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 the axis is:

Let the four corners be labeled such that:

Mass 1 is on the axis: $$r_1 = 0$$

Masses 2 and 3 are at adjacent corners, each at a perpendicular distance equal to half the diagonal: $$r_2 = r_3 = \frac{l}{\sqrt{2}}$$

Mass 4 is at the opposite corner, at a perpendicular distance equal to the full length of the diagonal: $$r_4 = \sqrt{2}l$$

$$I = m_1 r_1^2 + m_2 r_2^2 + m_3 r_3^2 + m_4 r_4^2$$

$$I = m(0)^2 + m\left(\frac{l}{\sqrt{2}}\right)^2 + m\left(\frac{l}{\sqrt{2}}\right)^2 + m\left(\sqrt{2}l\right)^2$$

$$I = 0 + \frac{1}{2}ml^2 + \frac{1}{2}ml^2 + 2ml^2 = 3ml^2$$

$$L = I\omega \implies L = 3ml^2\omega$$