Find the moment of inertia of a uniform annul | vedclass

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(A) The moment of inertia $I$ of an annular disc about an axis passing through its centre and perpendicular to its plane is given by the formula: $I =

Vedclass · Accepted Answer

$2.5 	imes 10^{-3} \, kg \cdot m^2$

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(A) The moment of inertia $I$ of an annular disc about an axis passing through its centre and perpendicular to its plane is given by the formula: $I = M \left[ \frac{R_1^2 + R_2^2}{2} ight]$ Given: Mass $M = 100 \, g = 0.1 \, kg$ Inner radius $R_1 = 10 \, cm = 0.1 \, m$ Outer radius $R_2 = 20 \, cm = 0.2 \, m$ Substituting the values into the formula: $I = 0.1 \left[ \frac{(0.1)^2 + (0.2)^2}{2} ight]$ $I = 0.1 \left[ \frac{0.01 + 0.04}{2} ight]$ $I = 0.1 \left[ \frac{0.05}{2} ight]$ $I = 0.1 	imes 0.025 = 0.0025 \, kg \cdot m^2$ $I = 2.5 	imes 10^{-3} \, kg \cdot m^2$

Find the moment of inertia of a uniform annular disc of mass $100 \, g$,having inner radius $10 \, cm$ and outer radius $20 \, cm$,about an axis passing through its centre and perpendicular to its plane.

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