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.