Question

A particle moving at speed $u$ on a circular path of radius $R$ begins to speed up at a constant rate. The magnitude of change in its velocity vector is $2u$, when it completes one fourth revolution. The magnitude of its radial acceleration after completing one-fourth revolution will be

• A. $\frac{u^2}{4R}$
• B. $\frac{3u^2}{R}$
• C. $\frac{2}{3}\frac{u^2}{R}$
• D. $\frac{5u^2}{4R}$

Answer 1

The correct option is B $\frac{3u^2}{R}$


Let that particle was initially at point $\text{A}$ with speed $u$ and now it is at point $\text{B}$ (as shown in figure) with speed $v$.

$\overrightarrow{\Delta v}=\overrightarrow{v_f}-\overrightarrow{v_i}$

$\Rightarrow \overrightarrow{\Delta v}=-v\hat{j}-u\hat{i}$

$\Rightarrow |\overrightarrow{\Delta v}|=\sqrt{v^2+u^2}$

Given that magnitude of change in velocity vector during one-quarter revolution is $2u$.

$\therefore |\overrightarrow{\Delta v}|=2u$

putting the value of $|\overrightarrow{\Delta v}|$,

$\Rightarrow \sqrt{(v^2+u^2)}=2u$

$\Rightarrow v^2+u^2=4u^2$

$\therefore v^2=3u^2$

Hence, radial acceleration at point $\text{B}$ is given by

$a_r=\frac{v^2}{R}$

$\therefore a_r=\frac{3u^2}{R}$

Hence, option $\left(b\right)$ is correct.