A particle of mass m is moving in a circular path of constant radius r such that its centripetal acc

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5.Work, Energy, Power and Collision

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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

$2\pi mk^2r^2$

$mk^2r^2t$

$\frac{{m{k^4}{r^2}{t^5}}}{3}$

Zero

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 $\quad P=F \times v$

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

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

Standard 11

Physics

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