The moment of inertia of a solid disc made of | vedclass

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(C) The moment of inertia $I$ of a body about an axis is defined as $I = \sum m_i r_i^2$,where $r_i$ is the perpendicular distance of the mass element

Vedclass · Accepted Answer

$\frac{M R^2}{4}$

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(C) The moment of inertia $I$ of a body about an axis is defined as $I = \sum m_i r_i^2$,where $r_i$ is the perpendicular distance of the mass element $m_i$ from the axis of rotation.When the disc is folded in half about its diameter,the mass distribution relative to that diameter remains unchanged. Every mass element $dm$ at a distance $r$ from the diameter in the original disc remains at the same distance $r$ from the diameter after folding.Since the mass $M$ and the perpendicular distances $r$ of all mass elements from the axis (the diameter) remain the same,the integral $I = \int r^2 dm$ does not change.Therefore,the moment of inertia about the diameter remains $\frac{M R^2}{4}$.

The moment of inertia of a solid disc made of thin metal of radius $R$ and mass $M$ about one of its diameters is given by $\frac{M R^2}{4}$. What will be the moment of inertia about this axis,if the disc is folded in half about this diameter?

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