Question

A man $A$ at radius $2\text{m}$ and other man $B$ at radius $3\text{m}$ perform uniform circular motion on the floor of a merry-go-round as the ride turns. They are on the same radial line. At one instant, the acceleration of the man $A$ is $(2\hat{i}+4\hat{j}){\text{m/s}}^2$. At that instant and in unit-vector notation, what is the acceleration of man $B$?

• A. $(6\hat{i}+3\hat{j}){\text{m/s}}^2$
• B. $(6\hat{i}-3\hat{j}){\text{m/s}}^2$
• C. $(3\hat{i}+6\hat{j}){\text{m/s}}^2$
• D. $(3\hat{i}-6\hat{j}){\text{m/s}}^2$

Answer 1

The correct option is C $(3\hat{i}+6\hat{j}){\text{m/s}}^2$

Given:
$r_A=2\text{m};r_B=3\text{m}$

As the man $A$ and $B$ are in uniform circular motion, thus, there is no tangential acceleration.

But centripetal or radial acceleration will be,

$a_c={\omega }^{2}r$

For the same rotating system,

$a_c\propto r$

So, $\frac{(a_c{)}_A}{(a_c{)}_B}=\frac{r_A}{r_B}$

$\Rightarrow (a_c{)}_B=\frac{r_B}{r_A}\times (a_c{)}_A$

$\Rightarrow (a_c{)}_B=\frac{3}{2}\times (a_c{)}_A.....\left(1\right)$

Here, $\overrightarrow{(a_c{)}_A}=(2\hat{i}+4\hat{j}){\text{m/s}}^2$

$\therefore (a_c{)}_A=\sqrt{2^2+4^2}=\sqrt{20}........(2)$

Let, $\overrightarrow{(a_c{)}_B}=(x\hat{i}+y\hat{j}){\text{m/s}}^2$

$\therefore (a_c{)}_B=\sqrt{x^2+y^2}........(3)$

Substituting $\left(2\right)$ and $\left(3\right)$ in $\left(1\right)$

$\sqrt{x^2+y^2}=\frac{3}{2}\left(\sqrt{20}\right)$

$\Rightarrow x^2+y^2=\frac{9}{4}\times 20$

$\Rightarrow x^2+y^2=45.......\left(4\right)$

Slope of $\overrightarrow{(a_c{)}_A}$ with $+x-$ axis,

$\tan \theta =\frac{4}{2}=2$

As the position of $B$ is not mentioned, $B$ can be on the same side of $A$ or on the opposite side.

So, direction of $\overrightarrow{(a_c{)}_B}$ with $+x-$ axis will be either $\theta$ or $\theta +\pi$.

But $\tan (\theta +\pi )=\tan \left(\theta \right)$

$\Rightarrow \tan \theta =\frac{y}{x}=2$

$\Rightarrow y=2x........\left(5\right)$

Substituting eq. $\left(5\right)$ in $\left(4\right)$ we get,

$x^2+4x^2=45$

$\Rightarrow 5x^2=45\Rightarrow x=\pm 3$

From $\left(5\right)$, $y=\pm 6$

$\therefore {\overrightarrow{\left(a_c\right)}}_B=(3\hat{i}+6\hat{j}){\text{m/s}}^2$ or $(-3\hat{i}-6\hat{j}){\text{m/s}}^2$

Hence, option (c) is the correct answer.

Why this question ? It tests your skill to apply knowledge of vectors, trigonometry and circular motion in a blend form. Such questions are generally asked in JEE.