Physics - Rotation Question with Solution | TestHub

Let $m$ be the mass and $R$ be the radius of the disc. $mg\sin \theta -f=m{a}_{cm}$ $\begin{array}{rl}& {\tau }_{cm}=I_{c{m}^{\alpha }}\\ & fR=\frac{m{R}^2}{2}\left(\frac{a_{cm}}{R}\right)\end{array}$ $f=\frac{m{a}_{cm}}{2}$ or $a_{cm}=\frac{2f}{m}$ So, $mg\sin \theta -f=m\left(\frac{2f}{m}\right)$ $f=\frac{mg\sin \theta }{3}=\frac{1}{2}\times \frac{10}{3}\times \frac{1}{\sqrt{2}}=\frac{5}{3\sqrt{2}}\mathrm{N}$