The moment of inertia of the system about an axis through the center of the square, perpendicular to the plane is 0.064 kg.m²
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Further explanation
Complete Question:
Four small spheres, each of which you can regard as a point of mass 0.200 kg, are arranged in a square 0.400 m on a side and connected by light rods. Find the moment of inertia of the system about an axis through the center of the square, perpendicular to the plane!
Given:
mass of sphere = m = 0.200 kg
length of side = x = 0.400 m
Asked:
net moment of inertia = ΣI = ?
Solution:
Firstly , let's find distance between center of the square and the sphere:
[tex]R = \frac{1}{2} \sqrt{x^2+x^2}[/tex]
[tex]R = \frac{1}{2} \sqrt{2x^2}[/tex]
[tex]R = \frac{1}{2}\sqrt{2} x[/tex]
[tex]R = \frac{1}{2} \sqrt{2} (0.400)[/tex]
[tex]R = 0.200\sqrt{2} \texttt{ m}[/tex]
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Next , we could find total moment of inertia as follows:
[tex]\Sigma I = mR^2 + mR^2 + mR^2 + mR^2[/tex]
[tex]\Sigma I = 4mR^2[/tex]
[tex]\Sigma I = 4(0.200)(0.200\sqrt{2})^2[/tex]
[tex]\Sigma I = 0.064 \texttt{ kgm}^2[/tex]
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Learn more
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Answer details
Grade: High School
Subject: Physics
Chapter: Circular Motion
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Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant
