Solved Problem on Moment of Inertia

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Derive the

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 2
\[
I=I_{CM}+2d\int r\;dm+d^{2}\int dm
\]

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.
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
*r*_{m} and *r*_{M} 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 (

Another system with three different masses with values

\[
\begin{split}
\sum_{i}r_{xi}m_{i} &=-r_{1x}m_{1}+(-r_{2x})m_{2}+r_{3x}m_{3}=\\
&=-r_{1x}m_{1}-r_{2x}m_{2}+r_{3x}m_{3}=0
\end{split}
\]

\[
\begin{split}
\sum_{i}r_{yi}m_{i} &=r_{1y}m_{1}+(-r_{2y})m_{2}+0.m_{3}=\\
&=r_{1y}m_{1}-r_{2y}m_{2}+0=0
\end{split}
\]

The distances from the masses to the center of mass are different, but the masses are also different, this makes

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

\[
\begin{gather}
\bbox[#FFCCCC,10px]
{I=I_{CM}+Md^{2}} \tag{Q.E.D.}
\end{gather}
\]

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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 .