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

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(B) Given centripetal acceleration $a_c = k^2 r t^2$.We know that $a_c = \frac{v^2}{r}$,where $v$ is the speed of the particle.Equating the two expres

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$m k^2 r^2 t$

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(B) Given centripetal acceleration $a_c = k^2 r t^2$.We know that $a_c = \frac{v^2}{r}$,where $v$ is the speed of the particle.Equating the two expressions: $\frac{v^2}{r} = k^2 r t^2$.This simplifies to $v^2 = k^2 r^2 t^2$,so $v = krt$.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 $F_t = m a_t = mkr$.The power $P$ delivered to the particle is the product of the tangential force and the velocity: $P = F_t \cdot v = (mkr) \cdot (krt) = m k^2 r^2 t$.

$A$ 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 $a_c = k^2 r t^2$. The power delivered to the particle by the forces acting on it is:

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