Physics - Rotation Question with Solution | TestHub

As moment of inertia of a disc about a diameter is$\frac{1}{2}\left(\frac{1}{2}{\text{mR}}^2\right)\text{,}$the moment of inertia of the disc about the chord PQ by 'theorem of parallel axes' will be

${\left({\text{I}}_{\text{D}}\right)}_{\text{PQ}}=\frac{1}{4}{\text{mR}}^2+\text{m}{\left(\frac{1}{4}\text{R}\right)}^2=\frac{5}{\text{16}}{\text{mR}}^2$

and as particle of mass m is at a distance [R + (R/4) = (5/4)R] from PQ, the moment of inertia of the system about PQ

$\text{I}={\left({\text{I}}_{\text{D}}\right)}_{\text{PQ}}+{\left({\text{I}}_{\text{P}}\right)}_{\text{PQ}}=\frac{5}{\text{16}}{\text{mR}}^2+\text{m}{\left(\frac{5}{4}\text{R}\right)}^2=\frac{15}{8}{\text{mR}}^2$

Now if ω is the angular speed of the system when A reaches the lowest point A' on rotation about the axis PQ, by 'conservation of mechanical energy',

$\frac{1}{2}\text{I}{\text{ω}}^2=\text{mg}\left({\text{AA}}^{\prime }\right)+{\text{mgCC}}^{\prime }=\text{mg}\left[\text{2AD}+\text{2CD}\right]$



$\text{i.e.,}\frac{1}{2}\times \frac{\text{15}}{8}\text{mR}{}^2{\text{ω}}^2=2\text{mg}\left[\left(\text{R}+\frac{1}{4}\text{R}\right)+\frac{1}{4}\text{R}\right]\text{,}$

$\text{i.e.,}\text{ω}=4\sqrt{\frac{\text{g}}{\text{5R}}}$

$\text{so}{\left(\text{v}\right)}_{{\text{A}}^{\prime }}=\text{r}\text{ω}=\left(\text{R}+\frac{1}{4}\text{R}\right)\times 4\sqrt{\frac{\text{g}}{\text{5R}}}=\sqrt{\text{5gR}}$