Answer:
The moment of inertia about an axis through the center and perpendicular to the plane of the square is
[tex]I_s = \frac{Ma^2}{3}[/tex]
Explanation:
From the question we are told that
The length of one side of the square is [tex]a[/tex]
The total mass of the square is [tex]M[/tex]
Generally the mass of one size of the square is mathematically evaluated as
[tex]m_1 = \frac{M}{4}[/tex]
Generally the moment of inertia of one side of the square is mathematically represented as
[tex]I_g = \frac{1}{12} * m_1 * a^2[/tex]
Generally given that [tex]m_1 = m_2 = m_3 = m_4 = m[/tex] it means that this moment inertia evaluated above apply to every side of the square
Now substituting for [tex]m_1[/tex]
So
[tex]I _g= \frac{1}{12} * \frac{M}{4} * a^2[/tex]
Now according to parallel-axis theorem the moment of inertia of one side of the square about an axis through the center and perpendicular to the plane of the square is mathematically represented as
[tex]I_a = I_g + m [\frac{q}{2} ]^2[/tex]
=> [tex]I_a = I_g + {\frac{M}{4} }* [\frac{q}{2} ]^2[/tex]
substituting for [tex]I_g[/tex]
=> [tex]I_a = \frac{1}{12} * \frac{M}{4} * a^2 + {\frac{M}{4} }* [\frac{q}{2} ]^2[/tex]
=> [tex]I_a = \frac{Ma^2}{48} + \frac{Ma^2}{16}[/tex]
=> [tex]I_a = \frac{Ma^2}{12}[/tex]
Generally the moment of inertia of the square about an axis through the center and perpendicular to the plane of the square is mathematically represented as
[tex]I_s = 4 * I_a[/tex]
=> [tex]I_s = 4 * \frac{Ma^2}{12}[/tex]
=> [tex]I_s = \frac{Ma^2}{3}[/tex]
The moment of inertia through the center of a square is [tex]I_s = \frac{1}{3} Ma^2[/tex]
The mass of one side of the square is given as;
[tex]m_1 = \frac{M}{4}[/tex]
The moment of inertia for one side of a square is given as;
[tex]I_g = \frac{1}{12} \times m_1 \times a^2\\\\I_g = \frac{1}{12} \times \frac{M}{4} \times a^2\\\\I_g = \frac{Ma^2}{48}[/tex]
Apply parallel axis theorem;
[tex]I_a = I_g \ + \ m_1(\frac{a}{2} )^2\\\\I_a = I_g \ + \ \frac{M}{4} (\frac{a^2}{4} )\\\\I_a = I_g \ + \ \frac{Ma^2}{16}[/tex]
[tex]I_a = \frac{Ma^2}{48} \ +\ \frac{Ma^2}{16} \\\\I_a = \frac{Ma^2 \ + \ 3Ma^2}{48} \\\\I_a = \frac{4Ma^2}{48} \\\\I_a = \frac{Ma^2}{12}[/tex]
The moment of inertia of the square about an axis through the center and perpendicular to the plane of the square is
[tex]I_s = 4 \times I_a\\\\I_s = 4 \times \frac{Ma^2}{12} \\\\I_s = \frac{Ma^2}{3}[/tex]
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