This document outlines the calculation of the moment of inertia for various common shapes and discusses fundamental concepts linking torque, moment of inertia, and angular momentum.

Key Relationships

1. Torque ($\tau$): The rotational equivalent of force. It is a measure of how much a force acting on an object causes that object to rotate. For a force $F$ applied at a position vector $r$ relative to the axis of rotation, the torque is $\vec{\tau} = \vec{r} \times \vec{F}$. Its magnitude is often $rF\sin\theta$, where $\theta$ is the angle between $r$ and $F$.
2. Moment of Inertia ($I$): The rotational equivalent of mass. It measures an object's resistance to changes in its rotational motion (i.e., resistance to angular acceleration). It depends on the object's mass distribution relative to the axis of rotation. For a system of point masses $m_i$ at distances $r_i$ from the axis, $I = \sum m_i r_i^2$. For a continuous body, it's calculated by integration:
   $$ I = \int r^2 \, dm $$ where $r$ is the perpendicular distance of the mass element $dm$ from the axis of rotation.
3. Angular Acceleration ($\alpha$): The rate of change of angular velocity ($\omega$). $\alpha = d\omega/dt$.
4. Newton's Second Law for Rotation: The net external torque acting on a rigid body is equal to the product of its moment of inertia and its angular acceleration about the axis of rotation.
   $$ \tau_{net} = I \alpha $$
5. Angular Momentum ($L$): The rotational equivalent of linear momentum. For a rigid body rotating with angular velocity $\omega$ about an axis of symmetry, its angular momentum is:
   $$ L = I \omega $$ More generally, for a point particle with linear momentum $p$ at position $r$ relative to the origin, $\vec{L} = \vec{r} \times \vec{p}$.
6. Torque and Angular Momentum: The net external torque acting on a system is equal to the rate of change of the system's angular momentum.
   $$ \vec{\tau}_{net} = \frac{d\vec{L}}{dt} $$ If the net external torque is zero, the total angular momentum of the system is conserved ($\vec{L} = \text{constant}$).

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1. Moment of Inertia of a Uniform Disk

Problem Statement

Calculate the moment of inertia $I$ of a uniform disk with mass $m$ and radius $R$, rotating about an axis passing through its center $O$ and perpendicular to the plane of the disk.

Solution using Integration

1. Choose an Infinitesimal Element: Consider the disk composed of many infinitesimal thin rings, each with radius $r$ ($0 \le r \le R$) and infinitesimal width $dr$.
2. Define Mass Density: The surface mass density is $\sigma = \frac{m}{\pi R^2}$.
3. Calculate Mass of the Ring ($dm$): The area $dA = (2\pi r) dr$. The mass $dm = \sigma \, dA = \left(\frac{m}{\pi R^2}\right) (2\pi r \, dr) = \frac{2m}{R^2} r \, dr$.
4. Calculate Moment of Inertia of the Ring ($dI$): The moment of inertia of the ring is $r^2 dm$.
   $$ dI = r^2 \left( \frac{2m}{R^2} r \, dr \right) = \frac{2m}{R^2} r^3 \, dr $$
5. Integrate to Find Total Moment of Inertia ($I$): Integrate $dI$ from $r=0$ to $r=R$:
   $$ I = \int_{0}^{R} \frac{2m}{R^2} r^3 \, dr = \frac{2m}{R^2} \left[ \frac{r^4}{4} \right]_{0}^{R} = \frac{2m}{R^2} \left( \frac{R^4}{4} \right) $$

Final Result

The moment of inertia of the uniform disk about its central perpendicular axis is:

$$ I = \frac{1}{2} m R^2 $$

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2. Moment of Inertia of a Uniform Spherical Shell

Problem Statement

A uniform spherical shell has mass $m$ and radius $R$. Find its moment of inertia $I$ when rotating about an axis passing through a diameter.

Solution using Integration

1. Strategy: Divide the shell into infinitesimal thin rings perpendicular to the axis of rotation (z-axis). Use spherical coordinates $(R, \theta, \phi)$.
2. Consider an Infinitesimal Ring: Take a band between angles $\theta$ and $\theta + d\theta$. The radius of this ring is $r = R \sin\theta$. Its width along the surface is $R \, d\theta$.
3. Calculate Mass of the Ring ($dm$): Surface density $\sigma = \frac{m}{4\pi R^2}$. Area of the ring $dA = (2\pi r) (R \, d\theta) = 2\pi R^2 \sin\theta \, d\theta$. Mass $dm = \sigma \, dA = \left(\frac{m}{4\pi R^2}\right) (2\pi R^2 \sin\theta \, d\theta) = \frac{m}{2} \sin\theta \, d\theta$.
4. Calculate Moment of Inertia of the Ring ($dI$): $dI = dm \cdot r^2 = dm \cdot (R \sin\theta)^2$
   $$ dI = \left(\frac{m}{2} \sin\theta \, d\theta\right) (R^2 \sin^2\theta) = \frac{m R^2}{2} \sin^3\theta \, d\theta $$
5. Integrate to Find Total Moment of Inertia ($I$): Integrate $dI$ as $\theta$ goes from $0$ to $\pi$:
   $$ I = \int_{0}^{\pi} \frac{m R^2}{2} \sin^3\theta \, d\theta = \frac{m R^2}{2} \int_{0}^{\pi} \sin^3\theta \, d\theta $$ The integral $\int_{0}^{\pi} \sin^3\theta \, d\theta = \frac{4}{3}$ (using substitution $u = \cos\theta$).
   $$ I = \frac{m R^2}{2} \times \frac{4}{3} $$

Final Result

The moment of inertia of the uniform spherical shell about a diameter is:

$$ I = \frac{2}{3} m R^2 $$

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3. Moment of Inertia of a Thin Uniform Rod

Consider a thin uniform rod of mass $m$ and length $L$.

A. Rotation About the Center

Calculate the moment of inertia $I_{cm}$ when the rod rotates about an axis passing through its center of mass (CM) and perpendicular to its length.

1. Strategy: Place the rod along the x-axis from $x = -L/2$ to $x = +L/2$. The axis of rotation is the y-axis (passing through $x=0$).
2. Mass Density: The linear mass density (mass per unit length) is $\lambda = \frac{m}{L}$.
3. Choose an Infinitesimal Element ($dm$): Consider a small segment of the rod of length $dx$ at position $x$. Its mass is $dm = \lambda \, dx = \frac{m}{L} \, dx$.
4. Moment of Inertia of the Element ($dI$): This element is at a distance $r = |x|$ from the axis of rotation. Its moment of inertia is $dI = r^2 dm = x^2 dm$.
   $$ dI = x^2 \left( \frac{m}{L} \, dx \right) = \frac{m}{L} x^2 \, dx $$
5. Integrate to Find Total Moment of Inertia ($I_{cm}$): Integrate $dI$ from $x = -L/2$ to $x = +L/2$:
   $$ I_{cm} = \int_{-L/2}^{+L/2} \frac{m}{L} x^2 \, dx = \frac{m}{L} \int_{-L/2}^{+L/2} x^2 \, dx $$ Evaluate the integral:
   $$ \int_{-L/2}^{+L/2} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{-L/2}^{+L/2} = \frac{1}{3} \left[ \left(\frac{L}{2}\right)^3 - \left(-\frac{L}{2}\right)^3 \right] $$ $$ = \frac{1}{3} \left[ \frac{L^3}{8} - \left(-\frac{L^3}{8}\right) \right] = \frac{1}{3} \left[ \frac{L^3}{8} + \frac{L^3}{8} \right] = \frac{1}{3} \left( \frac{2L^3}{8} \right) = \frac{L^3}{12} $$ Substitute back:
   $$ I_{cm} = \frac{m}{L} \left( \frac{L^3}{12} \right) $$

Final Result (Center)

The moment of inertia of a thin uniform rod about its center is:

$$ I_{cm} = \frac{1}{12} m L^2 $$

B. Rotation About One End

Calculate the moment of inertia $I_{end}$ when the rod rotates about an axis passing through one end and perpendicular to its length.

1. Strategy: Place the rod along the x-axis from $x = 0$ to $x = L$. The axis of rotation is the y-axis (passing through $x=0$).
2. Mass Density: $\lambda = \frac{m}{L}$.
3. Choose an Infinitesimal Element ($dm$): A small segment of length $dx$ at position $x$. Its mass is $dm = \lambda \, dx = \frac{m}{L} \, dx$.
4. Moment of Inertia of the Element ($dI$): The element is at a distance $r = x$ from the axis. $dI = r^2 dm = x^2 dm$.
   $$ dI = x^2 \left( \frac{m}{L} \, dx \right) = \frac{m}{L} x^2 \, dx $$
5. Integrate to Find Total Moment of Inertia ($I_{end}$): Integrate $dI$ from $x = 0$ to $x = L$:
   $$ I_{end} = \int_{0}^{L} \frac{m}{L} x^2 \, dx = \frac{m}{L} \int_{0}^{L} x^2 \, dx $$ Evaluate the integral:
   $$ \int_{0}^{L} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{L} = \frac{L^3}{3} - 0 = \frac{L^3}{3} $$ Substitute back:
   $$ I_{end} = \frac{m}{L} \left( \frac{L^3}{3} \right) $$

Final Result (End)

The moment of inertia of a thin uniform rod about one end is:

$$ I_{end} = \frac{1}{3} m L^2 $$

(Alternatively, the result for rotation about the end can be found using the Parallel Axis Theorem: $I = I_{cm} + Md^2$, where $d$ is the distance between the parallel axes. Here, $d = L/2$. $I_{end} = I_{cm} + m(L/2)^2 = \frac{1}{12}mL^2 + m\frac{L^2}{4} = \frac{1}{12}mL^2 + \frac{3}{12}mL^2 = \frac{4}{12}mL^2 = \frac{1}{3}mL^2$)

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