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

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

(A) r² ω

(B) r² θ

(C) r²ω/2

(D) rω²/2

Solution :

Areal velocity = dA/dt  ;

where A = area of the sector
=r²θ/2

$\large \frac{dA}{dt} = \frac{d}{dt}(\frac{r^2 \theta}{2}) $

$\large = \frac{1}{2}r^2  \frac{d\theta}{dt} = \frac{r^2 \omega}{2}$