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. Find the position on this rod through which the axis should pass in order that the work required to set the rod rotating with angular velocity \[{{\omega }_{0}}\] should be minimum.              

Answer:

                Let the axis of rotation be at a distance x from \[{{m}_{1}}\] Fig. 5(HT).5. K.E. of translation = 0 K.E. of rotation = \[\frac{1}{2}\left( {{I}_{1}}+{{I}_{2}} \right)\omega _{0}^{2}\] According to work energy principle, Work done = Increase in energy  \[W=\frac{1}{2}\left( {{I}_{1}}+{{I}_{2}} \right)\omega _{0}^{2}=\frac{1}{2}[{{m}_{1}}{{x}^{2}}+{{m}_{2}}{{\left( L-x \right)}^{2}}]\omega _{0}^{2}\]                                                                ??.. (i) For W to be minimum, \[\frac{dW}{dx}=0\] From (i), \[\frac{dW}{dx}=\frac{1}{2}{{m}_{1}}2x+\frac{1}{2}{{m}_{2}}\times 2(L-x)(-1)=0\] or    \[{{m}_{1}}x-{{m}_{2}}\left( L-x \right)=0\]   or    \[\left( {{m}_{1}}+{{m}_{2}} \right)x={{m}_{2}}L\] \[x={{m}_{2}}L/\left( {{m}_{1}}+{{m}_{2}} \right)\] This is the position of centre of mass of the rod from \[{{m}_{1}}\]. Hence the required axis should pass through centre of mass of the rod and perpendicular 10 the length of the rod as shown in Fig. 5(HT).5.