Question

Find (i) Radius of gyration (ii) Moment of inertia of a rod of mass ${{100g}}$ and length ${{100cm}}$ about an axis passing through its center and to it length.

Answer 1

Radius of gyration about a given axis is the perpendicular distance. From the axis to a point where if whole mass of the system/body is supposed to be concentrated the body hall have same moment of inertia as it has with actual distribution of mass and moment of inertia of a body is the quantity which measure rotational inertia of the body

Formula used:
${{{K}}_{{c}}}{{ = }}\dfrac{{{{{L}}^{{2}}}}}{{{{12}}}}$ , Where ${{{K}}_{{c}}}$ is radius of gyration about center of road, ${{L,}}$ is length of rod.
${{{I}}_{{c}}}{{ = }}\dfrac{{{{M}}{{{L}}^{{2}}}}}{{{{12}}}}$, Where ${{{I}}_{{c}}}$ is moment of inertia about the axis passing through its center.
${{{K}}_{{e}}}{{ = }}\dfrac{{{{{L}}^{{2}}}}}{{{3}}}{{,}}$ Where ${{{K}}_{{e}}}$ is radius of gyration of the rod about end, ${{L}}$ is length of rod.
$I_e$ = $\dfrac{ML^2}{3}$ Where ${{{I}}_{{e}}}$ is moment of inertia of rod about its length, ${{M}}$ is mass of rod, ${{L}}$ is length of rod.

Complete step by step solution:
In this question, it is given that,
Mass of rod ${{M = 0}}{{.1Kg}}$
Length of rod ${{L = 0}}{{.1m}}$
As we know, moment of inertia about the center is given by {{{I}}_{{c}}}{{ = }}\dfrac{{{{M}}{{{L}}^{{2}}}}}{{{{12}}}}{{\\_\\_}}\left( {{1}} \right)
By substituting the values of ${{M}}$ and ${{L}}$ in $\left( 1 \right)$ we get
${{{I}}_{{c}}}{{ = }}\dfrac{{{{0}}{{.1 \times }}{{\left( {{{0}}{{.1}}} \right)}^{{2}}}}}{{{{12}}}}{{ = }}{{8}}{{.3 \times 1}}{{{0}}^{{{ - 5}}}}{{Kg}}{{{m}}^{{2}}}$
Also we know Radius of gyration of a rod about its center is given as
${{{K}}_{{c}}}{{ = }}\dfrac{{{{{L}}^{{2}}}}}{{{{12}}}},$ by substituting values in this equation we get
${{{K}}_{{c}}}{{ = }}\dfrac{{{{\left( {{{0}}{{.1}}} \right)}^{{2}}}}}{{{{12}}}}{{ = }}{{8}}{{.3 \times 1}}{{{0}}^{{{ - 5}}}}{{{m}}^{{2}}}$
Further we also, know that radius of gyration about the end of rod is given as ${{{K}}_{{e}}}{{ = }}\dfrac{{{{{L}}^{{2}}}}}{{{3}}}$
By substituting we get, ${{{K}}_{{e}}}{{ = }}\dfrac{{{{\left( {{{0}}{{.1}}} \right)}^{{2}}}}}{{{3}}}{{ = }}{{3}}{{.3 \times 1}}{{{0}}^{{{ - 3}}}}{{{m}}^{{2}}}$
And moment of inertia about end of rod is given as ${{{I}}_{{e}}}{{ = }}\dfrac{{{{M}}{{{L}}^{{2}}}}}{{{3}}}{{ = }}\dfrac{{\left( {{{0}}{{.1}}} \right){{ \times }}{{\left( {{{0}}{{.1}}} \right)}^{{2}}}}}{{{3}}}{{ = }}{{3}}{{.3 \times 1}}{{{0}}^{{{ - 4}}}}{{Kg}}{{{m}}^{{2}}}$

Note: In this question the thickness of the rod is assumed to be negligible. Otherwise the results will be similar to that of a cylinder. The radius of gyration of a body is referred to as the radial distance from the rotational axis at which the entire body mass is supposed to be concentrated.