A hollow cylinder has mass $M$ outside radius $R_{2}$ and inside radius $R_{1}$ . Its moment of inertia about an axis parallel to its symmetry axis and tangential to the outer surface is equal to

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

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

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

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\frac{M}{2} (3 R_{2}^{2} + R_{1}^{2})

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Solution

The correct option is D $\frac{M}{2} (3 R_{2}^{2} + R_{1}^{2})$
Moment of Inertial of hollow cylinder about axis I
$I = \frac{M}{2} \times ({R_{1}}^{2} + {R_{2}}^{2}) . . . . . . . . . . . . . . . . . . (1)$
By parallel axis theorem
$I = I_{1} + M \times (R_{2})^{2}$

$I = [\frac{M}{2} \times ({R_{1}}^{2} + {R_{2}}^{2})]$ + $[M \times (R_{2})^{2}]$

$I = \frac{M}{2} \times ({R_{1}}^{2} + 3 {R_{2}}^{2})$

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