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

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Question

Here the tangential acceleration also exits which requires power.

Given that $\quad a_{c}=k^{2} r t^{2} \quad$ and     &nbsp

Vedclass · Accepted Answer

$mk^2r^2t$

Vedclass · Answer

Here the tangential acceleration also exits which requires power.

Given that $\quad a_{c}=k^{2} r t^{2} \quad$ and       $a_{c}=\frac{v^{2}}{r}$

$\frac{v^{2}}{r}=k^{2} r t^{2} 1$

$\mathrm{Or}$          $\mathrm{v}^{2}=\mathrm{k}^{2} \mathrm{r}^{2} \mathrm{t}^{2}$

$\mathrm{v}=\mathrm{krt}$

Tangential acceleration $\quad a=\frac{d v}{d t}=k r$

Now force $F=m a$

$=\mathrm{m} \mathrm{kr}$

So power $\quad P=F 	imes v$

$\mathrm{P}=(\mathrm{mkr}) 	imes(\mathrm{krt})$

$P=m k^{2} r^{2} t$

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