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# A particle is moving along a circle of radius$20/\pi$ with constant tangential acceleration. The velocity of the particle is $80m/s$ at the end of ${2^{nd}}$ revolution after the motion has begun. Then the tangential acceleration is.

Last updated date: 13th Sep 2024
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Hint Tangential acceleration can be defined as the rate of change of velocity along the tangent when the body is moving in a circular path. In a way, the tangential acceleration is nothing but linear acceleration that is viewed from a reference point that is tangential to the circle. Since the tangential acceleration is linear acceleration from a reference frame, the equations of motion hold true to a body executing tangential acceleration.
Formulas used: We will be using the formulas from the equations of motion. We have velocity and radius of the circular path, thus we can use the equation, ${v^2} - {u^2} = 2as$ where $v$ is the final velocity of the body, $u$ is the initial velocity of the body, $a$ is the tangential acceleration of the body, and $s$ is the displacement covered by the body.

We know that when a body executes circular motion along a circular path of a fixed radius, then the motion causes a tangential acceleration on the body. This tangential acceleration shows linear characteristics from a reference frame.
Thus, all the equations of motion corresponding to the linear motion will hold true for tangential acceleration. Now from the problem we can infer that the radius of the circular path is $r = \dfrac{{20}}{\pi }$ and the velocity of the particle at the end of the second revolution is $v = 80m/s$ .
Now we know that the circumference of the circle will be the distance travelled by the particle for each revolution. In the second revolution the distance will be, $s = 2\left( {2\pi r} \right)$
Substituting the known values, we get,
$s = 2\left( {2 \times \pi \times \dfrac{{20}}{\pi }} \right)$
$\Rightarrow s = 80m$
Now that we have the distance travelled, to find the tangential acceleration let us use the equation of motion, ${v^2} - {u^2} = 2as$ .
Let us consider the particle starts from rest, so the initial velocity $u = 0$ .
${\left( {80} \right)^2} - {\left( 0 \right)^2} = 2 \times a \times 80$
$a = \dfrac{{{{\left( {80} \right)}^2}}}{{2 \times 80}}$
Solving for $a$ we get,
$\Rightarrow a = 40m/{s^2}$

Thus, the tangential acceleration of the particle during the second revolution is $a = 40m/{s^2}$ .

Note: The tangential acceleration is closely related to angular acceleration and is given by,
${a_T} = \alpha r$ where $\alpha$ is the angular acceleration experienced by the body.