For the same total mass,which of the followin | vedclass

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(D) Let the total mass of each body be $m$. The moment of inertia $I$ about an axis passing through the centre of mass and perpendicular to the plane

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Four rods forming a square of side $2a$

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(D) Let the total mass of each body be $m$. The moment of inertia $I$ about an axis passing through the centre of mass and perpendicular to the plane is:$1$. For a disc of radius $a$: $I_{disc} = \frac{1}{2} m a^2 = 0.5 m a^2$$2$. For a ring of radius $a$: $I_{ring} = m a^2 = 1.0 m a^2$$3$. For a square lamina of side $L = 2a$: $I_{square} = \frac{m(L^2 + L^2)}{12} = \frac{m(4a^2 + 4a^2)}{12} = \frac{8 m a^2}{12} = \frac{2}{3} m a^2 \approx 0.67 m a^2$$4$. For four rods of length $2a$ forming a square: Each rod has mass $m/4$. For a single rod of length $L=2a$ rotating about an axis through its centre perpendicular to its length,$I_{rod} = \frac{(m/4)(2a)^2}{12} = \frac{m a^2}{12}$. Using the parallel axis theorem for each rod to shift the axis to the centre of the square (distance $d = a$): $I_{rod, shifted} = I_{rod} + (m/4)a^2 = \frac{m a^2}{12} + \frac{m a^2}{4} = \frac{m a^2 + 3 m a^2}{12} = \frac{4 m a^2}{12} = \frac{1}{3} m a^2$. Total $I = 4 	imes (\frac{1}{3} m a^2) = \frac{4}{3} m a^2 \approx 1.33 m a^2$.Comparing the values: $1.33 m a^2 > 1.0 m a^2 > 0.67 m a^2 > 0.5 m a^2$. Thus,the four rods have the largest moment of inertia.

For the same total mass,which of the following will have the largest moment of inertia about an axis passing through its centre of mass and perpendicular to the plane of the body?

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