A particle of mass 'm' moves along a circle of radius 'r' with constant tangential
acceleration. If kinetic energy 'E' of the particle becomes three times by the end of
third revolution after beginning of the motion then the magnitude of tangential
acceleration is

\(\mathrm{E}_{1}=\frac{1}{2} \mathrm{mv}^{2}=\frac{1}{2} \mathrm{mr}^{2} \omega_{1}^{2}\)
\(E_{2}=\frac{3}{2} m r^{2} \omega_{1}^{2}\)
\(\frac{E_{2}}{E_{1}}=3 \quad \therefore E_{2}=3 E_{1}\)
\(\frac{1}{2} \mathrm{mr}^{2} \omega^{2}=3 \frac{1}{2} \mathrm{~m} \omega_{0}^{2} \mathrm{r}^{2}\)
\(\omega^{2}=3 \omega_{0}^{2}\)
\(\frac{1}{2} m \omega_{0}^{2} r^{2}=E\)
\(\omega_{0}^{2}=\frac{2 \mathrm{E}}{\mathrm{mr}^{2}}\)
\(\omega^{2}=\omega_{0}^{2}+2 \alpha \theta\)
\(3 \omega_{0}^{2}=\omega_{0}^{2}+2 \alpha(3.2 \pi)\)
\(2 \omega_{0}^{2}=12 \alpha \pi\)
\(\alpha=\frac{\omega_{0}^{2}}{6 \pi}=\frac{2 \mathrm{E}}{\mathrm{mr}^{2}} \times \frac{1}{6 \pi}=\frac{\mathrm{E}}{3 \pi \mathrm{mr}^{2}}\)
But \(\mathrm{a}=r \alpha=r \times \frac{\mathrm{E}}{3 \pi \mathrm{mr}^{2}}=\frac{\mathrm{E}}{3 \pi \mathrm{mr}}\)