Download pdf - Lecture 2 - Rigid-Body Physics - Utrecht University 2 - … · · 2016-05-03Lecture II: Rigid-Body Physics 3 Rigid-Body Kinematics • Objects as sets of points. • Relative distances

Lecture II: Rigid-Body Physics

2Lecture II: Rigid-Body Physics

Rigid-Body Motion

• Previously: Point dimensionless objects moving through a trajectory.

• Today: Objects with dimensions, moving as one piece.


Rigid-Body Kinematics• Objects as sets of points.• Relative distances between all points are

invariant to rigid movement.• Free body movement: around the center of mass

(COM).• Movement has two components:

• Linear trajectory of a central point.• Relative rotation around the point.


Mass• The measure of the amount of matter in the

volume of an object:

𝑚 = #𝜌𝑑𝑉

(

• 𝜌 : the density of each point the object volume 𝑉.• 𝑑𝑉: the volume element.

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


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

volume:

𝑚 = ###𝜌(𝑥, 𝑦, 𝑧) 𝑑𝑥𝑑𝑦𝑑𝑧

• For uniform density (𝜌 constant):

𝑚 = 𝜌 ∗ 𝑉


Center of Mass• The center of mass (COM) is the “average” point

of the object, weighted by density:

𝐶𝑂𝑀 =1𝑚#𝜌 4 �⃗�𝑑𝑉

(• 𝑝: point coordinates.

• Point of balance for the object.• Uniform density: COM ó centroid.

COM


Center of Mass of System• Sets of bodies have a mutual center of mass:

𝐶𝑂𝑀 =1𝑚7𝑚8𝑝8

9

8:;• 𝑚8: mass of each body.• 𝑝8: location of individual COM.• 𝑚=∑ 𝑚8

98:; .

• Example: two spheres in 1D

𝑥=>? =𝑚;𝑥; +𝑚A𝑥A𝑚; +𝑚A

COM𝒎𝟏

𝒎𝟐

𝒙𝟐

𝒙𝟏𝒙𝑪𝑶𝑴


Center of Mass

• Quite easy to determine for primitive shapes

• What about complex surface based models?


Rotational Motion• 𝑃 a point on to the object. • 𝐶 is the center of rotation.• Distance vector �⃗� = 𝑃 − 𝐶.

• 𝑟 = 𝑟 : the distance.• Object rotates ó 𝑃

travels along a circular path.• Unit-length axis of rotation: 𝑢.

• Here, 𝑢=�̂� (“out” from the screen).• Rotation: counterclockwise.

• right-hand rule.

𝐶𝑃

𝑟

𝑃

𝜃

𝑠

𝑢

𝑣


Angular Displacement• Point 𝑝 covers linear distance 𝑠.

• 𝜃 is the angular displacement of the object:

𝜃 = 𝑠/𝑟

• 𝑠: arc length.

• Unit is radian (𝑟𝑎𝑑)

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

𝐶𝑃

𝑟

𝑃

𝜃

𝑠

𝑢

𝑣


Angular Velocity• Angular speed: the rate of change of the angular

displacement:

𝜔 =𝑑𝜃𝑑𝑡

• unit is UVW XYZ⁄ .

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

𝜔 = 𝜔𝑢


Angular Acceleration• Angular acceleration: the rate of change of the

angular velocity:

𝛼 =𝑑𝜔𝑑𝑡

• Paralleling definition of linear acceleration.

• Unit is 𝑟𝑎𝑑/𝑠A


Tangential and Angular Velocities• Every point moves with the same angular velocity.

• Direction of vector: 𝑢.

• Tangential velocity vector:

�⃗� = 𝜔×�⃗�Or:

𝜔 =�⃗�×�⃗�𝑟A

• 𝜔 = ^_ U⁄ (abs. values)• due to 𝑠 = 𝜃 4 𝑟.

• Only the tangential part matters!

𝐶 𝑃(𝑡)

𝑃(𝑡+ ∆𝑡)

𝑇(𝑡)

𝑉(𝑡)

𝐶𝑃

𝑟

𝑃

𝜃

𝑠

𝑢

𝑣


Dynamics• The centripetal force creates curved

motion.

• In the direction of (negative) �⃗�• Object is in orbit.

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


Tangential & Centripetal Accelerations• Tangential acceleration �⃗� holds:

�⃗� = �⃗�×𝑟

• cf. velocity equation𝑣 = 𝜔×�⃗�.

• The centripetal acceleration drives the rotational movement:

�⃗�9 =^b

U�̂� = −𝜔A�⃗�.

• What is the “centrifugal” force?


Angular Momentum• Linear motion è linear momentum: �⃗� = 𝑚�⃗�.• Rotational motion è angular momentum about

any fixed relative point (to which �⃗� is measured):

𝐿 = # �⃗�×�⃗� 𝜌𝑑𝑉(

• unit is 𝑁 4 𝑚 4 𝑠• 𝜌𝑑𝑉 = 𝑑𝑚 (mass element)

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

• Caveat: conserved w.r.t. the same point.


Angular Momentum• Plugging in angular velocity:

�⃗�×�⃗�𝑑𝑉 = �⃗�×�⃗� 𝑑𝑚 = �⃗�× 𝜔×�⃗� 𝑑𝑚

• Integrating, we get:

𝐿 = # �⃗�× 𝜔×�⃗� 𝑑𝑚

(

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


Moment of Inertia• Define: 𝑟 =

𝑥𝑦𝑧

and 𝜔 =𝜔f𝜔g𝜔h

.

• For a single rotating body: the angular velocity is constant.• We get:

𝐿 = #𝑟× 𝜔×𝑟 𝑑𝑚

(= #

𝑦A + 𝑧A 𝜔f − 𝑥𝑦𝜔g − 𝑥𝑧𝜔h−𝑦𝑥𝜔f + 𝑧A + 𝑥A 𝜔g − 𝑦𝑧𝜔h−𝑧𝑥𝜔f − 𝑧𝑦𝜔g + (𝑥A + 𝑦A)𝜔h

𝑑𝑚 =

𝐼ff −𝐼fg −𝐼fh−𝐼gf 𝐼gg −𝐼gh−𝐼hf −𝐼hg 𝐼hh

𝜔f𝜔g𝜔h

.

• Note: replacing integral with a (constant) matrix operating on a vector!


Momentum and Inertia• The inertia tensor 𝐼only depends on the geometry

of the object and the relative fixed point (often, COM):

𝐼fg = 𝐼gf = # 𝑥𝑦 𝑑𝑚

𝐼fh = 𝐼hf = # 𝑥𝑧 𝑑𝑚

𝐼gh = 𝐼hg = # 𝑦𝑧 𝑑𝑚

𝐼ff = # 𝑦A + 𝑧A 𝑑𝑚

𝐼gg = # 𝑧A + 𝑥A 𝑑𝑚

𝐼hh = # 𝑥A + 𝑦A 𝑑𝑚

𝑟 =𝑥𝑦𝑧


The Inertia Tensor• Compact form:

𝐼 =

# 𝑦A + 𝑧A 𝑑𝑚 −# 𝑥𝑦 𝑑𝑚 −# 𝑥𝑧 𝑑𝑚

−# 𝑥𝑦 𝑑𝑚 # 𝑧A + 𝑥A 𝑑𝑚 −# 𝑦𝑧 𝑑𝑚

−# 𝑥𝑧 𝑑𝑚 −# 𝑦𝑧 𝑑𝑚 # 𝑥A + 𝑦A 𝑑𝑚

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

• The off-diagonal elements are called products of inertia.


The Inertia Tensor• Equivalently, we separate mass elements to

density and volume elements:

𝐼 = #𝜌 𝑥, 𝑦, 𝑧𝑦A + 𝑧A −𝑥𝑦 −𝑥𝑧−𝑥𝑦 𝑧A + 𝑥A −𝑦𝑧−𝑥𝑧 −𝑦𝑧 𝑥A + 𝑦A

𝑑𝑥𝑑𝑦𝑑𝑧

(

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

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


Moment of Inertia• The moment of inertia 𝐼j, with respect to a

rotation axis 𝑢, measures how much the mass “spreads out”:

𝐼j = # 𝑟j A𝑑𝑚(

• 𝑟j: perpendicular distance to axis.• Through the central rotation origin point.

• Measures ability to resist change in rotational motion.• The angular equivalent to mass!


Moment and Tensor• We have: 𝑟j A = 𝑢×�⃗� A = 𝑢k𝐼(𝑞)𝑢 for any point 𝑞. (Remember: �⃗� is distance to origin).

• Thus:𝐼j = ∫ 𝑢k𝐼(𝑞)𝑢 𝑑𝑚

? =𝑢k𝐼𝑢

• The scalar angular momentum around the axis is then 𝐿j = 𝐼j𝜔.

• Reducible to a planar problem (axis as 𝑍axis).


Moment of Inertia• For a mass point:

𝐼 = 𝑚 4 𝑟jA

• For a collection of mass points:𝐼 = ∑ 𝑚8𝑟8A8

• For a continuous mass distribution on the plane:𝐼 = ∫ 𝑟jA? 𝑑𝑚

𝑟𝑚

𝑟;𝑚;

𝑟A 𝑚A

𝑟o 𝑚o

𝑟𝑑𝑚


Inertia of Primitive Shapes

• 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: 𝑥A + 𝑦A + 𝑧A = 𝑅A


Inertia of Primitive Shapes• Distance to axis of rotation is the radius of the disc at the

cross section along 𝑧: 𝑟A = 𝑥A + 𝑦A = 𝑅A − 𝑧A.

• Summing moments of inertia of small cylinders of inertia 𝐼q= Ubr

Aalong the z-axis:

𝑑𝐼q =12𝑟A𝑑𝑚 =

12𝑟A𝜌𝑑𝑉 =

12𝑟A𝜌𝜋𝑟A𝑑𝑧

• We get:

𝐼q =;A𝜌𝜋 ∫ 𝑟u𝑑𝑧v

wv = ;A𝜌𝜋 ∫ 𝑅A − 𝑧A A𝑑𝑧v

wv = ;A𝜌𝜋[𝑅u𝑧 − 2𝑅A 𝑧o 3⁄

+ 𝑧z 5⁄ ]wvv = 𝜌𝜋 1 − 2 3⁄ + 1 5⁄ 𝑅z.

• As 𝑚 = 𝜌 4 3⁄ 𝜋𝑅o, we finally obtain: 𝐼q =Az𝑚𝑅A.


Inertia of Primitive Shapes• Solid sphere, radius 𝑟 and mass 𝑚:

• Hollow sphere, radius 𝑟 and mass 𝑚:

𝐼 =

25𝑚𝑟

A 0 0

025𝑚𝑟

A 0

0 025𝑚𝑟

A

𝐼 =

23𝑚𝑟

A 0 0

023𝑚𝑟

A 0

0 023𝑚𝑟

A

xz

y


Inertia of Primitive Shapes• Solid ellipsoid, semi-axes 𝑎, 𝑏, 𝑐 and mass 𝑚:

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

𝐼 =

15𝑚(𝑏

A+𝑐A) 0 0

015𝑚(𝑎

A+𝑐A) 0

0 015𝑚(𝑎

A+𝑏A)

𝐼 =

112𝑚(ℎA+𝑑A) 0 0

0112𝑚(𝑤A+𝑑A) 0

0 0112𝑚(𝑤A+ℎA)

xz

y

wd

h


Inertia of Primitive Shapes• Solid cylinder, radius 𝑟, height ℎ and mass 𝑚:

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

𝐼 =

112𝑚(3𝑟

A+ℎA) 0 0

0112𝑚(3𝑟

A+ℎA) 0

0 012𝑚𝑟

A

𝐼 =

112𝑚(6𝑟

A+ℎA) 0 0

0112𝑚(6𝑟

A+ℎA) 0

0 0 𝑚𝑟A

h

h


Parallel-Axis Theorem• The object does not necessarily rotate around the

center of mass.• Some point can be fixed!

• parallel axis theorem:

𝐼 = 𝐼=>? +𝑚𝑑A

• 𝐼^: inertia around axis 𝑢.• 𝐼=>? inertia about a parallel axis through the COM.• 𝑑 is the distance between the axes.


Parallel-Axis Theorem• More generally, for point displacements:

𝑑f, 𝑑g, 𝑑h

𝐼ff = # 𝑦A + 𝑧A 𝑑𝑚+ 𝑚𝑑fA

𝐼gg = # 𝑧A + 𝑥A 𝑑𝑚+𝑚𝑑gA

𝐼hh = # 𝑥A + 𝑦A 𝑑𝑚+𝑚𝑑hA

𝐼fg = # 𝑥𝑦 𝑑𝑚+ 𝑚𝑑f𝑑g

𝐼fh = # 𝑥𝑧 𝑑𝑚 +𝑚𝑑f𝑑h

𝐼gh = # 𝑦𝑧 𝑑𝑚+𝑚𝑑g𝑑h


Perpendicular-Axis Theorem• 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:

𝑥

𝑦𝑧

𝐼h = 𝐼f + 𝐼gfor any planar object

𝑥

𝑦𝑧

𝐼h = 2𝐼f = 2𝐼gfor symmetrical objects


Reference Frame• The representation of the inertia tensor is coordinate

dependent.

• The physical effect should be invariant to coordinates!

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

𝐼��U�W = 𝑅 4 𝐼��Wg 4 𝑅k

• The moment of inertia is invariant!• Why?


Torque• A force �⃗� applied at a distance r from

a (fixed) center of mass.

• Tangential part causes tangential acceleration:

�⃗�� = 𝑚 4 �⃗��

• The torque 𝜏 is defined as:𝜏 = 𝑟×�⃗�

• So we get𝜏 = 𝑚 ∗ 𝑟 ∗ 𝛼 ∗ 𝑟 = 𝑚 ∗ 𝑟A ∗ 𝛼

• unit is 𝑁 ∗ 𝑚• rotates an object about its axis of

rotation.


Newton’s Second Law

• The law �⃗� = 𝑚 4 �⃗� has an equivalent with the inertia tensor and torque:

𝜏 = 𝐼�⃗�

• Force ó linear acceleration

• Torque ó angular acceleration


Torque and Angular Momentum• Reminder: in the linear case: �⃗� = W�⃗

W�(�⃗� is the linear

momentum).• Similarly with torque and angular momentum:

𝑑𝐿𝑑𝑡

=𝑑�⃗�𝑑𝑡×�⃗� + �⃗�×

𝑑�⃗�𝑑𝑡

= �⃗�×𝑚�⃗� + �⃗�×�⃗� = 0 + 𝜏

• Kinematics:𝑑𝐿𝑑𝑡

=𝑑(𝐼𝜔)𝑑𝑡

= 𝐼𝑑𝜔𝑑𝑡

= 𝐼�⃗� = 𝜏

• Force ó derivative of linear momentum.

• Torque ó derivative of angular momentum.


Rotational Kinetic Energy• Translating energy formulas to rotational motion.

• The rotational kinetic energy is defined as:

𝐸�U =12𝜔k ⋅ 𝐼 ⋅ 𝜔


Conservation of Mechanical Energy• Adding rotational kinetic energy

𝐸�� 𝑡 + ∆𝑡 + 𝐸� 𝑡 + ∆𝑡 + 𝐸�U 𝑡 + ∆𝑡= 𝐸�� 𝑡 + 𝐸� 𝑡 + 𝐸�U(𝑡) + 𝐸>

• 𝐸�� is the translational kinetic energy.• 𝐸� is the potential energy.• 𝐸�U is the rotational kinetic energy. • 𝐸> the “lost” energies (surface friction, air

resistance etc.).


Impulse• 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


Rigid Body Forces

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

40

𝑭

𝑪𝑶𝑴𝒂

𝜶


Position of An Object• 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.

41

𝑭


Complex Objects• 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.

42