A particle of mass m is revolving in a horizontal circle of radius r with constant angular speed $ω$ . The areal velocity of the particle is:

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$\frac{r ω^{2}}{2}$

$\frac{r^{2} ω}{2}$

$r^{2} θ$

$r^{2} ω$

answer is C.

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Detailed Solution

Sol : Areal velocity = dA/dt
Where, A=area of the arc $= \frac{r^{2} θ}{2}$
$∴ \frac{d A}{d t} = \frac{d}{d t} (\frac{r^{2} θ}{2}) = \frac{r^{2}}{2} \frac{d θ}{d t} = \frac{r^{2} ω}{2}$

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