Physics - Circular Motion Question with Solution | TestHub

Tangential acceleration $a_{\mathrm{t}}=r\mathrm{\alpha }=$ constant $=\mathrm{K}$

$\mathrm{\alpha }=\frac{\mathrm{K}}{r}$

At the end of second revoluation angular velocity is $\omega$ then

${\omega }^2-{\omega }_0^2=2\alpha \mathrm{\theta }$

${\omega }^2=2\left(\frac{\mathrm{K}}{r}\right)\left(4\pi \right)$

${\omega }^2=\frac{8\mathrm{\pi }\mathrm{K}}{r}$

K.E. of the particle is $=\mathrm{K}.\mathrm{E}.=\frac{1}{2}\mathrm{m}{\mathrm{v}}^2$

$\mathrm{K}.\mathrm{E}.=\frac{1}{2}m{r}^2{\omega }^2$

$\text{K}.\text{E}.=\frac{1}{2}\mathit{\text{m}}\left(r^2\right)\left(\frac{8\text{πK}}{r}\right)=\frac{1}{2}mr\left(8\pi \mathrm{K}\right)$

$8\times {10}^{-4}=\frac{1}{2}\times 10\times {10}^{-3}\times 6.4\times {10}^{-2}\times 8\times 3.14\times \text{K}$

$\mathrm{K}=\frac{2}{6.4\times 3.14}=0.1 \mathrm{m} {\mathrm{s}}^{-2}$