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

Hint: We know that tangential acceleration is an instantaneous acceleration. When particle performs circular motion a centripetal force acts along with centripetal acceleration, which is given by $\dfrac{m{{v}^{2}}}{r}$ and tangential acceleration is given by $\dfrac{dv}{dt}$. The condition given in question equates both the formula since the particle is under circular motion. Integrate equated equations find value of constant by taking the initial speed (u) condition. This way you will get the answer.

Complete answer:
Consider a particle revolving on the circumference of a circle. At any instant of time the tangential acceleration of particle performing circular motion is given by $\dfrac{dv}{dt}$ since it is moving in circle hence centripetal acceleration must have been applied centripetal acceleration is given by,
$a=\dfrac{{{v}^{2}}}{r}$
When particle is under circular motion then according to given condition that tangential and normal acceleration is equal i.e.
$\dfrac{dv}{dt}=\dfrac{{{v}^{2}}}{r}$
Use variable separable method, we get
$\dfrac{dv}{{{v}^{2}}}=\dfrac{1}{r}dt......(1)$
Integrate it we get
$\begin{align}
  & \int{\dfrac{dv}{{{v}^{2}}}=\int{\dfrac{1}{r}dt}} \\
 & -\dfrac{1}{v}=\dfrac{t}{r}+c......(2) \\
\end{align}$
Apply condition that when t=0 particle is at rest i.e. v=u
Therefore equation (2) implies,
$\begin{align}
  & -\dfrac{1}{u}=0+c \\
 & c=-\dfrac{1}{u} \\
\end{align}$
Put this value in equation (2), we get
$\begin{align}
  & -\dfrac{1}{v}=\dfrac{t}{r}-\dfrac{1}{u} \\
 & \dfrac{1}{v}=\dfrac{1}{u}+\dfrac{t}{r} \\
 & \dfrac{1}{v}=\dfrac{r+tu}{ur} \\
 & v=\dfrac{ur}{r+tu} \\
\end{align}$
Now we know that we want particles to come at rest at a certain time. If v=0 then t must be infinity
Because, if t=∞
$v=\dfrac{ur}{r+\infty }=\dfrac{ur}{\infty }=0$
i.e. velocity of particles is zero.
So clearly we can see that after infinity time, particles performing circular motion come to rest.

Additional information:
Instantaneous acceleration is defined as acceleration of a particle at a particular instant of time and it is denoted as$\overrightarrow{a}$.basically it is a limit of average acceleration as the time interval of particle will become infinitesimal.
$\Rightarrow \overrightarrow{a}=\underset{\Delta t\to 0}{\mathop{\lim }}\,\dfrac{\Delta \overrightarrow{v}}{\Delta t}=\dfrac{d\overrightarrow{v}}{dt}$
The particle is said to be accelerated if the velocity of the particle increases with time. Generally when term acceleration is used it is basically an instantaneous acceleration. Acceleration is zero, when velocity is constant.

Note:
When acceleration of a particle is uniform then instantaneous acceleration is equal to average acceleration. If velocity of particle decreases with time, then the particle is said to be decelerated or also called as retarded. This kind of acceleration is called negative acceleration.