Question

A particle moves with deceleration along the circle of radius R so that at any moment of the time its tangential and normal acceleration are equal in magnitude at the initial moment. Find time at which particle will come to u rest?

Answer 1

t any instant tangential acceleration is given by $=\frac{dV}{dt}$

but this is equal to centripetal acceleration which is given as $=\frac{m{V}^2}{r}$

so $\frac{dV}{dt}=\frac{V^2}{r}$

$\frac{dV}{V^2}=\frac{1}{r}dt$

$\int \frac{dV}{V^2}=\int \frac{1}{r}dt$

$\frac{-1}{V}=\frac{1t}{r}+C$

apply condition at $t=0V=u$

so $C=\frac{-1}{c}$

equation become $\frac{-1}{V}=\frac{1t}{r}+\frac{-1}{u}$

$\frac{1}{V}=\frac{1}{u}+\frac{t}{r}$

$\frac{1}{V}=\frac{r+tu}{ur}$

$V=\frac{ur}{r+tu}$

so clearly after infinite time particle will come to rest