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 $a_c$ is varying with time $t$ as, $a_c = k^2rt^2$, The power delivered to the particle by the forces acting on it is

A

$2\pi mk^2r^2t$

B

$mk^2r^2t$

C

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

D

Zero

Solution

Here the tangential acceleration also exits which requires power.

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

or $v^{2}=k^{2} r^{2} t^{2}$ or $v=k r t$

Tangential acceleration $a$ $=\frac{\mathrm{dv}}{\mathrm{dt}}=\mathrm{kr}$

Now force $F=m \times a=m k r$

So power $\mathrm{P}=\mathrm{F} \times \mathrm{v}=\mathrm{mkr} \times \mathrm{krt}=\mathrm{mk}^{2} \mathrm{r}^{2} \mathrm{t}$

Standard 11

Physics

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