Question

A particle is moving along a circle of radius $\dfrac{20}{\pi }$ with constant tangential acceleration. The velocity of the particle is $80m{{s}^{-1}}$ at the end of 2nd revolution after motion has begun. Then the tangential acceleration is

Answer 1

The particle is moving along a circular orbit with constant tangential acceleration. So we can apply the equation of motion for circular motion to the object to find out acceleration. Taking initial angular velocity to zero and final angular velocity $\omega$ we can calculate the angular acceleration $\alpha$. From angular acceleration and radius of orbit we can calculate tangential acceleration.

Formula used:
For a particle having final angular velocity $\omega$ and initial angular velocity ${{\omega }_{0}}$ with angular acceleration $\alpha$ and angular displacement $\theta$ the equation of motion is given by
${{\omega }^{2}}={{\omega }_{0}}^{2}+2\alpha \theta$
If the radius of the orbit is $r$and angular velocity is $\omega$ then tangential velocity is $v=\omega r\text{ or }\omega =\dfrac{v}{r}$
If $\alpha$ is the angular acceleration and tangential acceleration is $a=\alpha r\text{ or }\alpha =\dfrac{a}{r}$

Complete answer:
According to the question the particle is moving with a circular path with radius $\dfrac{20}{\pi }$ and constant tangential acceleration.
Let the radius be $r=\dfrac{20}{\pi }$ and the tangential acceleration is $a$.and angular acceleration is $\alpha$
Let the initial angular velocity is ${{\omega }_{0}}$.
Let after two complete revolutions the angular velocity of the particle is $\omega$. So after two revolutions the angular displacement of the particle is $\theta =4\pi$.
Applying the equation of motion ${{\omega }^{2}}={{\omega }_{0}}^{2}+2\alpha \theta$
Given after two revolution the velocity is $v=80m{{s}^{-1}}$
But $\omega =\dfrac{v}{r}$ and ${{\omega }_{0}}=\dfrac{{{v}_{0}}}{r}$
Also $\alpha =\dfrac{a}{r}$

Putting these values in above we get

& {{\left( \dfrac{v}{r} \right)}^{2}}={{\left( \dfrac{{{v}_{0}}}{r} \right)}^{2}}+2\dfrac{a}{r}\theta \\\ & \Rightarrow \dfrac{80\times 80}{{{r}^{2}}}=0+2\dfrac{a}{r}\times 4\pi \left( \because v=80m{{s}^{-1}}\text{ and }{{v}_{0}}=0\text{ and }\theta =4\pi \right) \\\ & \Rightarrow \dfrac{6400}{r}=8\pi a \\\ & \Rightarrow a=\dfrac{6400}{8\pi r}=\dfrac{6400}{8\pi \left( \dfrac{20}{\pi } \right)}=40m{{s}^{-2}} \\\ \end{aligned}$$ So we get the tangential acceleration after two revolution will be $$40m{{s}^{-2}}$$ **Note:** Just like the equation of motion for linear motion, the equation of motion for circular motion only valid for constant tangential acceleration. The equations are not valid for accelerated motion. The force due to which an object moves in a circular path is called centripetal force and it is always directed towards the centre.