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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William Yazzie · Accepted Answer

Take the square with four uniformly thin rods of length a.Each rod&#039;s center of mass is shown by the little blue circles in the centre of each rod.If we considere a perpendicular axis of rotation through each center of mass:&amp;nbsp;Icm=112(M4)a2Use the Parallel - Axis Theorem:&amp;nbsp;Ip=Icm+(M4)(a2)2=112(M4)a2+14(M4)a2=112Ma2The total moment of inertia is equal to four times the value f or each rod (the figure is symmetric):&amp;nbsp;Ip&amp;nbsp;total=4Ip=4(112)Ma2=13Ma2

Ralph Lester · Accepted Answer

Let a the length of one side of the squareLen M the total mass of the square&amp;nbsp;The mass of one size of the squarem1=M4&amp;nbsp;Typically, one side of a square&#039;s moment of inertia is mathematically described asIg=112⋆m1⋆a2&amp;nbsp;m1=m2=m3=m4=m&amp;nbsp;means that this moment inertia evaluated above apply to every side of the square&amp;nbsp;Substituting for&amp;nbsp;m1&amp;nbsp;Ig=112⋆M4⋆a2&amp;nbsp;The moment of inertia of one side of a square about an axis passing through its center and perpendicular to its plane is now theoretically stated as follows by the parallel-axis theorem:Ia=Ig+m[g2]2&amp;nbsp;⇒Ia=Ig+M4⋆[g2]2&amp;nbsp;substituting for&amp;nbsp;Ig&amp;nbsp;⇒Ia=12⋆M4⋆a2+M4⋆[g2]2&amp;nbsp;⇒Ia=Ma248+Ma216&amp;nbsp;⇒Ia=Ma212&amp;nbsp;Is=4⋆Ia&amp;nbsp;⇒Is=4⋆Ma212&amp;nbsp;⇒Is=Ma23

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

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