First slide

Moment of intertia

Question

If the moment of inertia of a rod of mass M and length L about an axis passing through its centre of mass and perpendicular to its length is (ML²/12), then moment of inertia of the same rod about an axis passing through one of its ends and parallel to the given axis will be:

Easy

A

(ML²/12)
B

(ML²/6)
C

(ML²/3)
D

(13/12)ML²

Solution

$\begin{array}{l} I = I_{CM} + {Md}^{2} = \frac{{ML}^{2}}{12} + M {(\frac{L}{2})}^{2} \\ = \frac{{ML}^{2}}{3} \end{array}$

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