\subsection{Moment of Inertia and Mass Distribution}

This subsection introduces the rotational analog of mass for fixed-axis motion. In AP mechanics, the key idea is that both the amount of mass and how far that mass lies from the chosen axis determine how strongly an object resists angular acceleration.

\dfn{Moment of inertia for discrete and continuous mass distributions}{Consider a rigid body about a chosen fixed axis. For a discrete collection of particles, let particle $i$ have mass $m_i$, let $\vec{r}_i$ denote its position vector relative to a point on the axis, and let $r_{\perp,i}$ denote its perpendicular distance to the axis. The \emph{moment of inertia} of the body about that axis is
\[
I=\sum_i m_i r_{\perp,i}^2.
\]
For a continuous mass distribution, let $dm$ denote a small mass element and let $r_\perp$ denote that element's perpendicular distance to the same axis. Then
\[
I=\int r_\perp^2\,dm.
\]
For a single point mass,
\[
I=mr_\perp^2.
\]
The SI unit of moment of inertia is $\mathrm{kg\cdot m^2}$. Because the distance to the axis is squared, mass farther from the axis contributes much more strongly to $I$.}

\wc{Moment of inertia does not depend on speed or rotation}{The moment of inertia $I$ is a geometric property of the mass distribution about an axis. It depends only on the object's mass, shape, and axis position --- \emph{not} on angular velocity, angular acceleration, or torque. A spinning top and a stationary top have the same $I$.}

\thm{Key fixed-axis relations and axis dependence}{Consider a rigid body rotating about a fixed axis with unit vector $\hat{k}$. Let $\vec{\alpha}=\alpha\hat{k}$ denote its angular acceleration, let $\vec{\tau}_{\mathrm{net}}=\tau_{\mathrm{net}}\hat{k}$ denote the net external torque about that axis, and let $I$ denote the moment of inertia about that same axis. Then
\[
\tau_{\mathrm{net}}=I\alpha,
\qquad
\vec{\tau}_{\mathrm{net}}=I\vec{\alpha}.
\]
Thus, for the same net torque, a larger moment of inertia gives a smaller angular acceleration.

Now let $M$ denote the total mass of the body, let $I_{\mathrm{cm}}$ denote the moment of inertia about an axis through the center of mass, and let $d$ denote the perpendicular distance from that axis to a second axis parallel to it. Then the parallel-axis theorem states
\[
I=I_{\mathrm{cm}}+Md^2.
\]
Therefore moment of inertia depends on the chosen axis as well as on the mass distribution. Moving the axis farther from the center of mass increases $I$.}

\nt{The parallel-axis theorem relates the moment of inertia about any axis to the moment of inertia about a parallel axis through the center of mass: $I_{\mathrm{any}}=I_{\mathrm{cm}}+Md^2$, where $d$ is the perpendicular distance between the axes. This is useful for computing $I$ about arbitrary pivot points. The perpendicular-axis theorem ($I_z=I_x+I_y$) applies only to flat planar objects and relates the moment of inertia about an axis perpendicular to the plane to the moments of inertia about two orthogonal in-plane axes.}

\ex{Illustrative example}{A light turntable carries two small clay balls, each of mass $m=0.40\,\mathrm{kg}$. In arrangement A, each ball is at distance $r_A=0.10\,\mathrm{m}$ from the axis. In arrangement B, each ball is at distance $r_B=0.20\,\mathrm{m}$ from the axis.

For arrangement A,
\[
I_A=2mr_A^2=2(0.40)(0.10)^2=8.0\times 10^{-3}\,\mathrm{kg\cdot m^2}.
\]
For arrangement B,
\[
I_B=2mr_B^2=2(0.40)(0.20)^2=3.2\times 10^{-2}\,\mathrm{kg\cdot m^2}.
\]
So doubling the distance from the axis makes the moment of inertia four times as large, even though the total mass is unchanged.}

\nt{Equal mass does not guarantee equal rotational response. Two objects can have the same total mass but different moments of inertia if their mass is distributed differently or if the axis is changed. In fixed-axis rotation, the object with larger $I$ has smaller angular acceleration for the same net torque.}

\qs{Worked example}{A light rigid rod of length $L=0.80\,\mathrm{m}$ lies on a horizontal frictionless table and can rotate about a vertical axle. A small mass $m_1=0.50\,\mathrm{kg}$ is attached at the left end, and a small mass $m_2=1.50\,\mathrm{kg}$ is attached at the right end. First, the axle passes through the center of the rod, so each mass is at distance $r=0.40\,\mathrm{m}$ from the axis. A string pulls perpendicularly on the right end with force magnitude $F=3.0\,\mathrm{N}$.

Find:
\begin{enumerate}[label=(\alph*)]
\item the moment of inertia $I$ about the center axle,
\item the torque magnitude $\tau$ due to the pull,
\item the angular acceleration magnitude $\alpha$, and
\item the new moment of inertia and angular acceleration if the axle is moved to the left end of the rod while the same perpendicular force is still applied at the right end.
\end{enumerate}}

\sol For parts (a)--(c), the axis passes through the center of the rod, so each mass is $r=0.40\,\mathrm{m}$ from the axis. Because the rod is light, treat its mass as negligible and include only the two point masses.

For part (a), the moment of inertia is
\[
I=\sum m_i r_i^2=m_1r^2+m_2r^2.
\]
Substitute the given values:
\[
I=(0.50)(0.40)^2+(1.50)(0.40)^2.
\]
Since $(0.40)^2=0.16$,
\[
I=(0.50)(0.16)+(1.50)(0.16)=0.08+0.24=0.32\,\mathrm{kg\cdot m^2}.
\]
So the moment of inertia about the center axle is
\[
I=0.32\,\mathrm{kg\cdot m^2}.
\]

For part (b), the pull is perpendicular to the rod, so the torque magnitude is
\[
\tau=rF.
\]
Here the force is applied at the right end, which is $0.40\,\mathrm{m}$ from the center axle. Therefore,
\[
\tau=(0.40\,\mathrm{m})(3.0\,\mathrm{N})=1.2\,\mathrm{N\cdot m}.
\]

For part (c), use the rotational form of Newton's second law:
\[
\tau=I\alpha.
\]
Hence,
\[
\alpha=\frac{\tau}{I}=\frac{1.2}{0.32}=3.75\,\mathrm{rad/s^2}.
\]
So for the center axle,
\[
\alpha=3.75\,\mathrm{rad/s^2}.
\]

For part (d), move the axle to the left end. Then the left mass is on the axis, so its distance is $r_1=0$, and the right mass is $r_2=L=0.80\,\mathrm{m}$ from the axis. The new moment of inertia is
\[
I' = m_1r_1^2+m_2r_2^2.
\]
Thus,
\[
I'=(0.50)(0)^2+(1.50)(0.80)^2.
\]
Since $(0.80)^2=0.64$,
\[
I'=0+(1.50)(0.64)=0.96\,\mathrm{kg\cdot m^2}.
\]

Now the same perpendicular force is applied at the right end, which is $0.80\,\mathrm{m}$ from the new axis, so the torque magnitude is
\[
\tau'=(0.80\,\mathrm{m})(3.0\,\mathrm{N})=2.4\,\mathrm{N\cdot m}.
\]
Then
\[
\alpha'=\frac{\tau'}{I'}=\frac{2.4}{0.96}=2.50\,\mathrm{rad/s^2}.
\]

Therefore, with the axle at the left end,
\[
I'=0.96\,\mathrm{kg\cdot m^2},
\qquad
\alpha'=2.50\,\mathrm{rad/s^2}.
\]
Even though the torque becomes larger, the angular acceleration becomes smaller because much more of the mass is farther from the axis, so the moment of inertia increases substantially.
