A particle of mass ' \(\mathrm{m}\) ' moves along a circle of radius ' \(r\) ' with constant tangential acceleration. If K.E. of the particle is ' \(E\) ' by the end of third revolution after beginning of the motion, then magnitude of tangential acceleration is

\(\begin{aligned} & \text { Using } 3^{\text {rd }} \text { equation of motion, } \\ & \mathrm{v}^2=\mathrm{u}^2+2 \mathrm{as} \\ & \therefore \quad \mathrm{v}^2=2 \mathrm{a}_{\mathrm{t}} \mathrm{s} \quad \ldots .(\text { For } \mathrm{u}=0)\end{aligned}\)
\(\therefore \quad \mathrm{a}_{\mathrm{t}}=\frac{\mathrm{v}^2}{2 \mathrm{~s}}\)
By the end of \(3^{\text {rd }}\) revolution, distance covered,
\(\begin{aligned}
& \mathrm{s}=3(2 \pi \mathrm{r})=6 \pi \mathrm{r} \\
\therefore \quad & \mathrm{a}_{\mathrm{t}}=\frac{\mathrm{v}^2}{2 \times 6 \pi \mathrm{r}}... (i)
\end{aligned}\)
\(\begin{aligned} & \text { Also, } \frac{1}{2} \mathrm{mv}^2=\mathrm{E} \\ \therefore \quad & \mathrm{v}^2=\frac{2 \mathrm{E}}{\mathrm{m}}... (ii)\end{aligned}\)
Substituting equation (ii) in equation (i),
\(a_t=\frac{2 E}{2 \times 6 \pi r \times m}=\frac{E}{6 \pi r m}\)