Question

A square is formed by joining four rods each of mass Mand length L. Its M about an axis PQ, in its plane
and passing through one of its corner is​

Answer 1

Answer:

First we shall try to find out the Moment of Inertia of that figure along it's centre of mass axis. Then we shall apply parallel law to Chang the axis to the given axis.

Refer to the attached photo to understand better

We can see that the above rods are at an angle 45° with the axis. And the lower rods are at 135° with axis.

Total moment of Inertia along axis AB will be :

$= \sum \bigg \{ \dfrac{m {l}^{2} }{3} { \cos}^{2} ( \theta) \bigg \}$

So putting the values :

$=2 \times \bigg \{ \dfrac{m {l}^{2} }{3} { \cos}^{2} ( 45 \degree) \bigg \} + 2 \times \bigg \{ \dfrac{m {l}^{2} }{3} { \cos}^{2} (135 \degree) \bigg \}$

$= \dfrac{m {l}^{2} }{3} + \dfrac{m {l}^{2} }{3}$

$= \dfrac{2m {l}^{2} }{3}$

Now we will shift the axis by a length of L/√2.

So, moment of Inertia along PQ :

$= \dfrac{2m {l}^{2} }{3} + 4 \times \{ m { (\dfrac{l}{ \sqrt{2} }) }^{2} \}$

$= \dfrac{2m {l}^{2} }{3} + 2m {l}^{2}$

$= \dfrac{8m {l}^{2} }{3}$

So final answer :

$\boxed{ \bold{ \huge{I_{net} = \dfrac{8m {l}^{2} }{3} }}}$

Answer 2

$</p><p>\tt{\huge{\purple{Hello!!! }}}$

$</p><p>\huge{\mathcal{\underline {\boxed{\boxed {\purple{Q} \green{U} \red{E} \purple{S} \orange{T} \blue{I} \red{O} \orange{N} \: - }}}}}$

• A square is formed by joining four rods each of mass Mand length L. Its M about an axis PQ, in its plane and passing through one of its corner is -

$</p><p>\tt{\red{\huge{\boxed{\boxed{\bullet{Solution \: - }}}}}}$

$</p><p>\tt{\orange{Step-By-Step-Explaination \: - }}$

The images from the figures above are self explanatory.

Anyway Explaining Again :

From the second figure we can note that :

For a rod of mass m and length L inclined to an axis at an angle $\phi$ is :

$</p><p>\tt{\red{\implies{\dfrac{M{L}^2}{3} Sin^2 (\phi)}}}$

The square has four such rods inclined at an angle of 45° to the y axis.

Hence,

$</p><p>\tt{\purple{\implies{I_{2} = 4(\dfrac{M{L}^2}{3} Sin^2 45° ) = \dfrac{2M{L}^2}{3}}}}$

$</p><p>\tt{\orange{I_{1</p><p>NET} = I_{2} + 4M{(\dfrac{\sqrt{2}L}{2})}^2 = \dfrac{8M{L}^2}{3} }.........[A]}$