Question

A particle is moving on a circular path of radius $1m$ with $2m/s$. If speed starts increasing at a rate of $2m/s^2$, then the acceleration of the particle is

• A. $2m/s^2$
• B. $4m/s^2$
• C. $2\sqrt{5}m/s^2$
• D. $8m/s^2$

Answer 1

The correct option is C $2\sqrt{5}m/s^2$

Radius, $r=1$ $m$

Velocity, $v=2$ $m/s=r\dot{\theta }$

Since, the particle moves in a circle:

$\dot{r}=\ddot{r}=0$

Acceleration is given by:

$\overrightarrow{a}=(\ddot{r}-r{\dot{\theta }}^2)\hat{r}+(r\ddot{\theta }+2\dot{r}\dot{\theta })\hat{\theta }$

Also $\ddot{\theta }=2$ $m/s^2$

$\therefore \overrightarrow{a}=[-1\times (2{)}^2]\hat{r}+[1\times 2]\hat{\theta }$

$⟹\overrightarrow{a}=-4\hat{r}+2\hat{\theta }$

$|\overrightarrow{a}|=\sqrt{(4{)}^2+(2{)}^2}$

$⟹|\overrightarrow{a}|=\sqrt{20}$

$⟹|\overrightarrow{a}|=2\sqrt{5}$ $m/s^2$