Question

In U. C. M (Uniform Circular Motion), prove the relation ${\displaystyle \overrightarrow{v}=\overrightarrow{w}\times \overrightarrow{r}}$, 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 centre O as shown.

Let

${\displaystyle \overrightarrow{\omega }}$= angular velocity of the particle

${\displaystyle \overrightarrow{v}}$= linear velocity of the particle

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

In the vector form, the linear dispalcement is

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

Dividing both sides by ${\displaystyle \delta t}$ we get

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

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

${\displaystyle \therefore \frac{\overrightarrow{dS}}{dt}=\frac{\overrightarrow{d\theta }}{\delta t}\times \overrightarrow{r}}$

but

${\displaystyle \frac{\overrightarrow{dS}}{dt}=\overrightarrow{v}}$ = Linear velocity

${\displaystyle \frac{\overrightarrow{d\theta }}{\delta t}=\overrightarrow{\omega }}$ = angular velocity

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