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3-2.Motion in Plane
hard
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 $\alpha=$_______________.
A$8$
B$5$
C$98$
D$45$
(JEE MAIN-2024)
Solution
$ \left|\overrightarrow{\mathrm{a}}_{\mathrm{c}}\right|=\left|\overrightarrow{\mathrm{a}}_{\mathrm{t}}\right| $
$ \frac{\mathrm{v}^2}{\mathrm{r}}=\frac{\mathrm{dv}}{\mathrm{dt}} $
$ \Rightarrow \int_4^{\mathrm{v}} \frac{\mathrm{dv}}{\mathrm{v}^2}=\int_0^{\mathrm{t}} \frac{\mathrm{dt}}{\mathrm{r}} $
$ \Rightarrow\left[\frac{-1}{\mathrm{v}}\right]_4^{\mathrm{v}}=\frac{\mathrm{t}}{\mathrm{r}} $
$ \Rightarrow \frac{-1}{\mathrm{v}}+\frac{1}{4}=2 \mathrm{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_0^5 \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 \mathrm{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, $\alpha=8$
$ \frac{\mathrm{v}^2}{\mathrm{r}}=\frac{\mathrm{dv}}{\mathrm{dt}} $
$ \Rightarrow \int_4^{\mathrm{v}} \frac{\mathrm{dv}}{\mathrm{v}^2}=\int_0^{\mathrm{t}} \frac{\mathrm{dt}}{\mathrm{r}} $
$ \Rightarrow\left[\frac{-1}{\mathrm{v}}\right]_4^{\mathrm{v}}=\frac{\mathrm{t}}{\mathrm{r}} $
$ \Rightarrow \frac{-1}{\mathrm{v}}+\frac{1}{4}=2 \mathrm{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_0^5 \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 \mathrm{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, $\alpha=8$
Standard 11
Physics
