91Ó°ÊÓ

The moment of inertia of a uniform rod,

Length \( = L\)

mass \( = M\)

Now, the rod ends and perpendicular rod will be,

\(I = \frac{1}{3}M'{L^2}\)----------(1)

Here we get half of the rod equation,

\(M' = \frac{1}{2}M\)

\(L' = \frac{1}{2}L\)

Put the value in equation (1)

\({I_1} = \frac{1}{3}\left( {\frac{1}{2}M} \right){\left( {\frac{1}{2}L} \right)^2}\)

\({I_1} = \frac{1}{6}M \times \frac{1}{4}{L^2}\)

If the uniform rode bent on the center

So, \({I_{{T_1}}} = 2{I_1}\)

\({I_{{T_1}}} = 2 \times \frac{1}{{24}}M{L^2}\)

\({I_{{T_1}}} = \frac{1}{{12}}M{L^2}\)

Here we will calculate the moment of inertia for one segment,

\(ac = \sqrt {{{\left( {\frac{1}{2}L} \right)}^2} + {{\left( {\frac{1}{2}L} \right)}^2}} \)

\( = \sqrt {\frac{1}{4}{L^2} + \frac{1}{4}{L^2}} \)

\( = \sqrt {\frac{1}{2}{L^2}} \)

\( = \frac{1}{{\sqrt 2 }}L\)

Now,

\(ad = \frac{1}{2}ac\)

\( = \frac{1}{{2\sqrt 2 }}L\)

\(Ed = \sqrt {{{\left( {ad} \right)}^2} - {{\left( {aE} \right)}^2}} \)

\( = \sqrt {{{\left( {\frac{1}{{2\sqrt 2 }}L} \right)}^2} - {{\left( {\frac{1}{4}L} \right)}^2}} \)

\( = \frac{1}{4}L\)

We know that \(I = \frac{1}{{12}}M'{L'^2}\)

Here we calculate half of the rod

\(M' = \frac{1}{2}M,\)

\(L' = \frac{1}{2}L\)

\({I_E} = \left( {\frac{1}{{12}}} \right) \times \left( {\frac{1}{2}M} \right) \times {\left( {\frac{1}{2}L} \right)^2}\)

\( = \frac{1}{{96}}M{L^2}\)

E is the midpoint of half rod ab.

Here we can use the parallel axis theorem.