Engineering Science (MENG10004)
Mechanics : Statics, Kinematics, Dynamics
2022/2023
Handout 9 – Angular Momentum
9.1
Angular Momentum of Particles
Meriam & Kraige, Dynamics: 3/10
The angular momentum of a particle is defined as the moment of momentum about a point O:
HO = r × p = r × mv
(9.1)
and is a vector, with units Nms. For planar problems the angular momentum has axis k and magnitude:
HO = mv d = mv r sin β
(9.2)
where r is the distance from O to the particle, and β the angle between the position and momentum vectors.
β
d
y
HO
O
p = mv
r
HO = r x p
x
The angular momentum of a particle depends on the choice of the point O about which it is calculated:
it might be positive, negative or zero. This is in contrast to linear momentum, whose magnitude does not
depend on the choice of the (inertial) reference frame.
The time derivative of the angular momentum ḢO is equal to the moment MO of the resultant force on the
particle around the same point O:
ḢO = MO
(9.3)
Angular impulse is the time integral of the moment MO , and is equal to the change in angular momentum:
Zt2
MO dt = HO2 − HO1 = ∆HO
(9.4)
t1
If the net force on the particle produces no moment around point O (which can be conveniently chosen), the
angular momentum is conserved and ∆HO = 0. This is a very powerful tool in solving dynamics problems.
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mvx
HO = r × mv
= m(rxvy - ryvx)
y
HO
mv
mvy
r
x
O
Fx
MO = r × F
= (rxFy - ryFx)
y
MO
r
Fy
F
x
O
proof: from the vector formulation for angular momentum
HO = r × mv
take the time derivative:
+ r × mv̇ = r × mv̇
mv
ḢO = ṙ × mv + r × mv̇ = v×
and note that v × mv is zero, as both vectors are parallel. Consider the moment due to the resultant force:
MO = r × F = r × mv̇
and substitute F = ma = mv̇. It is now evident that:
ḢO = MO
Example 9.1 – Orbital Mechanics
Consider a comet in a highly eccentric orbit around the Sun. The distance from the Sun to aphelion (furthest
point from the Sun) is ra = 6 · 109 km, and distance to perihelion (closest point) rp = 75 · 106 km. Its speed
at aphelion is va = 740 m/s.
va
Sun
vp
rp
ra
Q: What is the velocity vp at perihelion?
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The only significant force acting on the comet is the gravitational attraction, which is central (and thus passes
through the centre of the sun). Thus, the angular moment HO about the centre of the Sun is conserved:
mva ra = mvp rp
and therefore
vp =
ra
va = 59.200 m/s
rp
Example 9.2 – Spinning Desk Chair / Figure Skater
A first-year engineering student, during a brief moment of procrastination from revision, is swivelling on his
desk chair. He notices that his angular velocity increases when he tucks in his legs and decreases when he
swings them out. After taking the mechanics module, he is confident that he can explain what is happening!
For convenience, the problem is simplified to a massless rod with two equal masses m at a distance r1 , spinning
at an angular velocity ω1 . As the masses are moved inwards to radius r2 , the angular velocity increases to ω2 .
t = t1
t = t2
v1=ω1r1
v2=ω2r2
m
ω1
ω2
m
v1=ω1r1
r2
r1
v2=ω2r2
r1
The centripetal forces on the masses produce no net moment around the point of rotation. Thus, the angular
momentum is conserved (∆H = 0) between r1 and r2
HO 1 = HO 2
2m (ω1 r1 ) r1 = 2m (ω2 r2 ) r2
to give the angular velocity:
ω2 =
r1
r2
2
ω1
The change in kinetic energy:
1
1
∆T = T2 − T1 = mω22 r22 − mω12 r12
2
2
1
1
1
= mω12 r14
− 2
2
r22
r1
is positive for r1 > r2 , and the kinetic energy has increased.
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Q: Where did the required energy come from?
A: A centripetal force Fn = man is required to maintain the masses at a radius r, where an = ω 2 r. The work
done by the centripetal force over the change in radius:
Zr2
Zr2
Fn dr = −2
U12 = −2
r1
r1
Zr2
= −2
mω 2 rdr
m
r12
ω1
r2
2
rdr =
−2mω12 r14
r1
= mω12 r14
Zr2
1
dr
r3
r1
1
1
− 2
r22
r1
= ∆T
is equal to the change in kinetic energy.
Example 9.3 – Spinning and Dropping Mass
Particle A, with mass mA = 1 kg, slides on a smooth horizontal surface and is connected by a weightless string
to particle B, with mass mB = 2 kg. Particle B is released from rest at t = 0, when A has a circumferential
velocity of v0 = 6 m/s at a radius r0 = 0.5 m.
eθ
mAg
er
ez
T
v
eθ
r
er
ez
N
θ
eθ
er
T
z
mA
mBg
mB
Q: What are the equations of motion of the system?
A: The angular momentum of particle A around the central hole
HO = r × mA v = rer × mA ṙer + rθ̇eθ = mA r rθ̇ ez
is conserved, as neither the tension in the rope T or contact force N produce a moment. Therefore, the
angular velocity θ̇ at radius r is:
r 2
0
θ̇ =
θ̇0
r
where θ̇0 = v0 /r0 .
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From the FBD of particle A:
X
X
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−T = mA r̈ − rθ̇2
0 = mA θ̈r + 2ṙθ̇
Fr :
Fθ :
and particle B:
X
T − mB g = mB a
Fz :
The motion of particles A and B is linked:
a = r̈
Combine equations to eliminate T and substitute expression for angular velocity:
r̈ (mA + mB ) −
1
mA r02 v02 + mB g = 0
r3
Substitute values to give:
1
2
+ g=0
r3
3
This differential equation is solved numerically, to find the motion of the particle. The angular velocity can be
found from the conservation of angular momentum, and is integrated for angular position.
r̈ − 3
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Mechanics: Angular Momentum
9.2
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Angular Momentum of Rigid Bodies
Meriam & Kraige, Dynamics: 4/4, 6/2, 6/5, 6/8
9.2.1
Linear Momentum
As derived in Handout 8, the linear momentum of a rigid body is:
p = mvG
(9.5)
dp
= F = maG
dt
(9.6)
where vG is the velocity of the centre of mass.
The resultant force F is found as:
9.2.2
Angular Momentum
The angular momentum of a rigid body around a general point O is calculated as:
HO = HG + rG × p
(9.7)
= HG + rG × mvG = IG ω + rG × mvG
where rG is the position vector of the centre of mass, and p the linear momentum of the rigid body, IG the
moment of inertia tensor, and ω the angular velocity vector.
For planar problems:
HO = IG ω + mvG d
(9.8)
HO = HG + rG × mvG
O
rG
d
y
p = mvG
G
HG = IGω
x
HO = IGω + mvGd
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proof: consider the rigid body as a system of infinitesimal elements with mass dm, and take the vector sum
of their angular momentum around point O.
b
O
ω
d
y
β
G
vG
r
dm
x
A
vA/G = ωr
Taking the direction of velocity vG as reference axis, the moment of momentum HO about O is equal to:
Z
Z
HP = (vG + ωr sin β) (d + r sin β) dm − (ωr cos β) (b − r cos β) dm
Z
=
Z
=
vG d + vG r sin β + ωrd sin β + ωr2 sin2 β − ωrb cos β + ωr2 cos2 β dm
vG d + ωr2 dm + (vG + ωd)
Z
βdm
r sin
βdm − ωb r cos
Z
= IG ω + mvG d
where the last two terms cancel out by definition of the centre of mass, and IG =
7
R
r2 dm.
Mechanics: Angular Momentum
9.2.3
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Moment Equations
The resultant moment (as a result of the external forces on the body) about a point can be calculated from
the angular momentum about that point. The form these equations take depends on the choice of reference
point about which the angular momentum is calculated.
centre of mass:
taking the centre of mass G, where d = 0, results in:
MG = ḢG =
d
(IG ω) = IG θ̈
dt
(9.9)
This supports the assertion in Handout 5 that the general motion of a rigid body could be described by the
translation of the centre of mass, and rotation about the centre of mass.
fixed axis of rotation: for a rigid body with inertially fixed axis of rotation:
MO = IG θ̈ + maG d
= IG θ̈ + m θ̈d d
(9.10)
= IG + md2 θ̈
= IO θ̈
where IO is the moment of inertia around point O, which is found using the parallel axis theorem.
HO = IGω + mvGd
HG = IGω
y
x
general point:
G
d
ω
p = mvG = m(ωd)
O
for a general point P , with velocity and acceleration, the moment equation becomes:
MP = IG θ̈ + rG × maG
where rG and aG are the position and acceleration of the centre of mass.
Alternatively, if point P is fixed to the body, it can be rewritten as:
MP = IP θ̈ + rG × maP
where IP is the moment of inertia around point P , and aP is the acceleration of point P . For details of
derivation, the interested reader is referred to Meriam & Kraige (section 4/4 and 6/2).
This discussion highlights that in dynamics the point about which moments are taken is crucial in the correct
formulation of the equations of motion. This is unlike statics, where moments can be taken about any point!
note: in this unit only moment equations about centre of mass or a fixed axis of rotation are considered.
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Example 9.4 – Billiard Ball: Slide and Roll
A billiard ball is struck with the cue and gains velocity v0 . Initially the ball slides, before the friction with the
table initiates a rotation. After a while, the rotation has increased so that the ball is no longer sliding and is
purely rolling.
t=0
y
t
v0
t = t1
ω
ω1
v
mg
v1=ω1R
x
F
N
Q: At what time t1 does the ball stop sliding?
A: The billiard ball has radius R and mass m, and the moment of inertia for a sphere is IG = 52 mR2 .
From the FBD, write the equations of linear impulse/momentum:
−µN t1 = ∆p = mv1 − mv0
and angular impulse/momentum:
µN Rt1 = ∆HG = IG ω1 − 0
At time t = t1 the ball is purely rolling:
v1 = ω1 R1
The three equations are solved simultaneously:
v1 =
v0
IG
1 + mR
2
ω1 =
v1
R
t1 =
v0
µg 1 +
9
mR2
IG
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Example 9.5 – Tilting Block
A uniform rectangular block of mass m slides on a smooth horizontal surface with velocity v before hitting a
small step in the surface.
A
h
v
b
Q: assuming neglible rebound, (a) calculate the minimum value of v required to allow the block to pivot
around the edge and reach standing position with no velocity; (b) calculate the percentage energy loss ∆E/E
for the case of b = h.
A: (a) It is assumed that the block pivots around point O, and that the height of the step is neglible compared
to dimensions of the block. Furthermore, the angular impulse due to the weight of the block is neglected due
to the short time of impact. Thus, the conservation of momentum is taken to be valid.
The angular momentum of the block around O before impact is
HO = mv
b
2
v
ω
Rx
r
y
mg
x
O
Ry
The velocity of the centre of mass immediately after impact is v̄, with angular velocity ω = v̄/r. The angular
velocity directly after impact is given from the angular momentum:
HO = IO ω
1
where IO = IG + mr2 , with IG = 12
m b2 + h2 for a rectangle, and thus
"
2 2 !#
1
b
h
2
2
HO =
m b +h +m
+
ω
12
2
2
=
m 2
b + h2 ω
3
From conservation of angular momentum, ∆HO = 0:
mv
b
m 2
=
b + h2 ω
2
3
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and thus
ω=
3vb
+ h2 )
2 (b2
In order to tilt the block upright, the kinetic energy after impact must be as great as gain in potential energy:
∆T + ∆Vg = 0

s
2 2
h
b
1
b
+
− =0
0 − IO ω 2 + mg 
2
2
2
2
This allows to solve for required initial velocity v:
s h2 p 2
g
v=2
1+ 2
b + h2 − b
3
b
(b) the percentage loss in energy:
∆E
=
E
1
1
2
2
2 mv − 2 IO ω
1
2
2 mv
= ... = 1 −
3
4 1+
h2
b2
which for b = h results in an energy loss of 62.5%.
Example 9.6 – Boomerangs and Gyroscopic Precession (not examinable)
Using the concept of angular momentum, we are able to provide a simplified and qualitative analysis of the
flight of a boomerang. The turning motion relies on the concept of gyroscopic precession.
t
t + dt
H
H
θ
M
H1
Mdt
Consider a spinning object with angular momentum vector H. A moment M applied to the spinning object,
perpendicular to the angular momentum vector, produces an angular impulse. At time t + dt this results in a
new angular momentum vector H1 at angle θ.
Assuming H M dt, we find:
tan θ ≈ θ =
|M dt|
|H|
which results in a precession rate Ω of
Ω=
|M |
|H|
For more details of gyroscopic precession and nutation, see Meriam & Kraige: Dynamics, Section 7/11.
However, here be dragons: 3D kinematics and dynamics is significantly harder than 2D covered in this unit!
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Next, consider a cross-shaped boomerang which is thrown with a forward velocity v and angular velocity ω.
Due to the angular velocity, the top and bottom of the boomerang have different air speeds and thus generate
a different amount of lift. This results in a net lift force L with moment M .
v+ωa
ω
a
ω
L
v
v
M
v-ωa
v
L = mv
R
2
H = IGω
Ω
M
R
The moment M causes a gyroscopic precession which turns the boomerang at a rate
Ω=
M
IG ω
To maintain a constant angle to the direction of travel, the forward velocity v and precession rate ω are linked:
Ω=
v
R
The lift force generated also should be sufficient to generate the necesary centripetal acceleration:
L=
mv 2
R
The radius R turns out to be inherent to the design of the boomerang. The boomerang motion is a careful
balance between lift forces and gyroscopic effects, requiring a correct ratio of forward velocity and angular
velocity (the ‘flick of the wrist’) which explains why they are tricky to throw!
For more details: https://plus.maths.org/content/os/issue7/features/boomerangs/index
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Revision Objectives Handout 9:
Angular Impulse and Momentum of Particles
ˆ calculate angular momentum of particle (HO = r × p = r × mv, HO = mv d)
ˆ recall that change in angular momentum around and axis through point O equals the moment the
resultant forces on the particle produce around that axis (ḢO = MO )
ˆ recall that angular impulse results in a change in angular momentum
Angular Impulse and Momentum of Rigid Bodies
ˆ calculate angular momentum of a rigid body (HO = HG + rG × mvG , HO = ωIG + mvG d)
ˆ use dynamic moment equations for rotation around:
– centre of mass of the body (MG = IG θ̈)
– inertially fixed rotation axis (MO = IO θ̈)
ˆ recognise that, in contrast to moment equilibrium equations in statics, the axis about which a
moment is taken is of critical importance in dynamics
Apply the equations for angular momentum to solve problems in dynamics of particles and rigid bodies.
Note: gyroscopic precession (as seen in the boomerang) is not examinable.
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