P3 Dynamics
Mark Cannon
Hilary Term 2012
0-1
Lecture 7
Rigid Body Dynamics
7-1
Review: Circular motion
Velocity and acceleration components in polar co-ordinates
R̈ − R θ̇2 radially
R θ̈ + 2Ṙ θ̇ tangentially
Components of acceleration in circular motion (R = constant)
−R θ̇2 radially
R θ̈ tangentially
For a satellite orbiting point O:
? Moment of momentum about O is
conserved
m
v
u
ur = h = constant
r
M
0
? Total energy is conserved
R
1 2
gR 2
(u + v 2 ) −
= constant
2
r
7-2
Review: Moment of momentum of a particle
A particle of mass m and velocity Ṙ has:
? momentum vector
p = mṘ
? moment of momentum about O
H = R × mṘ
H gives the magnitude of the moment and the direction of its axis.
For planar motion: e.g. particle moving with speed u in the x direction
? the momentum has magnitude
u
m
p = mu
? the moment of momentum about O has
magnitude
H = −muy
7-3
y
H
0
x
Moment of momentum for rotation about a fixed point
How can we find the moment of momentum of a general rotating object?
e.g. bicycle wheel rotating about its axle
— treat as collection of particles rotating around a single point
7-4
Moment of momentum for rotation about a fixed point
Consider two masses rotating with equal angular velocity about O in the x-y
plane
m1
V2
V1
r1
m2
r2
w
0
? Kinematic relations:
V1 = ωr1
and
V2 = ωr2
? The moment of momentum about O has magnitude
H = r1 m1 V1 + r2 m2 V2
= (m1 r12 + m2 r22 )ω
= Iω,
? I is the moment of inertia about O
I has units kg m2
7-5
I = m1 r12 + m2 r22
Moment of momentum for rotation about a fixed point
Bicycle wheel, total mass m, rotating at angular velocity ω
dm
u dm 6
r
To find the magnitude H of the wheel’s moment
of momentum about its axle:
? an element of mass dm has moment of
inertia
dH = ru dm = r 2 ω dm
?
H, ω
? assume all mass is concentrated at the
rim, at radius r , then
H = mr 2 ω
I = mr 2
= Iω,
7-6
Moments of momentum and inertia of a rigid body
w
z
A rigid body, mass m, rotates about O
with angular velocity components ωx , ωy , ωz
about x, y and z axes.
v
dm
wz
u
y
wy
x
0
Consider a particle of mass dm
at position (x, y , z)
z
y
wx
x
? the body’s total mass is the integral of the masses of elements dm
Z
m=
dm
? the centre of mass of the body is defined as (xg , yg , zg ), where
 
 
Z x
xg
m yg  = y  dm
zg
z
7-7
Moments of momentum and inertia of a rigid body
w
z
A rigid body, mass m, rotates about O
with angular velocity components ωx , ωy , ωz
about x, y and z axes.
v
dm
wz
u
y
wy
z
x
0
Consider a particle of mass dm
at position (x, y , z)
y
wx
x
? from the diagram, the element has velocity components
u = zωy − y ωz
v = xωz − zωx
=⇒
w = y ωx − xωy

 
u
0
v  = −  z
w
−y
 
y
ωx


−x
ωy 
0
ωz
−z
0
x
7-7
Moments of momentum and inertia of a rigid body
w
z
A rigid body, mass m, rotates about O
with angular velocity components ωx , ωy , ωz
about x, y and z axes.
v
dm
wz
u
y
wy
x
0
Consider a particle of mass dm
at position (x, y , z)
z
y
wx
x
? from the diagram, the element has moment of momentum



wy − vz
0 −z



z
0
dH = dm uz − wx
=⇒
dH = dm
vx − uy
−y
x
7-7
 
y
u


−x
v
0
w
Moments of momentum and inertia of a rigid body
w
z
A rigid body, mass m, rotates about O
with angular velocity components ωx , ωy , ωz
about x, y and z axes.
v
dm
wz
u
y
wy
x
0
Consider a particle of mass dm
at position (x, y , z)
z
y
wx
x
? substituting for velocities in terms of angular velocities gives


 
0 −z
y
0 −z
y
ωx





z
0 −x
z
0 −x
ωy 
dH = − dm
−y
x
0
−y
x
0
ωz
 2



(y + z 2 )
−yx
−zx
ωx
2
2
(x + z )
−zy  ωy 
= dm  −xy
2
ωz
−xz
−yz
(x + y 2 )
7-7
Moments of momentum and inertia of a rigid body
Integrate dH over all the elements in the body, giving


 
Z (y 2 + z 2 )
−yx
−zx
ωx
2
2
(x + z )
−zy  dm ωy 
H =  −xy
2
ωz
−xz
−yz
(x + y 2 )
define the nine elements of the inertia matrix as:
moments of inertia
Z
Ixx = (y 2 + z 2 ) dm
Z
Iyy = (x 2 + z 2 ) dm
Z
Izz = (x 2 + y 2 ) dm
products of inertia
Z
Ixy = Iyx = xy dm
Z
Iyz = Izy = yz dm
Z
Izx = Ixz = zx dm
The moment of momentum for a body spinning about an arbitrary axis is then

Ixx

H = −Ixy
−Ixz
7-8
−Iyx
Iyy
−Iyz
 
−Izx
ωx


−Izy
ωy 
Izz
ωz
(HLT pp25-28 gives moments and products of inertia of common shapes)
Simple cases
If x, y , z are the principal axes of

Ixx
0

Iyy
H= 0
0
0
the body, then H simplifies to
  

0
ωx
Ixx ωx
0  ωy  = Iyy ωy 
Izz
ωz
Izz ωz
Often we are concerned with the rotation of a body about one of the
principal axes, so H simplifies further:

  

Ixx
0
0
0
0
Iyy
0 0= 0 
H= 0
0
0
Izz
ωz
Izz ωz
which can be written in scalar form as
Hz = Izz ωz
or
H = Iω
7-9
Kinetic energy of a body rotating about a fixed axis
w
z
v
dm
wz
Consider again an element dm of a rigid body
at position (x, y , z)
moving with velocity components u, v and w
u
y
wy
x
0
z
y
wx
x
? The element has kinetic energy
dT =
1
2
dm (u 2 + v 2 + w 2 )
where
 

u
0
v  = −  z
w
−y
−z
0
x
 
y
ωx


−x
ωy 
0
ωz
substituting for u, v and w in terms of angular velocities ωx , ωy and ωz :
1
dm (zωy − y ωz )2 + (xωz − zωx )2 + (y ωx − xωy )2
2
1
= dm (y 2 + z 2 )ωx2 + (z 2 + x 2 )ωy2 + (x 2 + y 2 )ωz2
2
− dm xy ωx ωy + yzωy ωz + zxωz ωx
dT =
7 - 10
Kinetic energy of a body rotating about a fixed axis
w
z
v
dm
wz
Consider again an element dm of a rigid body
at position (x, y , z)
moving with velocity components u, v and w
u
y
wy
x
0
z
y
wx
x
? Integrate over the entire element to give
T =
1
1
1
Ixx ωx2 + Iyy ωy2 + Izz ωz2 − Ixy ωx ωy − Iyz ωy ωz − Izx ωz ωx
2
2
2
? For x, y , z aligned with the principal axes and ωx = ωy = 0:
T =
1
Izz ωz2
2
or
T =
1
Iω 2
2
? Note that the origin O is not necessarily the mass centre of the body.
7 - 10
Moments of inertia for flat bodies
z
x
y
For a planar body in the x-y plane with
? negligible thickness (z = 0 throughout the body)
? mass per unit area ρ
the moments of inertia and products of inertia are
Z
Z
2
Ixx = ρ y dA
Ixy = ρ xy dA
Z
Iyy = ρ x 2 dA
Iyz = 0
Z
Izz = ρ (x 2 + y 2 ) dA
Izx = 0
7 - 11
Moments of inertia for flat bodies
z
y
x
For a planar body in the x-y plane with
? negligible thickness (z = 0 throughout the body)
? mass per unit area ρ
therefore
Izz = Ixx + Iyy
this is known as the perpendicular axis theorem
7 - 11
Perpendicular axis theorem
Example – Circular disc
y
dr
Calculate Ixx and Iyy for a thin flat circular disc with
mass per unit area ρ
r
a
G
x
Solution:
Either
(a) integrate across the plane of the disc to get Ixx and Iyy
or
(b) find the polar moment of inertia Izz using axial symmetry, then
use the perpendicular axis theorem
(b) is much easier...
7 - 12
Perpendicular axis theorem
Example – Circular disc
y
dr
Calculate Ixx and Iyy for a thin flat circular disc with
mass per unit area ρ
r
a
G
To find Izz
– consider a circular element at radius r
dIzz = (2πr dr ρ)r 2
– integrate over the whole disc:
Z a
ρπa4
1
2
Izz =
(2πr ρ)r dr =
= ma2
2
2
0
– then use the perpendicular axis theorem:
Ixx = Iyy =
1
ma2
4
7 - 12
Parallel axis theorem
y
Consider a planar body
with mass per unit area ρ
dA
y
G
B
h
G
xG
B
? Moment of inertia about GG through the centre of mass G:
Z
IGG = ρ y 2 dA
? Moment of inertia about BB parallel to GG, passing through y = −h:
Z
IBB = ρ (y + h)2 dA
Z
= ρ (y 2 + 2hy + h2 ) dA
7 - 13
x
Parallel axis theorem
y
Consider a planar body
with mass per unit area ρ
dA
y
G
h
B
? Hence
Z
IBB = ρ
G
2
xG
B
Z
y dA + 2h
y dA + h
2
Z
dA
Z
? but
y dA = 0
Z
ρ dA = m
since GG passes through G
by definition
therefore
IBB = IGG + mh2
this is the parallel axis theorem
Warning: GG must go through the centre of mass (see HLT p.25)
7 - 13
Parallel axis theorem
Example – Eccentric circular cam
A device for converting rotational into linear motion
follower
cam time
-
Cams are used to operate the intake and exhaust valves of i.c. engines
Camshaft from a car engine
7 - 14
Parallel axis theorem
Example – Eccentric circular cam
C
e
Find the moment of inertia of a circular cam with radius a and eccentricity e
about the axis of rotation through C
G
a
Solution:
? For a uniform circular disc about an axis perpendicular to the disc through
the centre of mass G:
1
IG = ma2
2
? For rotation about C, at distance e from G, we get
IC = IG + me 2
7 - 14
Parallel axis theorem
Example – Thin uniform rod
L
Find the moments of inertia of a thin rod,
mass m, length L, about G and about A
A
G
x
Solution:
Integrate to find IG and IA :
Z L/2
m 2
IG = 2
x dx
L
0
Z
and
IA =
0
L
m 2
x dx
L
giving
IG =
1
mL2
12
and
IA =
1
mL2
3
Check this by finding IA using the parallel axis theorem:
2
2
L
1
L
1
2
IA = IG + m
=
mL + m
= mL2
2
12
2
3
7 - 15
dx
Motion of a disc or flywheel
y
dq
dr
Determine from first principles how the
angular acceleration of a flywheel is related to the applied moment
r
a
q
x
G
? An element at radius r has mass
dm = ρr dθ dr
ρ = mass per unit area
? a force dF causes a constant tangential acceleration ω̇r
dF = dm ω̇r
the corresponding moment or torque acting on the element is
dM = r dF
? Integrating over the whole disc gives
Z a Z 2π
1
M = ρω̇
dθ r 3 dr = πa4 ρ ω̇
2
0
0
7 - 16
Motion of a disc or flywheel
y
dq
dr
Determine from first principles how the
angular acceleration of a flywheel is related to the applied moment
r
a
q
G
Compare M = 12 πa4 ρω̇ with the moment of inertia (HLT p.25):
)
I = 12 ma2
=⇒ I = 21 πa4 ρ
2
m = πa ρ
Hence the equation of motion is equivalent to
M = I ω̇
this is the rotational form of F = ma, Newton’s second law
7 - 16
x
Work done and kinetic energy of a flywheel
If φ is the angle of rotation of the flywheel, then
ω̇ =
dω
dω dφ
dω
=
=ω
dt
dφ dt
dφ
hence M = I ω̇ gives
therefore
M = Iω
dω
dφ
M dφ =
1
Iω 2
2
Z
i.e. the work done is equal to the change in kinetic energy
7 - 17
Motion of a disc or flywheel
Example – Spinning and slipping disc
r
w
mg
R
F
A spinning disc, of mass m and radius r , is placed on a flat surface
with coefficient of friction µ. If the disc’s initial angular velocity is ω1 ,
find:
a. the eventual angular velocity ω2 when slipping ceases
b. the time t during which slip occurs.
7 - 18
Motion of a disc or flywheel
Example – Spinning and slipping disc
w
r
mg
R
F
Solution:
? there is no vertical acceleration so R = mg and F = µmg
? equating the change in moment of momentum about the disc’s centre
while slipping occurs to the impulse of the moment due to F gives
Iω2 − Iω1 = −µmgr t
? the disc rolls with velocity r ω2 when slipping ceases
? so the change in the linear momentum of disc while it’s slipping is
mr ω2 − 0 = µmg t.
7 - 18
Motion of a disc or flywheel
Example – Spinning and slipping disc
w
r
mg
R
F
Solution contd:
? eliminating t from these two equations gives an expression for ω2 :
(I + mr 2 )ω2 = Iω1
=⇒
ω2 =
I
ω1
I + mr 2
but the moment of inertia of the disc about an axis through its centre is
I = 12 mr 2 , so
ω1
ω2 =
3
? and the time that the disc slips is:
t=
7 - 18
r
ω1 .
3µg
Summary
Moment of momentum for a body spinning about an arbitrary axis:

 
Ixx −Iyx −Izx
ωx



Iyy −Izy
ωy 
H = −Ixy
−Ixz −Iyz
Izz
ωz
the inertia matrix is diagonal if x, y , z are the principal axes of the body
For a planar body in the x-y plane, rotating about the z axis
? the moment of inertia is
Z
Izz =
(x 2 + y 2 ) dm
? the moment of momentum and the kinetic energy are
H = Izz ωz
and
T = 12 Izz ωz2
Perpendicular axis theorem (for planar bodies in the x-y plane):
Izz = Ixx + Iyy
Parallel axis theorem (for moment of inertia about BB):
IBB = IGG + mh2
7 - 19