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 ml^{2}w
Option: 2 2ml^{2}w
Option: 3 3ml^{2}w
Option: 4 4ml^{2}w

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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