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A particle of mass $m$ is rotating in a circle of radius $r$ with power $P = m K^{2} r^{2} t .$ Find the centripetal acceleration :

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$K^{2} r^{2} t$

$K^{2} r t^{2}$

$K r^{2} t^{2}$

$K r^{2} t$

answer is C.

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Detailed Solution

$P = m K^{2} r^{2} t = m a_{t} . v$

$a_{t} v = K^{2} r^{2} t, v \frac{d v}{d t} = K^{2} r^{2} t, \frac{v^{2}}{2} = \frac{K^{2} r^{2} t^{2}}{2} o r v = K r t$

$a_{t} = \frac{d v}{d t} = K r s o, a_{c} = \frac{v^{2}}{r} = K^{2} r t^{2}$

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