HRK Physics Notes: Rotational Dynamics - Torque, Moment of Inertia & Rotational Motion

House of Physics October 19, 2025

Rotational Dynamics: Complete Physics Guide

Mastering Torque, Moment of Inertia, Rotational Kinetic Energy, and Rotational Motion Equations
Rotational Dynamics Torque Moment of Inertia Rotational Kinetic Energy Parallel Axis Theorem Reading Time: 25 min

📋 Table of Contents

📜 Historical Background

Rotational dynamics developed as an extension of Newtonian mechanics to rotational motion:

  • Leonhard Euler (1707-1783): Developed Euler's equations for rigid body rotation
  • Isaac Newton (1643-1727): Established laws of motion that form the basis for rotational dynamics
  • Joseph-Louis Lagrange (1736-1813): Formulated Lagrangian mechanics for rotational systems
  • William Rowan Hamilton (1805-1865): Developed Hamiltonian mechanics unifying translational and rotational motion

These developments established the mathematical framework for understanding how objects rotate and the forces that cause rotational motion.

Introduction to Rotational Dynamics

🔬 What is Rotational Dynamics?

Rotational dynamics is the study of the motion of objects that rotate about an axis. While linear dynamics deals with straight-line motion, rotational dynamics focuses on circular or rotational motion.

For linear motion, we have the fundamental equation:

\[ F = ma \]

In rotational dynamics, we analyze the rotational motion of rigid bodies about a certain axis of rotation. The resulting motion depends not only on the magnitude of force but also on where and in what direction it's applied.

📝 Understanding Torque

Just as we regard force as a push or pull, we can regard torque as a twist. Torque takes into account both the magnitude of the force and its point of application relative to the axis of rotation.

The effort required to put a body into rotation depends on the distribution of mass about the axis of rotation. If mass is closer to the axis, it's easier to rotate the body, and vice versa.

⚡ The Rotational Analog of Newton's Second Law

The fundamental equation for rotational dynamics is:

\[ \tau = I\alpha \]

Where:

  • \(\tau\) = Torque (rotational analog of force)
  • \(I\) = Moment of inertia (rotational analog of mass)
  • \(\alpha\) = Angular acceleration (rotational analog of linear acceleration)

Unlike mass, moment of inertia is not an intrinsic property of a body - it depends on the axis of rotation and how mass is distributed relative to that axis.

Relationship Between Linear and Rotational Motion

📊 Linear vs. Rotational Motion Comparison

Linear Motion Rotational Motion
Linear Displacement \(x\) Angular Displacement \(\phi\)
Linear Velocity \(v = \frac{dx}{dt}\) Angular Velocity \(\omega = \frac{d\phi}{dt}\)
Linear Acceleration \(a = \frac{dv}{dt}\) Angular Acceleration \(\alpha = \frac{d\omega}{dt}\)
Force \(F = ma\) or \(F = \frac{dp}{dt}\) Torque \(\tau = I\alpha\) or \(\tau = \frac{dL}{dt}\)
Work \(W = \int F dx\) Work \(W = \int \tau d\phi\)
Kinetic Energy \(K = \frac{1}{2}mv^2\) Kinetic Energy \(K = \frac{1}{2}I\omega^2\)
Power \(P = Fv\) Power \(P = \tau\omega\)
Linear Momentum \(p = mv\) Angular Momentum \(L = I\omega\)
Linear Impulse = \(Ft\) Angular Impulse = \(\tau t\)
Mass (Translational Inertia) \(m\) Rotational Inertia \(I\)

💡 Practical Insight

The relationships between linear and rotational motion allow us to solve complex problems by drawing analogies. For example, just as a larger mass requires more force to accelerate, a larger moment of inertia requires more torque to achieve the same angular acceleration.

Rotational Kinetic Energy

🧮 Derivation of Rotational Kinetic Energy

Step 1: Consider a Rigid Body

Consider a rigid body spinning with uniform angular velocity \(\omega\) about an axis of rotation. The object consists of n particles with masses \(m_1, m_2, ..., m_n\) at distances \(r_1, r_2, ..., r_n\) from the axis.

Step 2: Kinetic Energy of Individual Particles

\[ \text{K.E.}_1 = \frac{1}{2} m_1 r_1^2 \omega_1^2 \]
\[ \text{K.E.}_2 = \frac{1}{2} m_2 r_2^2 \omega_2^2 \]
\[ \vdots \]
\[ \text{K.E.}_n = \frac{1}{2} m_n r_n^2 \omega_n^2 \]

Step 3: Total Rotational Kinetic Energy

\[ K.E._{\text{rot}} = K.E._1 + K.E._2 + \cdots + K.E._n \]
\[ = \frac{1}{2} m_1 r_1^2 \omega_1^2 + \frac{1}{2} m_2 r_2^2 \omega_2^2 + \cdots + \frac{1}{2} m_n r_n^2 \omega_n^2 \]

Step 4: Rigid Body Condition

Since the body is rigid, all masses rotate with the same angular velocity \(\omega\):

\[ K.E._{\text{rot}} = \frac{1}{2} (m_1 r_1^2 + m_2 r_2^2 + \cdots + m_n r_n^2) \omega^2 \]

Step 5: Introduce Moment of Inertia

\[ K.E._{\text{rot}} = \frac{1}{2} \left( \sum_{i=1}^n m_i r_i^2 \right) \omega^2 \]
\[ = \frac{1}{2} I \omega^2 \]

Where \(I = \sum_{i=1}^n m_i r_i^2\) is the moment of inertia of the rigid body.

⚡ Key Formula: Rotational Kinetic Energy

\[ K.E._{\text{rot}} = \frac{1}{2} I \omega^2 \]

This formula shows that rotational kinetic energy depends on both the moment of inertia and the square of the angular velocity, analogous to how linear kinetic energy depends on mass and the square of linear velocity.

Parallel Axis Theorem

🔬 Parallel Axis Theorem Definition

The parallel axis theorem states that the moment of inertia of a body about any axis is equal to the moment of inertia about a parallel axis through the center of mass plus the product of the mass of the body and the square of the distance between the two axes.

\[ I = I_{\text{cm}} + Mh^2 \]

Where:

  • \(I\) = Moment of inertia about the new axis
  • \(I_{\text{cm}}\) = Moment of inertia about the center of mass axis
  • \(M\) = Total mass of the body
  • \(h\) = Distance between the two parallel axes

Proof of Parallel Axis Theorem

🧮 Proof of Parallel Axis Theorem

Step 1: Set Up the Coordinate System

Consider a rigid body with its center of mass at the origin O. Let the axis through O be perpendicular to the plane of the figure. Consider another axis parallel to this axis and passing through point P, with OP = h.

Step 2: Consider a Mass Element

Consider a mass element \(m_i\) at point A, with position vector \(\vec{r}_i\) relative to O and \(\vec{r}_i'\) relative to P.

\[ \vec{r}_i' = \vec{r}_i - \vec{h} \]

Step 3: Moment of Inertia About P

\[ I_P = \sum m_i (r_i')^2 \]
\[ = \sum m_i (\vec{r}_i - \vec{h}) \cdot (\vec{r}_i - \vec{h}) \]
\[ = \sum m_i (r_i^2 - 2\vec{r}_i \cdot \vec{h} + h^2) \]

Step 4: Expand the Summation

\[ I_P = \sum m_i r_i^2 - 2\vec{h} \cdot \sum m_i \vec{r}_i + h^2 \sum m_i \]

Step 5: Use Center of Mass Properties

Since O is the center of mass:

\[ \sum m_i \vec{r}_i = 0 \]

Also:

\[ \sum m_i r_i^2 = I_{\text{cm}} \]
\[ \sum m_i = M \]

Step 6: Final Result

\[ I_P = I_{\text{cm}} - 2\vec{h} \cdot 0 + Mh^2 \]
\[ = I_{\text{cm}} + Mh^2 \]

💡 Practical Application

The parallel axis theorem is extremely useful because it allows us to calculate moments of inertia about any axis if we know the moment of inertia about a parallel axis through the center of mass. This saves us from having to perform complex integrations for every possible axis.

Calculating Moments of Inertia

🔬 Moment of Inertia Definition

The moment of inertia of a rigid body about a given axis is defined as the sum of the products of the masses of its particles and the squares of their respective distances from the axis of rotation.

\[ I = \sum_{i=1}^n m_i r_i^2 \]

For continuous mass distributions, this becomes:

\[ I = \int r^2 dm \]

The moment of inertia depends on:

  • The mass of the body
  • The distribution of mass relative to the axis of rotation
  • The position and orientation of the axis of rotation

Thin Uniform Rod (Center Axis)

🧮 Moment of Inertia: Thin Rod About Center

Step 1: Set Up the Problem

Consider a thin uniform rod of length L and mass M. We want to find the moment of inertia about an axis perpendicular to the rod through its center.

Step 2: Define Mass Element

Let the linear mass density be \(\lambda = \frac{M}{L}\). Consider a small element of length dx at a distance x from the center.

\[ dm = \lambda dx = \frac{M}{L} dx \]

Step 3: Moment of Inertia of Element

\[ dI = x^2 dm = x^2 \cdot \frac{M}{L} dx \]

Step 4: Integrate Over the Rod

\[ I = \int_{-L/2}^{L/2} x^2 \cdot \frac{M}{L} dx \]
\[ = \frac{M}{L} \int_{-L/2}^{L/2} x^2 dx \]
\[ = \frac{M}{L} \left[ \frac{x^3}{3} \right]_{-L/2}^{L/2} \]
\[ = \frac{M}{L} \left( \frac{(L/2)^3}{3} - \frac{(-L/2)^3}{3} \right) \]
\[ = \frac{M}{L} \left( \frac{L^3}{24} + \frac{L^3}{24} \right) \]
\[ = \frac{M}{L} \cdot \frac{L^3}{12} \]
\[ = \frac{1}{12} ML^2 \]

Thin Uniform Rod (End Axis)

🧮 Moment of Inertia: Thin Rod About End

Step 1: Set Up the Problem

Consider the same thin uniform rod of length L and mass M, but now we want the moment of inertia about an axis perpendicular to the rod through one end.

Step 2: Define Mass Element

Let the linear mass density be \(\lambda = \frac{M}{L}\). Consider a small element of length dx at a distance x from the end.

\[ dm = \lambda dx = \frac{M}{L} dx \]

Step 3: Moment of Inertia of Element

\[ dI = x^2 dm = x^2 \cdot \frac{M}{L} dx \]

Step 4: Integrate Over the Rod

\[ I = \int_{0}^{L} x^2 \cdot \frac{M}{L} dx \]
\[ = \frac{M}{L} \int_{0}^{L} x^2 dx \]
\[ = \frac{M}{L} \left[ \frac{x^3}{3} \right]_{0}^{L} \]
\[ = \frac{M}{L} \cdot \frac{L^3}{3} \]
\[ = \frac{1}{3} ML^2 \]

💡 Using Parallel Axis Theorem

We can verify this result using the parallel axis theorem:

\[ I_{\text{end}} = I_{\text{center}} + M\left(\frac{L}{2}\right)^2 \]
\[ = \frac{1}{12}ML^2 + \frac{1}{4}ML^2 \]
\[ = \frac{1}{3}ML^2 \]

This confirms our direct integration result.

Hollow and Solid Cylinders

🧮 Moment of Inertia: Hollow Cylinder

Step 1: Set Up the Problem

Consider a hollow cylinder of mass M, length L, with inner radius R₁ and outer radius R₂, rotating about its central axis.

Step 2: Define Mass Element

Consider a thin cylindrical shell at radius r with thickness dr.

\[ \text{Volume of shell} = 2\pi r L dr \]
\[ \text{Mass density} = \rho = \frac{M}{\pi (R_2^2 - R_1^2) L} \]
\[ dm = \rho \cdot 2\pi r L dr = \frac{2M r dr}{R_2^2 - R_1^2} \]

Step 3: Moment of Inertia of Element

\[ dI = r^2 dm = r^2 \cdot \frac{2M r dr}{R_2^2 - R_1^2} \]
\[ = \frac{2M r^3 dr}{R_2^2 - R_1^2} \]

Step 4: Integrate Over the Cylinder

\[ I = \int_{R_1}^{R_2} \frac{2M r^3 dr}{R_2^2 - R_1^2} \]
\[ = \frac{2M}{R_2^2 - R_1^2} \int_{R_1}^{R_2} r^3 dr \]
\[ = \frac{2M}{R_2^2 - R_1^2} \left[ \frac{r^4}{4} \right]_{R_1}^{R_2} \]
\[ = \frac{2M}{R_2^2 - R_1^2} \cdot \frac{R_2^4 - R_1^4}{4} \]
\[ = \frac{M}{2} \cdot \frac{R_2^4 - R_1^4}{R_2^2 - R_1^2} \]
\[ = \frac{M}{2} (R_1^2 + R_2^2) \]

🧮 Moment of Inertia: Solid Cylinder

Step 1: Special Case of Hollow Cylinder

For a solid cylinder, R₁ = 0 and R₂ = R.

Step 2: Apply the Formula

\[ I = \frac{M}{2} (0 + R^2) \]
\[ = \frac{1}{2} MR^2 \]

Disk and Hoop (Ring)

🧮 Moment of Inertia: Solid Disk

Step 1: Set Up the Problem

Consider a solid disk of mass M and radius R, rotating about an axis perpendicular to its plane through its center.

Step 2: Define Mass Element

Consider a thin ring at radius r with thickness dr.

\[ \text{Area of ring} = 2\pi r dr \]
\[ \text{Mass density} = \sigma = \frac{M}{\pi R^2} \]
\[ dm = \sigma \cdot 2\pi r dr = \frac{2M r dr}{R^2} \]

Step 3: Moment of Inertia of Element

\[ dI = r^2 dm = r^2 \cdot \frac{2M r dr}{R^2} \]
\[ = \frac{2M r^3 dr}{R^2} \]

Step 4: Integrate Over the Disk

\[ I = \int_{0}^{R} \frac{2M r^3 dr}{R^2} \]
\[ = \frac{2M}{R^2} \int_{0}^{R} r^3 dr \]
\[ = \frac{2M}{R^2} \left[ \frac{r^4}{4} \right]_{0}^{R} \]
\[ = \frac{2M}{R^2} \cdot \frac{R^4}{4} \]
\[ = \frac{1}{2} MR^2 \]

🧮 Moment of Inertia: Hoop or Ring

Step 1: Set Up the Problem

Consider a thin hoop or ring of mass M and radius R, rotating about an axis perpendicular to its plane through its center.

Step 2: All Mass at Distance R

For a thin hoop, all mass is essentially at distance R from the axis.

\[ I = \sum m_i r_i^2 = \sum m_i R^2 \]
\[ = R^2 \sum m_i \]
\[ = MR^2 \]

Spherical Shell

🧮 Moment of Inertia: Spherical Shell

Step 1: Set Up the Problem

Consider a thin spherical shell of mass M and radius R, rotating about an axis through its center.

Step 2: Define Mass Element

Consider a ring element at angle θ from the axis, with angular width dθ.

\[ \text{Radius of ring} = R \sin\theta \]
\[ \text{Circumference of ring} = 2\pi R \sin\theta \]
\[ \text{Width of ring} = R d\theta \]
\[ \text{Area of ring} = 2\pi R^2 \sin\theta d\theta \]
\[ \text{Mass density} = \sigma = \frac{M}{4\pi R^2} \]
\[ dm = \sigma \cdot 2\pi R^2 \sin\theta d\theta = \frac{M}{2} \sin\theta d\theta \]

Step 3: Moment of Inertia of Element

\[ dI = (\text{radius})^2 dm = (R \sin\theta)^2 \cdot \frac{M}{2} \sin\theta d\theta \]
\[ = \frac{1}{2} MR^2 \sin^3\theta d\theta \]

Step 4: Integrate Over the Sphere

\[ I = \int_{0}^{\pi} \frac{1}{2} MR^2 \sin^3\theta d\theta \]
\[ = \frac{1}{2} MR^2 \int_{0}^{\pi} \sin^3\theta d\theta \]
\[ = \frac{1}{2} MR^2 \int_{0}^{\pi} (1 - \cos^2\theta) \sin\theta d\theta \]
\[ = \frac{1}{2} MR^2 \left[ -\cos\theta + \frac{\cos^3\theta}{3} \right]_{0}^{\pi} \]
\[ = \frac{1}{2} MR^2 \left[ (1 - \frac{1}{3}) - (-1 + \frac{1}{3}) \right] \]
\[ = \frac{1}{2} MR^2 \cdot \frac{4}{3} \]
\[ = \frac{2}{3} MR^2 \]

Solid Sphere

🧮 Moment of Inertia: Solid Sphere

Step 1: Set Up the Problem

Consider a solid sphere of mass M and radius R, rotating about an axis through its center.

Step 2: Define Mass Element

Consider a spherical shell at radius r with thickness dr.

\[ \text{Volume of shell} = 4\pi r^2 dr \]
\[ \text{Mass density} = \rho = \frac{M}{\frac{4}{3}\pi R^3} \]
\[ dm = \rho \cdot 4\pi r^2 dr = \frac{3M r^2 dr}{R^3} \]

Step 3: Moment of Inertia of Element

From the spherical shell result, the moment of inertia of a thin shell is \(\frac{2}{3} mr^2\).

\[ dI = \frac{2}{3} r^2 dm = \frac{2}{3} r^2 \cdot \frac{3M r^2 dr}{R^3} \]
\[ = \frac{2M r^4 dr}{R^3} \]

Step 4: Integrate Over the Sphere

\[ I = \int_{0}^{R} \frac{2M r^4 dr}{R^3} \]
\[ = \frac{2M}{R^3} \int_{0}^{R} r^4 dr \]
\[ = \frac{2M}{R^3} \left[ \frac{r^5}{5} \right]_{0}^{R} \]
\[ = \frac{2M}{R^3} \cdot \frac{R^5}{5} \]
\[ = \frac{2}{5} MR^2 \]

📊 Common Moments of Inertia

Object Axis of Rotation Moment of Inertia
Thin rod, length L Through center, perpendicular to rod \(\frac{1}{12}ML^2\)
Thin rod, length L Through end, perpendicular to rod \(\frac{1}{3}ML^2\)
Hollow cylinder Central axis \(\frac{1}{2}M(R_1^2 + R_2^2)\)
Solid cylinder Central axis \(\frac{1}{2}MR^2\)
Solid disk Through center, perpendicular to plane \(\frac{1}{2}MR^2\)
Hoop or ring Through center, perpendicular to plane \(MR^2\)
Spherical shell Through center \(\frac{2}{3}MR^2\)
Solid sphere Through center \(\frac{2}{5}MR^2\)

Applications and Real-World Examples

🚗 Automotive Engineering

Flywheels in engines use rotational inertia to smooth out power delivery. The moment of inertia of the flywheel determines how effectively it can store rotational energy.

⚽ Sports Physics

The moment of inertia affects how objects rotate in sports. A figure skater spins faster when pulling arms in (reducing moment of inertia) due to conservation of angular momentum.

🏗️ Structural Engineering

Rotational dynamics principles are used in designing rotating machinery, bridges, and buildings to withstand rotational forces and vibrations.

🛰️ Aerospace Engineering

Satellites and spacecraft use reaction wheels with precisely calculated moments of inertia for attitude control and stabilization in space.

🧠 Problem-Solving Strategy

When solving rotational dynamics problems:

  1. Identify the axis of rotation
  2. Determine the appropriate moment of inertia formula
  3. Apply Newton's second law for rotation: \(\tau = I\alpha\)
  4. Use energy methods when appropriate: \(K.E._{\text{rot}} = \frac{1}{2}I\omega^2\)
  5. Apply conservation laws (energy, angular momentum) when applicable

Frequently Asked Questions

❓ Why is moment of inertia different from mass?

Mass is an intrinsic property that measures an object's resistance to linear acceleration. Moment of inertia, however, depends not only on mass but also on how that mass is distributed relative to the axis of rotation.

Key differences:

  • Mass is scalar and independent of axis
  • Moment of inertia depends on the axis of rotation
  • Mass measures resistance to linear acceleration
  • Moment of inertia measures resistance to angular acceleration
❓ Why does a figure skater spin faster when pulling arms in?

This phenomenon demonstrates conservation of angular momentum. When a figure skater pulls their arms in, they decrease their moment of inertia. Since angular momentum \(L = I\omega\) must remain constant (in the absence of external torques), the angular velocity \(\omega\) must increase to compensate.

\[ I_1\omega_1 = I_2\omega_2 \]
\[ \text{If } I_2 < I_1, \text{ then } \omega_2 > \omega_1 \]

The same principle applies to divers, gymnasts, and planetary systems.

❓ What is the physical significance of the parallel axis theorem?

The parallel axis theorem tells us that the moment of inertia about any axis is always greater than or equal to the moment of inertia about a parallel axis through the center of mass. The additional term \(Mh^2\) represents the moment of inertia that the object would have if all its mass were concentrated at the center of mass.

This theorem is extremely useful because:

  • It simplifies calculations of moments of inertia
  • It shows how moment of inertia increases as we move the axis further from the center of mass
  • It helps in understanding why objects tend to rotate about axes through their center of mass

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