Moment of Inertia: 4 Spheres Connected by Rods in Square

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SUMMARY

The moment of inertia for a system of four small spheres, each with a mass of 0.200 kg and arranged in a square with a side length of 0.400 m, was calculated for three different axes. For the axis through the center of the square, the moment of inertia is 0.0032 kg·m². For the axis bisecting two opposite sides, the moment of inertia is 0.0192 kg·m². Lastly, for the axis passing through the centers of the upper left and lower right spheres, the moment of inertia is 0.0128 kg·m².

PREREQUISITES
  • Understanding of moment of inertia calculations
  • Familiarity with the parallel axis theorem
  • Basic knowledge of physics concepts related to mass and distance
  • Ability to perform calculations involving squares and distances in physics
NEXT STEPS
  • Study the parallel axis theorem for further applications in moment of inertia
  • Explore the concept of rotational dynamics and its equations
  • Learn about the moment of inertia for different geometric shapes
  • Investigate the effects of mass distribution on moment of inertia
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Homework Statement


Four small spheres, each of which you can regard as a point of mass 0.200 , are arranged in a square 0.400 on a side and connected by light rods.

The picture has 4 spheres connected by rods in the shape of a square. there's a point O in the middle of the square and a horizontal line through the point O with point A on one end and point B on the other end.

a)Find the moment of inertia of the system about an axis through the center of the square, perpendicular to its plane (an axis through point O in the figure).

b)Find the moment of inertia of the system about an axis bisecting two opposite sides of the square (an axis along the line AB in the figure).

c)Find the moment of inertia of the system about an axis that passes through the centers of the upper left and lower right spheres and through point O.


I=m1(r1^2)+m2(r2^2)+...
 
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Homework Equations I=m1(r1^2)+m2(r2^2)+...The Attempt at a Solution a)I=m1(r1^2)+m2(r2^2)+m3(r3^2)+m4(r4^2)I=(0.200)(0.2^2)+(0.200)(0.2^2)+(0.200)(0.2^2)+(0.200)(0.2^2)=0.0032 kgm^2b)I=(0.200)(0.2^2)+(0.200)(0.2^2)+(0.200)(0.2^2)+(0.200)(0.2^2)+(0.200)(0.4^2)=0.0192 kgm^2c)I=(0.200)(0.2^2)+(0.200)(0.2^2)+(0.200)(0.4^2)=0.0128 kgm^2
 

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