Question

In circular motion if $\overline{v}$ is velocity vector, $\overline{a}$ is acceleration vector, $\overline{r}$ is instantaneous position vector, $\overline{p}$ is momentum vector and $\overline{\omega }$ is angular velocity of particle, then:

• A. $\overline{v}$, $\overline{\omega }$, $\overline{r}$ are mutually perpendicular
• B. $\overline{p}$, $\overline{v}$, $\overline{\omega }$ are mutually perpendicular
• C. $\overline{r}\times \overline{v}=0$ and $\overline{r}\times \overline{\omega }=0$
• D. $\overline{r}.\overline{v}=0$ and $\overline{r}.\overline{\omega }=0$

Answer 1

Answer: (A), (D)

The correct options are
A $\overline{v}$, $\overline{\omega }$, $\overline{r}$ are mutually perpendicular
D $\overline{r}.\overline{v}=0$ and $\overline{r}.\overline{\omega }=0$

$p$ in the direction of velocity
$\therefore \overrightarrow{v},\overrightarrow{w},\overrightarrow{r}$ are mutually tan
$\overrightarrow{r}.\overrightarrow{v}=0;\overrightarrow{r}.\overrightarrow{w}=0$
$\overrightarrow{v}\times \overrightarrow{p}=0$