k r)^2)`
=`kr sqrt(k^2 t^4 + 1)`.
(d) From (Fig. 148), `tan prop = (a_c)/(a_t) = (k^2 r t^2)/(k r) = k t^2`
`rArr prop = tan^-1 (kt^2)`.
.