Moment of inertia of a thin uniform rod – Rankers Physics

Moment of inertia of a thin uniform rod

The moment of inertia of a thin uniform rod of mass $M$ and length $L$ about an axis passing through its midpoint and perpendicular to its length is $I_{0}$. Its moment of inertia about an axis passing through one of its ends perpendicular to its length is (2011 Mains)

$I_{0} + ML^{2}/2$
$I_{0} + ML^{2}/4$
$I_{0} + 2ML^{2}$
$I_{0} + ML^{2}$

Solution:

Using the parallel axis theorem, $I = I_{cm} + Md^{2}$. Here, the center of mass moment of inertia is $I_{cm} = I_{0}$ and the distance to the parallel axis is $d = \frac{L}{2}$. Thus, $I = I_{0} + M(\frac{L}{2})^{2} = I_{0} + \frac{ML^{2}}{4}$.

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