Question

Four particles each of mass are placed at the four corners of a square having edge length "l".The moment of inertia of system about an axis along its one diagonal is​

Answer 1

Answer:

option (a) 4ml2

Explanation:

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Answer 2

Given:

Four particles each of mass are placed at the four corners of a square having edge length "l".

To find:

Moment of inertia of the system about an axis along one of its diagonal.

Calculation:

The two particles present along the axis will not contribute to Moment Of Inertia.

So, the perpendicular distance of the other two particles from the axis is:

$AB = \dfrac{1}{2} \bigg( \sqrt{ {l}^{2} + {l}^{2} } \bigg)$

$= > AB = \dfrac{1}{2} \bigg( \sqrt{ 2{l}^{2} } \bigg)$

$= > AB = \dfrac{1}{2} \bigg(l \sqrt{ 2 } \bigg)$

$= > AB = \dfrac{l}{ \sqrt{2} }$

Now , total moment of inertia of the system:

$I = m {r}^{2} + m {r}^{2}$

$= > I = m {( \dfrac{l}{ \sqrt{2} } )}^{2} + m {( \dfrac{l}{ \sqrt{2} } )}^{2}$

$= > I =2 \times m {( \dfrac{l}{ \sqrt{2} } )}^{2}$

$= > I =2 \times \dfrac{m {l}^{2} }{2}$

$= > I =m {l}^{2}$

So, final answer is:

$\boxed{ \bf{I =m {l}^{2} }}$