A particle of mass m is moving in a circular | vedclass

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(B) Given the centripetal acceleration $a_c = k^2rt^2$.Since $a_c = \frac{v^2}{r}$,we have $\frac{v^2}{r} = k^2rt^2$,which implies $v^2 = k^2r^2t^2$ o

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$mk^2r^2t$

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(B) Given the centripetal acceleration $a_c = k^2rt^2$.Since $a_c = \frac{v^2}{r}$,we have $\frac{v^2}{r} = k^2rt^2$,which implies $v^2 = k^2r^2t^2$ or $v = krt$.The tangential acceleration $a_t$ is given by $a_t = \frac{dv}{dt} = \frac{d}{dt}(krt) = kr$.The tangential force acting on the particle is $F_t = m a_t = m(kr) = mkr$.The power $P$ delivered to the particle is the product of the tangential force and the velocity: $P = F_t \cdot v = (mkr) \cdot (krt) = mk^2r^2t$.

$A$ particle of mass $m$ is moving in a circular path of constant radius $r$ such that its centripetal acceleration $a_c$ is varying with time $t$ as $a_c = k^2rt^2$. The power delivered to the particle by the forces acting on it is

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