04 - Rigid Body

TARGET DECK: MED::I::Medical Physics::04 - Rigid Body

Systems of Points

Foundations

Core Principles

The third law of dynamics and the conservation of momentum and energy are the foundations for studying systems composed of multiple point particles.

The various states of aggregation of matter can be described as systems of point particles with different characteristic interactions:

State	Interaction Type
Liquid	Short-range attraction (approximated)
Perfect gas	Elastic collisions between atoms and container walls; no inter-atomic attraction
Solid	Fixed (or almost fixed) elastic interactions between neighbouring atoms → oscillations and rotations around fixed positions

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Rigid Bodies

Definition

A rigid solid body is an ideal object where the internal distances between atoms do not change. It has a well-defined shape and volume and cannot be deformed.

Comparison of states:

Property	Rigid Solid	Liquid	Gas
Fixed volume	✓	✓	✗
Fixed shape	✓	✗	✗

Why the Rigid Body Approximation is Useful

To describe a rigid body we only need:

• The position of the center of mass
• How the body is rotated in space (3 additional numbers)

We do not need to specify the positions of all individual atoms.

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Properties of the Rigid Body

Some properties are direct extensions of the single point particle:

• Total mass of $N$ point particles: $M={\sum }_{i=1}^N{m}_i$

• Linear momentum: $\vec{Q}={\sum }_{i=1}^N{\vec{q}}_i$

• Total energy: sum of potential and kinetic energies of each point particle.

• Center of mass: the mass-weighted average position of all particles — moves in uniform rectilinear motion in an isolated system.

What is the total mass of a system of N point particles?

$M={\sum }_{i=1}^N{m}_i$

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The linear momentum of a rigid body is the {1:sum of all individual linear momenta} of each point particle: $\vec{Q}={\sum }_{i=1}^N{\vec{q}}_i$.

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Center of Mass vs. Center of Gravity

Any force applied on the body can be divided into two terms:

1. One that moves the center of mass without rotating the body
2. One that rotates the body around an axis through the center of mass without moving it

Center of Gravity

The center of gravity is the center of all forces applied on the body such that there is no net torque.

• In a uniform gravitational field: center of mass = center of gravity
• In a non-uniform field: they differ (e.g., in the Empire State Building, they differ by ~1 mm due to the gravity gradient between top and bottom floors)

When do the center of mass and center of gravity coincide?

In the approximation of a uniform gravitational field. In non-uniform fields they differ (e.g., ~1 mm difference in the Empire State Building).

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Center of Mass and the Fulcrum

• If a rigid bar is placed on a hinge point (the fulcrum):

  • Fulcrum under the CoM → no motion
  • Fulcrum off-center → rotation around the fulcrum (and translation of the CoM)

• For continuous matter (instead of discrete point particles), the CoM is determined by an integral over the mass distribution rather than a summation.

• For simple geometric shapes, the CoM can often be found using geometric considerations.

• Complex objects can be approximated as unions of simple shapes.

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Variations of the Center of Mass with Body Posture

Biomechanical Examples

• Standing: CoM is inside the body
• Bending over: CoM can move outside the body
• Fosbury flop (high jump): the athlete’s body bends over the bar such that the CoM does not need to clear the bar — enabling clearance of greater heights
• Hurdles race: this CoM-outside-body trick does not apply

Why does the Fosbury flop technique allow athletes to clear higher bars?

By bending the body over the bar, the center of mass (CoM) passes below the bar while the body arcs over it — meaning the CoM does not need to exceed the bar height.

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Translational and Rotational Motions

Rotational Analogues of Linear Quantities

Linear Quantity	Rotational Equivalent
Force $\vec{F}$	Torque $\vec{M}$
Linear velocity $\vec{v}$	Angular velocity $\omega$
Linear momentum $\vec{p}$	Angular momentum $\vec{L}$
Mass $m$	Rotational inertia $I$

Motion of the Rigid Body

The motion can be decomposed into:

1. Translation of the center of mass
2. Rotation around the center of mass (described by angular speed $\omega$, analogous to uniform circular motion)

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Angular Momentum

$\vec{L}=\vec{r}\times \vec{p}=\vec{r}\times m\vec{v}$

Direction determined by the right-hand rule:

• Counterclockwise rotation in the $xy$-plane → $\vec{L}$ points in the $+z$ direction
• Clockwise rotation → $\vec{L}$ points in the $-z$ direction

What is the formula for angular momentum?

$\vec{L}=\vec{r}\times \vec{p}=\vec{r}\times m\vec{v}$

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A counterclockwise rotation in the $xy$-plane generates an angular momentum vector pointing in the {1:$+z$ direction} (right-hand rule).

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Biomechanics

Axis of Rotation

Mozzi's Axis

At any moment in time, given the motion of a body, we can identify a single axis of rotation called Mozzi’s axis.

• In simple motions: stays constant in time
• In general: changes over time
• Valid even for motion with no applied force — angular momentum can be calculated relative to any chosen point

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Rotational Equilibrium of a Rigid Bar

If the fulcrum is not under the CoM, a force $F$ must be applied to prevent rotation.

Each force produces a torque equal to the force multiplied by its distance from the fulcrum, with a sign depending on the direction of rotation.

Equilibrium condition:

$F\cdot b-m_{T}g\cdot d=0⟹F=m_{T}g\cdot \frac{d}{b}$

What is the equilibrium condition for a rigid bar on a fulcrum?

The sum of all torques must be zero: $F\cdot b-m_{T}g\cdot d=0$, giving $F=m_{T}g\cdot \frac{d}{b}$.

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Levers

Class	Arrangement	Mechanical Advantage
First class	Fulcrum between the two forces	Can be neutral, advantageous, or disadvantageous
Second class	Weight between fulcrum and applied force	Always advantageous
Third class	Applied force between fulcrum and weight	Always disadvantageous

Mnemonic — Lever Classes

“F-W-E / F-E-W / E-F-W” → remembering the order of Fulcrum, Weight (resistance), and Effort (applied force):

• 1st: FWE → Fulcrum in the middle
• 2nd: WFE → Weight in the middle (advantageous)
• 3rd: WEF → Effort in the middle (disadvantageous)

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Levers in the Human Body: Joints

The three lever classes are all found in joints of the human body (I, II, and III class examples exist in human anatomy).

Which class of lever is always mechanically disadvantageous, and give a human body example?

Third class levers are always disadvantageous. Example: the biceps acting on the forearm — the muscle force is applied between the elbow joint (fulcrum) and the hand (load).

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Optimal Range of Motion at Joints

Torque and Joint Angle

The torque applied depends on the direction of force application relative to the lever arm.

• The effectiveness of a muscle changes with joint position.
• For the biceps: maximum effectiveness near 90° of elbow flexion; least effective when the arm is fully extended.
• This is why range of motion is as important as weight load (e.g., in muscle recovery exercises).

At what elbow angle is the biceps most effective, and why?

Around 90°, because the muscle force is applied most perpendicularly to the forearm lever arm, maximising the torque.

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Square-Cube Law

Square-Cube Law

• Muscle strength / bone resistance ∝ cross-sectional area → scales as $L^2$
• Weight of bone/muscle ∝ volume → scales as $L^3$

If all dimensions double:

• Resistance increases 4-fold ($2^2$)
• Weight increases 8-fold ($2^3$)

→ The larger the body, the harder it is to move relative to its own weight. → This is why elephants cannot jump.

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According to the square-cube law, if body size doubles, muscle strength increases {1:4}-fold while body weight increases {1:8}-fold.

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Torque and Rotations

Torque (General Definition)

$\vec{M}=\vec{r}\times \vec{F}$

Magnitude:

$|\vec{M}|=|\vec{r}|\cdot |\vec{F}|\cdot \sin (\theta )$

Dimensions: $F\cdot l=M{L}^2{T}^{-2}$

SI unit: $\text{N}\cdot \text{m}/\text{rad}$

What is the formula for torque and its SI unit?

$\vec{M}=\vec{r}\times \vec{F}$, with magnitude $|\vec{M}|=|\vec{r}||\vec{F}|\sin \theta$. SI unit: N·m/rad.

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Rigid Body Equilibrium

Equilibrium = absence of any motion (both translational and rotational).

${\sum }_{i=1}^N{\vec{F}}_i=0$ ${\sum }_{i=1}^N{\vec{M}}_i=0$

Both conditions must hold simultaneously for full equilibrium.

What are the two conditions for rigid body equilibrium?

1. The vector sum of all forces is zero: $\sum {\vec{F}}_i=0$
2. The vector sum of all torques is zero: $\sum {\vec{M}}_i=0$

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Rotational Inertia (Moment of Inertia)

Definition Rotational inertia

$I$ is the resistance a body exerts against changes to its state of rotation — the rotational analogue of mass.

It depends on the distribution of matter and the axis of rotation.

For a discrete system of point masses:

$I_{A.R.}={\sum }_{i=1}^n{m}_i{r}_i^2$

where $r_i$ is the perpendicular distance of mass $m_i$ from the axis of rotation.

Common Moments of Inertia

Object	Axis	$I$
Thin rod	Through center	$\frac{1}{12}m{L}^2$
Thin rod	Through end	$\frac{1}{3}m{L}^2$
Solid cylinder / disk	Central axis	$\frac{1}{2}m{r}^2$
Solid sphere	Through center	$\frac{2}{5}m{r}^2$

What is the moment of inertia of a thin rod rotating around one end?

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

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Rotational inertia is the rotational analogue of {1:mass}, and for a system of point masses it is $I=\sum m_{i}1:r_i^2$.

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Rotation Around a Fixed Axis

Analogous to Newton’s second law for translation:

$I\frac{d\omega }{dt}=M$

where $\omega$ is angular velocity and $\theta$ is the angle of rotation.

If the torque $M$ is constant, the resulting motion follows time laws equivalent to those of uniformly accelerated linear motion.

What is the rotational analogue of Newton's second law?

$I\frac{d\omega }{dt}=M$ where $I$ is the moment of inertia, $\omega$ is angular velocity, and $M$ is the net torque.

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Rotational Work and Energy

Rotational work:

$W={\int }_{{\theta }_1}^{{\theta }_2}\vec{M}\cdot d\theta$

When rotation and torque share the same axis:

$W={\int }_{{\theta }_1}^{{\theta }_2}|M|,d\theta$

Rotational kinetic energy:

$E_r=\frac{1}{2}I{\omega }^2$

What is the formula for rotational kinetic energy?

$E_r=\frac{1}{2}I{\omega }^2$

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Angular Momentum Conservation

$\vec{L}=I\omega$

Conservation Law

Angular momentum is conserved (like linear momentum). This arises from the third principle of dynamics.

Noether’s Theorem

Conservation laws arise from symmetries of the system:

Conserved Quantity	Corresponding Symmetry
Energy	Time translation invariance
Linear momentum	Spatial homogeneity
Angular momentum	Spatial isotropy

According to Noether's theorem, what symmetry corresponds to conservation of angular momentum?

Spatial isotropy (the laws of physics are the same in all directions).

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Angular momentum
$\vec{L}=1:I\omega$ is conserved as a consequence of {1:spatial isotropy}, according to Noether’s theorem.

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Exercises

Exercise 1 — Holding a Weight

Problem: A person holds a 500 N dumbbell in their right hand. The forearm and hand are stationary in the horizontal position (no rotation at the elbow joint).

• Forearm + hand weight: 17 N
• CoG of forearm/hand: 0.23 m from elbow joint
• CoG of dumbbell: 0.34 m from elbow joint
• Bicep muscle insertion: 0.05 m from elbow joint

How much muscle force is required to maintain static equilibrium?

Solution

For equilibrium: $\sum \vec{M}=0$ at the elbow.

Force	Value	Moment arm	Torque
Dumbbell weight	500 N	0.34 m	$-170.0$ N·m
Forearm/hand weight	17 N	0.23 m	$-3.9$ N·m
Bicep force	$F$	0.05 m	$+F\cdot 0.05$

Required counter-torque from bicep:

$M_{bicep}=-[(-170.0)+(-3.9)]=173.9\text{ Ncdotpm}$

Required muscle force:

$F=\frac{M}{r}=\frac{173.9}{0.05}\approx 3,478\text{ N}$

Key Insight

The bicep insertion is very close to the elbow (a third-class lever), so the muscle must exert a force nearly 7× greater than the total load it supports — a mechanically disadvantageous arrangement.

In the dumbbell holding exercise, why must the bicep exert ~3,478 N to hold a 500 N weight?

Because the bicep inserts only 0.05 m from the elbow joint (fulcrum), while the load acts at 0.34 m — a third-class lever arrangement that is mechanically disadvantageous. Force = torque / moment arm = 173.9 / 0.05 ≈ 3,478 N.

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Exercise 2 — Rotating the Shoulder

Problem: Consider the force applied by the shoulder muscles to swing the arm.

• Arm mass: 6 kg
• Arm length: 60 cm = 0.6 m
• Swing period: 1 second
• Muscle attachment distance from shoulder: 5 cm = 0.05 m
• Muscle assumed perpendicular to humerus
• Time to reach full speed from rest: 2 seconds

How much force must the shoulder muscles apply?

Solution

Step 1 — Moment of inertia (arm modelled as a bar rotating around one end):

$I=\frac{1}{3}m{L}^2=\frac{1}{3}\cdot 6\cdot (0.6{)}^2=0.72{\text{ kgcdotpm}}^2$

Step 2 — Angular velocity (one full rotation per second):

$\omega =2\pi \text{ rad/s}$

Step 3 — Average torque (from rest to $\omega$ in $\Delta t=2$ s):

$M=I\cdot \frac{\Delta \omega }{\Delta t}=0.72\cdot \frac{2\pi }{2}=0.72\pi \text{ Ncdotpm/rad}$

Step 4 — Required muscle force (muscle perpendicular to humerus, so $\cos \theta =1$):

$M=F\cdot r⟹F=\frac{M}{r}=\frac{0.72\pi }{0.05}\approx 45.2\text{ N}$

What moment of inertia formula applies to the arm swinging at the shoulder, and why?

$I=\frac{1}{3}m{L}^2$ — the arm is modelled as a uniform rod rotating around one end (the shoulder joint).

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Exercise 2 Extension — Throwing a Ball

Problem: How does the calculation change if the person holds a 0.5 kg ball at the end of their arm?

Solution:

Additional moment of inertia from ball (treated as a point mass at $r=0.6$ m):

$I_b=m_b\cdot r^2=0.5\cdot (0.6{)}^2=0.18{\text{ kgcdotpm}}^2$

This is a 25% increase over the original $I=0.72{\text{ kgcdotpm}}^2$.

Since required force $\propto$ torque $\propto$ rotational inertia:

$\Delta F=0.25\times 45.2\approx 11.3\text{ N additional force}$

Key Principle

Rotational inertia is additive. A point mass at distance $r$ contributes $I=m{r}^2$ to the total.

How does holding a 0.5 kg ball at arm's length affect the required shoulder muscle force when swinging the arm?

The ball adds $I_b=0.5\times 0.6^2=0.18$ kg·m², a 25% increase in rotational inertia. Since force ∝ torque ∝ inertia, the required force increases by 25%, i.e., approximately +11.3 N.

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TLDR — Rigid Body Dynamics

Systems of Points

• Total mass: $M=\sum m_i$; total momentum: $\vec{Q}=\sum {\vec{q}}_i$
• Rigid body: fixed internal distances; described by CoM position + 3 rotation angles
• Center of mass ≠ center of gravity in non-uniform fields (but approximately equal in most practical contexts)

Center of Mass

• Position of CoM determines rotation vs. translation when a force is applied
• CoM can lie outside the body (e.g., bending over); exploited in Fosbury flop

Biomechanics

• Mozzi’s axis: instantaneous axis of rotation at any moment
• Lever classes: 1st (fulcrum between forces), 2nd (load between fulcrum and effort — advantageous), 3rd (effort between fulcrum and load — disadvantageous)
• Joints act as all three lever types; biceps near 90° = maximum torque efficiency
• Square-cube law: strength ∝ $L^2$, weight ∝ $L^3$ → larger animals proportionally weaker relative to weight

Torque

• $\vec{M}=\vec{r}\times \vec{F}$; magnitude $=rF\sin \theta$; unit: N·m/rad
• Rigid body equilibrium: $\sum {\vec{F}}_i=0$ AND $\sum {\vec{M}}_i=0$

Rotational Inertia

• $I=\sum m_i{r}_i^2$ (discrete); depends on axis and mass distribution
• Common values: rod at end $=\frac{1}{3}m{L}^2$; disk $=\frac{1}{2}m{r}^2$; sphere $=\frac{2}{5}m{r}^2$
• Newton’s 2nd law for rotation: $I\frac{d\omega }{dt}=M$
• Rotational KE: $E_r=\frac{1}{2}I{\omega }^2$; rotational work: $W=\int M,d\theta$

Angular Momentum

• $\vec{L}=I\omega =\vec{r}\times m\vec{v}$; conserved quantity
• Noether’s theorem: energy ↔ time symmetry; linear momentum ↔ space homogeneity; angular momentum ↔ space isotropy

Exercises

• Dumbbell curl: bicep must exert ~3,478 N to hold 500 N load (3rd-class lever disadvantage)
• Shoulder swing: 45.2 N needed to accelerate arm to 2π rad/s in 2 s; adding a 0.5 kg ball increases this by ~25% (+11.3 N)