question_answer

Point masses \[{{m}_{1}}\] and \[{{m}_{2}}\] are placed at the opposite ends of a rigid rod of length L and negligible mass. The rod is to be set rotating about an axis perpendicular to it. The position of point P on this rod through which the axis should pass, so that the work required to set the rod rotating with angular velocity \[{{\omega }_{0}}\] is minimum, is given by

A)  \[x=\frac{{{m}_{1}}L}{{{m}_{1}}+{{m}_{2}}}\]                    

B)  \[x=\frac{{{m}_{1}}}{{{m}_{2}}}L\]         

C)         \[x=\frac{{{m}_{2}}}{{{m}_{1}}}L\]         

D)         \[x=\frac{{{m}_{2}}L}{{{m}_{1}}+{{m}_{2}}}\]

Correct Answer: D

Solution :

As two point masses \[{{m}_{1}}\] and \[{{m}_{2}}\] are placed at opposite ends of a rigid rod of length L and negligible mass as shown in figure. Total moment of inertia of the rod \[I={{m}_{1}}{{x}^{2}}+{{m}_{2}}{{(L-x)}^{2}}\] \[I={{m}_{1}}{{x}^{2}}+{{m}_{2}}{{L}^{2}}+{{m}_{2}}{{x}^{2}}-2{{m}_{2}}Lx\] As, \[I\] is minimum i.e. \[\frac{dI}{dx}\,=2{{m}_{1}}\,\,x+0+2x{{m}_{2}}-2{{m}_{2}}L=0\] \[\Rightarrow \]               \[x\,(2{{m}_{1}}\,+2{{m}_{2}})\,=2{{m}_{2}}L\] \[\Rightarrow \]               \[x=\frac{{{m}_{2}}L}{{{m}_{1}}+{{m}_{2}}}\] When \[I\] is minimum, then work done on rotating a rod 1/2 \[I{{\omega }^{2}}\] with angular velocity \[{{\omega }_{0}}\] will be minimum. Shortcut Way The position of point P on rod through which the axis should pass, so that the work required to set the rod rotating with minimum angular velocity \[{{\omega }_{0}}\] is their centre of mass, we have \[{{m}_{1}}x={{m}_{2}}(L-x)\Rightarrow x=\frac{{{m}_{2}}L}{{{m}_{1}}+{{m}_{2}}}\]