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

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, $a_{\text{tangential }}=\frac{dv}{dt}=\frac{d}{dt}=[K\sqrt{s}]$
or, $a_{\text{tangential }}=K\times \frac{1}{2\sqrt{s}}\times \frac{ds}{dt}$
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
$\tan \theta =\frac{K^{2}S}{\frac{R}{\frac{K^2}{2}}}$
$\Rightarrow \tan \theta =\frac{2s}{R}$