A particle of mass $$10\,g$$  moves along a circle of radius $$6.4\,cm$$  with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to $$8 \times {10^{ - 4}}J$$   by the end of the second revolution after the beginning of the motion ?

Solution :
Given: Mass of particle, $$M = 10\,g = \frac{{10}}{{1000}}kg$$
radius of circle $$R = 6.4\,cm$$
Kinetic energy $$E$$ of particle $$ = 8 \times {10^{ - 4}}J$$
acceleration $${a_t} = ?$$
$$\eqalign{ & \frac{1}{2}m{v^2} = E \Rightarrow \frac{1}{2}\left( {\frac{{10}}{{1000}}} \right){v^2} = 8 \times {10^{ - 4}} \cr & \Rightarrow {v^2} = 16 \times {10^{ - 2}} \Rightarrow v = 4 \times {10^{ - 1}} = 0.4\,m/s \cr} $$
Now, using
$$\eqalign{ & {v^2} = {u^2} + 2{a_t}s\,\,\left( {s = 4\pi R} \right) \cr & {\left( {0.4} \right)^2} = {0^2} + 2{a_t}\left( {4 \times \frac{{22}}{7} \times \frac{{6.4}}{{100}}} \right) \cr & \Rightarrow {a_t} = {\left( {0.4} \right)^2} \times \frac{{7 \times 100}}{{8 \times 22 \times 6.4}} = 0.1\,m/{s^2} \cr} $$