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Proof 30: Sphere’s moment-of-inertia

Solid sphere (ball) spinning about an axis through its centre-of-mass

Let’s cut the ball into many thin slices. Each slice has a small thickness $\mathrm dr$. Each slice is a circle or disc, and in fact, we can think of it as a very thin solid cylinder where $\mathrm dr$ is its length. A solid cylinder’s moment-of-inertia is:

$$I_\text{cylinder}=\frac12 d^2m\quad\Leftrightarrow\quad \mathrm dI_\text{cylinder}=\frac12 d^2\, \mathrm d m $$

where $d$ is the radius of the cylinder (the distance from sphere-edge to axis-of-rotation and not from edge to the centre). By adding up all moments-of-inertia of these infinitely many thin cylinder-slices, we will get the sphere’s moment-of-inertia. The tricky thing is just that those cylinders vary in radius $d$. We sum them all together with an integral:

$$I=\int \,\mathrm dI_\text{cylinder}= \int \frac 12 d^2\, \mathrm d m$$

Each slice carries a small piece $\mathrm dm$ of the total mass $m$ and covers a small volume $\mathrm dV$ of the total volume $V$. The mass is uniform (evenly spread), so $\mathrm dm$ covers the same amount of the total mass $m$ as $\mathrm d V$ covers of the total volume $V$. Meaning, they are both the same fraction of the total:

$$\frac{\mathrm dm}{m}= \frac{\mathrm dV}{V}\quad\Leftrightarrow \quad \mathrm dm= \frac{m}{V} \mathrm dV $$

Plug this in:

$$I= \int \frac 12 d^2\, \underbrace{\frac{m}{V} \mathrm dV }_{\mathrm d m}= \frac 12 \frac{m}{V} \int d^2\, \mathrm dV $$

Since the slices are very thin, we will ignore the tilted edge and consider them perfect cylinders. Their volumes $\mathrm dV$ will thus be area $A$ times the small thickness $\mathrm dr$:

$$\mathrm dV=A\mathrm dr=\pi d^2\, \mathrm dr $$

Plug this in as well as the formula for the volume $V$ for the full sphere, $V_\text{sphere}=\pi\frac 43 R^3$ (we here choose to symbolise the full sphere radius with an uppercase $R$ for now):

$$I= \frac 12 \frac{m}{V} \int d^2\, \underbrace{ \pi d^2\, \mathrm dr }_{\mathrm dV} = \frac 12 \cancel\pi \frac{m}{ \underbrace{\cancel \pi\frac 43 R^3}_V } \int d^4\, \mathrm dr= \frac 12 \frac 34 \frac{m}{ R^3} \int d^4\, \mathrm dr= \frac 38 \frac{m}{ R^3} \int d^4\, \mathrm dr $$

The cylinder radiuses $d$ depend on how far we are from the centre, $r$. What is their relationship? $r$ is the distance to a cylinder slice from the centre and must thus be perpendicular to that cyclinder. $d$ is parallel to the cylinder, reaching from cylinder-centre to sphere-edge. From that point on the edge there is then the full sphere radius $R$ straight back to the sphere-centre. These three constitute a right-angled triangle in which we’ll use Pythagoras’ Theorem:

$$R^2=r^2+d^2\quad\Leftrightarrow \quad d^2=R^2-r^2$$

Plug this in:

$$ I= \frac 38 \frac{m}{ R^3} \int d^4\, \mathrm dr = \frac 38 \frac{m}{ R^3} \int (d^2)^2\, \mathrm dr = \frac 38 \frac{m}{ R^3} \int (R^2-r^2)^2\, \mathrm dr= \frac 38 \frac{m}{ R^3} \int R^4+r^4-2R^2r^2\, \mathrm dr $$

Integrate over the whole axis-of-rotation, meaning over the full radius on either side of the centre; so from $-R$ to $R$:

$$\begin{align}
I&=\frac 38 \frac{m}{ R^3} \int_{-R}^R R^4+r^4-2R^2r^2\, \mathrm dr\\
&= \frac 38 \frac{m}{ R^3} \left[ rR^4+\frac15 r^5-2R^2\frac 13 r^3\right]_{-R}^R \\
&= \frac 38 \frac{m}{ R^3} \left( \left(RR^4+\frac15 R^5-2R^2\frac 13 R^3\right)-\left( (-R)R^4+\frac15 (-R)^5-2R^2\frac 13 (-R)^3 \right)\right)\\
&= \frac 38 \frac{m}{ R^3} \left( \left(R^5+\frac15 R^5- \frac 23R^5\right)-\left( -R^5-\frac15 R^5+\frac 23R^5 \right)\right)\\
&= \frac 38 \frac{m}{ R^3} \left( R^5+\frac15 R^5- \frac 23R^5+R^5+\frac15 R^5-\frac 23R^5\right)\\
&= \frac 38 \frac{m}{ R^3} \left( 2R^5+\frac25 R^5- \frac 43R^5\right)\\
&= \frac 38 \frac{m}{ R^3} \left( \frac{60}{30}R^5+ \frac{12}{30}R^5- \frac{40}{30} R^5\right)\\
&= \frac 38 \frac{m}{ R^3} \frac{60+12-40}{30}R^5= \frac 38 \frac{m}{ R^3} \frac{32}{30}R^5 = \frac 38 \frac{m}{ R^3} \frac{16}{15}R^5\\
&= \frac 38 \frac{16}{15} \frac{m}{ R^3}R^5= \frac{48}{120} mR^2= \frac25 mR^2\\
\end{align} $$

Let’s rename the full radius to $r$ and then we are done:

$$I= \frac 25 mr^2\qquad\blacksquare $$

Hollow shell (thin-walled empty ball) spinning about an axis through its centre-of-mass

We can repeat the above derivation but start with the moment-of-inertia of a hollow, thin-walled cylinder instead of a solid cylinder:

$$I_\text{hollow cylinder}=d^2m\quad\Leftrightarrow\quad \mathrm dI_\text{hollow cylinder}=d^2\, \mathrm d m$$

This time all mass is carried by the thin surface area $A$. The thin ring that the hollow cylinder constitutes has a small area $\mathrm dA$, which again is the same ratio of the full surface area $A$ as its small mass $\mathrm dm$ is the the total mass $m$:

$$\frac{\mathrm dm}{m}= \frac{\mathrm dA}{A}\quad\Leftrightarrow \quad \mathrm dm= \frac{m}{A} \mathrm dA $$

We’ll integrate over all of these rings and plug that in:

$$I=\int dI_\text{hollow cylinder} =\int d^2\, \mathrm d m =\int d^2\,\frac{m}{A} \mathrm dA= \frac{m}{A} \int d^2\, \mathrm dA $$

The very thin ring with area $\mathrm dA$ tilts slightly. We will ignore the slight curvature that causes either side of the ring to have slightly different lengths. When assuming same lengths, this small area is thus similar to a curved rectangle area which is length around, meaning the circle circumference $2\pi d$, times its width $\mathrm ds$. This tiny width $\mathrm ds$ is simply a geometric bond: $\mathrm ds=R\,\mathrm d\theta$, where $ \mathrm d\theta $ is a tiny angle (a tiny portion of the trip around this ring).

$$\mathrm dA=lw=2\pi d\cdot \mathrm ds= 2\pi d\cdot R\,\mathrm d\theta$$

Plugging this in:

$$I=\frac{m}{A} \int d^2\, \mathrm dA = \frac{m}{A} \int d^2\, 2\pi d R\,\mathrm d\theta = 2\pi R \frac{m}{A} \int d^3\,\mathrm d\theta $$

To do this integral, we need to know how $d$ depends on $\mathrm d\theta$. As used above, $d$, $R$ and $r$ constitute a right-angled triangle. The angle in this triangle is $\theta$. Let’s use the sine relation giving:

$$\sin(\theta)=\frac{\text{opposite side}}{\text{hypotenuse}}=\frac dR\quad\Leftrightarrow\quad d=R\sin(\theta)$$

Plug this in and integrate until the axis-of-rotation is covered. It is covered, when we have integrated through half a round of the sphere, so from $0$ to $\pi$:

$$\begin{align}
I&= 2\pi R \frac{m}{A} \int_0^{\pi} d^3\,\mathrm d\theta\\
&= 2\pi R \frac{m}{A} \int _0^{\pi} \left( R\sin(\theta) \right)^3\,\mathrm d\theta\\
&= 2\pi R \frac{m}{A} \int _0^{\pi} R^3\sin(\theta)^3\,\mathrm d\theta\\
&= R^32\pi R \frac{m}{A} \int _0^{\pi} \sin(\theta)^3\,\mathrm d\theta\\
&= 2\pi R^4 \frac{m}{A} \int _0^{\pi} \sin(\theta)^2\sin(\theta)\,\mathrm d\theta\\
&= 2\pi R^4 \frac{m}{A} \int _0^{\pi} \underbrace{(1-\cos(\theta)^2)}_{\sin(\theta)^2}\sin(\theta)\,\mathrm d\theta\quad\leftarrow \cos(\theta)^2+ \sin(\theta)^2=1^2\\
\end{align}$$

The last step is possible due to Pythagoras’ Theorem in a unit circle. To continue we’ll now use the method of integration by substitution. Let’s choose to substitute $\cos(\theta)=u$. This thus means that $\frac{\mathrm du}{\mathrm d\theta}= \frac{\mathrm d}{\mathrm d\theta}\cos(\theta)=-\sin(\theta) \quad\Leftrightarrow\quad \mathrm d \theta= \frac 1{-\sin(\theta)}\mathrm du$. Let’s substitute these two portions in – note how the range on the integral also must adjust to this new substituted variable $u$ by being inserted in $u$ (meaning inserted in $u=\cos(\theta)$):

$$\begin{align}
I&= 2\pi R^4 \frac{m}{A} \int _0^{\pi} (1-u^2)\sin(\theta)\,\underbrace{ \frac 1{-\sin(\theta)} \mathrm du }_{\mathrm d\theta}\\
&=- 2\pi R^4 \frac{m}{A} \int _{\cos(0)}^{\cos(\pi)} (1-u^2)\cancel{\sin(\theta)}\, \frac 1{\cancel{\sin(\theta)}} \mathrm du\\
&= -2\pi R^4 \frac{m}{A} \int _{1}^{-1} 1-u^2\,\mathrm du \\
&= -2\pi R^4 \frac{m}{A} \left[u-\frac 13u^3\right] _{1}^{-1} \\
&= -2\pi R^4 \frac{m}{A} \left((-1)-\frac 13(-1)^3)-( (1)-\frac 13(1)^3 )\right) \\
&= -2\pi R^4 \frac{m}{A} \left(-1+\frac 13-1+\frac 13\right) = -2\pi R^4 \frac{m}{A} \left(-2+\frac 23\right) \\
&= -2\pi R^4 \frac{m}{A} \left(-\frac 63+\frac 23\right) = -2\pi R^4 \frac{m}{A} \left(-\frac 43\right)\\
&= 2\pi R^4 \frac{m}{A} \frac 43 = \frac 83 \pi R^4 \frac{m}{A}
\end{align} $$

Finally, we plug in the formula for a sphere’s surface area, $A_\text{sphere surface}=4\pi R^2$:

$$I=\frac 83 \pi R^4 \frac{m}{A}= \frac 83 \cancel \pi R^4 \frac{m}{ \underbrace{4\cancel \pi R^2 }_A}=\frac 23m R^2 $$

And again, let’s rename the full radius as $r$ instead of $R$:

$$I= \frac 23m r^2 \qquad\blacksquare$$

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