Question

A particle moves along a circle of radius $ r $ with constant tangential acceleration. If the velocity of the particle is $ v $ at the end of second revolution, after the revolution has started, then the tangential acceleration is:
A. $ \dfrac{{{v}^{2}}}{8\pi r} $
B. $ \dfrac{{{v}^{2}}}{6\pi r} $
C. $ \dfrac{{{v}^{2}}}{4\pi r} $
D. $ \dfrac{{{v}^{2}}}{2\pi r} $

Answer 1

Hint: First we have to write the equation of velocity with respect to distance and acceleration. Now as in the question the particle moves in a circular motion, so the acceleration due to gravity will be termed as angular acceleration, the distance now will be termed as the angular distance. After figuring out all the properties of the particle, like velocity, replace those values with the values of the velocity formula with respect to distance and time.

Complete step-by-step answer:
In the equilibrium, final velocity(v) is,
 $ {{v}^{2}}={{u}^{2}}+2as, $ ……(i)
v is the final velocity,
u is the initial velocity,
a is the acceleration,
s is the distance traveled.
Now, we have a= tangential acceleration,
And for the circular motion,
 $ s=r\theta $ and $ u=0 $ ,
Hence when we replace these values in equation (i) we have,
 $ {{v}^{2}}=2ar\theta $ ……..(ii)
But we know that,
 $ \theta =4\pi $ in two revolutions.
Replacing the value of $ \theta $ in equation (ii),we have
 $ {{v}^{2}}=2ar(4\pi )=8\pi ar $
From here we can find the value of tangential acceleration,
 $ \therefore a=\dfrac{{{v}^{2}}}{8\pi r} $ .
Therefore, option A is the correct option.

Additional Information:
The concept of tangential acceleration is used to measure the change in the tangential velocity at a point with a specific radius with the change in time.
Tangential acceleration is defined as the rate of change of tangential velocity of the matter in the circular path.
Equilibrium is the condition of a system when neither it is in a state of motion nor its internal energy state tends to change with time.
The radius of a circle or sphere is any of the line segments from its center point to its perimeter.

Note: Remember that $ \theta =4\pi $ is already given so when swapping the values for theta be a bit careful. Try to remember the relations and the formula for the velocity comprising distance and acceleration. Always remember that for circular motion Initial velocity is always equal to zero.