M S ODELLING AND CONTROL OF MULTIBODY MECHANICAL SYSTEMS AMPLE EXAM PAPER

M ODELLING AND CONTROL OF MULTIBODY MECHANICAL SYSTEMS
S AMPLE EXAM PAPER
1.
Consider the (simplified and normalised) equations describing the motion of a vertical
take-off and landing aircraft moving in a horizontal plane, namely
x¨ = −(sin θ )v1 + ε (cos θ )v2
z¨ = (cos θ )v1 + ε (sin θ )v2 − g
θ¨ = v2 ,
where (x, z) describes the position of the centre of mass of the aircraft in a vertical
plane, θ the roll angle, g the gravity acceleration, v1 and v2 the control actions and
ε > 0 a parameter that captures the effect of the “slopped” wings and induces a coupling
between the vertical and the roll dynamics.
z
ε
v2
v1
θ
g
centre of mass
x
a)
Show that the system, with v1 = v2 = 0, can be written as an Hamiltonian system with M = I. In particular, write the internal Hamiltonian H0 (q, p), where
[ 4 marks ]
q = (x, y, θ ) and p are the corresponding momenta.
b)
Verify that if v1 = v2 = 0 then H˙ 0 (q, p) = 0.
c)
Show that the system is not a simple Hamiltonian system, i.e. it is not possible
to define a Hamiltonian function H(q, p, u) = H0 (q, p) − H1 (q)v1 − H2 (q)v2 ,
with H0 (q, p) as in part a), such that
∂H ′
∂H ′
q˙ =
p˙ = −
.
∂p
∂q
[ 2 marks ]
[ 4 marks ]
d)
Let v1 and v2 be constant. Compute the equilibrium points of the system. Give
a physical interpretation for the obtained result.
[ 4 marks ]
e)
Show that the point (q, p) = (0, 0) is an equilibrium of the system. What is the
value of the input signals associated to this equilibrium?
[ 2 marks ]
f)
Compute the linearization of the system around the equilibrium (q, p) = (0, 0).
Show that this equilibrium can be locally asymptotically stabilised by a state
feedback control law exploiting Lyapunov first method.
Finally, suppose that it is possible to measure only y = q. Is the equilibrium
(q, p) = (0, 0) locally asymptotically stabilizable by a dynamic output feedback
control law?
[ 4 marks ]
Modelling and control of multibody mechanical systems
1/6
2.
Consider a simple Hamiltonian system with
H0 (q, p) =
1
1
1 2
p (1 + α q2 ) + q2 − qn ,
2
2
n
α > 0, n > 2 and even, and H1 (q) = q.
a)
Write the Hamiltonian equations of motion.
[ 4 marks ]
b)
Find the equilibria of the system for u = 0.
[ 2 marks ]
c)
Study the local stability properties of each equilibrium.
[ 2 marks ]
d)
Asymptotically stabilise the system around the equilibrium (0, 0) using Lyapunov first method.
[ 4 marks ]
e)
Using the shaping function method find a control law which globally asymptotic stabilises the equilibrium (0, 0).
[ 8 marks ]
Modelling and control of multibody mechanical systems
2/6
3.
A circular disk of mass m and radius a rolls without sliding on a horizontal plane as
shown in Figure 3.1 (the disk in the figure has non-zero width for illustration purposes).
The plane of the disk remains always vertical. The moment of inertia of the disk about
its spin axis is Iyy and about a diameter is Izz .
i
O
i′
j′
j
j
ψ
r
−θ˙
C
Figure 3.1 A disk rolling on a horizontal plane.
A moving Cartesian coordinate system with unit vectors i ′ and j ′ is used to analyse the
motion of the object. This coordinate system has a fixed origin O but it rotates by an
angle ψ so that it has the same orientation as the body fixed axes (shown with dashed
lines on the object).
a)
The coordinates of the centre of mass, C, in the moving reference frame are
(x′ , y′ ). Give the kinetic energy of the disk in terms of the four generalised
coordinates x′ , y′ , ψ , θ .
[5]
b)
Write the equations of the rolling constraint.
[3]
c)
Hence derive the equations of motion of the object.
[8]
d)
What are the forces that maintain the rolling constraint?
[4]
Modelling and control of multibody mechanical systems
3/6
4.
a)
A helicopter blade of mass m is attached onto a massless rotor that rotates with
a fixed angular speed ω . The blade has a lagging freedom relative to the rotor
described by the angle γ as shown in Figure 4.1.
e
rotor
γ
f
C
blade
ω
Figure 4.1 Plan view of a helicopter rotor with one blade.
The radius of the rotor is e, and the distance along the blade from the blade attachment point to the centre of mass of the blade, C, is f . The moment of inertia
of the blade about the axis passing through the centre of mass and which is normal to the plane of the diagram is Izz . A damping moment of magnitude −Dγ˙
opposes the motion of the blade relative to the rotor, where D is the damping
coefficient.
i)
Write an expression for the kinetic energy of the system.
ii)
Show that the lagging equation of motion is
[4]
(m f 2 + Izz )γ¨ + Dγ˙ + mω 2 e f sin γ = 0.
[6]
b)
Prove that for a general rigid body motion about a fixed point the rate of change
of the kinetic energy T is given by
dT
= Ω · N,
dt
where Ω is the angular velocity vector of the body and N is the external torque.
[ 10 ]
Modelling and control of multibody mechanical systems
4/6
5.
A cylindrical bar is made to rotate uniformly with an angular speed of ω about an axis
passing through the centre of the bar and making an angle θ with the bar as shown in
Figure 5.1.
ω
θ
i
k
Figure 5.1 Cylindrical bar.
a)
Find the angular velocity vector of the bar expressed in a body fixed reference
frame.
[2]
b)
Find the angular momentum vector of the bar expressed in the same body fixed
reference frame.
[2]
c)
Find the magnitude and direction of the torque driving the bar.
d)
If in addition to ω the bar rotates about its axis of symmetry with a speed of φ˙
then
[6]
i)
what are the new angular velocity and angular momentum vectors?
[5]
ii)
what is the extra torque that is needed to drive the bar?
Modelling and control of multibody mechanical systems
[5]
5/6
6.
A particle of mass m slides without friction on a wedge of angle α and mass M that can
move without friction on a smooth horizontal surface, as shown in Figure 6.1.
m
M
α
Figure 6.1 A particle slides on a wedge. The wedge slides on the horizontal surface.
a)
Treating the constraint of the particle on the wedge by the method of Lagrange
multipliers, find the equations of motion for the particle and wedge.
[ 15 ]
b)
Also obtain an expression for the forces of constraint between the particle and
the wedge.
[5]
Modelling and control of multibody mechanical systems
6/6
Modelling and control of multibody mechanical systems
Sample exam paper - Model answers
Question 1
a) Setting v1 = v2 = 0 yields
x
¨=0
z¨ = −g
θ = 0.
˙ Then
Let q = (x, z, θ) and M = I, hence p = (x,
˙ z,
˙ θ).


0


p˙ =  −g  .
0
q˙ = p
The above system is a simple Hamiltonian system with internal Hamiltonian
H0 (q, p) =
b) Note that
1 ′
p p + gz.
2
H˙ 0 = p′ p˙ + gz˙ = −zg
˙ + gz˙ = 0.
c) The system can be written as
q˙ = p






0
− sin θ
ǫ cos θ


 


p˙ =  −g  +  cos θ  v1 +  ǫ sin θ  v2 .
0
0
1
The system can be written as a simple Hamiltonian system provided there exist H1 (q)
and H2 (q) such that


′
− sin θ
∂H1


=  cos θ 
∂q
0
and
∂H2
∂q
From the first equation we have that
′


ǫ cos θ


=  ǫ sin θ  .
1
∂H1
=0
∂θ
implying that H1 (q) is not a function of θ. However, this contradicts the equation
∂H1
= sin θ.
∂x
As a result the system cannot be written as a simple Hamiltonian system.
1
d) Setting q˙ = p˙ = 0 yields p = 0 and
0 = − sin θv1 + ǫ cos θv2
0 = cos θv1 + ǫ sin θv2 − g
0 = v2 .
This implies v2 = 0 and
0 = − sin θv1
0 = cos θv1 − g.
Note that we do not have any constraint on x and z. As a result, θ = 0 + 2kπ, yielding
v1 = g, or θ = π + 2kπ, yielding v1 = −g, with k integer. In summary we have two
types of equilibria:
q = (⋆, ⋆, 0)
p = (0, 0, 0)
(v1 , v2 ) = (g, 0)
and
q = (⋆, ⋆, π)
p = (0, 0, 0)
(v1 , v2 ) = (−g, 0).
Both equilibria correspond to the aircraft ’floating’ at some fixed location with zero roll
angle or with roll angle equal to π, i.e. the aircraft is upside down. The corresponding
input signals are such that the aircraft does not roll: v2 = 0, and the effect of gravity
is compensated: v1 = ±g.
e) From the above discussion we have that q = p = 0 is an equilibrium associated to v1 = g
and v2 = 0.
f) The system linearised around the zero equilibrium is described by
"
δ˙q
δ˙q
with
#
=
"
#"
0 I
−G 0
δq
δq
#
+
"
0
B
#"
v1
v2
#


0 0 −g


0 
G= 0 0
0 0
0
and


0 ǫ


B =  1 0 .
0 1
The zero equilibrium may be stabilised by a state feedback control law if the reduced
system
x˙ = Gx + Bu
is controllable. The controllability matrix of this system is
C=
h
B GB G2 B
i


0 ǫ 0 −g 0 0


0 0 0 
= 1 0 0
0 1 0
0 0 0
and this has rank three, hence the system is controllable.
Finally, the zero equilibrium may be stabilised by an output feedback control law if the
system with output y = q is observable. A submatrix of the observability matrix is
˜=
O
"
I 0
0 I
#
which has rank six, hence the system is observable.
2
Question 2
a) The Hamiltonian equations of motion are
p˙ = −αqp2 − q + q n−1 + u.
q˙ = p(1 + αq 2 )
b) The equilibria of the system for u = 0 are obtained solving the equation
−q + q n−1 = q(−1 + q n−2 ) = 0.
Since n > 2, and even, we have three solutions: q = 0, q = 1 and q = −1. Therefore,
for u = 0 we have three equilibria
(q, p) = (0, 0)
(q, p) = (1, 0)
(q, p) = (−1, 0).
c) The equilibrium (0, 0) is associated to a local minimum of the potential energy, hence
it is a stable equilibrium. The other two equilibria are associated to local maxima of
the potential energy, hence they are unstable equilibria.
d) The linearised system around the (0, 0) equilibrium is described by the equation
δ˙x =
"
0 1
−1 0
#
δx +
"
0
1
#
δu .
Setting δ = Kδx , with K = [K1 , K2 ] yields a closed-loop system which is described by
δ˙x =
"
0
1
−1 + K1 K2
#
δx .
This system is asymptotically stable if K2 < 0 and −1 + K1 < 0.
e) To obtain a control law which globally asymptotically stabilises the zero equilibrium
using the shaping function method we first have to modify the potential energy of the
system. To this end, consider a modified Hamiltonian
Hm (q, p) = H0 (q, p) − qu + Vd (q),
with Vd (q) such that
1 2 1 n
q − q + Vd (q)
2
n
has a global minimum at q = 0. For example, select Vd (q) = n1 q n . Note now that
H˙ 0 + V˙ d = p(1 + αq 2 )(u + q n−1 ).
Hence, selecting
u = −q n−1 − p
yields
H˙ 0 + V˙ d = −p2 .
This implies that p converges to zero. The system restricted to the set p = 0 is described
by
q˙ = 0
0 = p˙ = −q,
which implies that the equilibrium (0, 0) is globally asymptotically stable.
3
Question 3
a) The position vector of the centre of mass is given by
r = x′ i ′ + y ′ j ′ ,
and therefore the velocity vector by differentiation is
˙ ′ + (y˙ ′ + x′ ψ)j
˙ ′.
r˙ = (x˙ ′ − y ′ ψ)i
Hence the kinetic energy of the system is
1 ˙ 2 + (y˙ ′ + x′ ψ)
˙ 2 + 1 Izz ψ˙ 2 + 1 Iyy θ˙ 2 .
T = m (x˙ ′ − y ′ ψ)
2
2
2
b) The equations of the rolling constraint are
x˙ ′ − y ′ ψ˙ + aθ˙ = 0,
y˙ ′ + x′ ψ˙ = 0.
(1)
(2)
c) The Lagrangian function is L = T −V = T . The Lagrangian equation for the generalised
coordinate x′ is
∂L
d ∂L
− ′ + λ1 = 0,
′
dt ∂ x˙
∂x
or
d ˙ − m(y˙ ′ + x′ ψ)
˙ ψ˙ + λ1 = 0,
m(x˙ ′ − y ′ ψ)
dt
or by making use of Equations 1, 2,
d −maθ˙ + λ1 = 0,
dt
or
¨
λ1 = maθ.
The Lagrangian equation for the generalised coordinate y ′ is
d
dt
∂L
∂ y˙ ′
−
∂L
+ λ2 = 0,
∂y ′
or
d ˙ + m(x˙ ′ − y ′ ψ)
˙ ψ˙ + λ2 = 0,
m(y˙ ′ + x′ ψ)
dt
or by making use of Equations 1, 2,
˙
λ2 = maθ˙ ψ.
The Lagrangian equation for the generalised coordinate ψ is
d
dt
or
∂L
∂ ψ˙
!
−
∂L
− λ1 y ′ + λ2 x′ = 0,
∂ψ
d ˙ ′ + (y˙ ′ + x′ ψ)x
˙ ′ ) + Izz ψ˙ − λ1 y ′ + λ2 x′ = 0,
m(−(x˙ ′ − y ′ ψ)y
dt
4
or by making use of Equations 1, 2 and substituting the λ’s from above we get
or
d ˙ ′ + Izz ψ˙ − maθy
¨ ′ + maθ˙ψx
˙ ′ = 0,
maθy
dt
¨ ′ + maθ(
˙ y˙ ′ + ψx
˙ ′ ) + Izz ψ¨ − maθy
¨ ′ = 0,
maθy
or by making use of Equation 2
Izz ψ¨ = 0.
The Lagrangian equation for the generalised coordinate θ is
d
dt
or
∂L
∂ θ˙
−
∂L
+ λ1 a = 0,
∂θ
d Iyy θ˙ + λ1 a = 0,
dt
or by substituting λ1 by maθ¨ we get
(Iyy + ma2 )θ¨ = 0.
d) The force which maintains the longitudinal rolling constraint is
−λ1 = −maθ¨ = 0.
The force which prevents side-slipping of the object is
˙
−λ2 = −maθ˙ ψ.
5
Question 4
a) We attach two rotating Cartesian coordinate systems, one on the rotor and one on the
blade as shown in Figure 1.
e
rotor
i
i′
j
γ
j′
C
f
blade
ω
Figure 1: Plan view of a helicopter rotor with one blade.
i) The position vector of the blade centre of mass is
r = ei + f i′,
and therefore the velocity vector is
r˙ = eωj + f (ω + γ)j
˙ ′ = −f (ω + γ)
˙ sin γi + (eω + f (ω + γ)
˙ cos γ)j.
The kinetic energy is thus
T =
or
1
1 2
m f (ω + γ)
˙ 2 sin2 γ + (eω + f (ω + γ)
˙ cos γ)2 + Izz (ω + γ)
˙ 2,
2
2
1
1 T = m e2 ω 2 + 2ef ω(ω + γ)
˙ cos γ + f 2 (ω + γ)
˙ 2 + Izz (ω + γ)
˙ 2.
2
2
ii) We find the lagging equation of motion using the Lagrangian approach. The
Lagrangian function is L = T − V = T and therefore,
d
dt
∂L
∂ γ˙
−
∂L
= −Dγ,
˙
∂γ
or
or
d m(ef ω cos γ + f 2 (ω + γ))
˙ + Izz (ω + γ)
˙ + mef ω(ω + γ)
˙ sin γ = −Dγ,
˙
dt
(mf 2 + Izz )¨
γ + Dγ˙ + mω 2 ef sin γ = 0.
b) The kinetic energy is given by
1
T = Ω · H.
2
The rate of change of T is
d
dT
=
dt
dt
1
Ω·H
2
1
=
2
6
dΩ
dH
·H +Ω·
dt
dt
We can make use of a rectangular coordinate system that is fixed onto the rigid body
and rotates with it. The derivatives of the vectors appearing in the above expression can
be expressed in terms of the derivatives observed inside the rotating reference frame.
Thus,
1
2
dH
dΩ
·H +Ω·
dt
dt
1
=
2
d′ Ω
Ω×Ω+
dt
d′ H
·H +Ω· Ω×H +
dt
The scalar triple product identity gives us
(Ω × Ω) · H = (Ω × H) · Ω.
If we express Ω in terms of its components along the body fixed axes then

The vector H in similarly given by

Ωx


Ω =  Ωy  .
Ωz


Ωx


H = I  Ωy  ,
Ωz
where I is the inertia tensor of the rigid body. The derivatives of Ω and H observed
in the moving reference frame are
˙x
Ω
 ˙ 
=  Ωy  ,
dt
˙z
Ω

d′ Ω
and

˙x
Ω
d′ H
 ˙ 
= I  Ωy  .
dt
˙z
Ω

The term

h
i
d′ Ω
˙x Ω
˙y Ω
˙ z IΩ,
·H = Ω
dt
and the term
Ω·
′
h
i
d′ H
d′ H
˙y Ω
˙ z I T Ω = d Ω · H,
=
· Ω = Ω˙ x Ω
dt
dt
dt
since I T = I. The rate of change of kinetic energy becomes
dT
1
d′ H
d′ H
=
·Ω = Ω×H +
2(Ω × H) · Ω + 2
dt
2
dt
dt
7
·Ω=
dH
· Ω = N · Ω.
dt
Question 5
a) The angular velocity vector of the bar in body fixed axes is
Ω = −ω sin θi + ω cos θk.
b) The angular momentum vector in body fixed axes is
H = −Ixx ω sin θi + Izz ω cos θk,
since the chosen axes are principal. Izz is the moment of inertia about the axis of
symmetry and Ixx is the moment of inertia about a diameter passing through the
centre of mass.
c) The torque driving the bar is given by the rate of change of the angular momentum:
N=
d′ H
dH
=Ω×H +
,
dt
dt
when the derivative is observed in the body fixed reference frame. This gives
Ω×H +
d′ H
= (−ω sin θi + ω cos θk) × (−Ixx ω sin θi + Izz ω cos θk) + 0,
dt
since θ and ω are not changing. We therefore get
N = (Izz − Ixx )ω 2 sin θ cos θj.
The direction j is always perpendicular to both the axis of rotation and the axis of
symmetry of the bar.
d)
i) The new angular velocity vector is given by
˙
Ω = −ω sin θ cos φi + ω sin θ sin φj + (ω cos θ + φ)k.
The new angular momentum vector is given by
˙
H = −Ixx ω sin θ cos φi + Iyy ω sin θ sin φj + Izz (ω cos θ + φ)k,
where Iyy is the moment of inertia about the direction of j and is equal to Ixx due
to symmetry.
ii) The torque that is needed to drive the bar is
N=
dH
d′ H
=Ω×H +
.
dt
dt
The first term on the right-hand side is
˙ + (Izz − Ixx )ω sin θ cos φ(ω cos θ + φ)j
˙ +
Ω × H = (Izz − Iyy )ω sin θ sin φ(ω cos θ + φ)i
+(Ixx − Iyy )ω 2 sin2 θ sin φ cos φk,
or
˙
Ω × H = (Izz − Ixx )ω sin θ(ω cos θ + φ)(sin
φi + cos φj),
8
since Ixx = Iyy . The last term on the right-hand side is
d′ H
= Ixx ω φ˙ sin θ sin φi + Iyy ω φ˙ sin θ cos φj = Ixx ω φ˙ sin θ(sin φi + cos φj).
dt
The torque is therefore
N = (Izz − Ixx )ω cos θ + Izz φ˙ ω sin θ(sin φi + cos φj).
This torque is always in the direction normal to the axis of symmetry of the bar
and the axis of rotation with speed ω (same as in part c) above). The magnitude
of the additional torque needed due to the extra rotation with speed φ˙ is
˙ sin θ.
Izz φω
9
Question 6
a) We make use of a fixed Cartesian coordinate system as shown in Figure 2.
i
k
m
M
α
Figure 2: A particle slides on a wedge. The wedge slides on the horizontal surface.
Let’s assume that
rm = xm i + zm k
and
rM = xM i + zM k
are the position vectors of the particle and the highest vertex of the wedge respectively.
The system configuration can be characterised by three generalised coordinates xM , xm
and zm . zM is a constant. The constraint that the particle stays on the wedge can be
expressed by a dot product as follows:
f = (rm − rM ) · (sin αi + cos αk) = 0,
or
f = (xm − xM ) sin α + (zm − zM ) cos α = 0.
The kinetic energy of the system is
T =
1
1
2
M x˙ 2M + m(x˙ 2m + z˙m
),
2
2
and the potential energy is
V = −mgzm .
The Lagrangian function is
1
1
2
L = T − V = M x˙ 2M + m(x˙ 2m + z˙m
) + mgzm .
2
2
The Lagrangian equation corresponding to the generalised coordinate xM is
d
dt
or
∂L
∂ x˙ M
−
∂f
∂L
+λ
= 0,
∂xM
∂xM
d
(M x˙ M ) − λ sin α = 0,
dt
10
(3)
or
Mx
¨M − λ sin α = 0.
(4)
The Lagrangian equation corresponding to the generalised coordinate xm is
d
dt
or
∂L
∂ x˙ m
−
∂f
∂L
+λ
= 0,
∂xm
∂xm
d
(mx˙ m ) + λ sin α = 0,
dt
or
m¨
xm + λ sin α = 0.
(5)
The Lagrangian equation corresponding to the generalised coordinate zm is
d
dt
or
∂L
∂ z˙m
−
∂f
∂L
+λ
= 0,
∂zm
∂zm
d
(mz˙m ) − mg + λ cos α = 0,
dt
or
m¨
zM − mg + λ cos α = 0.
From Equations 4, 5 we get
x
¨M = −
(6)
m
x
¨m .
M
From Equation 5, 6 we get
m¨
xm cos α − m¨
zm sin α + mg sin α = 0,
and by differentiating Equation 3 and substituting x¨M we get
z¨m = − 1 +
m
M
tan α¨
xm .
If we substitute the value of z¨m into Equation 7 we get
x
¨m = −
By substitution
z¨m =
m
(1 + M
)g sin2 α
,
m
sin2 α
1+ M
and
x
¨M =
and finally
λ=−
g sin α cos α
.
m
1+ M
sin2 α
m
M g sin α cos α
,
m
1+ M
sin2 α
m¨
xm
mg cos α
=
.
m
sin α
1+ M
sin2 α
11
(7)
b) The force of constraint on the particle in the horizontal direction is −λ sin α or
−
mg cos α sin α
,
m
1+ M
sin2 α
and in the vertical direction it is −λ cos α or
−
mg cos2 α
.
m
1+ M
sin2 α
The force of constraint on the wedge exerted by the particle is given by λ sin α or
mg cos α sin α
.
m
sin2 α
1+ M
12