A thin, uniform rod is bent into a square of side length a. If the total mass is

Edmund Adams 2021-11-21 Answered
A thin, uniform rod is bent into a square of side length a. If the total mass is M, find the moment of inertia about an axis through the center and perpendicular to the plane of the square.

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Expert Answer

William Yazzie
Answered 2021-11-22 Author has 291 answers

Consider the square made of 4 thin uniform rods of length a.
The blue small circles in the middle of each rod represents its center of mass.
If we considere a perpendicular axis of rotation through each center of mass: \(\displaystyle{I}_{{{c}{m}}}={\frac{{{1}}}{{{12}}}}{\left({\frac{{{M}}}{{{4}}}}\right)}{a}^{{{2}}}\)
To find the moment of inertia of each rod about an axis through the center of the square and perpendicular to its plane (point P in the figure), use the Parallel - Axis Theorem: \(\displaystyle{I}_{{{p}}}={I}_{{{c}{m}}}+{\left({\frac{{{M}}}{{{4}}}}\right)}{\left({\frac{{{a}}}{{{2}}}}\right)}^{{{2}}}={\frac{{{1}}}{{{12}}}}{\left({\frac{{{M}}}{{{4}}}}\right)}{a}^{{{2}}}+{\frac{{{1}}}{{{4}}}}{\left({\frac{{{M}}}{{{4}}}}\right)}{a}^{{{2}}}={\frac{{{1}}}{{{12}}}}{M}{a}^{{{2}}}\)
The total moment of inertia is equal to four times the value f or each rod (the figure is symmetric): \(\displaystyle{I}_{{{p}\ to{t}{a}{l}}}={4}{I}_{{{p}}}={4}{\left({\frac{{{1}}}{{{12}}}}\right)}{M}{a}^{{{2}}}={\frac{{{1}}}{{{3}}}}{M}{a}^{{{2}}}\)

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Ralph Lester
Answered 2021-11-23 Author has 130 answers
The moment of inertia about an axis through the center and perpendicular to the plane of the square is
\(\displaystyle{I}_{{{s}}}={\frac{{{M}{a}^{{{2}}}}}{{{3}}}}\)
Explanation:
From the question we are told that
The length of one side of the square is a
The total mass of the square is M
Generally the mass of one size of the square is mathematically evaluated as
\(\displaystyle{m}_{{{1}}}={\frac{{{M}}}{{{4}}}}\)
Generally the moment of inertia of one side of the square is mathematically represented as
\(\displaystyle{I}_{{{g}}}={\frac{{{1}}}{{{12}}}}\star{m}_{{{1}}}\star{a}^{{{2}}}\)
Generally given that \(\displaystyle{m}_{{{1}}}={m}_{{{2}}}={m}_{{{3}}}={m}_{{{4}}}={m}\) means that this moment inertia evaluated above apply to every side of the square
Now substituting for \(\displaystyle{m}_{{{1}}}\)
So
\(\displaystyle{I}_{{{g}}}={\frac{{{1}}}{{{12}}}}\star{\frac{{{M}}}{{{4}}}}\star{a}^{{{2}}}\)
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
\(\displaystyle{I}_{{{a}}}={I}_{{{g}}}+{m}{\left[{\frac{{{g}}}{{{2}}}}\right]}^{{{2}}}\)
\(\displaystyle\Rightarrow{I}_{{{a}}}={I}_{{{g}}}+{\frac{{{M}}}{{{4}}}}\star{\left[{\frac{{{g}}}{{{2}}}}\right]}^{{{2}}}\)
substituting for \(\displaystyle{I}_{{{g}}}\)
\(\displaystyle\Rightarrow{I}_{{{a}}}={\frac{{{1}}}{{{2}}}}\star{\frac{{{M}}}{{{4}}}}\star{a}^{{{2}}}+{\frac{{{M}}}{{{4}}}}\star{\left[{\frac{{{g}}}{{{2}}}}\right]}^{{{2}}}\)
\(\displaystyle\Rightarrow{I}_{{{a}}}={\frac{{{M}{a}^{{{2}}}}}{{{48}}}}+{\frac{{{M}{a}^{{{2}}}}}{{{16}}}}\)
\(\displaystyle\Rightarrow{I}_{{{a}}}={\frac{{{M}{a}^{{{2}}}}}{{{12}}}}\)
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
\(\displaystyle{I}_{{{s}}}={4}\star{I}_{{{a}}}\)
\(\displaystyle\Rightarrow{I}_{{{s}}}={4}\star{\frac{{{M}{a}^{{{2}}}}}{{{12}}}}\)
\(\displaystyle\Rightarrow{I}_{{{s}}}={\frac{{{M}{a}^{{{2}}}}}{{{3}}}}\)
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