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Standard XII

Physics

Center of Mass as an Average Point

Two rods OA a...

Question

Two rods OA and OB of equal length and mass are lying in xy plane as shown in figure. Let $I_{x}, I_{y}$ and $I_{z}$ be the moment of inertia of both the rods about x, y and z axis respectively. Then?

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Solution

$I_{x} = I_{y} = 2 [\frac{{m l}^{2}}{3} {s i n}^{2} 45^{0}] = \frac{{m l}^{2}}{3}$

$I_{z} = 2 [\frac{{m l}^{2}}{3}] = \frac{2 {m l}^{2}}{3} I_{x} = I_{y} < I_{z}$

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Two rods OA and OB of equal length and mass are lying in xy plane as shown in figure. Let Ix,Iy and Iz be the moment of inertia of both the rods about x, y and z axis respectively. Then?

Ix=Iy=2[ml23sin2450]=ml23Iz=2[ml23]=2ml23Ix=Iy<Iz

Two rods OA and OB of equal length and mass are lying in xy plane as shown in figure. Let $I_{x}, I_{y}$ and $I_{z}$ be the moment of inertia of both the rods about x, y and z axis respectively. Then?

$I_{x} = I_{y} = 2 [\frac{{m l}^{2}}{3} {s i n}^{2} 45^{0}] = \frac{{m l}^{2}}{3}$

$I_{z} = 2 [\frac{{m l}^{2}}{3}] = \frac{2 {m l}^{2}}{3} I_{x} = I_{y} < I_{z}$