Circular and General Curvilinear Motion
49
5
5.1
Circular and General Curvilinear Motion
5.2 Polar Coordinates
50
Here the first term is due to the change in the magnitude of the velocity vector
in time dt, whereas the second is due to the change in direction as V moves
to V� .
et det
Normal-tangential Coordinates
It is often convenient to describe curvilinear motion of a particle using path
et
en
variables, i.e. measurements made along the tangent t and normal n to the
path. The axes can be pictured as a right-angled bracket moving along with
the particle. The t-axis always points in the direction of travel, while the naxis points towards the centre of curvature. The unit vectors et and en are
shown in figure 35.
d!
Figure 36: Incremental change in the tangential unit vector
Figure 36 shows how the derivative of the unit vector et is found: its end
rotates angle d β in the direction of en . Dividing this increment by dt we get
path
et
ėt = β̇ en .
V
Substituting this into (5.2) and using (5.1) leads to the following expressions
for acceleration in normal-tangential coordinates
C
en A
d!
"
et
V
a = at et + an en
en
at = V̇ = ρβ̈ + ρ̇β̇
A
an = ρβ̇ 2 =
V2
= V β̇
ρ
Figure 35: Normal-tangential coordinates along a path
Suppose that the particle moves from A to A� in time dt. The incremental
change in the distance s moved along the path in time dt is
5.2
Polar Coordinates
y
..
ds = ρ dβ
..
(5.1)
The acceleration vector can be found by differentiating: a = dV/dt. To
perform the differentiation, just apply the usual product rule to equation (5.1):
a = V̇ et + V ėt
(5.2)
R
R
. ..
#$$#$$#
tangent to the path:
V = ρ β̇
.2
R ! R#
where ρ is the radius of curvature of the path. The particle’s speed is then
V = ds/dt = ρ dβ/dt = ρ β̇, and the velocity vector is, of course, always
V = V et
..
2R# " R#
x
0
Figure 37: General motion in polar coordinates
A particle moving at radius R around a fixed centre O in the x-y plane,
as shown in figure 37, has an instantaneous position vector R whose three
Circular and General Curvilinear Motion
51
5.2 Polar Coordinates
52
y
components are given by


cos θ


R = R  sin θ 
0
.2
(5.3)
. ..
"##"##"
Figure 38: Motion in a circle




cos θ
− sin θ




R̈ = −R θ̇2  sin θ  + R θ̈  cos θ 
0
0
0
Motion in a Circle
If the radius is fixed then Ṙ = 0 and R̈ = 0, as shown in figure 38. Equations
5.4 and 5.5 become


− sin θ


Ṙ = R θ̇  cos θ 
0
5.2.2
Example – Bicycle on a Curve
Problem: A bicycle travels at a steady speed of 10 m s−1 round a curve of
radius r = 20 m. By what angle α does the rider need to lean inwards?
Solution:
Consider the free-body diagram shown in figure 39. The angular
..
)
Collecting the terms, we can identify a radial component R̈ − R θ̇2 and a
tangential component 2Ṙ θ̇ + R θ̈ so that the acceleration vector can be written
as




cos θ
− sin θ
�
�
�
�




(5.5)
R̈ = R̈ − R θ̇2  sin θ  + 2Ṙ θ̇ + R θ̈  cos θ 
x
0
0
radial direction, and the other has magnitude R θ̇ in the direction tangential
to the circle. Differentiating a second time yields the acceleration R̈ of the
particle








cos θ
− sin θ
− sin θ
cos θ








R̈ = R̈  sin θ  + 2Ṙ θ̇  cos θ  + R θ̈  cos θ  − R θ̇2  sin θ 
0
0
0
0
5.2.1
R
R
Thus the velocity vector has two components. One has magnitude Ṙ in the
0
R"
! R"
Differentiating both terms in this product with respect to time, the velocity Ṙ
of the particle is




cos θ
− sin θ




Ṙ = Ṙ  sin θ  + R θ̇  cos θ 
(5.4)
0
..
.2
r"
mg
RH
! RV
Figure 39: Bicycle on a curve
velocity is θ̇ = 10/20 = 0.5 rad s−1 . Hence the inward radial acceleration is
r θ̇2 = 20 × 0.52 = 5 m s−2
Using Newton’s second law ΣF = ma, the radial equation of motion is
RH = mr θ̇2
and for vertical equilibrium ΣV = 0 gives
RV = mg
Circular and General Curvilinear Motion
53
For equilibrium in roll, the resultant of the normal reaction RV and the tangential reaction RH must pass through the mass centre. This defines the angle
of bank. Hence
RH
5
tan α =
= 0.510
=
RV 9.81
5.3 Moment of Momentum
54
which simplifies to
dT
= − ρA r θ̇2
dr
Separate variables and integrate to give
⇒ α = 27.0o
T =−
1
ρAθ̇2 r 2 + C
2
where C is the constant of integration. At the tip radius r = R2 , T = 0 hence
5.2.3
Example – Tension in Turbine Blade
C=
Problem: Find the greatest tensile stress (tension/cross-section area) in a
gas turbine blade made from steel and having a uniform cross-section of
1
ρAθ̇2 R22
2
so that at any radius r the stress is
area A, a root radius of R1 = 0.7 m and a tip radius of R2 = 0.8 m, at 10000
rpm.
T =
1
ρAθ̇2 (R22 − r 2 )
2
Clearly the greatest tension T and the greatest stress T /A occur at the root
where the radius has its smallest value r = R1 . Substituting the given numer-
__
r
R2
T ! dT dr
dr
σ=
dr
blade
ical values together with the density of steel ρ = 7800 kg m−3 (HLT page 41),
we find the root stress σ is
T
R1
T
2π
= 0.5×7800×(10000× )2 ×(0.82 −0.72 ) = 0.641 kN mm−2 = 641 MPa
A
60
Comment: Is there a steel strong enough for this duty?
disc
shaft
Figure 40: Tension in turbine blade
Solution: Figure 40 illustrates the blade and the forces acting on a small
element of thickness dr at a radius r . If the density of steel is ρ, then an
element dr has mass dm = ρA dr . The radial force causing the centripetal
acceleration of this mass element arises from the gradient of the tension T
in the blade. The tension on the inner face of the element is T , and on the
outer face is T + (dT /dr ) dr . Thus the equation of motion for the element is
T − (T +
dT
dr ) = ρA dr r θ̇2
dr
5.3
5.3.1
Moment of Momentum
Definition
The moment of momentum H about a point O is defined for a particle of mass
m moving with velocity Ṙ, illustrated in figure 41, as the vector product of the
linear momentum mṘ and the distance R from O.
H = R × m Ṙ
Moment of momentum is also known as angular momentum.
(5.6)
Circular and General Curvilinear Motion
55
5.3 Moment of Momentum
56
y
w
z
v
m
Hz
v
u
R
x
y
u
m
Hy
x
0
z
y
H
y
x
0
Hx
Figure 42: Moment of momentum of a particle in planar motion
x
Figure 41: Moment of momentum of a particle
5.3.2
5.3.3
General Moment Equation
Illustration of Components
Differentiating equation 5.6 using the product rule gives
Let


Hx
 
H =  Hy 
Hz
 
x
 
R = y 
z
 
u
 
Ṙ =  v 
w
The vector product can be accomplished easily by rewriting the first vector
as a matrix (HLT page 19) giving

 


0 −z y
u
wy − vz

 


H = R × m Ṙ = m  z
0 −x   v  = m uz − wx 
−y x
0
w
vx − uy
In the first-year course, we only need consider circular motion in two dimensions, called planar motion. If we choose the x-y plane, then z = 0 throughout, and the moment of momentum for a particle simplifies to
H = m (vx − uy) k
Note that the moment of momentum vector is in the z or k direction, perpendicular to the x-y plane. As we will be working in two dimensions, we can
write this in scalar form as
H = m (vx − uy)
It can be seen from figure 42 that this result is simply the sum of the x and y
components of velocity times their respective moment arms.
dH
= R × m R̈ + Ṙ × m Ṙ
dt
and since Ṙ × Ṙ = 0 this becomes
dH
= R × m R̈
dt
By Newton’s second law Σ F = ma, the vector acceleration R̈ of the mass m
is caused by a force vector F
F = m R̈
so that
dH
=R×F
dt
From section 3.6.3 we recognise R × F as the moment M (or torque) of F
about O, or
M=R×F
so we have proved that
M=
dH
dt
or in words
torque = rate of change of moment of momentum
Circular and General Curvilinear Motion
57
5.3.4
6
Example – Mass on a String
Problem: A mass m rotates on a light inelastic string of length R about a
fixed point O as shown in figure 43(a). The angular velocity is ω1 . The length
of the string is now halved by pulling it through an aperture at O as in figure
43(b). What is the new angular velocity ω2 ?
(a)
!1
0
58
(b)
m
!2 m
R
R/2
0
Figure 43: Mass on a string
Solution: There is no applied torque as the only force is the string tension
passing through O. Therefore the moment of momentum about O is unchanged, hence
� �2
R
2
mR ω1 = m
ω2
2
ω2
⇒
=4
ω1
That is, the new angular velocity is ω2 = 4ω1 .
Question: What has happened to the kinetic energy ? Has it been conserved? If not, where did the additional energy come from?
Gravity and Satellite Orbits
6.1
Introduction
The word “orbit” is so familiar that we probably do not stop to think exactly
what it means. The earth is in orbit round the sun. A spacecraft is in orbit
round the earth. The simplest orbit is circular, but there are many that are
not. Halley’s comet passes close to the earth and away again. An ICBM
has an orbit which actually intersects the earth. A space-probe to another
planet has an orbit which never comes back. For the present study, we shall
define an orbit as the path of an object whose motion is subject only to a
force towards a fixed point.
6.1.1
Types of Orbit
All satellite orbits are conic sections. It is beyond the limits of the present
syllabus to prove this, although a brief analysis is given later in this chapter.
Here we shall simply sketch and name them. Our primary interest at this
stage is in circular and elliptical orbits.
hyperbola
parabola
ellipse
circle
Figure 44: Conic sections
The circle, ellipse, parabola and hyperbola, illustrated in figure 44, are
known in mathematics as conic sections. The equations defining these shapes
are given in HLT page 20.
Gravity and Satellite Orbits
59
6.1.2
6.2 Newton’s Law of Gravitation
60
Apogee radius
Kepler’s Laws
ra = a + OF = a +
Kepler’s laws for planetary motion are stated in HLT page 139.
Enclosed area
�
a2 − b 2
A = πab
6.1.3
Equations for the Elliptical Orbit
Kepler’s second law (HLT page 139)
ur = h = constant
a
v
u
r
b
!
L perigee
F
apogee
0
rp
Period P of orbit (from rate of sweeping area)
� P
1
1
A=
ur dt = urP
or
2
2
0
P=
2A
h
Comment: These equations are given in or easily derived from HLT page 20
and may be used to solve a number of simple orbit problems. However, they
are presented without proof and we need now to examine orbital motion from
ra
first principles in order to understand them.
Figure 45: Elliptical orbit
Figure 45 shows some of the terminology for an elliptical orbit (HLT page
20).
x2 y2
+
=1
a2 b 2
Polar equation (origin F)
r=
L
2(1 − � cos θ)
�=
Latus rectum
Focal distance
Newton’s Law of Gravitation
m1
m2
F
Cartesian equation (origin O)
Eccentricity
6.2
�
b2
1− 2
a
The force of gravitational attraction F between any two bodies having
masses m1 and m2 and whose centres of mass are separated by a distance
r as in figure 46 is given by
Gm1 m2
r2
(6.1)
where G is called the gravitational constant. On the surface of the earth,
which has mass M and radius R, the weight mg of any body of mass m is
governed by the same law so that
OF = a�
rp = a − OF = a −
Figure 46: Gravitational attraction
F =
2b2
L=
a
Perigee radius
r
�
a2 − b 2
mg =
GMm
R2
(6.2)
Gravity and Satellite Orbits
61
6.4 Moment of Momentum and Energy
Experimental values: G = 6.673 × 10−11 N m2 kg−2 , M = 5.976 × 1024 kg,
R = 6.371 × 106 m. Equation (6.2) then gives g = 9.824 m s−2 . This value is
m
r
M
0
R
Figure 47: Satellite orbiting the earth
Potential Energy
At any height above the surface of the earth, defined by a radius r , the gravitational force F on a mass m is less than mg. Substituting (6.2) into (6.1) we
obtain:
v
u
the mean absolute acceleration due to gravity. The actual value varies with
latitude because the earth is not truly spherical. It is different from the familiar
9.81 m s−2 because the latter includes the centripetal acceleration due to the
rotation of the earth, which makes all bodies seem lighter.
6.3
62
� �2
R
F = mg
r
detail, there are two simple statements which can be made.
(a) the moment of momentum is constant
(6.3)
To raise a body away from the earth, work must be done against the variable
force F and the potential energy U of the body is increased according to
dU
=F
dr
Because the force F acts through O, the moment of momentum mur about
O cannot change. Thus, for a fixed mass
ur = h = constant
(6.5)
It also follows geometrically that the radius vector r sweeps out area at a
constant rate 12 h. This is Kepler’s second law (HLT page 139). It is useful for
calculating the period of an orbit, provided that we know the area of the conic
Substituting for F and integrating gives
�
1
mgR 2
U = mgR 2
dr
=
−
+C
r2
r
section. (HLT page 20 gives the area of an ellipse as πab.)
The usual boundary condition is U = 0 when r = ∞. The resulting values for
U are all negative but this is not a problem.
mgR 2
U=−
r
direction and u normal to the radius vector. Without analysing the motion in
(b) the total energy is constant
The potential energy at any instant is
U=−
(6.4)
6.4 Moment of Momentum and Energy
Consider a satellite of mass m which is negligible compared with the mass
M of the earth as shown in figure 47. The motion then has a centre O at
the centre of the earth. However, the orbit is not necessarily circular. At
any instant, the radius is r and the velocity components are v in the radial
mgR 2
r
and the kinetic energy is
1
m (u 2 + v 2 )
2
Since no work external to the system comprising earth plus satellite is done,
the total energy U + T is constant. Thus
T =
1 2
gR 2
(u + v 2 ) −
= constant
2
r
(6.6)
Gravity and Satellite Orbits
63
6.5
Calculations Without Full Analysis
With three unknowns u, v and r , equations (6.5) and (6.6) are not sufficient
to determine the orbit. For a complete analysis, we need the differential
equations of motion set out in section 6.6.
6.5 Calculations Without Full Analysis
64
(d) For how long, and in which directions, should the rocket be fired, to
initiate and terminate the transfer orbit?
(e) If P and Q are in conjunction when t = 0, when should the transfer orbit
be initiated?
For the present course, we must be prepared to accept without proof that
orbits are circular , elliptical, parabolic or hyperbolic. Then there are a few
simple calculations that can be done using conic section data from HLT page
20 together with equations (6.5) and (6.6). These usually involve the symmetrical points on an orbit (perigee and apogee) where v = 0.
6.5.1
Example – Space Stations
Q
rq
P
rp
Solution:
(a) Calculate the radii rp and rq of the two circular orbits, and the corresponding moments of momentum hp and hq per unit mass.
For a circular orbit with period T , equilibrium exists when acceleration due
to gravity matches the centripetal acceleration.
� �2
u2
R
=g
r
r
Substituting
u=
2πr
T
and solving for r gives
r 3 = gR 2
The moment of momentum is
Figure 48: Space stations
Problem: Two space stations P and Q are in circular orbits in the same plane,
moving in the same direction round the earth, as illustrated in figure 48. P
has a period of 12 hours and Q has a period of 24 hours so that, once a day,
P and Q are in conjunction vertically above the same point on the earth. A
small vehicle with a rocket motor producing an acceleration of 3g is required
to leave station P and rendezvous with station Q. Take g = 9.824 m s−2 and
R = 6371 km, so that gR 2 = 3.9875 × 1014 m3 s−2 .
(a) Calculate the radii rp and rq of the two circular orbits, and the corresponding moments of momentum hp and hq per unit mass.
(b) Find the moment of momentum h per unit mass in the transfer orbit.
(c) Calculate the duration of the transfer orbit.
�
h = ur =
T
2π
�2
2πr 2
T
Thus for orbit P with 12 hour period, by equation (6.7)
�
�
�2 �1/ 3
12 × 3600
14
rp = 3.9875 × 10 ×
= 2.6614 × 107 m
2π
and by equation (6.8)
hp = 2π
(2.6614 × 107 )2
= 1.030 × 1011 m2 s−1
12 × 3600
Similarly for orbit Q with 24 hour period we get
rq = 4.2246 × 107 m
and
hq = 1.2979 × 1011 m2 s−1
(6.7)
(6.8)
Gravity and Satellite Orbits
65
(b) Find the moment of momentum h per unit mass, in the transfer orbit.
The transfer orbit is an ellipse with perigee radius equal to rp and apogee
radius equal to rq . By conservation of energy, as in equation (6.6) applied at
the perigee and the apogee where v = 0, we get
1 2 gR 2 1 2 gR 2
u −
= uq −
2 p
rp
2
rq
(6.9)
and
uq =
a=
(6.10)
(6.14)
1
(rq + rp ) = 3.443 × 107 m
2
so that
�
a2 − b 2 =
1
(rq − rp ) = 0.782 × 107 m
2
b = 3.354 × 107 m
rq = 4.2246 × 107 m
and from equation (6.12)
Using this data, we can solve equations (6.9) and (6.10) to find
�
2gR 2 rq
= 4288.2 m s−1
up =
(rp + rq )rp
�
(6.13)
From equation (6.14) minus equation (6.13), we get
We were given that gR 2 = 3.9875×1014 m3 s−2 while from question (a) above
rp = 2.6614 × 107 m
The perigee and apogee radii are
�
rp = a − a2 − b2 = 2.6614 × 107 m
�
rq = a + a2 − b2 = 4.2246 × 107 m
66
From equation (6.13) plus equation (6.14), we get
while, by conservation of moment of momentum as in equation (6.5)
up rp = uq rq
6.5 Calculations Without Full Analysis
A = πab = 3.627 × 1015 m2
The duration of transfer is T /2 where T = 2A/h hence
T 1 2A
3.627 × 1015
=
=
= 8.83 hours
2 2 h
1.1411 × 1011 × 3600
2gR 2 rp
= 2701.1 m s−1
(rp + rq )rq
Then finally we get
(d) For how long, and in which directions, should the rocket be fired, to
h = up rp = uq rq = 1.1411 × 1011 m2 s−1
initiate and terminate the transfer orbit?
To insert the vehicle from the circular P orbit into the elliptical transfer
orbit, the moment of momentum h per unit mass must be increased from
(c) Calculate the duration of the transfer orbit.
Using the equations and notation in HLT page 20, and noting that the
centre of the earth is at the focus F of the ellipse, we get the orbit period
P=
2A
h
(6.11)
1.0302 × 1011 to 1.1411 × 1011 m2 s−1 , by an impulse at radius 2.6614 × 107
m. With the rocket thrust aligned normal to the radius vector , this impulse is
equivalent to a 3g acceleration of duration t1 where
3 × 9.824 × t1 × 2.6614 × 107 = (1.1411 − 1.0302) × 1011
which gives t1 = 14.14 seconds.
where the area of the ellipse is
The point at which this first burn occurs will become the perigee of the
A = πab
(6.12)
transfer orbit. At the apogee, to insert the vehicle into the outer Q orbit, a
Gravity and Satellite Orbits
67
further increase in h is required from 1.1411 × 1011 to 1.2979 × 1011 m2 s−1
at radius 4.2246 × 107 m. With the rocket thrust again aligned normal to the
radius vector , this impulse is equivalent to a 3g acceleration of duration t2
where
3 × 9.824 × t2 × 4.2246 × 107 = (1.2979 − 1.1411) × 1011
giving t2 = 12.59 seconds.
6.6 Full Analysis of Satellite Orbits
68
where i and j are unit vectors in the radial and tangential directions.
Equation of Tangential Motion
Since the only force acting is radial, there is no acceleration normal to the
radius vector, so the j component in equation 6.15 is zero, giving
r θ̈ + 2ṙ θ̇ = 0
(e) If P and Q are in conjunction when t = 0, when should the transfer orbit
be initiated?
This is a problem you might like to try by yourselves! At the end of the
transfer orbit, Q must be 180o ahead of where P was at the time of the first
burn. During the 8.83 hours of the transfer orbit, P travels 265o and Q travels
132o . (Answer: t = −3.17 hours.)
6.6
6.6.1
Multiplying both sides by r gives
r 2 θ̈ + 2r ṙ θ̇ =
dr 2 θ̇
=0
dt
hence
r 2 θ̇ = h = constant
Of course we also have that r θ̇ = u, the tangential velocity component. Threfore equation 6.16 can be written as
Full Analysis of Satellite Orbits
ur = h
Differential Equations of Motion
. ..
j 2r.# " r#
i
.2
r ! r#
..
(6.16)
(6.17)
This equation is identical to equation 6.5, and h may be recognised once
again as the moment of momentum per unit mass. It also follows geometrically that the radius vector r sweeps out area at a constant rate 12 h. This is
Kepler’s second law (HLT page 139).
r
. ..
#$$#$$#
Equation of Radial Motion
Figure 49: Radial and tangential components of acceleration
The acceleration in the radial direction is caused by gravitational attraction so
In equation (5.5), back in section 5.2, we found that the acceleration of a
particle in general motion in polar coordinates could be represented as the
vector sum of radial and tangential components of acceleration. Describing
applying Newton’s second law ΣF = ma to the i component of acceleration
in equation 6.15 gives
� �2
R
−mg
= m(r̈ − r θ̇2 )
r
the position of the satellite by the polar coordinates r and θ as shown in
figure 49, we can rewrite equation 5.5 for the acceleration vector a for general
motion as
a = i (r̈ − r θ̇2 ) + j (r θ̈ + 2ṙ θ̇)
(6.15)
hence
r 3 θ̇2 − r 2 r̈ = gR 2
(6.18)
Gravity and Satellite Orbits
69
6.6.2
Solution to Find the Shapes of Orbits
6.6.3
Although (6.18) is a nonlinear differential equation, it is easy to obtain the
shapes of orbits from this by using equations (6.16) and (6.17) to eliminate r
and θ̇, and to change the independent variable from t to θ.
Thus, from (6.16) and (6.17) we obtain
r=
h
u
and
θ̇ =
6.6 Full Analysis of Satellite Orbits
u u2
= ,
r
h
Facts About the Solution
Conic Section
Equation (6.20) may be rearranged in the form
� 2 � �
�
h
Ah
r=
/
1
+
cos
θ
gR 2
gR 2
This is the polar equation for an ellipse seen in section 6.1.3 if
Ah
= −�
gR 2
so the first term in equation (6.18) can be re-written in terms of u and h as
r 3 θ̇2 =
h3 u 4
= uh.
u 3 h2
and, by differentiating again, we obtain the second term in (6.18) as
dθ2
θ̇ = −
d 2u
h.
dθ2
With these substitutions in (6.18) we get the linear differential equation
uh +
or equivalently
d 2u
h = gR 2
dθ2
(6.19)
The solution to equation (6.19) must be in the form u = A cos θ + B. Substituting this into equation (6.19) gives B = gR 2 /h so that the tangential velocity
component u, or the radius r , is given at any polar angle θ by
h
gR
= A cos θ +
r
h
Since A is negative, the maximum u or minimum r are found when θ = π in
equation (6.20). This is the perigee of the orbit and since r is a minimum
dr/dθ = v = 0 and the motion is purely normal to the radius vector. At the
perigee
gR 2
h
up =
−A
and
rp =
h
up
Apogee and Escape Condition
d 2u
gR 2
+u =
.
2
dθ
h
u=
2h2
=L
gR 2
Perigee
h � du � u
h � du � u 2 du
h � du �
θ̇ = − 2
=− 2
=
ṙ = − 2
u dθ
u dθ r
u dθ h
dθ
� d 2u �
and
For an elliptical orbit 0 < � < 1 so 0 > A > −gR 2 /h. That is, A is negative.
Also, differentiating r = h/u using the chain-rule gives
r 2 r̈ = −r 2
70
2
The minimum u and maximum r occur when θ = 0 in equation (6.20). This is
the apogee, where the radial velocity v is again zero. At the apogee
ua =
gR 2
+A
h
ra =
h
ua
Clearly, at the apogee the velocity becomes zero and the radius infinite when
A=−
(6.20)
and
gR 2
h
or
�=1
If this happens, the apogee ceases to exist, the orbit becomes a parabola
and the satellite escapes from the gravitational field of the earth.
Gravity and Satellite Orbits
71
72
Total Energy
7
The total energy E, which is the same at all points on an orbit, is obtained
conveniently from the perigee condition where v = 0
7.1
E 1 2 gR 2
= u −
m 2 p
rp
By substituting for rp and up we get
E 1
=
m 2
�
gR 2
−A
h
�2
gR 2
−
h
�
�
�
� 2 �2 �
gR 2
1
gR
2
−A =
A −
h
2
h
Rigid Bodies
Moment of Momentum of a Particle
If a particle of mass m has a velocity Ṙ, then that particle has a momentum G
which is a vector in the direction of motion. In section 5.3 we also introduced
the concept of moment of momentum H. This arises when a particle of mass
m moves past a reference point with velocity Ṙ at a distance R.
G = mṘ
and
H = R × mṘ
2
The energy is zero when A = −gR /h. This is the limiting escape condition
already noted above. When the energy is negative, the orbit is closed. When
it is zero or positive, the satellite escapes.
7.1.1
Moment of Momentum for an Arbitrary Moving Particle
Circular Orbits
u
m
If Ah/gR 2 = � = 0, then equation (6.20) becomes independent of θ. u and
r are then constant and the orbit is circular . Equation (6.20) is then precisely equivalent to the simple Newtonian statement of equilibrium between
centripetal acceleration and gravitational attraction
u 2 gR 2
= 2
r
r
Hyperbolic Trajectories
Beyond the parabola when A < −gR 2 /h, orbits are hyperbolic. Halley’s
comet has an elliptical orbit which is very close to parabolic, while comets
which pass only once and never reappear are in hyperbolic orbits.
y
H
0
x
Figure 50: Moment of momentum for an arbitrary moving particle
A particle of mass m moves with speed u in the x direction in the x-y plane
at a distance y from the origin O as illustrated in figure 50. The magnitude of
the momentum G is
G = mu
The momentum is actually a vector in the x direction aligned with u. The
magnitude of the moment of momentum H about O is
H = − muy
By convention, positive H is drawn anticlockwise about an axis normal to the
paper through O.
Rigid Bodies
73
7.1.2
Moment of Momentum for Rotation about a Fixed Point
Instead of arbitrary particle velocities we now consider the motion of two
masses m1 and m2 at radii r1 and r2 which are rotating in the same x-y plane
with the same angular velocity ω about O, as shown in figure 51.
m1
V2
V1
r1
m2
r2
7.2 Moments of Momentum and Inertia of a Rigid Body
7.2
74
Moments of Momentum and Inertia of a Rigid Body
A rigid body can be thought of as a group of particles that are permanently
and rigidly connected to each other. We can therefore determine the moment
of momentum of a rigid body by summing the contribution from each of its
constituent particles. For example, if a lightweight bicycle wheel with radius
r is modelled as having all of its mass m distributed around its rim, then the
moment of momentum for rotation about its axle with angular velocity ω is
simply H = mr 2 ω. This approach can be extended to general rigid bodies
using integration.
!
0
7.2.1
Analysis for a Mass Element
Figure 51: Moment of momentum for rotation about a fixed point
w
z
Kinematic relations are
v
dm
!z
V1 = ωr1
and
u
V2 = ωr2
y
!y
The masses have momentum m1 V1 and m2 V2 each in the direction indicated.
The vector sum of these is non-zero, so the system is unbalanced and would
shake about. This is because O is not the centre of mass. However, our
primary concern is the moment of momentum H about O. The moment of
momentum is
H = r1 m1 V1 + r2 m2 V2
Eliminating V1 and V2 gives
x
0
z
y
!x
x
Figure 52: Moment of momentum of a rigid body
Consider a particle of mass dm located at vector position (x, y , z) as
H = (m1 r12 + m2 r22 ) ω
These equations describe the magnitude of the moment of momentum H.
The direction of the vector H is out of the paper through O, the k direction.
This is the same as the direction of the angular velocity vector ω . Notice that
this result has been achieved by summing the products mass × square of
radius. This introduces the definition of moment of inertia, which we explore
rigorously in the next section.
shown in figure 52. It is part of a rigid body of total mass m, which rotates
about the origin O with angular velocity components ωx , ωy , ωz about the x,
y and z axes respectively.
The mass of the body is the integral of the masses of the elements dm or
�
m=
dm
The centre of mass of the body is defined by co-ordinates (xg , yg , zg ) given
Rigid Bodies
75
by
7.2 Moments of Momentum and Inertia of a Rigid Body
It does not matter which of the two multiplications we perform first, so we
leave the angular velocity vector untouched and multiply the two 3×3 position
matrices to give
 
 
xg
� x
 
 
m yg  = y  dm
zg
z

u = zωy − yωz
7.2.2
v = xωz − zωx
w = y ωx − xωy
Noting the rule of matrix multiplication, these three identities may be written
Again we see by inspection of the diagram that the moment of momentum
dH of the element dm about O is a vector with components about the x, y


wy − vz


dH = dm uz − wx 
vx − uy
Again recalling the standard matrix multiplication procedure, this is equivalent
to

 
0 −z y
u

 
dH = dm  z
(7.2)
0 −x   v 
−y
x
0
w
Substituting for linear velocities in terms of angular velocities using equation
7.1

 

0 −z y
ωx
0 −z y

 

dH = − dm  z
0 −x   z
0 −x  ωy 
−y x
0
−y x
0
ωz

−zx
ωx

 
(x 2 + z 2 )
−zy  ωy 
−yz
(x 2 + y 2 )
ωz
First, we integrate to include all the elements in the body, giving
H=
(7.1)
−yx
Definition – Moments and Products of Inertia
as a single equation.
 

 
u
0 −z y
ωx
 

 
0 −x  ωy 
v  = −  z
w
−y x
0
ωz
(y 2 + z 2 )

dH = dm  −xy
−xz
By inspection of figure 52, the velocity components u, v and w for any
element may be written in terms of the position co-ordinates and the angular
velocity components.
and z axes given by
76
�

(y 2 + z 2 )

dm  −xy
−xz
−yx

−zx
ωx

 
(x 2 + z 2 )
−zy  ωy 
−yz
(x 2 + y 2 )
ωz
We now name each of the nine terms in the inertia matrix as
moments
of inertia
�
2
Ixx = (y + z 2 ) dm
�
Iyy = (x 2 + z 2 ) dm
�
Izz = (x 2 + y 2 ) dm
products of�inertia
Ixy = Iyx =
Iyz = Izy =
Izx = Ixz =
xy dm
�
�
yz dm
zx dm
Moments and products of inertia for a selection of common shapes are given
in HLT pages 25-28. Using these definitions, the moment of momentum for
a body spinning about an arbitrary axis is

 
Ixx −Iyx −Izx
ωx

 
H = −Ixy Iyy −Izy  ωy 
−Ixz −Iyz Izz
ωz
Rigid Bodies
77
7.2.3
7.4 Inertia Properties for Thin Flat Bodies
Using these components to substitute individually for u, v and w in the kinetic
energy gives
Simple Cases for the Prelim Syllabus
Symmetrical Bodies
1
dm [(zωy − y ωz )2 + (xωz − zωx )2 + (y ωx − xωy )2 ]
2
Multiplying out and collecting terms gives
dT =
If we choose x, y and z to be principal axes of the body, we find that the
products of inertia become zero, so the equation simplifies to


  
Ixx 0 0
Ixx ωx
ωx


  
H =  0 Iyy 0  ωy  = Iyy ωy 
0
0
Izz
ωz
Izz ωz
1
dT = 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 ]
and integrating over the whole rigid body and recalling the definitions of moments and products of inertia produces
Rotation about a Single Axis
T =
Much of the time we are concerned with the rotation of a body about a single
axis, for example the z axis, which is also one of the principal axes of the
body, so the equation simplifies further to

  

Ixx 0 0
0
0

  

H =  0 Iyy 0   0  =  0 
0 0 Izz
ωz
Izz ωz
which can be written in scalar form as
Hz = Izz ωz
7.3
78
or
H = Iω
Kinetic Energy of Body Rotating about Fixed Axis
For an element dm of a rigid body, moving with velocity components u, v and
w, the kinetic energy dT is given by
dT =
1
dm (u 2 + v 2 + w 2 )
2
But as noted in equation 7.1, if the rigid body rotates about the axes Ox, Oy
and Oz with angular velocity components ωx , ωy and ωz respectively then

  

 
0 −z y
ωx
zωy − y ωz
u

  

 
0 −x  ωy  = xωz − zωx 
v  = −  z
w
−y x
0
ωz
y ωx − xωy
1
1
1
Ixx ωx2 + Iyy ωy2 + Izz ωz2 − Ixy ωx ωy − Iyz ωy ωz − Izx ωz ωx
2
2
2
In most cases the bodies considered will be symmetrical with products of
inertia equal to zero, and they will probably also rotate about only one axis.
Thus the form in which we will expect to use kinetic energy will be
T =
1
Izz ωz2
2
or
T =
1
Iω 2
2
Notice that the chosen fixed origin O is not necessarily the mass centre of
the body.
7.4
Inertia Properties for Thin Flat Bodies
Clearly, since moments of inertia are required for the calculation both of moment of momentum and for kinetic energy of rotating rigid bodies, we need
to be able to calculate them.
For a thin flat body, say in the x-y plane, these definitions are simplified
because z = 0. Then, taking ρ to be the mass per unit area, we may write
�
�
Ixx = ρ y 2 dA
Ixy = ρ xy dA
�
Iyy = ρ x 2 dA
Iyz = 0
(7.3)
�
Izz = ρ (x 2 + y 2 ) dA
Izx = 0
Rigid Bodies
79
7.4.1
7.4 Inertia Properties for Thin Flat Bodies
80
about the z axis. The integration for this is a simple summation of circular
rings as shown in figure 54 giving
� a
ρπa4 1
Izz =
(2πr dr ρ)r 2 =
= ma2
2
2
0
Perpendicular Axis Theorem
z
Then, by the perpendicular axis theorem, we get
y
x
Figure 53: Perpendicular axis theorem
From the definitions given in equation 7.3, we observe that, for any thin
flat body in the x-y plane, as shown in figure 53, the moment of inertia about
the z axis is the sum of the moments of inertia about the x and y axes, or
Ixx = Iyy =
1
ma2
4
as given in HLT on page 23.
7.4.3
Parallel Axis Theorem
y
Izz = Ixx + Iyy
dA
y
G
This sometimes provides a very convenient way to avoid difficult integrals
when calculating moments of inertia as illustrated in the following example.
B
h
G
xG
B
Figure 55: Parallel axis theorem
7.4.2
Example – Circular Disc
Consider the moment of inertia of the planar body of mass per unit area
Problem: Calculate Ixx and Iyy for a thin flat circular disc of mass per unit
area ρ.
y
ρ, illustrated in figure 55, calculated about the line GG through the centre of
mass G, and also about a parallel line BB at y = −h. For a small element of
area dA we see that
�
IGG = ρ
dr
hence
r
a
G
x
Figure 54: Moment of inertia of a circular disc
Solution: Rather than integrating across the plane of the disc to get Ixx and
Iyy , it is quicker to find the polar moment of inertia Izz using axial symmetry
y 2 dA
�
(y + h)2 dA = ρ (y 2 + 2hy + h2 ) dA
��
�
�
�
=ρ
y 2 dA + 2h y dA + h2 dA
IBB = ρ
�
But since GG passes through the centre of mass, the second term makes no
contribution to the integral because
�
y dA = 0
Rigid Bodies
81
Also we can simplify the third term because
�
ρ dA = m
giving
IBB = IGG + mh2
Note that this is only true if GG is an axis through the centre of mass. See
HLT page 25.
Example – Eccentric Circular Cam
IG =
7.5.1
a
1
mL2
12
and
IA =
1
mL2
3
The application of the parallel axis theorem is trivial here, but the agreement
is reassuring!
� �2
� �2
L
1
L
1
IA = IG + m
=
mL2 + m
= mL2
2
12
2
3
7.5
C
e
G
82
We can integrate to find the moments of inertia about the centre of mass
G, and the end A, of the rod, giving
� L/2
� L
m 2
m 2
IG = 2
x dx
and
IA =
x dx
L
0
0 L
Thus finally we conclude that
7.4.4
7.5 Further Examples
Further Examples
Motion of a Heavy Disc or Flywheel
y
Figure 56: Eccentric circular cam
For a uniform circular disc, about an axis perpendicular to the disc through
r
a
the centre of mass we get
IG =
d!
dr
!
G
x
1
ma2
2
For rotation with an eccentric radius e, as shown in figure 56, we get
IC = IG + me2
Figure 58: Flywheel
In figure 58, the mass of the element at radius r is
dm = ρr dθ dr
7.4.5
Example – Thin Uniform Rod
where ρ is the mass per unit area. Consider the force dF on this mass
element to cause a constant tangential acceleration ω̇r at radius r
L
A
G
x
dx
Figure 57: Thin uniform rod
dF = dm ω̇r
The corresponding moment or torque dM is
dM = r dF
Rigid Bodies
83
Integrating for the whole disc gives
� a � 2π
1
M = ρω̇
dθ r 3 dr = πa4 ρ ω̇
2
0
0
For comparison, the moment of inertia for the disc may be found in HLT on
page 25.
1
I = ma2
where
m = πa2 ρ
2
Thus we recognise that the equation of motion above is equivalent to
M = I ω̇
This is the rotational form of F = ma, Newton’s second law.
7.5.2
Work Done on a Flywheel and Kinetic Energy
Note that if φ is the angular movement
ω̇ =
dω dω dφ
dω
=
=ω
dt
dφ dt
dφ
Then
and therefore
�
M = Iω
dω
dφ
M dφ =
1
Iω 2
2
7.6 Combined rotation and translation
84
!
P
P
r
!r
r
⇐⇒
G
G
!
Figure 59: Equivalent representations of rigid body rotation and translation
[You have already encountered a similar relationship between forces and
moments in statics: a force F acting through P is equivalent to a force acting
through G and a moment Fr about an axis through G.]
Therefore any combination of rotation and translation may be represented
as a pure rotation with no translation about some axis. The point on a planar
rigid body through which this axis passes is known as the instantaneous
centre of rotation. This is the point of intersection of the lines perpendicular
to the instantaneous velocities of points on the body (point C in figure 60).
Note that the instantaneous centre may be a point outside the body. Also
note that if the lines perpendicular to the directions of velocities VP and VQ
are coincident, then the position of C on this line depends on the ratio VP /VQ .
VP
That is, the work done is equal to the change in kinetic energy.
P
VQ
7.6 Combined rotation and translation
7.6.1
Instantaneous Centre of Rotation
General planar rigid body motion can be represented as a rotation about any
point plus a suitable translation of the rigid body. For example, pure rotation
with angular velocity ω about point P in figure 59 is equivalent to rotation at
the same angular velocity ω about G plus translation at velocity ωr .
Q
C
Figure 60: The instantaneous centre of rotation
Rigid Bodies
85
7.6.2
7.6 Combined rotation and translation
86
result:
Moment of momentum for general planar motion
Consider a planar rigid body rotating with angular velocity ω, and suppose
that a point P (not necessarily the centre of mass) has instantaneous velocity
VP , as shown in figure 61.



 

moment of momentum 

 
 moment about P of 
HP = about G for rotation about G + the linear momentum





 

with angular velocity ω
of G
(7.6)
Here “the linear momentum of G” is the momentum of the body’s mass concentrated at G and moving at the instantaneous velocity of G.
VP
!
P
VP
VP
!
!
P
G
P
"
r
VP
"
r
r!
VP
G
G
VP cos"
Figure 61: General planar motion of a planar rigid body
The moment of momentum of the body about P can be decomposed as:
 

moment of momentum 
 
 moment about P of the 

HP = about P for rotation about P + linear momentum due to
(7.4)




 


translation at velocity Vp
with angular velocity ω



The linear momentum of the rigid body, mVP , acts through G (recall that for
translational motion, the rigid body can be considered as a particle with all
its mass concentrated at G). From figure 62(a) we therefore have
HP = IP ω + mVP r cos θ
(7.5)
An alternative expression for HP can be obtained by writing IP = IG + mr 2
using the parallel axis theorem. Then equation 7.5 can be expressed
HP = IG ω + mr (r ω + VP cos θ)
and since r ω +VP cos θ is the component of the velocity of G that acts perpendicular to the line joining P and G (see figure 62(b)), this gives the general
(a)
(b)
Figure 62: General planar motion
Equation 7.6 is easier to use than equation 7.4 in situations where the
instantaneous velocity of the centre of mass G is known, but at the same time
it is convenient to consider moments about some other point P. Examples of
the use of equation 7.6 are given in section 8.
Special cases are:
• HP = IP ω if P is the instantaneous centre of rotation
(since v = r ω so HP = IG ω + mr ω 2 = IP ω by the parallel axis theorem)
• HG = I G ω
(since r = 0 in this case).
Examples II
87
8
Examples II
8.1
8.2.1
88
and final velocities U1 and U2 of the mass centre
U1 = hω1 + UH
Introduction
U2 = hω2 + UH
In the previous chapter we set out the essential analysis which defines and
explains the application of momentum, moment of momentum and energy
equations to rigid bodies. This final chapter is devoted to applications and
examples.
8.2
8.2 Examples Using Moment of Momentum
Examples Using Moment of Momentum
Impulse Q reduces the momentum of mass centre
−Q = mU2 − mU1
Impulse Q offset by distance a reduces the moment of momentum about the
mass centre
−Qa = IG ω2 − IG ω1
Eliminating Q, U1 and U2 leaves
Centre of Percussion of a Cricket Bat
so that
UH
!
H
IG
mh
This is the right place to hit a ball. It is called the centre of percussion and
is a function only of the inertia characteristics of the bat. It is not affected by
the velocity or mass of the ball or by the way the bat is swung.
a=
h
G
mha(ω1 − ω2 ) = IG (ω1 − ω2 )
U
.
Q
a
8.2.2
Figure 63: Cricket bat
Problem: A cricket bat has mass m and moment of inertia IG about the mass
centre G. It is instantaneously vertical and pivots freely, both before and after
impact with the ball, about a point H (the batsman’s hands) at a height h
above G as illustrated in figure 63. The initial angular velocity of the bat
about H is ω1 at the instant before the ball strikes with impulse Q, and ω2
afterwards. There is no impulse at H (the batsman’s hands feel no shock)
and H has the same forward velocity UH both before and after impact. Find
the distance a below G where the ball strikes the bat.
Solution: The pivot constraint provides two kinematic equations for the initial
Impact of Rolling Disc against a Step
before impact
a
at impact
!1
G
a!1
Qv
after impact
!2
a!2
Qh
P
h
Figure 64: Impact of a rolling disc against a step
Problem: A uniform disc having mass m, radius of gyration k (where Izz =
mk 2 ) and radius a rolls along a flat surface with angular velocity ω1 until
it encounters a step of height h as shown in figure 64. Find the angular
Examples II
89
velocity ω2 of the disc immediately after impact with the step if it does not slip
or rebound.
Solution: The impulse acting at the corner of the step has components Qh
horizontally and Qv vertically, as shown in figure 64. This causes a sudden
change in linear momentum and in moment of momentum. To illustrate the
choices that are available, this problem will be solved twice with two different
locations for the fixed origin O.
(a) Choose O at the centre of mass G of the disc.
horizontal momentum
vertical momentum
−Qh = m(a − h)ω2 − maω1
�
+Qv = m a2 − (a − h)2 ω2
When considering moments about G, the moment arms for Qh and Qv are evident from the diagram. This is the easiest approach to understand because
G is the mass centre. Therefore the moment of momentum is calculated for
pure rotation about G, ignoring the linear motion, as
�
Qh (a − h) − Qv a2 − (a − h)2 = mk 2 (ω2 − ω1 )
8.3 Examples Using Conservation of Energy
of momentum about P cannot be written this way. As already discussed in
section 7.6.2, it must therefore be considered as
�
� �
�
moment of momentum
moment about P of
HP =
+
about centre of mass G
linear momentum of G
8.3
8.3.1
Examples Using Conservation of Energy
Example – Rolling Wheel
Problem: A wheel of mass m has radius a and moment of inertia IG about
its centre G. Find the kinetic energy when it rolls without slipping along a flat
surface with angular velocity ω.
a
!
m
2
G
a!
A
Eliminating Qh and Qv from these three equations leaves the solution
2
90
Figure 65: Rolling wheel
2
(k + a(a − h))ω1 = (k + a )ω2
(b) Choose O at the corner P of the step. In this case Qh and Qv do not
appear in the moment equation. Thus there are no unwanted unknowns to
eliminate and there is no need to use the linear momentum equations. Take
moments about P. The moment of momentum is the same, before and after
the impact, giving
Solution: A wheel rolling along a flat surface without slipping may be seen
as rotating instantaneously about the point of contact A shown in figure 65.
Using the parallel axis theorem, the moment of inertia about A is
IA = IG + ma2
Consider the kinetic energy during rotation about A
mk 2 ω1 + maω1 (a − h) = mk 2 ω2 + maω2 a
That is
(k 2 + a(a − h))ω1 = (k 2 + a2 )ω2
This result is identical to the first, but the algebra is very much simpler. Notice
that the final moment of momentum is m(k 2 + a2 )ω2 which is identical to IP ω2 .
However, because P is not the centre of rotation initially, the initial moment
T =
1
IA ω 2
2
Notice however that this could be written
T =
1
1
1
(IG + ma2 ) ω 2 = IG ω 2 + m(aω)2
2
2
2
Hence the kinetic energy could be found equally well by considering rotation
about G and adding the kinetic energy of the mass m moving at speed aω.
Examples II
91
8.3.2
8.3 Examples Using Conservation of Energy
92
That is
Example – Rolling Sphere
g cos θ = (a + b)θ̇2
a
(8.2)
Eliminating (a + b)θ̇2 from equations 8.1 and 8.2 leaves
cos θ =
! b
8.3.3
10
17
or
θ = 53.97o
Example – Rolling and Slipping Sphere
Figure 66: Rolling sphere
Problem: The above example is unrealistic because unless the coefficient
Problem: A solid sphere of uniform density ρ and radius a is placed exactly
on the top of a fixed sphere of radius b. It is displaced slightly and rolls
off without slipping. Find the angular position θ, measured from the upward
of friction µ is infinite, the moving sphere will start to slide at some angle θ1
and the non-slipping analysis will be invalid between this angle and the final
separation angle θ2 . Find θ1 and suggest a procedure for determining θ2 for
a given value of µ.
vertical, where it leaves the surface.
Solution: From HLT page 27, the mass and moment of inertia for a sphere
are
4
2
m = πρa3
and
I = ma2
3
5
Let θ be the angular displacement of the moving sphere from the top of the
fixed sphere, as illustrated in figure 66. At angle θ, we can write the linear
velocity of the centre of the moving sphere as (a+b)θ̇ and the angular velocity
of the moving sphere about its own centre as (a + b)θ̇/a. Conservation of
energy, namely that the loss of potential energy = the gain in kinetic energy,
then yields the equation
�
�2
1
1
(a + b)θ̇
mg(a + b)(1 − cos θ) = m((a + b)θ̇)2 + I
2
2
a
Substituting for I and simplifying gives
10g(1 − cos θ) = 7(a + b)θ̇2
(8.1)
The moving sphere leaves the surface of the fixed sphere when the radial reaction between them is zero and the inward component of the sphere weight
force is exactly sufficient to maintain the inward centripetal acceleration
�
�2
(a + b)θ̇
mg cos θ = m
a+b
"R
b
! R
mg cos!
a
mg sin!
Figure 67: Rolling and sliding sphere
Solution: Figure 67 shows the forces acting on the sphere. For limiting friction
at angle θ1 , using energy as before gives
10g(1 − cos θ) = 7(a + b)θ̇2
The radial equation of motion for the moving sphere gives
�
�2
(a + b)θ̇
mg cos θ − R = m
a+b
The equation of motion for the rotation of the moving sphere gives
2
a+b
ma2
θ̈
5
a
and the tangential equation of motion of the centre of mass gives
µRa =
mg sin θ − µR = m(a + b)θ̈
(8.3)
(8.4)
(8.5)
(8.6)
Examples II
93
8.3 Examples Using Conservation of Energy
94
Eliminate (a + b)θ̇2 from equations 8.3 and 8.4 to get
R
17 cos θ − 10
=
mg
17
(8.7)
G
Eliminate m(a + b)θ̈ from equations 8.5 and 8.6 gives
R
2 sin θ
=
mg
7µ
(8.8)
Eliminate R/mg from equations 8.7 and 8.8 to get
17 cos θ − 10 17
=
2 sin θ
7µ
Note that when µ = ∞ then cos θ = 10/17 as before.
Now we have to find the separation angle θ2 . For the motion beyond θ1 ,
energy is lost in sliding friction and the energy equation is not valid. Also,
as the sphere is now sliding as well as rotating, equation 8.5 is not valid
either. For the remainder of the solution to separation we need a numerical
integration of the tangential equation of motion 8.6.
The tangential motion of the centre of mass is
mg sin θ − µR = m(a + b)θ̈
with R given by equation 8.4 and subject to the initial condition
17 cos θ1 − 10 17
=
2 sin θ1
7µ
8.3.4
mg
!
Example – Falling ladder
Problem: A ladder of length L and mass m, which is initially vertical, is given
a gentle push and subsequently falls to the ground. Find the coefficient of
friction µ if the base of the ladder begins to slip at an angle θ0 . Which way
does it slip?
Solution: For small θ (while the angular velocity θ̇ is still small), the friction
force F on the base of the ladder in figure 68 acts to the right against the
couple formed by R and mg. As θ increases, θ̇ also increases, and F must
R
F
0
Figure 68: Falling ladder
act to the left to provide the required centripetal acceleration. Hence the
base initially slips to the left if µ is small, and to the right if µ exceeds a
critical value.
Before the ladder begins to slip energy is conserved. Since the moment
of inertia about O is IO = 13 mL2 , we therefore have
2 2
1
6 mL θ̇
+ mg(L/2) cos θ = mg(L/2).
(8.9)
Setting the moment about O equal to IO θ̈:
2
1
6 mL θ̈
= mg(L/2) sin θ
�
Resolving forces along the ladder and applying F = ma:
m(L/2)θ̇2 = mg cos θ + F sin θ − R cos θ
also, resolving forces perpendicular to the ladder and using
m(L/2)θ̈ = mg sin θ − F cos θ − R sin θ
(8.10)
(8.11)
�
F = ma gives
(8.12)
The friction force F and reaction R can then be found in terms of θ by eliminating θ̇ and θ̈ from equations 8.11 and 8.12 using equations 8.9 and 8.10.
Solving the resulting pair of equations for F and R gives
F = 14 mg sin θ(6 − 9 cos θ)
R = 14 mg(1 − 6 cos θ + 9 cos2 θ)
If the ladder begins to slip at angle θ0 , then from F = µR, the value of µ is
sin θ0 (6 − 9 cos θ0 )
µ=
.
1 − 6 cos θ0 + 9 cos2 θ0
mark.cannon@eng.ox.ac.uk