Physics - Rotation Question with Solution | TestHub

$\frac{{\text{mv}}^2}{\left(\text{R}+\text{r}\right)}=mg\; cos\theta$

$\text{mgh}=\frac{1}{2}{\text{mv}}^2+\frac{1}{2}\text{I}{\omega }^2$

$\text{mg}\left(\text{R}+\text{r}\right)\left(1-cos\theta \right)=\frac{1}{2}{\text{mv}}^2+\frac{1}{5}{\text{mv}}^2=\frac{7}{10}{\text{mv}}^2$

         

$\frac{10}{7}\text{mg}\left(1-cos\theta \right)=\text{mg}cos\theta$

${\text{mv}}^2=\frac{10}{7}\text{mg}\left(\text{R}+\text{r}\right)\left(1-\text{cos}\theta \right)\frac{10}{7}=\frac{17}{7}\text{cos}\theta$

or $cos\theta =\frac{10}{17}$

$\text{v}=\sqrt{\text{g}\left(\text{R}+\text{r}\right)cos\theta }=\sqrt{\frac{10}{17}\text{g}\left(\text{R}+\text{r}\right)}$

and $\omega =\frac{\text{v}}{\text{r}}=\sqrt{\frac{10\text{g}\left(\text{R}+\text{r}\right)}{17{\text{r}}^2}}=\sqrt{\frac{110\text{g}}{17\text{r}}}$

$\Rightarrow p+q=127$