Question

A stone is revolving in a circular path of radius $\frac{4}{\pi }m$, with speed $v=2t$ (in $m/s$). Initially if the stone is at rest, the magnitude and direction of total acceleration of the stone after it covers half of the circle are

• A. $2\pi \sqrt{1+{\left(\frac{1}{2\pi }\right)}^2};{81}^{\circ }$ with tangental acceleration
• B. $2\pi \sqrt{1+{\left(\frac{1}{2\pi }\right)}^2};{81}^{\circ }$ with centripetal acceleration
• C. $4\pi \sqrt{1+{\left(\frac{1}{2\pi }\right)}^2};{81}^{\circ }$ with tangental acceleration
• D. $4\pi \sqrt{1+{\left(\frac{1}{2\pi }\right)}^2};{81}^{\circ }$ with centripetal acceleration

Answer 1

The correct option is C $4\pi \sqrt{1+{\left(\frac{1}{2\pi }\right)}^2};{81}^{\circ }$ with tangental acceleration

Given, $v=2t$
$\Rightarrow a_t=\frac{dv}{dt}=\frac{d}{dt}\left(2t\right)=2m/s^2$
$\because a_t$ is uniform, we use
$s=ut+\frac{1}{2}{a}_t{t}^2$ for half circle distance
$\Rightarrow \pi r=\frac{1}{2}{a}_t{t}^2$
$\Rightarrow t^2=\frac{2\pi \times \frac{4}{\pi }}{2}=4$
$\Rightarrow t=2s$
$a_c=\frac{v^2}{r}=\frac{(2t{)}^2}{\frac{4}{\pi }}$ (at $t=2s$)
$=\frac{\pi }{4}\times (4{)}^2=4\pi m/s^2$


$\tan \theta =\frac{a_c}{a_t}=\frac{4\pi }{2}=2\pi$
$\Rightarrow \theta ={\tan }^{-1}2\pi ={81}^{\circ }$ with tangental acceleration

Total acceleration
$\overrightarrow{a}=\overrightarrow{a_c}+\overrightarrow{a_t}$
$\Rightarrow |\overline{a}|=\sqrt{(a_c{)}^2+(a_t{)}^2}=\sqrt{(4\pi {)}^2+(2{)}^2}=\sqrt{16{\pi }^2+4}=4\pi \sqrt{1+\frac{1}{(2\pi {)}^2}}$