question_answer

Two masses m and \[\frac{m}{2}\] are connected at the two ends of a massless rigid rod of length \[l.\] The rod is suspended by a thin wire of torsional constant k at the Centre of mass of the rod-mass system (see figure). Because of torsional constant k, the restoring torque is \[\tau \,=k\theta \] for angular displacement 0. If the rod is rotated by \[{{\theta }_{0}}\]and released, the tension in it when it passes through its mean position will be:

A) \[\frac{3k\theta _{0}^{2}}{l}\]

B) \[\frac{k\theta _{0}^{2}}{2l}\]

C) \[\frac{2k\theta _{0}^{2}}{2l}\]

D) \[\frac{k\theta _{0}^{2}}{l}\]

Correct Answer: D

Solution :

\[\omega \,=\,\sqrt{\frac{k}{\text{I}}},\,\omega \,=\,\sqrt{\frac{3k}{m{{\ell }^{2}}}}\]

\[\Omega \,=\,\omega {{\theta }_{0}}\,=\,\]average velocity

\[\text{T}\,=\,m{{\Omega }^{2}}{{r}_{1}}\]

\[\text{T}\,=\,m{{\Omega }^{2}}\frac{\ell }{3}=\,m{{\omega }^{2}}\theta _{0}^{2}\frac{\ell }{3}\]

\[\frac{{{r}_{1}}}{{{r}_{2}}}\,=\,\frac{1}{2}\]\[\Rightarrow \]\[{{r}_{1}}=\,\frac{1}{3}\]

\[=\,m\,\frac{3k}{m{{\ell }^{2}}}\theta _{0}^{2}\frac{\ell }{3}\,=\,\frac{\text{K}\theta _{0}^{2}}{\ell }\]

\[\text{I}\,=\,m\ell =\,\frac{\frac{{{m}^{2}}}{2}}{\frac{3m}{2}}{{\ell }^{2}}=\,\frac{m{{\ell }^{2}}}{3}.\]