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.
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