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ME 407 Advanced Dynamics
We will learn to model systems that can be viewed as collec;ons of rigid bodies
Common mechanical systems
Robots
Various wheeled vehicles
The focus will be on engineering applica;ons
Divers and gymnasts
1
I’m open to applica.ons you all care about
2
I expect you to be comfortable with mathema;cs and abstract thinking in general
even though our applica;ons will be concrete
I expect you to be familiar with
geometry
trigonometry
linear algebra
systems of ordinary differen;al equa;ons
vectors
Prerequisites
and some basic physics
3
YOU NEEDTO INTERRUPT ME IF YOU DON’T KNOW WHAT IS GOING ON
THIS IS IMPORTANT
Boilerplate
There’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 Simula7on available in preprint form from Jill in the department office.
Weekly problems sets
Probably two midterms
Meirovitch and/or Goldstein will be useful at the beginning both on two hour reserve in Carlson
Office hours Tuesday‐Thursday 2 – 4 or by appointment.
4
5
We will go from very fundamental to very applied
conserva;on of momentum and angular momentum
What is a rigid body?
Moments of iner;a
internal and external forces and torques
work and energy
geometry of three dimensional mo;on
angular velocity and angular momentum
coordinate systems
6
We will go from very fundamental to very applied
Hamilton’s principle
The Euler‐Lagrange equa;ons
Hamilton’s equa;ons
Kane’s method
The null‐space method
Computa;onal tricks: the method of Zs
7
We will go from very fundamental to very applied
engineering mechanisms: linkages, gears, etc.
robots and their rela;ves
wheeled vehicles of different sorts
I’m open to applica.ons you all care about
8
Let me show you a couple of hard problems so you can see where we are going
9
10
11
We will also need mathema;cal and computa;onal tools
We need nota;on to understand ourselves beker
Most of the interes;ng problems are wildly nonlinear and we’ll need to integrate differen;al equa;ons numerically
I’m perfectly happy to use commercial code to do this but you do need to have an idea of what to expect so you can figure out if it’s right.
You will find Mathema7ca 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 informa;on than you need.
hkp://www.me.rochester.edu/courses/ME201/websom/somw.html
Mathema7ca
12
13
A liDle bit about nota.on
vectors will be lower case bold face
matrices will be upper case bold face
“Vector nota;on”
Matrix/linear algebra nota;on
vectors will be column vectors, their transposes row vectors
Indicial nota;on
vectors have one superscript, their transposes have one subscript
“real matrices” have one superscript and one subscript deno;ng row and column respec;vely
14
€
a = ai =
a1
a2
M
aN
, aT = ai = a1 a2 L aN{ }
€
A = A. ji =
A11 A2
1 L AN1
A21 A2
2 M M
M M O M
AN1 L L AN
N
Matrices do not have to be square.
Examples of the nota.ons
15
Vector‐matrix mul.plica.on
€
Ax = A. ji x j =
A11 A2
1 L AN1
A21 A2
2 M M
M M O M
AN1 L L AN
N
x1
x 2
M
xN
€
a ⋅b = aibii=1
N
∑ = aibi
i=1
N
∑
€
ab = abT = aib j =
a1b1 a1b2 L a1bNa1b2 a2b2 M M
M M O M
a1bN L L aNbN
16
Summa;on conven;on
€
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
17
??
18
The iner;al 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 iner.al coordinate system
Start from the very basic: “f = ma” and consider a single par;cle/ point mass — moments of iner;a all zero
€
v = ˙ r ⇔ v i = ˙ r i, a = ˙ v = ˙ ̇ r ⇔ ai = ˙ v i = ˙ ̇ r i
€
f = m˙ ̇ r ⇔ f i = m˙ ̇ r i
Conserva;on of momentum
19
€
˙ x = dxdt
, ˙ r = drdt
, ˙ A = dAdt
L
20
Angular momentum
This doesn’t mean much for a par;cle, but we might as well start here
€
l = r ×p = mr × v
This angular momentum is defined wrt the iner;al origin, but any reference will do — different reference, different angular momentum
€
l* = r − r0( ) ×p = mr *×v
21
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
22
Example: a par;cle falling under gravity
€
x = x0, y = v0t, z = w0t −12gt 2
i
j
k
€
p = mv0j+ m w0 − gt( )k
€
l = x0i + v0tj+ w0t −12gt 2
k
× mv0j+ m w0 − gt( )k( )
€
l = − 12gt 2mv0i −m w0 − gt( )x0j+ mx0v0k
€
τ = ˙ l = −gtmv0i + mgx0j
23
??
24
WORK AND ENERGY
work = force ;mes distance, so
€
dW = f ⋅ ds = f ⋅ dr
€
dWdt
= f ⋅ drdt
= f ⋅ v
€
f = m˙ v ⇒ dWdt
= m˙ v ⋅ v =ddt
12mv ⋅ v
=
dTdt
€
T =12mv ⋅ v
The kine.c energy of a par;cle
25
€
dWdt
dt =W2 −W1 =t1
t2∫ dTdtdt =
t1
t2∫ T2 −T1⇔ΔW = ΔT
and we can go back to the beginning and note that
€
ΔT = f ⋅ dss1
s2∫
26
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 conserva.ve
27
Conserva;ve forces come from poten;als
A force is conserva;ve iff
€
f = −∇V r( )
Poten;als can be ;me‐dependent; we will not deal with ;me‐dependent poten;als
There’s a discussion of poten;als in the text, and I’ll do a likle on the board
Bokom line
The total energy, T + V, is conserved for a single par.cle under conserva.ve forces
28
An aside regarding poten.als
M
m
€
f = −G mMr2er
€
V = −G mMr
⇒ f = −∇V = −G mMr2er
29
For celes;al mechanics we do not include the m in the poten;al We associate the poten;al with the gravita;ng body
€
V = −G Mr⇒ f = −m∇V = −G mM
r2er
There are several simple orbital examples in the text.
30
SYSTEMS OF PARTICLES
€
f1 = m˙ ̇ r 1
€
f4 = m˙ ̇ r 4
€
f5 = m˙ ̇ r 5
€
f2 = m˙ ̇ r 2
€
f3 = m˙ ̇ r 3
31
The par;cles can interact — including ac;on 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 interac;on 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
32
€
˙ 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 ac;on and reac;on
This is called
The weak law of ac.on and reac.on
33
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 system under the weak law of ac.on and reac.on
34
What is the momentum of a system?
€
p =ddt
mirii=1
N
∑
write
€
M = mii=1
N
∑ then
€
p = M ddt
miriMi=1
N
∑ = M drCMdt
⇒ rCM =miriMi=1
N
∑
€
M˙ ̇ r CM = fi(e )
i=1
N
∑ = F
35
If the sum of the external forces ac;ng on a system is zero, the momentum of the system is conserved
For example: the contents of a shotgun shell fired in a vacuum
36
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( )
37
€
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 connec;ng the two par;cles €
f21 = −f12
€
r1 − r2( ) × f12
38
reference point
r2 r1
r1 – r2
€
r1 − r2( ) × f12 = 0
if f12 is parallel to r1 – r2
Gravity works this way, as does electrosta;cs
Not all internal forces work this way, but all the ones we care about do
39
That is the strong law of ac.on and reac.on
I will assume that throughout.
We have the following for systems
€
˙ p = fi(e )
i−1
N
∑
€
˙ l = ri × fi(e )
i−1
N
∑
40
The angular momentum of a system can be wriken
€
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.
41
€
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
42
??
43
Kine.c Energy
€
T =12
mivi ⋅ vii−1
N
∑
€
vi = vCM + ′ v i
€
T =12
mi vCM + ′ v i( ) ⋅ vCM + ′ v i( )i−1
N
∑ =12M mi
MvCM + ′ v i( ) ⋅ vCM + ′ v i( )
i−1
N
∑
€
T =12M mi
MvCM ⋅ vCM
i−1
N
∑ + M mi
MvCM ⋅ ′ v i
i−1
N
∑ +12M mi
M′ v i ⋅ ′ v i
i−1
N
∑
€
T =12MvCM ⋅ vCM +
12
mi ′ v i ⋅ ′ v ii−1
N
∑
44
€
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
∑ =rCMM
mii−1
N
∑ +1M
mi ′ r ii−1
N
∑
these are equal this is
zero
so the kine;c energy is as on the previous slide
45
€
T =12MvCM ⋅ vCM +
12
mi ′ v i ⋅ ′ v ii−1
N
∑
kine;c energy of the center of mass
internal kine;c energy
46
Let’s try to summarize today’s beginning
€
rCM =miriMi−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
47
??