Moment Of Inertia Of A Rod MCQ - Practice Questions & Answers

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The moment of inertia of a uniform cylinder of length $l$ and radius $R$ about its perpendicular bisector is $I$ . What is the ratio $\frac{l}{R}$ such that the moment of inertia is minimum?

$\sqrt{\frac{3}{2}}$

$\frac{\sqrt{3}}{2}$

$1$

$\frac{3}{\sqrt{2}}$

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Moment of inertia of a Rod

Let I=Moment of inertia of a ROD about an axis through its centre and perpendicular to it

To calculate I (Moment of inertia of rod)

Consider a uniform straight rod of length L, mass M and having centre C

mass per unit length of the rod = $\lambda = \frac{M}{L}$

Take a small element of mass dm with length dx at a distance x from the point C.

$dm = \lambda.dx = \frac{M}{L}.dx$

$\Rightarrow dI= x^2dm$

Now integrate this dI between the limits $x = -\frac{L}{2}\ to \ \frac{L}{2}$

$I=\int dI=\int x^2dm=\int_{\frac{-L}{2}}^{\frac{L}{2}} \frac{M}{L}x^2*dx= \frac{M}{L}\int_{\frac{-L}{2}}^{\frac{L}{2}}x^2dx=\frac{ML^2}{12}$

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Moment Of Inertia Of A Rod - Practice Questions & MCQ

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