JEE MAIN - Physics (2020 - 6th September Morning Slot - No. 17)
Four point masses, each of mass m, are fixed 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 corners of the square
and parallel to its diagonal, as shown in the figure. The angular momentum of the square about this
axis is :
_6th_September_Morning_Slot_en_17_1.png)
3m$$l$$2$$\omega $$
4m$$l$$2$$\omega $$
m$$l$$2$$\omega $$
2m$$l$$2$$\omega $$
Explanation
_6th_September_Morning_Slot_en_17_2.png)
I = $$m\left( {{{{l^2}} \over 2}} \right) \times 2 + m \times {\left( {\sqrt 2 l} \right)^2}$$
= $$3m{l^2}$$
$$ \therefore $$ L = I$$\omega $$ = $$3m{l^2}\omega $$
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