Solved Problem on Moment of Inertia
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Derive the Parallel Axis Theorem.
Problem diagram:
In Figure 1, dm is an element of mass of the rod, r is the distance from the mass element
to the axis perpendicular to the center of mass of the rod, s is the distance from the mass
element to an axis parallel to the axis passing through the center of mass and d is the distance
between the two axes.
Solution
From Figure 1, we can write
\[
\begin{gather}
s=r+d \tag{I}
\end{gather}
\]
The moment of inertia relative to the axis passing through the center of mass is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{I_{CM}=\int r^{2}\;dm} \tag{II}
\end{gather}
\]
The moment of inertia relative to an axis parallel to the axis passing through the center of mass is given by
\[
\begin{gather}
I=\int s^{2}\;dm \tag{III}
\end{gather}
\]
substituting the expression (I) into expression (III)
\[
I=\int (r+d)^{2}\;dm
\]
Expanding the binomial in the integral
\( (a+b)^{2}=a^{2}+2 ab+b^{2} \)
\[
I=\int r^{2}+2 rd+d^{2}\;dm
\]
the integral of the sum of functions is the sum of the integrals
\[
I=\int r^{2}\;dm+\int 2 rd\;dm+\int d^{2}\;dm
\]
The first integral represents the moment of inertia relative to the axis that passes through the center of
mass given by the expression (I). In the second integral the term 2d is constant and in the third
integral d2 is constant, moving these terms out of the integral
\[
I=I_{CM}+2d\int r\;dm+d^{2}\int dm
\]
In the integral
\( \int r\;dm=0 \),
all elements of mass multiplied by distance and added are equal to zero, relative to the center of
mass of the body.
Note:
\( \int r\;dm=0 \):
Figure 2
Figure 3
The distances of the masses to the center of mass are different (rm > rM), but the masses are also different (m < M), this makes the products rmm e rMM equal and the adds zero.
Another system with three different masses with values m1, m2, and m3 and placed at distances r1 r2 and, r3 from the center of mass of the system. We choose a frame of reference in the center of mass. Decomposing the position vectors in x and y directions, and calculating the product of the masses by the distance to the center of mass in the x and y directions and adding the results (figures 4-A and 4-B)
Figure 4
The distances from the masses to the center of mass are different, but the masses are also different, this makes rimi products in the x and y directions equal and adds zero.
Figure 5
We have a system with two equal masses m and placed the same distance r from the center
of mass of the system. We choose a frame of reference in the center of mass. Calculating the product
of the masses by the distance to the center of mass and adding the results (Figure 2)
Figure 2
\[
\sum_{i}r_{i}m_{1}=rm+(-r)m=rm-rm=0
\]
We have another system with two different masses with m and M and placed at distances
rm and rM from the center of mass of the system. We choose a frame
of reference in the center of mass. Calculating the product of the masses by the distance to the
center of mass and adding the results (Figure 3)
Figure 3
\[
\sum_{i}r_{i}m_{1}=r_{m}m+(-r_{M})M=r_{m}m-r_{M}M=0
\]
The distances of the masses to the center of mass are different (rm > rM), but the masses are also different (m < M), this makes the products rmm e rMM equal and the adds zero.
Another system with three different masses with values m1, m2, and m3 and placed at distances r1 r2 and, r3 from the center of mass of the system. We choose a frame of reference in the center of mass. Decomposing the position vectors in x and y directions, and calculating the product of the masses by the distance to the center of mass in the x and y directions and adding the results (figures 4-A and 4-B)
\[
\begin{split}
\sum_{i}r_{xi}m_{i} &=-r_{1x}m_{1}+(-r_{2x})m_{2}+r_{3x}m_{3}=\\
&=-r_{1x}m_{1}-r_{2x}m_{2}+r_{3x}m_{3}=0
\end{split}
\]
\[
\begin{split}
\sum_{i}r_{yi}m_{i} &=r_{1y}m_{1}+(-r_{2y})m_{2}+0.m_{3}=\\
&=r_{1y}m_{1}-r_{2y}m_{2}+0=0
\end{split}
\]
The distances from the masses to the center of mass are different, but the masses are also different, this makes rimi products in the x and y directions equal and adds zero.
For a rigid body of mass M, we consider mass element dm given by the position vector
r relative to the center of mass of the system. We choose a frame of reference in the center of
mass. As we have a continuous mass distribution we change from the sum to integral. Integration on all
mass elements will be zero (Figure 5)
\[
\int r\;dm=0
\]
Figure 5
The integral \( \int_{0}^{M}dm=M \), represents the total mass of the body.
The moment of inertia of the body relative to an axis parallel to the axis passing through the center of mass will be
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{I=I_{CM}+Md^{2}} \tag{Q.E.D.}
\end{gather}
\]
Note: Q.E.D is the abbreviation of the Latin expression Quod Erat Demonstrandum
which means Which was to be Demonstrated.
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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 .