Grade 12Mechanics

Q.What is the moment of inertia of a cube about its main diagonal and how to find it?

sunindra kanghujam

14 Years agoGrade 12

3 Answers

Aman Bansal

14 Years ago

Dear Sunindra

FIrst lets find moment Jz of a³ cube around its z-axis:

dJz = r² dm = (x² + y²) ρ dx dy dz
dJz = (x² + y²) aρ dx dy
Jz = integral (x² + y²) aρ dx dy
Jz = 2 integral x² aρ dx dy
Jz = 2a²ρ integral x² dx
Jz = 2/3 a²ρ [x³ from -a/2 to a/2]
Jz = 1/6 ma²

Since Jx=Jx=Jz, and x,y,z are obviously main axes of
tensor Jij of inertia, the elliopsoid of inertia of a cube
is perfectly degerate spherical.

Answer:
Moment of inertia of a cube around any axis
passing throug it center of mass is ma²/6

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Aman Bansal

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Gravemind

8 Years ago

Divide cube into 8 equal cubes. Let moment of inertia of each be I. Then moment of inertia of bigger cube is 32I since mass becomes 8 times and side length becomes 2 times. Applying parallel axes theorem, we get the equation 32I = 8I + 6(2/3 ML^2) solving which, we get I = ML^2/6 , which is the required value.

ankit singh

6 Years ago

ivide cube into 8 equal cubes. Let moment of inertia of each be I. Then moment of inertia of bigger cube is 32I since mass becomes 8 times and side length becomes 2 times. Applying parallel axes theorem, we get the equation 32I = 8I + 6(2/3 ML^2) solving which, we get I = ML^2/6 , which is the required value.