Calculate the ratio of moments of inertia of a circular lamina of radius 400 mm to that of 200 mm about their respective centroid axes.

Concept:

Area moment of inertia is given by, I = A × k2

Where A is area of section and k is radius of gyration of the section.

For circular section, k = D/4

\(\therefore {\rm{I}} = {\rm{A}} \times {{\rm{k}}^2} = \frac{{\rm{\pi }}}{{64}}{{\rm{D}}^4}\)

Calculation:

Given:

D₁ = 400 mm and D₂ = 200 mm

So, the ratio of MOI of both is given by

\({I_1\over I _2} ={ {\pi\times D^4_1\over 64} \over {\pi\times D^4_2\over 64}}\)

\({I_1 \over I_2 } = {{D_1^4} \over {D_2^4} }\)

\({I_1 \over I_2 } = {{400^4} \over {200^4} }\)

\({I_1 \over I_2 } = 16\)

Important Points

Polar Moment of Inertia (J) = 2I = \(\frac{{\rm{\pi }}}{{32}}{{\rm{D}}^4}\)