Calculate the moment of Inertia of an annular disc of inner radius $R_{1}$ and outer radius $R_{2}$ about an axis perpendicular to the plane in which the disc lies and passing through $O$ .

$\frac{M}{6} {R_{2}^{2} + R_{1}^{2}}$

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\frac{M}{2} {R_{1}^{2} + R_{2}^{2}}

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\frac{M}{4} {R_{1}^{2} + R_{2}^{2}}

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\frac{M}{2} {R_{1} + R_{2}}_{1}^{2}

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Solution

The correct option is B $\frac{M}{2} {R_{1}^{2} + R_{2}^{2}}$

Choose a ring at a distance X from O and of thickness dx

Mass per unit Area of annular

disc = $\frac{M}{π (R_{2}^{2} - R_{1}^{2})}$ = $, ρ$

Let $, ρ$ = mass per unit Area

Note that the disc in uniform and mass distribution is uniform too.

Now mass of area element considered, = $d m$

$d m$ = $\frac{M}{π (R_{2}^{2} - R_{1}^{2})} \times 2 π x d x$ = $\frac{2 M x d x}{(R_{2}^{2} - R_{1}^{2})}$

Moment of Inertia of this small ring element about $O$ = $d m . x^{2}$

Total moment of Inertia of annular disc = $R_{2} \int R_{1} x^{2} d m$

$I$ = $R_{2} \int R_{1} \frac{x^{2} .2 M x d x}{(R_{2}^{2} - R_{1}^{2})}$

$I$ = $\frac{2 M}{(R_{2}^{2} - R_{1}^{2})} R_{2} \int R_{1} x^{3} d x$

= $\frac{M ((R_{2}^{2} - R_{1}^{2}) (R_{2}^{2} + R_{1}^{2}))}{2 (R_{2} - R_{1}) (R_{2} + R_{1})}$

= $\frac{M (R_{2} - R_{1}) (R_{2} + R_{1}) (R_{2}^{2} + R_{1}^{2})}{2 (R_{2} - R_{1}) (R_{2} + R_{1})}$

$I = \frac{M}{2} {R_{1}^{2} + R_{2}^{2}}$

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Calculate the moment of Inertia of an annular disc of inner radius R1 and outer radius R2 about an axis perpendicular to the plane in which the disc lies and passing through O.

Calculate the moment of Inertia of an annular disc of inner radius $R_{1}$ and outer radius $R_{2}$ about an axis perpendicular to the plane in which the disc lies and passing through $O$ .