# Four point masses each of  mass $'m'$, are proved at the corners of a square of side $'l'$. The square is rotating with angular frequency $'\omega '$, about an axis passing through one of the corner of the square and parallel to its diagonal , as shown in figure. The angular momentum of the square about this axis is :- Option: 1 Option: 2 Option: 3 Option: 4

Answers (1)

Since, BD is parallel to axis.

$r_1 = r_3$ Half of the diagonal square $\frac{l}{\sqrt{2}}$

Also,

$r_2$ = Equal to the diagonal of square = ${\sqrt{2}}l$

$So, \\ \left(MO I\right)=\sum M_ir_{i}^{2} \Rightarrow mr_1^{2}+m r_{2}^{2}+m r_{3}^{2}$

Because all the masses are same  = m

By putting the value of r1, r2 , r3

We get

$I = 3 ml^2$

So, Angular momentum =  $I\omega$

Angular momentum = $3ml^2\omega$

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