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 $a_{C}=k^{2} r t^{2}$ and $a_{C}=\frac{v^{2}}{r} 	herefore \frac{v^{2

Vedclass · Accepted Answer

$mk^2r^2t$

Vedclass · Answer

Here the tangential acceleration also exits which requires power.

Given that $a_{C}=k^{2} r t^{2}$ and $a_{C}=\frac{v^{2}}{r} 	herefore \frac{v^{2}}{r}=k^{2} r t^{2}$

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 $F=m 	imes a=m k r$

So power $\mathrm{P}=\mathrm{F} 	imes \mathrm{v}=\mathrm{mkr} 	imes \mathrm{krt}=\mathrm{mk}^{2} \mathrm{r}^{2} \mathrm{t}$

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

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