ME 407 Advanced Dynamics
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Transcript of ME 407 Advanced Dynamics
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ME 407 Advanced Dynamics
We will learn to model systems that can be viewed as collections of rigid bodies
Common mechanical systems
Robots
Various wheeled vehicles
The focus will be on engineering applications
Divers and gymnasts
I’m open to applications you all care about
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I expect you to be comfortable with mathematics and abstract thinking in general
even though our applications will be concrete
I expect you to be familiar with
geometry
trigonometry
linear algebra
systems of ordinary differential equations
vectors
Prerequisites
and some basic physics
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YOU NEEDTO INTERRUPT MEIF YOU DON’T KNOW WHAT IS GOING ON
THIS IS IMPORTANT
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BoilerplateThere’s a web site: www.me.rochester.edu/courses/ME407
(NOT UP TO DATE — STAY TUNED)
My email, which I read regularly: [email protected]
Text: Engineering Dynamics: From the Lagrangian to Simulationavailable in preprint form from Jill in the department office.
Weekly problems sets
Probably two midterms
Meirovitch and/or Goldstein will be useful at the beginningboth on two hour reserve in Carlson
Office hours Tuesday-Thursday 2 – 4 or by appointment.
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We will go from very fundamental to very applied
conservation of momentum and angular momentum
What is a rigid body?
Moments of inertia
internal and external forces and torques
work and energy
geometry of three dimensional motion
angular velocity and angular momentum
coordinate systems
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We will go from very fundamental to very applied
Hamilton’s principle
The Euler-Lagrange equations
Hamilton’s equations
Kane’s method
The null-space method
Computational tricks: the method of Zs
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We will go from very fundamental to very applied
engineering mechanisms: linkages, gears, etc.
robots and their relatives
wheeled vehicles of different sorts
I’m open to applications you all care about
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Let me show you a couple of hard problemsso you can see where we are going
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We will also need mathematical and computational tools
We need notation to understand ourselves better
Most of the interesting problems are wildly nonlinearand we’ll need to integrate differential equations numerically
I’m perfectly happy to use commercial code to do thisbut you do need to have an idea of what to expect
so you can figure out if it’s right.
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You will find Mathematica very useful. It’s available on many UR computers.
We can take part of a class to deal with this if necessary.The following link will get you to more information than you need.
http://www.me.rochester.edu/courses/ME201/websoft/softw.html
Mathematica
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A little bit about notation
vectors will be lower case bold face
matrices will be upper case bold face
“Vector notation”
Matrix/linear algebra notation
vectors will be column vectors, their transposes row vectors
Indicial notation
vectors have one superscript, their transposes have one subscript
“real matrices” have one superscript and one subscriptdenoting row and column respectively
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€
a = ai =
a1
a2
MaN
⎧
⎨ ⎪ ⎪
⎩ ⎪ ⎪
⎫
⎬ ⎪ ⎪
⎭ ⎪ ⎪
, aT = ai = a1 a2 L aN{ }
€
A = A. ji =
A11 A2
1 L AN1
A21 A2
2 M MM M O M
AN1 L L AN
N
⎧
⎨ ⎪ ⎪
⎩ ⎪ ⎪
⎫
⎬ ⎪ ⎪
⎭ ⎪ ⎪
Matrices do not have to be square.
Examples of the notations
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Vector-matrix multiplication
€
Ax = A. ji x j =
A11 A2
1 L AN1
A21 A2
2 M MM M O M
AN1 L L AN
N
⎧
⎨ ⎪ ⎪
⎩ ⎪ ⎪
⎫
⎬ ⎪ ⎪
⎭ ⎪ ⎪
x1
x 2
Mx N
⎧
⎨ ⎪ ⎪
⎩ ⎪ ⎪
⎫
⎬ ⎪ ⎪
⎭ ⎪ ⎪
€
a ⋅b = aibii=1
N
∑ = aibi
i=1
N
∑
€
ab = abT = aib j =
a1b1 a1b2 L a1bN
a1b2 a2b2 M MM M O M
a1bN L L aNbN
⎧
⎨ ⎪ ⎪
⎩ ⎪ ⎪
⎫
⎬ ⎪ ⎪
⎭ ⎪ ⎪
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Summation convention
€
a ⋅b = aibii=1
N
∑ = aibi
i=1
N
∑ ⇒ aibi
“Metric tensor”
€
gij =1,i = j0,i ≠ j ⎧ ⎨ ⎩
, gij =1,i = j0,i ≠ j ⎧ ⎨ ⎩
⇒ a ⋅b = gijaib j , a ⋅b = gijaib j
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??
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The inertial coordinate system: coordinates x, y, z; unit vectors i, j, k
i
j
k
€
r = xi + yj+ zk
We will also have body coordinates, but not today
We have to do physics in the inertial coordinate system
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Start from the very basic: “f = ma” and consider a single particle/point mass — moments of inertia all zero
€
v = ˙ r ⇔ v i = ˙ r i, a = ˙ v = ˙ ̇ r ⇔ ai = ˙ v i = ˙ ̇ r i
€
f = m˙ ̇ r ⇔ f i = m˙ ̇ r i
Conservation of momentum
€
˙ x = dxdt
, ˙ r = drdt
, ˙ A = dAdt
L
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Angular momentum
This doesn’t mean much for a particle, but we might as well start here
€
l = r × p = mr × v
This angular momentum is defined wrt the inertial origin, but any reference will do —
different reference, different angular momentum
€
l* = r − r0( ) × p = mr *×v
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Its rate of change
€
˙ l * = m ˙ r − ˙ r 0( ) × v + m r − r0( ) × ˙ v
€
˙ r 0 = 0 = ˙ r × v⇒ ˙ l * = m r − r0( ) × ˙ v
which we call the torque.
The torque depends on the point of reference — remember this
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Example: a particle falling under gravity
€
x = x0, y = v0t, z = w0t − 12
gt 2
i
j
k
€
p = mv0j+ m w0 − gt( )k
€
l = x0i + v0tj + w0t − 12
gt 2 ⎛ ⎝ ⎜
⎞ ⎠ ⎟k
⎛ ⎝ ⎜
⎞ ⎠ ⎟× mv0j+ m w0 − gt( )k( )
€
l = − 12
gt 2mv0i − m w0 − gt( )x0j+ mx0v0k
€
τ =˙ l = −gtmv0i + mgx0j
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??
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WORK AND ENERGY
work = force times distance, so
€
dW = f ⋅ds = f ⋅dr
€
dWdt
= f ⋅ drdt
= f ⋅v
€
f = m˙ v ⇒ dWdt
= m˙ v ⋅v = ddt
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mv ⋅v ⎛ ⎝ ⎜
⎞ ⎠ ⎟= dT
dt
€
T = 12
mv ⋅v
The kinetic energy of a particle
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€
dWdt
dt = W2 −W1 =t1
t2∫ dTdt
dt =t1
t2∫ T2 − T1 ⇔ ΔW = ΔT
and we can go back to the beginning and note that
€
ΔT = f ⋅dss1
s2∫
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i
j
k
1
2
€
ΔT = f ⋅dss1
s2∫
In general the integral
will be different for the red path and the blue path
€
ΔT = f ⋅ds = 0∫
If the integral is the same for all paths, we’ll have
and the force is conservative
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Conservative forces come from potentials
A force is conservative iff
€
f = −∇V r( )
Potentials can be time-dependent; we will not deal with time-dependent potentials
There’s a discussion of potentials in the text, and I’ll do a little on the board
Bottom line
The total energy, T + V, is conserved for a single particle under conservative forces
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An aside regarding potentials
M
m
€
f = −G mMr2 er
€
V = −G mMr
⇒ f = −∇V = −G mMr2 er
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For celestial mechanics we do not include the m in the potentialWe associate the potential with the gravitating body
€
V = −G Mr⇒ f = −m∇V = −G mM
r2 er
There are several simple orbital examples in the text.
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SYSTEMS OF PARTICLES
€
f1 = m˙ ̇ r 1
€
f4 = m˙ ̇ r 4
€
f5 = m˙ ̇ r 5
€
f2 = m˙ ̇ r 2
€
f3 = m˙ ̇ r 3
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The particles can interact — including action at a distance
€
f1 = f1(e ) + f21 + f31 + f41 + f51
f2 = f2(e ) + f12 + f32 + f43 + f53
M
Split each force into an external part and an interaction part, within the system
momentum of the system
€
p = p1 + p2 + p3 + p4 + p5
the rate of change is equal to the force, so we have
€
˙ p = ˙ p 1 + ˙ p 2 + ˙ p 3 + ˙ p 4 + ˙ p 5= f1
(e ) + f21 + f31 + f41 + f51 + f2(e ) + f12 + f32 + f43 + f53 +L
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€
˙ p = ˙ p 1 + ˙ p 2 + ˙ p 3 + ˙ p 4 + ˙ p 5= f1
(e ) + f21 + f31 + f41 + f51 + f2(e ) + f12 + f32 + f43 + f53 +L
cancel
All such pairs cancel by Newton’s third law of action and reaction
This is called
The weak law of action and reaction
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from which we deduce
€
˙ p = f1(e ) + f2
(e ) + f3(e ) + f4
(e ) + f5(e )
or, more generally
€
˙ p = fi(e )
i−1
N
∑
Only the external forces change the momentum of a systemunder the weak law of action and reaction
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What is the momentum of a system?
€
p = ddt
mirii=1
N
∑
write
€
M = mii=1
N
∑ then
€
p = M ddt
miri
Mi=1
N
∑ = M drCM
dt⇒ rCM = miri
Mi=1
N
∑
€
M˙ ̇ r CM = fi(e )
i=1
N
∑ = F
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If the sum of the external forces acting on a system is zero, the momentum of the system is conserved
For example: the contents of a shotgun shell fired in a vacuum
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We can do the same thing for torque and angular momentum,and we’ll find we need a new law
€
l = r1 × p1 + r2 × p2 + r3 × p3 + r4 × p4 + r5 × p5
€
˙ l = τ = m2r1 × ˙ v 1 + m2r2 × ˙ v 2 + m3r3 × ˙ v 3 + m4r4 × ˙ v 4 + m5r5 × ˙ v 5 = r1 × f1 + r2 × f2 + r3 × f3 + r4 × f4 + r5 × f5
Look at a pair for simplicity’s sake
€
r1 × f1(e ) + f12( ) + r2 × f2
(e ) + f21( )
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€
r1 × f1(e ) + f12( ) + r2 × f2
(e ) + f21( )
€
r1 × f1(e ) + r2 × f2
(e ) + r1 × f12 + r2 × f21
The internal torques will cancel if the forces are parallel to a line connecting the two particles€
f21 = −f12
€
r1 − r2( ) × f12
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reference point
r2
r1
r1 – r2
€
r1 − r2( ) × f12 = 0
if f12 is parallel to r1 – r2
Gravity works this way, as does electrostatics
Not all internal forces work this way, but all the ones we care about do
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That is the strong law of action and reaction
I will assume that throughout.
We have the following for systems
€
˙ p = fi(e )
i−1
N
∑
€
˙ l = ri × fi(e )
i−1
N
∑
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The angular momentum of a system can be written
€
l = MrCM × vCM + mi ′ r i × ′ v ii−1
N
∑
where
€
′ r i = ri − rCM , ′ v i = ˙ r i − ˙ r CM = vi − vCM
You can establish this for homework. It’s not hard and it’s a good exercise.
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€
l = MrCM × vCM + mi ′ r i × ′ v ii−1
N
∑
angular momentum of the system wrt the reference
angular momentum of the system wrt the CM
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??
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Kinetic Energy
€
T = 12
mivi ⋅vii−1
N
∑
€
vi = vCM + ′ v i
€
T = 12
mi vCM + ′ v i( ) ⋅ vCM + ′ v i( )i−1
N
∑ = 12
M mi
MvCM + ′ v i( ) ⋅ vCM + ′ v i( )
i−1
N
∑
€
T = 12
M mi
MvCM ⋅vCM
i−1
N
∑ + M mi
MvCM ⋅ ′ v i
i−1
N
∑ + 12
M mi
M′ v i ⋅ ′ v i
i−1
N
∑
€
T = 12
MvCM ⋅vCM + 12
mi ′ v i ⋅ ′ v ii−1
N
∑
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€
M mi
MvCM ⋅ ′ v i
i−1
N
∑ = MvCM ⋅ mi
M′ v i
i−1
N
∑ = MvCM ⋅ ddt
mi
M′ r i
i−1
N
∑
€
rCM = mi
Mri
i−1
N
∑ = mi
MrCM + ′ r i( )
i−1
N
∑ = rCM
Mmi
i−1
N
∑ + 1M
mi ′ r ii−1
N
∑
these are equal thisis
zero
so the kinetic energy is as on the previous slide
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€
T = 12
MvCM ⋅vCM + 12
mi ′ v i ⋅ ′ v ii−1
N
∑
kinetic energy of the center of mass
internal kinetic energy
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Let’s try to summarize today’s beginning
€
rCM = miri
Mi−1
N
∑
€
p = M˙ r CM
€
˙ p = M˙ ̇ r CM = fi(e )
i=1
N
∑ = f
€
l = ri × pii−1
N
∑
€
˙ l = ri × ˙ p ii−1
N
∑ = ri × fi(e )
i−1
N
∑ = τ
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??