Question

A particle moving is a circle of radius R in such a way that at any instant the normal and tangential components of its acceleration are equal. If its speed at t $= 0$ is ${v_0}$ , the time taken to complete the first revolution is $-$

Answer 1

Hint : We have to know about radial, tangential and linear acceleration. When any particle moves in a circular way with an acceleration that is called radial or circular acceleration. This acceleration lies on the radius of the circle. Tangential acceleration is meant to measure the acceleration of a specific point with a constant or specific radius with the change in time.

Complete Step By Step Answer:
This is a circular motion, so we are assuming that the radial acceleration is ${a_r}$ which is equal to $\dfrac{{{v^2}}}{R}$
Now, we can assume that the tangential acceleration is ${a_t}$ which is equal to $\dfrac{{dv}}{{dt}}$
According to the question these two accelerations are equal. So, $\dfrac{{dv}}{{dt}} = \dfrac{{{v^2}}}{R}$
Or, $\left( {\dfrac{R}{{{v^2}}}} \right)dv = dt$
Now, integrating both sides we will get,
$\int {\dfrac{R}{{{v^2}}}} dv = \int {dt}$
$\dfrac{R}{v} = t + c$
Now, we have to calculate the value of the integrating constant C.
Putting the values of v and t we will get $C = - \dfrac{R}{{{v_0}}}$
Now, the relation between v and t is, $t = R\left[ {\dfrac{{v - {v_0}}}{{{v_0}.v}}} \right]$
Now, v is equal to $\dfrac{{ds}}{{dt}}$
Here, s is the length of the arc covered by the particle
$\therefore t = \dfrac{R}{{{v_0}}} - R.\dfrac{{dt}}{{ds}}$
$ds(\dfrac{R}{{{v_0}}} - t) = Rdt$
$\dfrac{{ds}}{R} = \dfrac{{dt}}{{\left( {\dfrac{R}{{{v_0}}} - t} \right)}}$
Again, we are going to integrate
$\int {\dfrac{{ds}}{R}} = \int {\dfrac{{dt}}{{\left( {\dfrac{R}{{{v_0}}} - t} \right)}}}$
$\dfrac{s}{R} = - \ln \left( {\dfrac{R}{{{v_0} - t}}} \right) + C$
Putting the values at t $= 0\& S = 0$
C $= \dfrac{{\ln R}}{{{v_0}}}$
Therefore,
$\dfrac{s}{R} = - \ln \left( {\dfrac{R}{{{v_0} - t}}} \right) + \dfrac{{\ln R}}{{{v_o}}} = \ln \left( {\dfrac{R}{{R - {v_0}t}}} \right)$
Now, finally for complete revolution, s is equal to $2\pi R$ and $t = T$
$\dfrac{{2\pi R}}{R} = \ln \left( {\dfrac{R}{{R - {v_0}T}}} \right)$
$T = \dfrac{R}{{{v_0}}}(1 - {e^{ - 2\pi }})$
This is the final answer.

Note :
we can get confused between linear acceleration, radial acceleration and tangential acceleration. But there is a huge difference between them. we have to clearly know what those acceleration means. Not only about the accelerations but also about the relation between force, time and acceleration.