A particle moves along a circle of radius $$\left( {\frac{{20}}{\pi }} \right)m$$   with constant tangential acceleration. If the velocity of the particle is $$80\,m/s$$  at the end of the second revolution after motion has begin, the tangential acceleration is

Solution :
The tangential acceleration is given by
$${a_T} = r\alpha \,......\left( {\text{i}} \right)$$
From 2nd equation of motion for rotational motion,
$${\omega ^2} = \omega _0^2 + 2\alpha \theta $$
Here, initial angular velocity, $${\omega _0} = 0,$$
Final angular velocity,
$$\eqalign{ & \omega = \frac{v}{I} = \frac{{80}}{{\frac{{20}}{\pi }}} \cr & = 4\pi \,rad/s \cr & \theta = 2 \times 2\pi \,rad \cr} $$
So, angular acceleration
$$\alpha = \frac{{{\omega ^2}}}{{2\theta }} = \frac{{{{\left( {4\pi } \right)}^2}}}{{2 \times \left( {2 \times 2\pi } \right)}} = \frac{{16{\pi ^2}}}{{8\pi }} = 2\pi $$
Hence, from Eq. (i), we have
$${a_T} = r\alpha = \frac{{20}}{\pi } \times 2\pi = 40\;m/{s^2}$$
Alternative
Initial velocity, $$u = 0$$
Final velocity, $$v = 80\,m/s$$
Radius of circle $$r = \left( {\frac{{20}}{{11}}} \right)m$$
Distance travelled, $$S = 2 \times \left( {2\pi r} \right) = 2 \times \left( {2\pi \times \frac{{20}}{\pi }} \right)$$
$$ = 80\,m$$
Now, by applying third equation of motion,
$$\eqalign{ & {v^2} = {u^2} + 2as \cr & {\left( {80} \right)^2} = 0 + 2 \times {a_f} \times 80 \cr & {a_f} = 40\,m/{s^2} \cr} $$