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(B) The moment of inertia of a cylinder of mass $M$,radius $R$,and length $L$ about an axis passing through its center of mass and perpendicular to it

Vedclass · Accepted Answer

$M \left[ \frac{D^2}{16} + \frac{L^2}{12} \right]$

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(B) The moment of inertia of a cylinder of mass $M$,radius $R$,and length $L$ about an axis passing through its center of mass and perpendicular to its length is given by the formula: $I = M \left( \frac{L^2}{12} + \frac{R^2}{4} \right)$ Given that the diameter is $D$,the radius $R = \frac{D}{2}$. Substituting $R$ into the formula: $I = M \left[ \frac{L^2}{12} + \frac{(D/2)^2}{4} \right]$ $I = M \left[ \frac{L^2}{12} + \frac{D^2/4}{4} \right]$ $I = M \left[ \frac{L^2}{12} + \frac{D^2}{16} \right]$

What is the moment of inertia of a cylinder of diameter $D$ and length $L$ about an axis passing through its center of gravity and perpendicular to its length?

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