Question 162: Work, Energy and Power & Circular Motion

Question

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 it's acceleration are equal. If its speed at $$\mathrm{t}=0$$ is $$4 \mathrm{~m} / \mathrm{s}$$, the time taken to complete the first revolution will be $$\frac{1}{\alpha}\left[1-e^{-2 \pi}\right] \mathrm{s}$$, where $$\alpha=$$ _________.

Accepted Answer

$$\begin{aligned}
& \left|\vec{a}_cight|=\left|\vec{a}_tight| \
& \frac{v^2}{r}=\frac{d v}{d t} \
& \Rightarrow \int_\limits4^v \frac{d v}{v^2}=\int_\limits0^t \frac{d t}{r} \
& \Rightarrow\left[\frac{-1}{v}ight]_4^v=\frac{t}{r} \
& \Rightarrow \frac{-1}{v}+\frac{1}{4}=2 t
\end{aligned}$$
$$\begin{aligned}
& \Rightarrow \mathrm{v}=\frac{4}{1-8 \mathrm{t}}=\frac{\mathrm{ds}}{\mathrm{dt}} \
& 4 \int_0^{\mathrm{t}} \frac{\mathrm{dt}}{1-8 \mathrm{t}}=\int_0^{\mathrm{s}} \mathrm{ds} \
& (\mathrm{r}=0.5 \mathrm{~m} \
& \mathrm{s}=2 \pi \mathrm{r}=\pi) \
& 4 	imes \frac{[\ell \mathrm{n}(1-8 \mathrm{t})]_0^{\mathrm{t}}}{-8}=\pi \
& \ell \mathrm{n}(1-8 \mathrm{t})=-2 \pi \
& 1-8 \mathrm{t}=\mathrm{e}^{-2 \pi} \
& \mathrm{t}=\left(1-\mathrm{e}^{-2 \pi}ight) \frac{1}{8} \mathrm{~s}
\end{aligned}$$
So, $$\alpha=8$$

Work, Energy and Power & Circular Motion

Integer / Numeric Answer Type