Physics - Rotation Question with Solution | TestHub

From the law of conservation of mechanical energy,

Final energy = Translational kinetic energy + Rotational kinetic energy

Hence, Final energy =$\frac{1}{2}{\mathrm{mv}}^2+\frac{1}{2}{\mathrm{I\omega }}^2$

$=\frac{1}{2}{\mathrm{mv}}^2+\frac{1}{2}\times \frac{2}{5}{\mathrm{mR}}^2\times {\left(\frac{\mathrm{v}}{\mathrm{R}}\right)}^2=\frac{7}{10}{\mathrm{mv}}^2$

Hence, we finally obtain,$\text{mg}\left(\text{R - r}\right)=\frac{7}{10}{\text{mv}}^2$.......(1)

and$\text{N}-\text{mg}=\frac{{\text{mv}}^2}{\text{R - r}}$..........(2)

Solving Eqs. (1) and (2), we get
$\text{N}=\frac{17}{7}\text{mg}$