Question

In U. C. M. (Uniform Circular Motion), prove the relation $\overrightarrow{v}=\overrightarrow{\omega }\times \overrightarrow{r}$, where symbols have their usual meanings.

Answer 1

Analytical method: Consider a particle revolving in the anticlockwise sense along the circumference of a circle of radius $r$ with center O as shown figure.

Let $\overrightarrow{v}=$ Linear velocity of the particle

$\overrightarrow{\omega }=$ Angular velocity of the particle

$\overrightarrow{r}=$ radius vector of the particle

In the vector from, the linear displacement is

$\overrightarrow{\delta }=\overrightarrow{\delta \theta }\times \overrightarrow{r}$

Dividing both side by $\delta t$ we get

$\frac{\overrightarrow{\delta s}}{\overrightarrow{\delta t}}=\frac{\overrightarrow{\delta \theta }}{\overrightarrow{\delta t}}\times \overrightarrow{r}$

${lim}_{\delta t\to 0}\frac{\overrightarrow{\delta s}}{\overrightarrow{\delta t}}={lim}_{\delta t\to 0}\frac{\overrightarrow{\delta \theta }}{\overrightarrow{\delta t}}\times \overrightarrow{r}$

$\frac{\overrightarrow{ds}}{\overrightarrow{\delta t}}=\frac{\overrightarrow{\delta \theta }}{\overrightarrow{\delta t}}\times \overrightarrow{r}$

But $\frac{\overrightarrow{ds}}{\overrightarrow{\delta t}}=\overrightarrow{r}$ Linear velocity and $\frac{\overrightarrow{\delta \theta }}{\overrightarrow{\delta t}}=\omega$ Angular velocity

$\therefore$ $\overrightarrow{v}=\overrightarrow{\omega }\times \overrightarrow{r}$