Motion In One Dimension Question 9
Question 9 - 2024 (29 Jan Shift 2)
A particle is moving in a circle of radius $50 \mathrm{~cm}$ 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 $4 \mathrm{~m} / \mathrm{s}$, the time taken to complete the first revolution will be $\frac{1}{\alpha}\left[1-\mathrm{e}^{-2 \pi}\right] \mathrm{s}$, where $\alpha=$
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Answer: (8)
Solution:
$\left|\vec{a}{c}\right|=\left|\vec{a}{t}\right|$
$\frac{v^{2}}{r}=\frac{d v}{d t}$
$\Rightarrow \int_{4}^{v} \frac{d v}{v^{2}}=\int_{0}^{t} \frac{d t}{r}$
$\Rightarrow\left[\frac{-1}{v}\right]_{4}^{v}=\frac{t}{r}$
$\Rightarrow \frac{-1}{v}+\frac{1}{4}=2 t$
$\Rightarrow \mathrm{v}=\frac{4}{1-8 \mathrm{t}}=\frac{\mathrm{ds}}{\mathrm{dt}}$
$4 \int_{0}^{\mathrm{t}} \frac{\mathrm{dt}}{1-8 \mathrm{t}}=\int_{0}^{\mathrm{s}} \mathrm{ds}$
( $\mathrm{r}=0.5 \mathrm{~m}$
$\mathrm{s}=2 \pi \mathrm{r}=\pi)$
$4 \times \frac{[\ln (1-8 \mathrm{t})]_{0}^{\mathrm{t}}}{-8}=\pi$
$\ln (1-8 t)=-2 \pi$
$1-8 \mathrm{t}=\mathrm{e}^{-2 \pi}$
$\mathrm{t}=\left(1-\mathrm{e}^{-2 \pi}\right) \frac{1}{8} \mathrm{~s}$
So, $\alpha=8$
