Mechanics Rigid Body Motion 2, Lecture Notes - Physics, Study notes of Mechanics

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MechanicsPhysics 151^ Lecture 9Rigid Body Motion(Chapter 4, 5)

What We Did Last Time „^ Discussed 3-dimensional rotation^ „^ Preparation for rigid body motion^ „^ Movement in 3-d + Rotation in 3-d = 6 coordinates „^ Looked for ways to describe 3-d rotation^ „^ Euler angles one of the many possibilities^ „^ Euler’s theorem „^ Defined infinitesimal rotation

d Ω

„^ Commutative (unlike finite rotation) „^ Behaves as an axial vector (like angular momentum)

d^ d =^ Φ Ω^ n d^ d =^ × r^ r^ Ω

Body Coordinates „^ Consider a rotating rigid body^ „^ Define body coordinates (

x’ ,^ y’ ,^ z’ )

„^ Between^ t^

and^ t^ +^ dt , the bodycoordinates rotate by „ The direction is CCW because weare talking about the coordinate axes „ Observed in the space coordinates,any point^ r^ on the body moves by

x^ ′ x z ′ y^ y ′ z d^ d^

d = Φ × = × r n r Ω^ r

n^ d Φ^ d r r d^ d =^ Φ Ω^ n Sign opposite from last lecturebecause the rotation is CCW

General Vectors „^ Now consider a general vector

G

„^ How does it move in space/body coordinates? „^ i.e. what’s the time derivative

d G / dt^?

„^ Movement

d G^ differs in space and body coordinatesbecause of the rotation of the latter ( ) (^ )^ (^ „ If G is fixed to the body

) space^ body^

rot d^ d^

d = + G^ G^

G^ Difference is due to rotation (^0) d = G (^) ( )body d^ d =^ ( )space

× G Ω G

and d d =^ × G Ω^ G^ ( )rot

Generally true

Coriolis Effect „^ Imagine a particle observed in a rotating system^ „^ e.g. watching an object’s motion on Earth^ „^ Velocity:^ „^ Acceleration:^ „^ Newton’s equation works in the space (inertial) system,i.e.

d^ d ⎛ ⎞^ ⎛^ ⎞=^ +^ × ω ⎜ ⎟^ ⎜^ ⎟ dt^ dt ⎝ ⎠^ ⎝^ ⎠^^ s^ r =^ +^ × v v^ ω^ r s r^2 s^ s (^ ) s^

s d^ d ⎛ ⎞^ ⎛^ ⎞= =^ +^ dt^ dts^ r r^ r

×

⎜^ ⎟^ ⎜^

⎝^ ⎠^ ⎝^

=^ +^ ×^ +

v^ v^ ×^ × a^

ω^ v a^ ω^ v^ ω

ω^ r m = F a^ s^2 (^ eff^

)^ (^ )

r^

r m^

m^ m = = − ×^ −^

×^ ×

a^ F^ F^

ω^ v^ ω^

ω^ r Object appears to move according to this force

Coriolis Effecteff^ „^ Last term is centrifugal force^ „^ Middle term is Coriolis effect^ „^ Occurs when object is moving in therotating frame^ „^ Moving objects on Earth appears to deflect toward right inthe northern hemisphere^ „^ Hurricane wind pattern^ „^ Foucault pendulum

2 (^ )^

(^ )

r^

r m^

m^ m = = − ×^ −^

×^ ×

a^ F^ F^

ω^ v^ ω^

ω^ r^

ω v^ r − × ω v^ r L

Euler Angles^ „^ Use angular velocity

ω^ to calculate particles’ velocities

„^ Use Euler angles to describe the rotation of rigid bodies „ How are they connected? „^ Infinitesimal rotations can be added like vectors^ z^ ζ^ η y^ x^ ξ

z ′^ ηζ ′ y x ′^ ξ^ φ

θ

′ y z ′ z ′^ xy^ ψ x ^ ^ φ^ θ^ ψ= +^ + ω n n^ n ′ z z^ ξ n^ z

n^^ ξ

n ′ z

Euler Angles „^ Let’s express

ω^ in^ x’-y’-z’ „ Doing this in^ x-y-z^ equally easy (or difficult) „ Must express^ n ,^ n ,^ n in^ x’-y’-z’z^ ξ z’^ „ n Æ An ,^ n Æ^ Bn ( n z z^ ξ^ ξ^ z’

is OK) z^ ζ^ η y x^ ξ

z ′^ η′ζ y x ′^ ξ^ φ

θ

′ y z ′ z ′^ xy^ ψ x

n^ z

n^^ ξ

n ′ z

D^

C^

B

Euler Angles^ „^ Remember: this is in the

x’-y’-z’^ system

„^ Now we know how to express velocities in terms oftime-derivatives of Euler angles^ „^ We can write down the Lagrangian

sin^ sin^ coscos^ sin^ sin cos z^ z^ ξ

φ^ ψ^ θ^ θ

φ^ θ^ ψ^

φ^ ψ^ θ^ θ

⎡^ ′

⎢^

=^ +^ +^

=^

ω^ n^ n^

^ n

^ ^

^  ^ 

Kinetic Energy „^ Kinetic energy of multi-particle system is^ „^ If we define the body axis from the center of mass^ „^ T’^ depends only on the angular velocity^ „^ Must be a 2

1 12 2 ′ T Mv^ m v = + i^ i^2 2 nd^ order homogeneous function Motion of CoM^

Motion around CoM 1 2 2 2 (^ )^

( ,^ ,^ )

T^ M x^ y^ z^2

′ T^ φ^ θ ψ

=^ +^

 ^ + +

^ ^ 

Remember Einstein convention

Rotational Motion „^ We concentrate on the rotational part^ „^ Translational part same as a single particle

Æ^ Easy

„^ Consider total angular momentum^ „^ v is given by the rotation i^

ω^ as

m = × L r^ v^ i^ i^ i = × v ω r i i 2 2 2

2 2 2

2

(^ )

(^ ) ( ) (^

) (^ )

i^ i^ i

i^ i^ i^

i^ i^ i^ i^

i^ i

i^ i^ i^ i^

i^ i^ i^ i^ i^

i^ i^ i^ i i^ i^ i^ i^

i^ i^ i^ i^

i

m

m^ r^ x^

m x y^ m x z m^ r^

m y x^ m^ r^

y^ m y z m z x^ m z y

m^ r^ z

=^ ×^ ×

⎡^

−^ −^

⎢^

⎡^

=^ −^

⋅^ =^ −^

−^ −

⎢^

⎣^

⎦^ ⎢^

−^ −^

⎣^

L^ r^ ω^ r^ ω^ r r

ω^

ω

Inertia tensor^ I

Inertia Tensor „^ Diagonal components are familiarmoment of inertia „^ What are the off-diagonal components?^ „^ I produces yx^

L when the object is turned around y^

x^ axis (^2) „ Imagine turning something like: (^2 2 2) ( )^ sin I m r x m r = − =^ Θ xx i i i^ i i

r^ x^ Θ Balanced^

Unbalanced

Unbalanced one has non-zerooff-diagonal components,which represents “wobbliness”of rotation

Kinetic Energy „^ Kinetic energy due to rotation is^ „^ Using^ „^ Using^ „^ Defining

n^ as the unit vector in the direction of

ω

„^ NB:^ n^ moves with time „^ I = I ( t ) must change accordingly with time

(^1) T m = ⋅ v^ v i^ i^ i 2 =^ × v ω^ r i i 1 (^ )^

(^ )

i^ i^ i^

i^ i^ i T^ m^

m =^ ⋅^ ×^

=^ ⋅^ ×^

ω^ ω =^ ⋅ v^ ω^ r^

r^ v^ L = L Iω^

⋅ ω Iω T = 2

1 ⋅ n In 2 2 where T I ω ω= = 2 2

2 2 (^ ) i i i I^ m^ ⎡^ ⎤ r = ⋅ = −^ ⋅ n In r^ n ⎣^ ⎦ Moment of inertia about axis (^2) ( )^ n jk^ i^ i^ jk^

ij^ ik I^ m^ r^

x x δ= − ω= ω n

Shifting Origin „^ Origin of body axes does not have to be at the CoM^ „^ It’s convenient – Separates translational/rotational motion „^ If it isn’t,^ I

can be easily translated^ i^

′= + r R r i from origin^

from CoM 2

2 (^ )^ [(^ )^2

]

(^ )^ (^

)^2 (^ ) (

i^ i^ i^

i i i i^

i I^ m^

′ m = × = + × ′ (^) M m m

=^ ×^ +^

×^ +^ ×

⋅^ ×

r^ n^ R

r^ n R^ n^ r^

n^ R^ n

r^ n I from CoM I^ of CoM