Static friction

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Question

AB is a thin rod of length l. Its moment of inertia about axis (1) is $I_{1}$ and that about axis (2) is $I_{2}$ . If $I_{1} > I_{2}$ , and c be the position of centre of mass of the rod, then

Moderate

Solution

By parallel axis Theorem,
$\begin{array}{l} I_{1} = I_{C} + m {(AC)}^{2} \\ I_{2} = I_{C} + m {(BC)}^{2} \end{array}$
Since, $I_{1} > I_{2},$ we conclude AC > BC.
Also AC + BC = l.
$∴ AC > \frac{l}{2} .$

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