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 [NEET 2015 (Re)] |
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 |
\[{{m}_{1}}x={{m}_{2}}(L-x)\Rightarrow x=\frac{{{m}_{2}}L}{{{m}_{1}}+{{m}_{2}}}\] |
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