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Question

(A) The moment of inertia of a solid cube of mass $m$ and side $l$ about an axis passing through its center of mass and parallel to an edge is $I_{cm}

Vedclass · Accepted Answer

$2\pi \sqrt {\frac{2\sqrt{2}}{3}\frac{l}{g}} $

Vedclass · Answer

(A) The moment of inertia of a solid cube of mass $m$ and side $l$ about an axis passing through its center of mass and parallel to an edge is $I_{cm} = \frac{1}{6}ml^2$. Using the parallel axis theorem,the moment of inertia $I$ about an edge is $I = I_{cm} + md^2$,where $d$ is the distance from the center of mass to the edge. For a cube,$d = \sqrt{(\frac{l}{2})^2 + (\frac{l}{2})^2} = \frac{l}{\sqrt{2}}$. Thus,$I = \frac{1}{6}ml^2 + m(\frac{l}{\sqrt{2}})^2 = \frac{1}{6}ml^2 + \frac{1}{2}ml^2 = \frac{2}{3}ml^2$. The time period of a physical pendulum is given by $T = 2\pi \sqrt{\frac{I}{mgR}}$,where $R$ is the distance from the pivot to the center of mass,which is $R = \frac{l}{\sqrt{2}}$. Substituting the values,$T = 2\pi \sqrt{\frac{\frac{2}{3}ml^2}{mg(\frac{l}{\sqrt{2}})}} = 2\pi \sqrt{\frac{2}{3} \cdot \sqrt{2} \cdot \frac{l}{g}} = 2\pi \sqrt{\frac{2\sqrt{2}}{3} \frac{l}{g}}$.

$A$ solid cube of side $l$ is made to oscillate about a horizontal axis passing through one of its edges. Its time period will be

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