Question

A thin rod is hinged at one end $O$ and it is in an unstable equilibrium position. It falls under gravity due to a slight disturbance. It makes angles ${60}^{\circ },{90}^{\circ }$ and ${180}^{\circ }$ with vertical in positions $\left(B\right),\left(C\right)$ and $\left(D\right)$ respectively. If ${\omega }_2,{\omega }_3,{\omega }_4$ are angular velocities at these positions, then : -

• A. ${\omega }_4=2{\omega }_3$
• B. ${\omega }_4=2{\omega }_2$
• C. ${\omega }_4=1.5{\omega }_2$
• D. ${\omega }_4=\sqrt{2}{\omega }_2$

Answer 1

The correct option is B ${\omega }_4=2{\omega }_2$

$\begin{array}{c}\text{Le the mass of the Rod be(M) and Length (L)}\\ \text{we have cosque acting on the rod equal to}\end{array}$

$\begin{array}{cc}\tau =F\cdot h& F\to \text{Perpendicular For}a\\ & r\to \text{distanu from Hinge}\\ t=\left(\mathrm{mg}\sin \theta \right)\cdot \frac{L}{2}& \end{array}$

$\begin{array}{c}\Rightarrow \tau =\frac{mgL}{2}\sin \theta =I\alpha I\to \text{moment of Inertia}\\ \Rightarrow I=\frac{mgL}{2}\sin \theta =\frac{m{L}^2}{3}\alpha \\ \Rightarrow \alpha =\frac{3g}{2L}\sin \theta =\omega \cdot \frac{d\omega }{d\theta }\\ \Rightarrow {\int }_0^{\omega }\omega \cdot d\omega ={\int }_0^{\theta }\frac{3g}{2L}\sin \theta \cdot d\theta \\ \Rightarrow \frac{{\omega }^2}{2}=\frac{3g}{2L}[1-\cos \theta ]\end{array}$

$\Rightarrow \omega =\sqrt{\frac{3g}{L}\left[1\cos \theta \right]}$

$\begin{array}{c}\ast \text{For}\theta ={60}^{\circ }\\ \begin{array}{cc}{\omega }_2& =\sqrt{\frac{3g}{L}[1-\frac{1}{2}]}=\sqrt{\frac{3y}{2L}}\\ \ast \text{For}\theta & ={90}^{\circ }\\ {\omega }_3& =\sqrt{\frac{3g}{L}[1-0]}=\sqrt{\frac{3g}{L}}\\ \ast \text{For}\theta & ={180}^{\circ }\\ {\omega }_4& =\sqrt{\frac{3g}{L}\cdot [1+1]}=\sqrt{\frac{6g}{L}}\\ \Rightarrow & {\omega }_4=2{\omega }_2\end{array}\\ \text{Henu, option (B) is correct.}\end{array}$