The ultimate goal is still to arrive at the Euler equations for rigid body dynamics. I often find that equations are neatly derived in books and in class but they often prove to be almost useless to the student because students are not comfortable in the actual calculation of the quantities in the equations. We so far have introduced rotational motion and gone through it in some detail. Thus the idea of jumping into the rotational reference frame in order to use these equations should not be foreign. The next step is to learn how to find center of mass, and moments of inertia of various shapes. The goal is to try and make the derivations as straight forward as possible, which means using brute force methods. We shall leave the trickery and elegance for pretentious physics professors and authors. I am a simple minded man and I hope you are too.
Our first goal is to be comfortable finding the center of mass, then we shall move to finding the moment of inertia. Along this journey we shall meet interesting concepts like the parallel axis theorem and the less known perpendicular axis theorem.
First let's begin simply, we shall find the center of mass for a 1D line, cylinder, sphere and a cone. The equation is easily recalled namely: \(\int_{V}\vec{r}dm\). The important thing to remember is to take \(\vec{r}\) seriously. This is the mistake I have often made in the past, because \(\vec{r}\) is not in cartesian co-ordinates, before we take the integral this will have to be changed to Cartesian co-ordinates. This point becomes very important when we go to 3D dimensions.
1) 1D line.
\begin{align}
&\frac{\rho}{M}\left( \int_0^L x dx \right) \\
& \rho \frac{x^2}{2}|_{0}^L = \rho\left(\frac{L^2}{2}\right) = L/2\\
&\rho \text{ here is }\frac{M}{L}
\end{align}
2) Cylinder
\begin{align}
&\frac{\rho}{M}\left( \int x \, dV \, \hat{x} + \int y\, dV\, \hat{y}+ \int z \, dV \, \hat{z} \right)\\
&\frac{\rho}{M}\left(\int r\,\cos\phi\, r d\phi dr dz\, \hat{x} + \int r\,\sin\phi\, rd\phi dr dz\,\hat{y}+ \int_0^L \int_0^R\, z\,r\,dr d\phi dz\, \hat{z} \right)
\end{align}
The first two integrals will go to zero since the integral of \(\sin \phi\) and \(\cos \phi \) over 2\(\pi\) radians is zero. So all we need is to concentrate on the last integral.
\begin{align}
&\frac{\rho}{M}\left( \frac{R^2}{2} 2\left(\pi\right) \frac{L^2}{2}\right)\\
&\text{put in the expression for the density of a cylinder to get} \frac{L}{2}
\end{align}
With the cylindrical example it is now clear what people mean when they say,"by symmetry, the center of mass is on this or that axis". It is important that one knows what occurs when options are eliminated by reasons of symmetry. This is why I am doing all the integrals, one has to justify one's intuitions however right they may be.
3) Spherical Shell
\begin{align}
&\frac{\rho}{M}\left( \int R\,\cos \phi \,R^2\, (\ sin\theta)^2\, d\phi d\theta\, \hat{x} + \int \, R\, \sin \phi \, (\sin\theta)^2 \, R^2 d\theta d\phi \hat{y} + \int R \cos \, R^2 \, \sin \theta\, d\theta d\phi\, \hat{z}\right)
\end{align}
I have already plugged in what x, y and z are in spherical co-ordinates to get the above expression
Again, the first two integrals go to zero and doing the third integral also gives you zero, i.e the center of mass is at the origin.
4)Solid Cone
\begin{align}
\frac{\rho}{M}\left( \int \, x \, r d\phi dr dz \, \hat{x}+ \int y \, r d\phi dr dz \, \hat{y}+ \int z\, r\, drd\phi dz \, \hat{z}\right)
\end{align}
Again the first two integrals disappear (could have argued this from the beginning by using symmetry arguments). Now r depends on z the integrals over r and z can't be done completely independently, we will need to write r in terms of z. This will be done by considering the limits. If the cone is balance on its pointed edge then r and z are related in the following manner \(r = \frac{zR}{L}\) . Thus the last integral is the following $$ \frac{\rho}{M}2\pi \left(\int_0^L z\, dz \int_0 ^{\frac{zR}{L}}r\, dr \right)$$ The integral gives \( \frac{\rho \pi R^2 L^2}{4M}\) and plugging for the density gives \(\frac{3L}{4}\)
All other shapes usually found in text books can be attacked from these examples by playing around with the limits.
Moments of Inertia
Again it is quite easy to remember the general equation which is \(\int r^2 \, dm \) but as with most thing s the devil is in the details. The above form is slightly misleading if you have had vector calculus. One is led to be believe that r is from the point dm to the origin but this is wrong. The r in the equation is from the dm to axis of rotation. Do not get the impression that calculation for moment of inertia is like the calculation for center of mass except the r is squared that could not be further from the truth. This point will be illustrated when we calculate the moment of inertia for a solid sphere.
Secondly, it matters what our axis of rotation is, you have to know that before you begin writing down integrals. This point will be illustrated by deriving the parallel axis theorem.
Let \(r_i\) be the distance from some origin to a \(m_i\) and \(r_i^{'}\) be the distance from the center of mass to that same mass and lastly let \(d\) be the distance from the center of mass to the fore mentioned origin. Then the moment of inertia about this origin is clearly, \(\sum r_i^2 m_i\) which is equal to \(\sum (r_i^{'}+d)^2 m_i\) since r = r' + d.
\begin{align}
\sum r_i^2 m_i &= \sum r{'}^2 m_i + \sum d^2 m_i + 2d \cdot \left( \sum r{'}m_i\right) \\
&= \sum r{'}^2 m_i +d^2M + 2d \cdot \left( \sum r' m_i \right)
\end{align}
Look at the last term on the right in parentheses. It is weighted term defined in order that it be zero. Look at the definition of center of mass and put all the terms on one side and you will see what I mean. We thus arrive at the parallel axis theorem. Whenever you find yourself trying to calculate the moment of inertia about some weird axis. STOP! do not just write down integrals. It might be easier to do the integrals about the axis that goes through the center of mass and then use the parallel axis theorem.
1) I of line about the end and about center of mass
\begin{align}
I &= \int r^2 dm = \rho \int r^2 dV \\
&= \rho \int_0^L x^2 dx = \rho\frac{L^3}{3}\\
&= \frac{ML^2}{3}
\end{align}
Now suppose we want to find the moment of inertia about the center of mass, do we re-do the integrals. NO! we apply the parallel axis theorem: \(I = I_{cm}+ I_d\) we just calculated I and the end is separated from the center of mass by length d. Plug and chug to get \(\frac{ML^2}{12}\).
2) I of cylinder about axis
\begin{align}
I &= \int r^2 dm = \rho \int r^2 dV \\
&= \rho \int r^3 \, dr d\phi dz\\
&= \frac{\rho \pi L R^4}{2}\\
&= \frac{MR^2}{2}
\end{align}
3) I of solid sphere about axis (any really)
\begin{align}
I &= \rho \int (r \sin \theta)^2 \, r^2 \sin \theta \, dr d\phi d\theta \\
&= \rho \int (\sin \theta)^3 r^4 \, d\theta d\phi \\
& = \frac{\rho R^5 8\pi}{15}\\
& = \frac{2MR^2}{5}
\end{align}
Notice the \(r \sin\theta\) in the first integral which would not have been there if r was the usual r in spherical co-ordinates and also I have not broken up r into its components x, y and z as in the center of mass case.
4) I of a Spherical shell
There is a rigorous way of assuming it has a finite but appreciable thickness , finding the moment of inertia then taking the limit of the thickness dropping to zero. I will not do that instead I shall I assume it has a finite mass in an infinitesimally thin space. Hence,
\begin{align}
I &= \rho \int r^2 dA\\
&= \rho \int (R \sin \theta)^2\, R^2 \sin \theta \,d\phi d\theta \\
&= \rho \int (\sin \theta)^3 R^4 \, d\theta d\phi\\
I &= \frac{2MR^2}{3}\\
\text { remember} \rho \text{ is }\frac{M}{4\pi R^2}
\end{align}
Our first goal is to be comfortable finding the center of mass, then we shall move to finding the moment of inertia. Along this journey we shall meet interesting concepts like the parallel axis theorem and the less known perpendicular axis theorem.
First let's begin simply, we shall find the center of mass for a 1D line, cylinder, sphere and a cone. The equation is easily recalled namely: \(\int_{V}\vec{r}dm\). The important thing to remember is to take \(\vec{r}\) seriously. This is the mistake I have often made in the past, because \(\vec{r}\) is not in cartesian co-ordinates, before we take the integral this will have to be changed to Cartesian co-ordinates. This point becomes very important when we go to 3D dimensions.
1) 1D line.
\begin{align}
&\frac{\rho}{M}\left( \int_0^L x dx \right) \\
& \rho \frac{x^2}{2}|_{0}^L = \rho\left(\frac{L^2}{2}\right) = L/2\\
&\rho \text{ here is }\frac{M}{L}
\end{align}
2) Cylinder
\begin{align}
&\frac{\rho}{M}\left( \int x \, dV \, \hat{x} + \int y\, dV\, \hat{y}+ \int z \, dV \, \hat{z} \right)\\
&\frac{\rho}{M}\left(\int r\,\cos\phi\, r d\phi dr dz\, \hat{x} + \int r\,\sin\phi\, rd\phi dr dz\,\hat{y}+ \int_0^L \int_0^R\, z\,r\,dr d\phi dz\, \hat{z} \right)
\end{align}
The first two integrals will go to zero since the integral of \(\sin \phi\) and \(\cos \phi \) over 2\(\pi\) radians is zero. So all we need is to concentrate on the last integral.
\begin{align}
&\frac{\rho}{M}\left( \frac{R^2}{2} 2\left(\pi\right) \frac{L^2}{2}\right)\\
&\text{put in the expression for the density of a cylinder to get} \frac{L}{2}
\end{align}
With the cylindrical example it is now clear what people mean when they say,"by symmetry, the center of mass is on this or that axis". It is important that one knows what occurs when options are eliminated by reasons of symmetry. This is why I am doing all the integrals, one has to justify one's intuitions however right they may be.
3) Spherical Shell
\begin{align}
&\frac{\rho}{M}\left( \int R\,\cos \phi \,R^2\, (\ sin\theta)^2\, d\phi d\theta\, \hat{x} + \int \, R\, \sin \phi \, (\sin\theta)^2 \, R^2 d\theta d\phi \hat{y} + \int R \cos \, R^2 \, \sin \theta\, d\theta d\phi\, \hat{z}\right)
\end{align}
I have already plugged in what x, y and z are in spherical co-ordinates to get the above expression
Again, the first two integrals go to zero and doing the third integral also gives you zero, i.e the center of mass is at the origin.
4)Solid Cone
\begin{align}
\frac{\rho}{M}\left( \int \, x \, r d\phi dr dz \, \hat{x}+ \int y \, r d\phi dr dz \, \hat{y}+ \int z\, r\, drd\phi dz \, \hat{z}\right)
\end{align}
Again the first two integrals disappear (could have argued this from the beginning by using symmetry arguments). Now r depends on z the integrals over r and z can't be done completely independently, we will need to write r in terms of z. This will be done by considering the limits. If the cone is balance on its pointed edge then r and z are related in the following manner \(r = \frac{zR}{L}\) . Thus the last integral is the following $$ \frac{\rho}{M}2\pi \left(\int_0^L z\, dz \int_0 ^{\frac{zR}{L}}r\, dr \right)$$ The integral gives \( \frac{\rho \pi R^2 L^2}{4M}\) and plugging for the density gives \(\frac{3L}{4}\)
All other shapes usually found in text books can be attacked from these examples by playing around with the limits.
Moments of Inertia
Again it is quite easy to remember the general equation which is \(\int r^2 \, dm \) but as with most thing s the devil is in the details. The above form is slightly misleading if you have had vector calculus. One is led to be believe that r is from the point dm to the origin but this is wrong. The r in the equation is from the dm to axis of rotation. Do not get the impression that calculation for moment of inertia is like the calculation for center of mass except the r is squared that could not be further from the truth. This point will be illustrated when we calculate the moment of inertia for a solid sphere.
Secondly, it matters what our axis of rotation is, you have to know that before you begin writing down integrals. This point will be illustrated by deriving the parallel axis theorem.
Let \(r_i\) be the distance from some origin to a \(m_i\) and \(r_i^{'}\) be the distance from the center of mass to that same mass and lastly let \(d\) be the distance from the center of mass to the fore mentioned origin. Then the moment of inertia about this origin is clearly, \(\sum r_i^2 m_i\) which is equal to \(\sum (r_i^{'}+d)^2 m_i\) since r = r' + d.
\begin{align}
\sum r_i^2 m_i &= \sum r{'}^2 m_i + \sum d^2 m_i + 2d \cdot \left( \sum r{'}m_i\right) \\
&= \sum r{'}^2 m_i +d^2M + 2d \cdot \left( \sum r' m_i \right)
\end{align}
Look at the last term on the right in parentheses. It is weighted term defined in order that it be zero. Look at the definition of center of mass and put all the terms on one side and you will see what I mean. We thus arrive at the parallel axis theorem. Whenever you find yourself trying to calculate the moment of inertia about some weird axis. STOP! do not just write down integrals. It might be easier to do the integrals about the axis that goes through the center of mass and then use the parallel axis theorem.
1) I of line about the end and about center of mass
\begin{align}
I &= \int r^2 dm = \rho \int r^2 dV \\
&= \rho \int_0^L x^2 dx = \rho\frac{L^3}{3}\\
&= \frac{ML^2}{3}
\end{align}
Now suppose we want to find the moment of inertia about the center of mass, do we re-do the integrals. NO! we apply the parallel axis theorem: \(I = I_{cm}+ I_d\) we just calculated I and the end is separated from the center of mass by length d. Plug and chug to get \(\frac{ML^2}{12}\).
2) I of cylinder about axis
\begin{align}
I &= \int r^2 dm = \rho \int r^2 dV \\
&= \rho \int r^3 \, dr d\phi dz\\
&= \frac{\rho \pi L R^4}{2}\\
&= \frac{MR^2}{2}
\end{align}
3) I of solid sphere about axis (any really)
\begin{align}
I &= \rho \int (r \sin \theta)^2 \, r^2 \sin \theta \, dr d\phi d\theta \\
&= \rho \int (\sin \theta)^3 r^4 \, d\theta d\phi \\
& = \frac{\rho R^5 8\pi}{15}\\
& = \frac{2MR^2}{5}
\end{align}
Notice the \(r \sin\theta\) in the first integral which would not have been there if r was the usual r in spherical co-ordinates and also I have not broken up r into its components x, y and z as in the center of mass case.
4) I of a Spherical shell
There is a rigorous way of assuming it has a finite but appreciable thickness , finding the moment of inertia then taking the limit of the thickness dropping to zero. I will not do that instead I shall I assume it has a finite mass in an infinitesimally thin space. Hence,
\begin{align}
I &= \rho \int r^2 dA\\
&= \rho \int (R \sin \theta)^2\, R^2 \sin \theta \,d\phi d\theta \\
&= \rho \int (\sin \theta)^3 R^4 \, d\theta d\phi\\
I &= \frac{2MR^2}{3}\\
\text { remember} \rho \text{ is }\frac{M}{4\pi R^2}
\end{align}
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