Three uniform similar rods of length L and ma | vedclass

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(C) Let the three rods be $A$,$B$,and $C$. The axis passes through the top vertex perpendicular to the plane.For rod $A$ and rod $B$,the axis passes t

Vedclass · Accepted Answer

$\frac{L}{\sqrt{2}}$

Vedclass · Answer

(C) Let the three rods be $A$,$B$,and $C$. The axis passes through the top vertex perpendicular to the plane.For rod $A$ and rod $B$,the axis passes through one end,so their moment of inertia is $I_A = I_B = \frac{mL^2}{3}$.For rod $C$,the distance of its center of mass from the axis is $d = L \cos(30^\circ) = \frac{\sqrt{3}L}{2}$.Using the parallel axis theorem,the moment of inertia of rod $C$ about the axis is $I_C = I_{cm} + md^2 = \frac{mL^2}{12} + m\left(\frac{\sqrt{3}L}{2}\right)^2 = \frac{mL^2}{12} + \frac{3mL^2}{4} = \frac{mL^2 + 9mL^2}{12} = \frac{10mL^2}{12} = \frac{5mL^2}{6}$.The total moment of inertia $I_{total} = I_A + I_B + I_C = \frac{mL^2}{3} + \frac{mL^2}{3} + \frac{5mL^2}{6} = \frac{2mL^2 + 2mL^2 + 5mL^2}{6} = \frac{9mL^2}{6} = \frac{3mL^2}{2}$.The total mass of the system is $M = 3m$. The radius of gyration $K$ is defined by $I_{total} = MK^2$.$\frac{3mL^2}{2} = (3m)K^2 \implies K^2 = \frac{L^2}{2} \implies K = \frac{L}{\sqrt{2}}$.

Three uniform similar rods of length $L$ and mass $m$ are joined to form an equilateral triangle. Find the radius of gyration about an axis passing through one corner and perpendicular to the plane of the triangle.

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