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One error that occurs when calculating moments of inertia has to do with the parallel axes theorem. The error is to change the moment of inertia of the axis passing through the center of mass with the parallel to it when applying the parallel axes theorem. Realizing that the subscript "CM" in the parallel axes theorem means "center of mass" helps to avoid this mistake. Also checking the response to ensure that the value of the moment of inertia about the center of mass axis is less than the other moment of inertia will detect the error.

## Massezentrum

Consider two particles with the same mass m, each at a different position on the x-axis of a Cartesian coordinate system.

Common sense dictates that the average position of the material making up the two particles is halfway between the two particles. Common sense is correct. We call the mean position of the material that makes up a distribution the name "center of mass," and the center of mass of a pair of particles of equal mass is actually midway between the two particles.

What if one of the particles is more massive than the other? You would expect the center of mass to be closer to the more massive particle, and again you'd be right. In such a case, to determine the position of the center of mass of the matter distribution, we calculate a weighted sum of the positions of the particles in the distribution, where the weighting factor for a given particle is that fraction of the total mass, that is the mass of the particulate matter itself. Thus, for two particles on the \(x\)-axis, one with mass \(m_1\) at \(x_1\) and the other with mass \(m_2\) at \(x_2\),

the position \(\bar{x}\) of the center of mass is given by

\[\bar{x}=\frac{m_1}{m_1+m_2} x_1+ \frac{m_2}{m_1+m_2} x_2 \label{22-1} \]

Note that each weight is a true fraction and the sum of the weights is always 1. Also note that if, for example, \(m_1\) is greater than \(m_2\), the rank \(x_1\ ) of particle 1 will count more in the sum, ensuring that the center of mass is closer to the more massive particle lies (as we know it should be). Also note that if \(m_1=m_2\), each weighting factor is \(\frac{1}{2}\), as is evident if we divide m by \(m_1\) and \(m_2\) replace in equation \(\ref{22-1}\):

\[\bar{x}=\dfrac{m}{m+m} x_1+ \frac{m}{m+m} x_2 \nonúmero\]

\[\bar{x}=\dfrac{1}{2}x_1+\frac{1}{2}x_2 \nonúmero\]

\[\bar{x}=\dfrac{x_1+x_2}{2} \nonumber\]

The center of gravity is halfway between the two particles, exactly where common sense tells us to.

## The center of mass of a thin rod

When asked about the location of the center of mass of a particle distribution, the particle distribution is often the group of particles that form a rigid body. The simplest rigid body to calculate the center of mass is the thin rod, since it only extends in one dimension. (Here we are talking about an ideal thin bar. A physical thin bar must have a non-zero diameter. However, the ideal thin bar is a good approximation of the physical thin bar, so long as the bar's diameter is small compared to its longitude. )

In the simplest case, the calculation of the position of the center of mass is trivial. In the simplest case, it is a uniformly thin bar. A uniformly thin rod is one in which the linear mass density \(\mu\), the mass per length of the rod, has the same value at all points on the rod. The center of mass of a uniform beam is at the center of the beam. For example, the centroid of a uniform bar extending along the x-axis from \(x=0\) to \(x=L\) is at (L/2, 0).

The linear mass density \(\mu\), often also called linear density when the connection is clear, is a measure of how close together the elementary particles that make up the bar are. Where the linear density is high, the particles are close together.

To visualize what a non-uniform rod means, a rod whose linear density is a function of position, imagine a thin rod made of an alloy composed of lead and aluminum. Also imagine that the amount of lead in the rod varies steplessly from 0% at one end of the rod to 100% at the other end. The linear density of such a bar would be a function of position along the bar. A one millimeter segment of the rod in one position would have a different mass than a one millimeter segment of the rod in another position.

People with some exposure to calculus understand more easily what linear density is than people without calculus, since linear density is only the ratio of the mass in a rod segment to the length of the segment, while it is final. Boundary when the boundary segment length goes to zero. Imagine a bar extending from \(0\) to \(L\) along the \(x\) axis. Now suppose that \(m_s(x)\) is the mass of the bar segment extending from \(0\) to \(x\), where \(x\ge0\) but \(x<L \ ) . Then the linear density of the bar at any point x along the bar is simply calculated \(\frac{dm_s}{dx}\) with the \(x\) value in question.

Now that you have a good idea of what we mean by linear mass density, let's use an example to illustrate how you can determine the position of the center of mass of a non-uniform thin rod.

##### Find the position of the center of mass of a thin bar extending from \(0\) to \(.890\)m along the \(x\) axis of a Cartesian coordinate system and having a linear density given by \ ( µ(x)=0.650\frac{kg}{m^3}x^2\).**Solution**

In order to determine the location of the center of mass of a rod of given length and linear density as a function of position, one must first know the mass of that rod. To do this, one might be tempted to use a method that only works for the special case of a uniform bar, ie try \(m=\mu L\), where \(L\) is the length of the bar is. The problem with this is that \(\mu\) varies along the stem. What value would be used for \(\mu\)? One might be tempted to evaluate and use the given \(\mu\) at \(x=L\), but that would act as if the linear density were constant at \(\mu=\mu(L)\ ). It is not. In fact, in the present case \(\mu(L)\) is the maximum linear density of the bar; it only has this value at one point on the bar.

What we can do is say that the infinitesimal mass \(dm\) in a segment \(dx\) of the rod is \(\mu dx\). Here we say that at a position \(x\) on the beam, the amount of mass in the infinitesimal length \(dx\) of the beam is the value of \(\mu\) at that point \(x\). value , multiplied by the infinitesimal length \(dx\). We don't have to worry about \(\mu\) changing with position here, since the segment \(dx\) is infinitely long, which essentially means it has zero length, so the entire segment essentially at is a position \(x\) and so the value of \(\mu\) at that location \(x\) is good for the entire segment \(dx\).

(Video) A thin uniform rod of mass M and length L is bent at its center \[dm=\mu (x) dx \label{22-2} \]

Well, that's true for any value of \(x\), but it only covers an infinitesimally small segment of the bar at \(x\). To get the mass of the whole beam we need to add all these contributions to the mass. Of course, since each \(dm\) corresponds to an infinitesimal length of the bar, we have an infinite number of terms in the sum of all \(dm\). An infinite sum of infinitely small terms is an integral.

\[\int dm=\int_{0}{L} \mu (x) dx \label{22-3} \]

where the values of \(x\) must be between \(0\) and \(L\) to cover the length of the stem, hence the limits on the right. Now, mathematicians have provided us with a rich set of algorithms for evaluating integrals, and indeed we will need access to this toolbox to evaluate the integral on the right-hand side, but to evaluate the integral on the left-hand side, can, may and we will not resort to the algorithm. . Instead, we'll use common sense and our conceptual understanding of what the left integral means. In the context of the present problem, \(\int dm\) means "the sum of all infinitesimal masses that make up the beam". Now if you add up all the infinitesimal pieces of mass that make up the beam, you get the mass of the beam. So \(\int dm\) is just the mass of the beam, which we call \(m\). The equation \(\ref{22-3}\) then becomes

\[m=\int_{0}{L} \mu(x) dx \label{22-4} \]

If I substitute \(\mu (x)\) for the given expression for the linear density \(\mu=0.650 \frac{kg}{m^3} x^2\), I decide that \(\ mu= bx ^ 2 \) with \(b\) defined by \(b=0.650 \frac{kg}{m^3}\) we get

\[m=\int_{0}{L}bx^2dx \nonnumber\]

Factoring constant returns

\[m=b\int_{0}{L}x^2dx \nonúmero\]

When integrating the variable of integration raised to a power, simply increase the power by one and divide by the new power. Is everything

\[m=b\frac{x^3}{3} \Big |_0^L \nonumber \]

Evaluate this using the lower and upper marginal returns

\[ m=b(\frac{L^3}{3}-\frac{0^3}{3}) \nonúmero\]

\[m=\frac{bL^3}{3} \nnúmero\]

The value of \(L\) is given as \(0.890 m\) and we define \(b\) as the constant \(0.650 \frac{kg}{m^3}\) in which for \ ( \ mu \), \(\mu=0.650\frac{kg}{m^3} x^2\), then

\[ m=\frac{0,650 \frac{kg}{m^3} (0,890m)^3}{3} \nonúmero\]

\[m=0,1527kg \nonúmero\]

This is a value useful in calculating the position of the center of mass. Now, when we calculate the center of mass of a set of discrete particles (where a discrete particle is one that exists on its own, as opposed to say a rigid body), we simply perform a weighted sum where each term was the position of a particle by its weight factor, and the weight factor was that fraction of the total mass represented by the mass of the particle. We perform a similar procedure for a continuous mass distribution as constituted by the beam in question. Let's start by writing a single addition term. Consider an infinitesimal length \(dx\) of the bar at a position \(x\) along the bar. As said, the position is \(x\), and the weighting factor is the fraction of the total mass \(m\) of the beam, which represents the mass \(dm\) with infinitesimal length \( dx\). . This means that the weighting factor is \(\frac{dm}{m}\), so one term in our weighted sum of positions looks like this:

\[\frac{dm}{m} x \nonúmero \]

Now \(dm\) can be expressed as \(\mu\)\(dx\), so our expression for the term in the weighted sum can be written as

\[ \frac{\mu dx}{m} x \number \]

This is a term in the weighted sum of positions, the sum that gives the position of the center of mass. The problem is that since the value of \(x\) is not specified, this term is good for any infinitely small segment of the bar. Each term in the sum looks like this. So we have an expression for each term in the sum. Since the expression is valid for an infinitesimal length \(dx\) of the bar, there are obviously an infinite number of terms in the sum. Again we have an infinite sum of infinitesimal terms. That means we have an integral again. Our expression for the location of the center of mass is:

\[\bar{x}=\int_{0}{L} \frac{\mu dx}{m} x \nonumber \]

Let's replace the given expression \(\mu (x)=0.650\frac{kg}{m^3} x^2\) by \(\mu\), which in turn we write as \(\mu=bx^2 \ ) with \(b\) defined by \(b=0.650\frac{kg}{m^3} \), generated

\[\bar{x}=\int_{0}{L} \frac{bx^2 dx}{m} x \nonúmero \]

Rearranging and factoring the constants, we get

(Video) Physics - Application of the Moment of Inertia (6 of 11) Acceleration=? When Pulley Has Mass (mu=0) \[\bar{x}=\frac{b}{m} \int_{0}{L} x^3 dx \nonumber \]

Then we do the integration

\[\bar{x}=\frac{b}{m} \frac{x^4}{4} \Big |_0^L \nonumber \]

\[\bar{x}=\frac{b}{m} (\frac{L^4}{4}-\frac{0^4}{4}) \nonúmero\]

\[\bar{x}=\frac{bL^4}{4m} \nonumber\]

Now we replace values with units; the mass m of the beam that we found earlier, the constant \(b\) that we defined to simplify the look of the linear density function, and the given length \(L\) of the beam:

\[\bar{x}=\frac{(0,650\frac{kg}{m^3})(0,890m)^4}{4(0,1527kg)} \nonúmero\]

\[ \bar{x}=0,668m \nonúmero\]

This is our final answer for the location of the center of mass. Note that it's closer to the denser end of the stem than you'd expect. The reader may also be interested to know that if we had substituted the expression \(m=\frac{bL^3}{3}\) for the mass, which we derived, instead of the value we would have obtained by evaluating this expression, would simplify our expression for \(\bar{x}\) to \(\frac{3}{4} L\), yielding \(\bar{x}=0.668m\), the same result as before.

**Solution**

## Moment of inertia: also called rotational inertia

You already know that the moment of inertia of a rigid object about a given axis of rotation depends on the mass of that object and how that mass is distributed about the axis of rotation. In fact, you know that the object has a lower moment of inertia when the mass is compressed close to the axis of rotation than if the same mass were expanded further around the axis of rotation. Let's quantify these ideas. (Quantify in this context means bring into equation form.)

We begin by building in our minds an idealized object whose mass is concentrated in a single location that is not on the axis of rotation: imagine a massless disk spinning with angular velocity w about an axis defined by the center of the axis runs disk. and perpendicular to their faces. Suppose a particle of mass m is embedded in the disk at a distance \(r\) from the axis of rotation. This is how it looks from the axis of rotation at some distance from the disk:

where the axis of rotation is marked with \(O\). Since the disk has no mass, we call the moment of inertia of the structure the moment of inertia of a particle, related to the rotation about an axis from which the particle is at a distance \(r\).

Knowing that the velocity of the particle can be expressed as \(v=r\omega\), one can show how \(I\) must be defined so that the expression for the kinetic energy \(K=\frac {1} { 2 } I \omega^2\) for the object seen as a rotating rigid body, equal to the kinetic energy term \(K=\frac{1}{2} m v^2\) for moving through space moving particles in a circle. Each view is valid, so both views must produce the same kinetic energy. Go ahead and derive what \(I\) should be, and then go back and read the derivation below.

Here the derivation:

Since \(K=\frac{1}{2} m v^2\), we replace \(v\) with \(r\omega\).

This results in \(K=\frac{1}{2} m (r\omega)^2\)

which can be written as

\[K=\frac{1}{2} (mr^2) \omega^2 \nonumber\]

For this to be equivalent

\[ K=\frac{1}{2} I \omega^2 \nonúmero\]

we must have

\[ me=sir^2 \label{22-5} \]

This is our result for the moment of inertia of a particle of mass \(m\) relative to an axis of rotation from which the particle is at a distance \(r\).

Now suppose we have embedded two particles in our massless disk, one with mass \(m_1\) at a distance \(r_1\) from the axis of rotation and another with mass \(m_2\) at a distance \(r_2 \) the axis of rotation.

The moment of inertia of the first would be only

\[I_1=m_1 r_1^2 \innumber\]

and the moment of inertia of the second particle would be alone

\[I_2=m_2 r_2^2 \sinnzahl\]

The total moment of inertia of the two particles embedded in the massless disc is simply the sum of the two individual moments of inertia.

\[I=I_1+I_2 \no number\]

\[I=m_1 r_1^2+m_2 r_2^2 \innumber\]

This concept can be extended to any number of particles. For each additional particle, simply add another term \(m_i r_i^2\) to the sum, where \(m_i\) is the mass of the additional particle and \(r_i\) is the distance between the additional particle and the axis of Rotation. In the case of a rigid object, we subdivide the object into an infinite set of elements with infinitesimal mass \(dm\). Each mass element contributes a moment of inertia

\[dI=r^2dm\label{22-6}\]

to the moment of inertia of the object, where \(r\) is the distance that the respective mass element has from the axis of rotation.

##### Determine the moment of inertia of the rod in the example \(\ref{22-1}\) in relation to the rotation around the z-axis.**Solution**

In example \(\ref{22-1}\), the linear density of the beam was given as \(\mu=0.650 \frac{kg}{m^3}x^2\). To reduce the number of times we need to write the value in this expression, we write it as \(\mu=bx^2\) with \(b\) defined as \(b=0.650 \frac{kg} {m^3}\).

The total moment of inertia of the bar is the infinite sum of the infinitesimal contributions

\[dI=r^2 dm \label{22-6}\]

of each element of mass dm forming the bar.

In the diagram we have specified an infinitesimal element \(dx\) of the bar at an arbitrary position on the bar. The z-axis, the axis of rotation, looks like a dot on the diagram and the distance \(r\) in \(dI=r^2 dm\), the distance between the considered mass bit and the axis of rotation, is simply the abscissa x the position of the mass element. Therefore, the equation \(\ref{22-6}\) for the case in question can be written as

\[dI=x^2 dm \no number\]

what are we copying here

\[dI=x^2 dm \not a number\]

By defining the linear mass density \(\mu\), the infinitesimal mass \(dm\) can be expressed as \(dm = \mu dx\). Plugging this into our expression for \(dI\) yields

\[dI=x^2 \mu dx \nonúmero \]

Now \(\mu\) was given as \(bx^2\) (where \(b\) is actually the symbol I chose to express the given constant \(0.650 \frac{kg}{m^3 ) to represent } \)). Substituting \(bx^2\) for \(\mu\) in our expression for \(dI\) yields

\[dI=x^2(bx^2)dx \unnumbered\]

\[dI=bx^4dx \uncountable\]

(Video) H. C. Verma Solutions - Chapter 10, Question 22 This expression for the contribution of a member \(dx\) of the member to the total moment of inertia of the member is valid for each member \(dx\) of the member. The infinite sum of all these infinitesimal contributions is therefore the integral

\[\int dI=\int_{0}{L} bx^4 dx \nonúmero\]

As with our last integration on the left, we again don't care about the limits of integration: the infinite sum of all infinitesimal contributions to the moment of inertia is simply the total moment of inertia.

\[I=\int_{0}{L}bx^4 dx \nonúmero\]

On the right, we use the limits of integration \(0\) to \(L\) to include every element of the root that extends from \(x=0\) to \(x=L\), where L is given as \ (0.890 m\). Factoring the constant \(b\) yields

\[I=b \int_{0}{L}x^4 dx \nonúmero \]

Now we perform the integration:

\[I=b\frac{x^5}{5} \Big|_0^L \nonumber \]

\[I=b(\frac{L^5}{5}-\frac{0^5}{5}) \nonúmero\]

\[I=b\frac{L^5}{5} \nonúmero\]

Substituting the given values of \(b\) and \(L\) gives:

\[I=0,650 \frac{kg}{m^3} \frac{(0,890m)^5}{5} \nonúmero\]

\[I=0,0726 kg\cdot m^2 \nonúmero\]

**Solution**

## parallel axis set

We formulate the parallel axis theorem without proof:

\[I=I_{cm}+md^2 \label{22-7} \]

in which:

- \(I\) is the moment of inertia of a body about an axis from which the center of mass of the body is a distance \(d\).
- \(I_{cm}\) is the moment of inertia of the body about an axis parallel to the first axis and passing through the center of mass.
- \(m\) is the mass of the object
- \(d\) is the distance between the two axes.

The parallel axes theorem relates the moment of inertia \(I_{CM}\) of a body with respect to an axis that runs through the center of mass of the body to the moment of inertia I of the same body to with respect to an axis that is parallel to the axis through the center of mass and has a distance d from the axis through the center of mass.

A conceptual assertion of the parallel axes theorem is one that could probably have been reached using common sense, namely that an object's moment of inertia about an axis through the center of mass is less than the moment of inertia. about each axis parallel to it. As you know, the closer the "packed" mass is to the axis of rotation, the lower the moment of inertia; Is; For a given object, by definition of the center of mass, the mass is compressed closer to the axis of rotation when the axis of rotation passes through the center of mass.

##### Find the moment of inertia of the bar of the examples \(\ref{22-1}\) and \(\ref{22-2}\) about an axis perpendicular to the bar and passing through the center of stem mass.**Solution**

Recall that the bar in question extends along the \(x\)-axis from \(x=0\) to \(x=L\) with \(L=0.890 m\) and that the bar has a density which is defined by \(\mu=bL^2\) with \(b=0.650 \frac{kg}{m^3} x^2\).

The relevant axis can be chosen parallel to the z-axis, the axis around which we move the moment of inertia \(I= 0.0726kg \cpoint m^2 \) when solving the example \(\ref{22-2}\). . If we solve the example \(\ref{22-1}\), we find that the mass of the rod is \(m=0.1527 kg\) and the center of gravity of the rod is at a distance of \(d=0.668 m lies \ ) away from the z-axis. Here we present the solution to the problem:

\[I=I_{cm}+md^2 \nonúmero\]

\[I_{cm}=I-md^2 \nonúmero \]

(Video) Physics - Ch 22A Test Your Knowledge: Thermal Expansion (13 of 20) Real Pendulum \[I_{cm}=0,0726kg\cdot m^2-0,1527kg (0,668m)^2 \nonumber\]

\[I_{cm}=0,0047 kg\cdot m^2 \nonumber\]

**Solution**