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?

Open in App

Solution

t any instant tangential acceleration is given by $= \frac{d V}{d t}$

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

so $\frac{d V}{d t} = \frac{V^{2}}{r}$

$\frac{d V}{V^{2}} = \frac{1}{r} d t$

$\int \frac{d V}{V^{2}} = \int \frac{1}{r} d t$

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

apply condition at $t = 0 V = u$

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

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

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

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

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

so clearly after infinite time particle will come to rest

Suggest Corrections

Similar questions

R

t = 0

v_{0}

R

t = 0

v_{0}

S

R

t = 0

v_{0}

R

t = 0

v_{0}

S

\begin{matrix} A body moves in a circular path of radius R with deceleration so that at any moment of time its tangential and normal accelerations are equal in magnitude. At the initial moment t = 0, the velocity of body is v_{0} then the velocity of body will be- \end{matrix}

Join BYJU'S Learning Program