Physics - Rotation Question with Solution | TestHub

From third equation of angular motion,

${\mathrm{\omega }}^2={\omega }_0^2-2\mathrm{\alpha }\mathrm{ }\mathrm{ }(\mathrm{H}\mathrm{e}\mathrm{r}\mathrm{e},\mathrm{\omega }=\frac{{\mathrm{\omega }}_0}{2},\mathrm{\theta }=36\times 2\mathrm{\pi })$

$\therefore$${\left(\frac{{\omega }_0}{2}\right)}^2={\omega }_0^2-2\alpha \times 36\times 2\pi$

or$4\times 36\pi \alpha ={\omega }_0^2-\frac{{\omega }_0^2}{4}$

or$4\times 36\pi \alpha =\frac{3{\omega }_0^2}{4}$

or$\alpha =\frac{{\omega }_0^2}{16\times 12\pi }$...(i)

According to question again applying the third equation of angular motion

${\omega }^2={\omega }_0^2-2\alpha \theta (\mathrm{H}\mathrm{e}\mathrm{r}\mathrm{e}, \omega =0)$

$\therefore$$0={\left(\frac{{\omega }_0}{2}\right)}^2-2\times \frac{{\mathrm{\omega }}_0^2.\mathrm{\theta }}{16\times 12\mathrm{\pi }}$

or$\mathrm{\theta }=24\mathrm{\pi }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{o}\mathrm{r}\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{\theta }=12\times 2\mathrm{\pi }$

But$2\pi =1$cycle

So,$\mathrm{\theta }=12\mathrm{ }\mathrm{c}\mathrm{y}\mathrm{c}\mathrm{l}\mathrm{e}$