Solved Problem on Moment of Inertia
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A system with four point bodies connected by bars of negligible mass, located at the corners of a square
of side R. Calculate the moment of inertia about an axis passing through the center of the square
and perpendicular to the plane containing the masses at the following cases:
a) The four bodies have masses equal to M;
b) The bodies have masses equal to 1 kg, 2 kg, 3 kg, 4 kg, and R = 2 m.
a) The four bodies have masses equal to M;
b) The bodies have masses equal to 1 kg, 2 kg, 3 kg, 4 kg, and R = 2 m.
Problem data:
- Distance between bodies: R.
a) The distance r from one of the bodies to the center will be half of the diagonal d of the square of side R. Applying the Pythagorean Theorem (Figure 1)
\[
\begin{gather}
d^{2}=R^{2}+R^{2}\\
d^{2}=2R^{2}\\
d=\sqrt{2R^{2}\;}\\
d=R\sqrt{2\;}
\end{gather}
\]
\[
\begin{gather}
r=\frac{R\sqrt{2\;}}{2} \tag{I}
\end{gather}
\]
The moment of inertia about the axis is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{I=\sum _{i=1}^{n}m_{i}r_{i}^{2}} \tag{II}
\end{gather}
\]
as all bodies have the same mass and are the same distance from the center
\[
\begin{gather}
I=M\left(\frac{R\sqrt{2}\;}{2}\right)^{2}+M\left(\frac{R\sqrt{2}\;}{2}\right)^{2}+M\left(\frac{R\sqrt{2}\;}{2}\right)^{2}+M\left(\frac{R\sqrt{2}\;}{2}\right)^{2}\\
I=4M\frac{R^{2}2}{4}
\end{gather}
\]
\[ \bbox[#FFCCCC,10px]
{I=2MR^{2}}
\]
b) Substituting the given masses and the given axis distance by expression (I) into expression (II)
\[
\begin{gather}
I=1\times\left(\frac{2\sqrt{2}\;}{2}\right)^{2}+2\times\left(\frac{2\sqrt{2}\;}{2}\right)^{2}+3\times\left(\frac{2\sqrt{2}\;}{2}\right)^{2}+4\times\left(\frac{2\sqrt{2}\;}{2}\right)^{2}\\
I=1\times2+2\times2+3\times2+4\times2\\
I=2+4+6+8
\end{gather}
\]
\[ \bbox[#FFCCCC,10px]
{I=20\;\text{kg.m}^{2}}
\]
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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .