Motion In One Dimension Question 9
Question 9 - 2024 (29 Jan Shift 2)
A particle is moving in a circle of radius $50 cm$ in such a way that at any instant the normal and tangential components of its acceleration are equal. If its speed at $t=0$ is $4 m / s$, the time taken to complete the first revolution will be $\frac{1}{\alpha}\left[1-e^{-2 \pi}\right] s$, where $\alpha=$
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Answer: (8)
Solution:
$\left|\vec{a} _c\right|=\left|\vec{a} _t\right|$
$\frac{v^{2}}{r}=\frac{d v}{d t}$
$\Rightarrow \int _4^{v} \frac{d v}{v^{2}}=\int _0^{t} \frac{d t}{r}$
$\Rightarrow\left[\frac{-1}{v}\right] _4^{v}=\frac{t}{r}$
$\Rightarrow \frac{-1}{v}+\frac{1}{4}=2 t$
$\Rightarrow v=\frac{4}{1-8 t}=\frac{ds}{dt}$
$4 \int _0^{t} \frac{dt}{1-8 t}=\int _0^{s} ds$
( $r=0.5 m$
$s=2 \pi r=\pi)$
$4 \times \frac{[\ln (1-8 t)] _0^{t}}{-8}=\pi$
$\ln (1-8 t)=-2 \pi$
$1-8 t=e^{-2 \pi}$
$t=\left(1-e^{-2 \pi}\right) \frac{1}{8} s$
So, $\alpha=8$
