Question

A particle of mass $m$ describes a circle of radius $r$. The centripetal accelaration of the particle is $\frac{4}{r^2}$.What will be the momentum of the particle?

• A. $\frac{4m}{\sqrt{r}}$
• B. $\frac{2m}{r}$
• C. $\frac{2m}{\sqrt{r}}$
• D. None of these

Answer 1

The correct option is C $\frac{2m}{\sqrt{r}}$

Let the speed of the particle be $v$.

So, its centripetal acceleration , $a_c=\frac{v^2}{r}$

But from the given equation,
$a_c=\frac{4}{r^2}$

$\therefore \frac{v^2}{r}=\frac{4}{r^2}$

$\Rightarrow v^2=\frac{4}{r}$

$\therefore v=\frac{2}{\sqrt{r}}$

Momentum of the particle, $p=mv$
$\therefore p=\frac{2m}{\sqrt{r}}$

Hence, option (b) is the correct answer.