Question

A point object moves along an arc of a circle of radius $R .$ Its velocity depends upon the distance covered $s$ as $v=K \sqrt{s}$, where $K$ is a constant. If $\theta$ is the angle between the total acceleration and tangential acceleration, then

• A. $\tan \theta=\sqrt{\frac{s}{R}}$
• B. $\tan \theta=\sqrt{\frac{s}{2 R}}$
• C. $\tan \theta=\frac{s}{2 R}$
• D. $\tan \theta=\frac{2 s}{R}$

Answer 1

Answer: (D)

Given, velocity depends upon distance covered $s$ as $V=k \sqrt{s} .$
$\because \tan \theta=\frac{a_{\text {radial }}}{a_{\text {tangential }}}=\frac{V^{2} / R}{a_{\text {tangential }}}$
$\Rightarrow \tan \theta=\frac{1}{a_{\text {tangential }}}\left[\frac{1}{R} \times K^{2} S\right]$
$\Rightarrow \tan \theta=\frac{1}{a_{\text {tangential }}}\left[\frac{K^{2} S}{R}\right]\,\,\,...(i)$
Now, $a_{\text {tangential }} =\frac{d v}{d t}=\frac{d}{d t}=[K \sqrt{s}] $
or, $ a_{\text {tangential }} =K \times \frac{1}{2 \sqrt{s}} \times \frac{d s}{d t} $
or, $ a_{\text {tangential }} =\frac{K}{2 \sqrt{s}} \times V $
$=\frac{K}{2 \sqrt{s}} \times K \sqrt{S}=\frac{K^{2}}{2}\,\,\,...(ii)$
From Eqs. (i) and (ii), we get
$\tan \theta=\frac{K^{2} S }{\frac{ R }{\frac{K^{2}}{2}}} $
$\Rightarrow \tan \theta=\frac{2\, s }{R}$