Question

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

• A. $r^2\omega$
• B. $r^2\theta$
• C. $\frac{r^2\omega }{2}$
• D. $\frac{r{\omega }^2}{2}$

Answer 1

Answer: (C)

$\frac{r^2\omega }{2}$

Let the angular displacement of the particle in time $\Delta t$ be $\Delta \theta$.

$\therefore$ $\Delta \theta =w\Delta t$

The area covered in $\Delta t$ time $\Delta A=\frac{\Delta \theta }{2\pi }\left(\pi r^2\right)=\frac{w\Delta t}{\left(2\pi \right)}\left(\pi r^2\right)$

$\therefore$ We get $\Delta A=\frac{w{r}^2}{2}\Delta t$

$\Rightarrow$ Areal velocity of particle $v_{areal}=\frac{\Delta A}{\Delta t}=\frac{\frac{w{r}^2}{2}\Delta t}{\Delta t}$$=\frac{w{r}^2}{2}$