Question

Moment of inertia of a uniform rod of length L and mass M, about an axis passing through L/4 from one end and perpendicular to its length is

Answer 1

$\textbf{\large{Answer -}}$

Length = $\frac{L}{4}$

Let the axis at length L/4 be AB.

We know that the moment of inertia of uniform rod with an axis passing through its middle is $\frac{ {ML}^{2}}{12}.$

So, $I_{AB} = I_{CD} + {ML}^{2}$

$I_{AB}= \frac{ {ML}^{2}}{12} +\frac{ {ML}^{2}}{16}$

$I_{AB}= \frac { {4ML}^{2} + {3ML}^{2}}{48}$

$\boxed {I_{AB}= \frac {{7ML}^{2}}{48}}$

Answer 2

Answer:

Explanation:

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