Question

Two identical discs of mass M and radius R are connected by a light rod. The assembly rests at a corner, the vertical wall is smooth and there is sufficient friction on the floor to ensure pure rolling of disc B. The system starts from position $\theta =0$. Find the velocity of the midpoint of the rod when $\theta ={60}^o$.

Answer 1

$\begin{array}{c}{V}_A\cos \theta =V_B\sin \theta --\left(1\right)\\ \Rightarrow V_A=V_B\sin \theta \\ Now,\\ wall=\Delta k\\ \Rightarrow mg\left(l-l\cos \theta \right)=\frac{1}{2}m{V}_A^2+\frac{1}{2}m{V}_B^{{}^2}+\frac{1}{2}\left[\frac{m{R}^2}{2}\right]\times \frac{V_B^2}{R^2}\\ at\theta ={60}^{\circ }\\ \Rightarrow mg\frac{l}{2}=\frac{1}{2}m{V}_A^2+\frac{1}{2}m{V}_B^2+\frac{m{V}_B^2}{4}\\ \Rightarrow g\frac{l}{2}=\frac{V_A^2}{2}+\frac{3}{4}{V}_B^2\\ \Rightarrow gl=\frac{V_B^2{\tan }^2\theta }{2}+\frac{3}{4}{V}_A^2\\ \Rightarrow gl=\frac{V_B^2{r}^3}{2}+\frac{3}{4}{V}_B^2\\ \Rightarrow gl=\frac{9}{4}{V}_B^2\\ \Rightarrow \frac{4}{9}\times 10\times \frac{60}{100}=V_B^2\\ \Rightarrow V_B=\sqrt{\frac{24}{9}}=\frac{\sqrt{24}}{3}\\ \Rightarrow V_a=\frac{\sqrt{24}}{3}\times \sqrt{3}=\sqrt{\frac{24}{3}}\\ Now,\\ V_{acidpoint}=\sqrt{{\left(\frac{V_A}{2}\right)}^2+{\left(\frac{V_B}{2}\right)}^2}\\ =\sqrt{\frac{24}{4\times 3}+\frac{24}{4\times 9}}\\ =\sqrt{\frac{6}{3}+\frac{6}{9}}\\ =\sqrt{2+\frac{2}{3}}\\ =\sqrt{\frac{2}{3}}\end{array}$

Which is the required answer.