calculate the moment of inertia for the system. the rods of length l are massless. axis

Explanation:

Step1: Recall moment - of - inertia formula

The moment of inertia of a point mass $m$ at a distance $r$ from the axis of rotation is $I = mr^{2}$.

Step2: Calculate $I$ for each mass

For mass $m$ at distance $L$ from the axis: $I_1=mL^{2}$. For mass $2m$ at distance $\sqrt{L^{2}+L^{2}}=\sqrt{2}L$ from the axis: $I_2 = 2m(\sqrt{2}L)^{2}=4mL^{2}$. For mass $3m$ at distance $L$ from the axis: $I_3=3mL^{2}$. For mass $4m$ at distance $\sqrt{L^{2}+L^{2}}=\sqrt{2}L$ from the axis: $I_4 = 4m(\sqrt{2}L)^{2}=8mL^{2}$.

Step3: Sum up the moments of inertia

$I = I_1+I_2+I_3+I_4=mL^{2}+4mL^{2}+3mL^{2}+8mL^{2}$. $I=(1 + 4+3 + 8)mL^{2}=16mL^{2}$.

Answer:

$16mL^{2}$