Rotational & Rigid-Body Mechanics

Lectures 3+4

• So far: point objects moving through a trajectory.

• Next: moving actual dimensional objects and rotating them.

2

Rotational Motion

• 𝐶 is the center of rotation.

• 𝑃 a point on to the object.

• 𝑟 is the distance vector 𝑃 − 𝐶

• 𝑟 = 𝑟 the distance between 𝐶 and 𝑃.• Object rotates 𝑃

travels along a circular path

• after ∆𝑡, 𝑃 covered distance 𝑠, and angle 𝜃.

• Unit-length axis of rotation: 𝑢.• In this example, the Z axis going “out” from the screen.

• Rotation: counterclockwise (right-hand rule).

3

Circular Motion - Definitions

𝐶𝑃

𝑟

𝑃

𝜃

𝑠

• This angle 𝜃 represents the rotation of the object:

𝜃 = 𝑠/𝑟

where 𝑠 is the arc length and 𝑟 the radius• unit is radian (𝑟𝑎𝑑)

• 1 radian=angle for arc length 1 at a distance 1.

4

Angular Displacement

𝐶𝑃

𝑟

𝑃

𝜃

𝑠

𝜔

𝑣

• Angular velocity: the rate of change of the angular displacement:

𝜔 =∆𝜃

∆𝑡=𝜃 𝑡 + ∆𝑡 − 𝜃(𝑡)

∆𝑡

• unit is 𝑟𝑎𝑑/𝑠.

• The angular velocity vector is collinear with the rotation axis:

𝜔 = 𝜔𝑢

5

Angular Velocity

• Angular acceleration: the rate of change of the angular velocity:

𝛼 =∆𝜔

∆𝑡=𝜔 𝑡 + ∆𝑡 − 𝜔(𝑡)

∆𝑡

• Exactly like acceleration is to velocity in a trajectory.

• unit is 𝑟𝑎𝑑/𝑠2

6

Angular Acceleration

• Deferring velocity from position and acceleration:

𝜔(𝑡 + ∆𝑡) = 𝜔(𝑡) + 𝛼∆𝑡

𝜔 =𝜔(𝑡 + ∆𝑡) + 𝜔(𝑡)

2

∆𝜃 = 1 2 𝜔(𝑡 + ∆𝑡) + 𝜔(𝑡) ∆𝑡

∆𝜃 = 𝜔(𝑡)∆𝑡 + 1 2𝛼 ∆𝑡2

𝜔(𝑡 + ∆𝑡)2 = 𝜔(𝑡)2 + 2𝛼∆𝜃

7

Equations of Motion

• The centripetal force creates curved motion.

• Orthogonal to the velocity of the object.• Object is in orbit.

• Constant force circular rotation with constant tangential velocity.• Why?

8

Kinetics of Rotational Motion

𝐶 𝑃(𝑡)

𝑃(𝑡 + ∆𝑡)

𝑇(𝑡)

𝑉(𝑡)

• Every point on a rigid body moves with the same angularvelocity.

• Different points on a rigid body can have different tangential velocities.• Different radii

9

Tangential and Angular Velocities

• Angular velocity vector: direction of rotation axis

• Tangential velocity vector: direction of movement direction

• Relation: 𝑣 = 𝜔 × 𝑟

Or:

𝜔 = 𝑟 × 𝑣

𝑟2

• Note that 𝑣 = 𝜔 ∗ 𝑟 (abs. values), due to 𝑠 = 𝜃 ∗ 𝑟.• Only the tangential part of any velocity matters for rotation!

10

Tangential Velocity

𝐶𝑃

𝑟

𝑃

𝜃

𝑠

𝜔

𝑣

• Tangential acceleration defined the same way

𝑎 = 𝛼 ∗ 𝑟

where 𝛼 is the angular acceleration.

• Similarly to the velocity equation 𝑣 = 𝜔 ∗ 𝑟

11

Tangential Acceleration

• The centripetal acceleration, orthogonal to the velocity (=towards the axis of rotation), drives the rotational movement:

𝑎𝑛 =𝑣𝑡2

𝑟= 𝑟𝜔2

• What is the centrifugal force?

12

Centripetal Acceleration

Rigid-Body Kinematics• Rigid bodies have dimensions, as opposed to single points.

• Relative distances between all points are invariant.

• Movement can be decomposed into two components:• Linear trajectory of any single points

• Relative rotation around the point

• Free body movement: around the center of mass (COM).

• The measure of the amount of matter in the volume of an object:

𝑚 = 𝑉

𝜌 𝑑𝑉

where 𝜌 is the density at each location in the object volume 𝑉.

• Equivalently: a measure of resistance to motion or change in motion.

14

Mass

• For a 3D object, mass is the integral over its volume:

𝑚 = 𝜌(𝑥, 𝑦, 𝑧) 𝑑𝑥 𝑑𝑦 𝑑𝑧

• For uniform density objects (simple rigid bodies):𝑚 = 𝜌 ∗ 𝑉

where 𝜌 is the density of the object and 𝑉 is its volume.

15

Mass

• The center of mass (COM) is the point at which all the mass can be considered to be ‘concentrated’• obtained from the first moment, i.e. mass times distance.

• point of ‘balance’ of the object

• if uniform density, COM is also the centroid

16

Center of Mass

COM

• Coordinate of the COM:

𝐶𝑂𝑀 =1

𝑚 𝑉

𝜌 𝑝 ∗ 𝑝 𝑑𝑉

where 𝑝 is the position at each location in 𝑉.

• For a set of bodies:

𝐶𝑂𝑀 =1

𝑚

𝑖=1

𝑛

𝑚𝑖𝑝𝑖

where 𝑚𝑖 is the mass of each COM 𝑝𝑖 in every body.

17

Center of Mass

• Example for a body made of two spheres in 1D

𝑥𝐶𝑂𝑀 =𝑚1𝑥1 +𝑚2𝑥2𝑚1 +𝑚2

18

Center of Mass

COM𝒎𝟏

𝒎𝟐

𝒙𝟐

𝒙𝟏

𝒙𝑪𝑶𝑴

• Quite easy to determine for primitive shapes

• But what about complex surface based models?

19

Center of Mass

• Remember linear momentum: 𝑝 = 𝑚 𝑣.

• Rotational motion also produces angular momentum of a any point on the object about the center of mass (or any relative point):

𝐿 =

𝑉

𝑟 × 𝑝 𝑑𝑉

• unit is 𝑁 ∗ 𝑚 ∗ 𝑠

• Angular momentum is a conserved quantity!• Just like the linear momentum.

• Remember: measured around a point.

20

Angular Momentum

• Plugging in angular velocity:

𝑟 × 𝑝 = 𝑟 × 𝑣 𝑑𝑚 = 𝑟 × 𝜔 × 𝑟 𝑑𝑚

• Integrating, we get:

𝐿 = 𝑀

𝑟 × 𝜔 × 𝑟 𝑑𝑚

• Note: The angular momentum and the angular velocity are not generally collinear!

21

Angular Momentum

Moment of Inertia

• Defining: 𝑟 =𝑥𝑦𝑧

and 𝜔 =𝜔𝑥𝜔𝑦𝜔𝑧

.

• Remember that the angular velocity is constant throughout the body.

• We get:

𝐿 =

𝑦2 + 𝑧2 𝜔𝑥 − 𝑥𝑦𝜔𝑦 − 𝑥𝑧𝜔𝑧

−𝑦𝑥𝜔𝑥 + 𝑧2 + 𝑥2 𝜔𝑦 − 𝑦𝑧𝜔𝑧

−𝑧𝑥𝜔𝑥 − 𝑧𝑦𝜔𝑦 + (𝑥2 + 𝑦2)𝜔𝑧

𝑑𝑚 =

𝐼𝑥𝑥 −𝐼𝑥𝑦 −𝐼𝑥𝑧−𝐼𝑦𝑥 𝐼𝑦𝑦 −𝐼𝑦𝑧−𝐼𝑧𝑥 −𝐼𝑧𝑦 𝐼𝑧𝑧

𝜔𝑥𝜔𝑦𝜔𝑧

.

• The inertia tensor only depends on the geometry of the object and the relative point (often, COM):

𝐼𝑥𝑥 = 𝑦2 + 𝑧2 𝑑𝑚

𝐼𝑦𝑦 = 𝑧2 + 𝑥2 𝑑𝑚

𝐼𝑧𝑧 = 𝑥2 + 𝑦2 𝑑𝑚

23

Momentum and Inertia

𝐼𝑥𝑦 = 𝐼𝑦𝑥 = 𝑥𝑦 𝑑𝑚

𝐼𝑥𝑧 = 𝐼𝑧𝑥 = 𝑥𝑧 𝑑𝑚

𝐼𝑦𝑧 = 𝐼𝑧𝑦 = 𝑦𝑧 𝑑𝑚

• Finally the inertia can be expressed as the matrix

𝐼 =

𝑦2 + 𝑧2 𝑑𝑚 − 𝑥𝑦 𝑑𝑚 − 𝑥𝑧 𝑑𝑚

− 𝑥𝑦 𝑑𝑚 𝑧2 + 𝑥2 𝑑𝑚 − 𝑦𝑧 𝑑𝑚

− 𝑥𝑧 𝑑𝑚 − 𝑦𝑧 𝑑𝑚 𝑥2 + 𝑦2 𝑑𝑚

• The diagonal elements are called the (principal) moment of inertia

• The off-diagonal elements are called products of inertia

24

The Inertia Tensor

• Equivalently, we separate mass elements to density and volume elements:

𝐼 =

𝑉

𝜌 𝑥, 𝑦, 𝑧

𝑦2 + 𝑧2 −𝑥𝑦 −𝑥𝑧

−𝑥𝑦 𝑧2 + 𝑥2 −𝑦𝑧

−𝑥𝑧 −𝑦𝑧 𝑥2 + 𝑦2𝑑𝑥 𝑑𝑦 𝑑𝑧

• The diagonal elements: distances to the respective principal axes.

• The non-diagonal elements: products of the perpendicular distances to the respective planes.

25

The Inertia Tensor

• The moment of inertia of a rigid body is a measure of how much the mass of the body is spread out.

• A measure of the ability to resist change in rotational motion

• Defined with respect to a specific rotation axis 𝑢.• Through the central rotation origin point.

• We have that: 𝐼𝑢 = 𝑉 𝑟𝑢2𝑑𝑚

26

Moment of Inertia

Moment and Tensor

• We have: 𝑟𝑢2 = 𝑢 × 𝑟 2 = 𝑢𝑇𝐼(𝑞)𝑢 for any point 𝑞. (Remember:

𝑟 is distance to origin).

• Thus:

𝐼𝑢 = 𝑀 𝑢𝑇𝐼(𝑞)𝑢 𝑑𝑚=𝑢𝑇𝐼𝑢

• The scalar angular momentum around the axis is then 𝐿𝑢 = 𝐼𝑢𝜔.

• Reducible to a planar problem (axis is the new z axis).

• For a mass point:𝐼 = 𝑚 ∗ 𝑟𝑢

2

• For a collection of mass points:𝐼 = 𝑖𝑚𝑖𝑟𝑖

2

• For a continuous mass distribution on the plane:𝐼 = 𝑀 𝑟𝑢

2 𝑑𝑚

28

Moment of Inertia

𝑟𝑚

𝑟1𝑚1

𝑟2𝑚2

𝑟3 𝑚3

𝑟𝑑𝑚

• For primitive shapes, the inertia can be expressed with the parameters of the shape

• Illustration on a solid sphere• Calculating inertia by integration of thin discs along

one axis (e.g. 𝑧).

• Surface equation: 𝑥2 + 𝑦2 + 𝑧2 = 𝑅2

29

Inertia of Primitive Shapes

• Distance to axis of rotation is the radius of the disc at the cross section along 𝑧: 𝑟2

= 𝑥2 + 𝑦2 = 𝑅2 − 𝑧2.

• Summing moments of inertia of small cylinders of inertia 𝐼𝑍 =𝑟2𝑚

2along the z-axis:

𝑑𝐼𝑍 =1

2𝑟2𝑑𝑚 =

1

2𝑟2𝜌𝑑𝑉 =

1

2𝑟2𝜌𝜋𝑟2𝑑𝑧

• We get:

𝐼𝑍 =1

2𝜌𝜋 −𝑅𝑅𝑟4𝑑𝑧 =

1

2𝜌𝜋 −𝑅𝑅𝑅2 − 𝑧2 2𝑑𝑧 =

1

2𝜌𝜋 𝑅4𝑧 − 2𝑅2 𝑧3 3 + 𝑧5 5 −𝑅

𝑅

= 𝜌𝜋 1 − 2 3 + 1 5 𝑅5.

• as 𝑚 = 𝜌 4 3 𝜋𝑅3, we finally obtain: 𝐼𝑍 =2

5𝑚𝑅2.

30

Inertia of Primitive Shapes

• Solid sphere, radius 𝑟 and mass 𝑚:

• Hollow sphere, radius 𝑟 and mass 𝑚:

31

Inertia of Primitive Shapes

𝐼 =

2

5𝑚𝑟2 0 0

02

5𝑚𝑟2 0

0 02

5𝑚𝑟2

𝐼 =

2

3𝑚𝑟2 0 0

02

3𝑚𝑟2 0

0 02

3𝑚𝑟2

xz

y

• Solid ellipsoid, semi-axes 𝑎, 𝑏, 𝑐 and mass 𝑚:

• Solid box, width 𝑤, height ℎ, depth 𝑑 and mass 𝑚:

32

Inertia of Primitive Shapes

𝐼 =

1

5𝑚(𝑏2+𝑐2) 0 0

01

5𝑚(𝑎2+𝑐2) 0

0 01

5𝑚(𝑎2+𝑏2)

𝐼 =

1

12𝑚(ℎ2+𝑑2) 0 0

01

12𝑚(𝑤2+𝑑2) 0

0 01

12𝑚(𝑤2+ℎ2)

xz

y

wd

h

• Solid cylinder, radius 𝑟, height ℎ and mass 𝑚:

• Hollow cylinder, radius 𝑟, height ℎ and mass 𝑚:

33

Inertia of Primitive Shapes

𝐼 =

1

12𝑚(3𝑟2+ℎ2) 0 0

01

12𝑚(3𝑟2+ℎ2) 0

0 01

2𝑚𝑟2

𝐼 =

1

12𝑚(6𝑟2+ℎ2) 0 0

01

12𝑚(6𝑟2+ℎ2) 0

0 0 𝑚𝑟2

h

h

• The object does not necessarily rotate around the center of mass.• Some point can be fixed!

• parallel axis theorem:

𝐼𝑣 = 𝐼𝐶𝑂𝑀 +𝑚𝑑2

Where:

𝐼𝑣: inertia around axis 𝑢.

𝐼𝐶𝑂𝑀 inertia about a parallel axis through the COM.

𝑑 is the distance between the axes.

34

Parallel-Axis Theorem

• More generally, for point displacements: 𝑑𝑥 , 𝑑𝑦 , 𝑑𝑧

35

Parallel-Axis Theorem

𝐼𝑥𝑥 = 𝑦2 + 𝑧2 𝑑𝑚 +𝑚𝑑𝑥

2

𝐼𝑦𝑦 = 𝑧2 + 𝑥2 𝑑𝑚 +𝑚𝑑𝑦

2

𝐼𝑧𝑧 = 𝑥2 + 𝑦2 𝑑𝑚 +𝑚𝑑𝑧

2

𝐼𝑥𝑦 = 𝑥𝑦 𝑑𝑚 +𝑚𝑑𝑥𝑑𝑦

𝐼𝑥𝑧 = 𝑥𝑧 𝑑𝑚 +𝑚𝑑𝑥𝑑𝑧

𝐼𝑦𝑧 = 𝑦𝑧 𝑑𝑚 +𝑚𝑑𝑦𝑑𝑧

• For a planar 2D object, the moment of inertia about an axis perpendicular to the plane is the sum of the moments of inertia of two perpendicular axes through the same point in the plane:

36

Perpendicular-Axis Theorem

𝑥

𝑦𝑧

𝐼𝑧 = 𝐼𝑥 + 𝐼𝑦for any planar object

𝑥

𝑦𝑧

𝐼𝑧 = 2𝐼𝑥 = 2𝐼𝑦for symmetrical objects

• The inertia tensor is coordinate dependent.

• If 𝑅 changes bases from body to world coordinate, the inertia tensor in world space is:

𝐼𝑤𝑜𝑟𝑙𝑑 = 𝑅 ∗ 𝐼𝑏𝑜𝑑𝑦 ∗ 𝑅𝑇

37

Reference Frame

Rigid-Body Dynamics

• The torque is a force F applied at a distance r from a (held) center of mass.

• Tangential part causes tangential acceleration:

𝐹𝑡 = 𝑚 ∗ 𝑎𝑡• Multiplying by the distance, the torque is:

𝜏 = 𝐹𝑡 ∗ 𝑟 = 𝑚 ∗ 𝑎𝑡 ∗ 𝑟

• We know that 𝑎𝑡 = 𝑟 ∗ 𝛼

• So we have 𝜏 = 𝑚 ∗ 𝑟 ∗ 𝛼 ∗ 𝑟 = 𝑚 ∗ 𝑟2 ∗ 𝛼• unit is 𝑁 ∗ 𝑚• rotates an object about its axis of rotation through the center of mass.

39

Torque

• A force in general is not applied in the direction of the tangent.

• The torque 𝜏 is then defined as: 𝜏 = 𝑟 × 𝐹

• The direction of the torque is perpendicular to both 𝐹 and 𝑟.

40

Torque

• The law 𝐹 = 𝑚 ∗ 𝑎 has an equivalent with inertia tensor and torque:

𝜏 = 𝐼 𝛼

• Force linear acceleration

• Torque angular acceleration

41

Newton’s Second Law

• Translating energy formulas to rotational motion.

• The rotational kinetic energy is defined as:

𝐸𝐾𝑟 =1

2𝜔𝑇 ∗ 𝐼 ∗ 𝜔

42

Rotational Kinetic Energy

• Adding rotational kinetic energy

𝐸𝐾𝑡 𝑡 + ∆𝑡 + 𝐸𝑃 𝑡 + ∆𝑡 + 𝐸𝐾𝑟 𝑡 + ∆𝑡= 𝐸𝐾𝑡 𝑡 + 𝐸𝑃 𝑡 + 𝐸𝐾𝑟(𝑡) + 𝐸𝑂

• 𝐸𝐾𝑡 is the translational kinetic energy.

• 𝐸𝑃 is the potential energy.

• 𝐸𝐾𝑟 is the rotational kinetic energy.

• 𝐸𝑂 the “lost” energies (surface friction, air resistance etc.).

43

Conservation of Mechanical Energy

Torque and Angular Momentum

• Remember, in the linear case: 𝐹 =𝑑 𝑝

𝑑𝑡( 𝑝 is the linear momentum).

• Similarly with torque and angular momentum:𝑑𝐿

𝑑𝑡=𝑑 𝑟

𝑑𝑡× 𝑝 + 𝑟 ×

𝑑 𝑝

𝑑𝑡= 𝑣 × 𝑚 𝑣 + 𝑟 × 𝐹 = 0 + 𝜏

• Force derivative of linear momentum.

• Torque derivative of angular momentum.

• We may apply off-center forces for a very short amount of time.

• Such ‘angular’ impulse results in a change in angular momentum, i.e.in angular velocity:

𝜏∆𝑡 = ∆𝐿

45

Impulse

• A force can be applied anywhere on the object, producing also a rotational motion.

46

Rigid Body Forces

𝑭

𝑪𝑶𝑴

𝒂

𝜶

• Remember: the object moves linearly as the COM moves.

• Rotation: the movement for all points relatively to the COM.

• Total motion: sum of the two motions.

47

Position of An Object

𝑭

• Most simple instance of a physics system• Each object (body) is a particle

• Each particle have forces acting upon it• Constant, e.g. gravity

• Position dependent, e.g. force fields

• Velocity dependent, e.g. drag forces

• Event based, e.g. collision forces

• Restrictive, e.g. joint constraint

• So net force is a function 𝐹 𝑝𝑜, 𝑣, 𝑎,𝑚, 𝑡, …

• Discretization: e.g., 𝑉 𝑓 𝑞 𝑑𝑚 becomes a sum: 𝑖=1𝑛 𝑓 𝑞𝑖 𝑚𝑖

48

Particle System

• Use the equations of motion to find the position of each particle at each frame.

• At the start of each frame:• Sum up all of the forces for each particle.

• From these forces compute the acceleration.

• Integrate into velocity and position.

• Rigid body: all particles receive the same rotation and translation.

49

Particle System

• When an object consists of multiple primitive shapes:• Calculate the individual inertia of each shape.• Use parallel axis theorem to transform to inertia about an

axis through the COM of the object.• Add the inertia matrices together.

50

Complex Objects

• A rigid body may not be free to move ‘on its own’.

• We wish to constrain its movement:• wheels on a chair

• human body parts

• trigger of a gun

• opening door

• actually almost anything you can think of in a game...

51

Motion Constraints

• To describe how a body can move in space, specify its degrees of freedom (DOF):• Translational

• Rotational

52

Degree of freedom

• A kinematic pair is a connection between two bodies that imposes constraints on their relative movement• Lower pair, constraint on a point, line or plane:

• Revolute pair, or hinged joint: 1 rotational DOF.

• Prismatic joint, or slider: 1 translational DOF.

• Screw pair: 1 coordinated rotation/translation DOF.

• Cylindrical pair: 1 translational + 1 rotational DOF.

• Spherical pair, or ball-and-socket joint: 3 rotational DOF.

• Planar pair: 3 translational DOF.

• Higher pair, constraint on a curve or surface.

53

Kinematic pair