Solved Problem on Moment of Inertia
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Derive the Perpendicular Axis Theorem.
Problem diagram:
In Figure 1, dm is a mass element of the body, and r is the distance from the mass element
to a perpendicular axis to the body (not necessarily passing through the Center of Mass).
Solution
In Figure 1, the z-axis is perpendicular to the plane containing the body, and the x and y axes are in the same plane as the body.
The moment of inertia relative to the z-axis is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{I_{z}=\int r^{2}\;dm} \tag{I}
\end{gather}
\]
Writing the moment of inertia for the two axes n the same plane as the body
\[
\begin{gather}
I_{x}=\int x^{2}\;dm \tag{II-a}
\end{gather}
\]
\[
\begin{gather}
I_{y}=\int y^{2}\;dm \tag{II-b}
\end{gather}
\]
Applying the Pythagorean Theorem (Figure 1)
\[
\begin{gather}
r^{2}=x^{2}+y^{2} \tag{III}
\end{gather}
\]
substituting the expression (II) into expression (I)
\[
\begin{gather}
I=\int \left(x^{2}+y^{2}\right)\;dm \tag{IV}
\end{gather}
\]
the integral of the sum of functions is the sum of the integrals
\[
\begin{gather}
I=\int x^{2}\;dm+\int y^{2}\;dm \tag{V}
\end{gather}
\]
substituting the expressions (II-a) and (II-B) into expression (V)
\[ \bbox[#FFCCCC,10px]
{I=I_{x}+I_{y}} \tag{Q.E.D.}
\]
Observação: Q.E.D é a abreviação da expressão em latim Quod Erat Demonstrandum que
significa Como Queríamos Demonstrar.
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