A particle is moving in a circle of radius 50 cm in such a way that at any instant the normal and ta

Question

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 it's 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 $\frac{1}{\alpha}\left[1-e^{-2 \pi}ight] \mathrm{s}$, where $\alpha=$ $\qquad$ .\

Professor PC Thomas Classes · Answer

(8) : Given,\ $r=50 \mathrm{~cm}, v=4 \mathrm{~m} / \mathrm{s}$ at $t=0$\ $\left|\vec{a}_{c}ight|=\left|\vec{a}_{r}ight| ; \frac{v^{2}}{r}=\frac{d v}{d t}$;\ $\int_{4}^{v} \frac{d v}{v^{2}}=\int_{0}^{t} \frac{d t}{r} ;\left[-v^{-1}ight]_{4}^{v}=\frac{1}{r}[t]_{0}^{t}$\ $-\frac{1}{v}+\frac{1}{4}=\frac{t}{50 / 100} \Rightarrow \frac{1}{4}-\frac{1}{v}=2 t ;$\ $v=\frac{4}{1-8 t}=\frac{d s}{d t}$\ $4 \int_{0}^{t} \frac{d t}{1-8 t}=\int_{0}^{s} d s \quad 	herefore \quad r=0.5 \mathrm{~m}, s=2 \pi r=\pi$\ $\left.4 \frac{\ln (1-8 t)}{-8}ight|_{0} ^{t}=\pi \ln (1-8 t)=-2 \pi ; 1-8 t=e^{-2 \pi}$ so, $\alpha=8$\