J. Peraire
16.07 Dynamics
Fall 2004
Version 1.2
Lecture D10 - Angular Impulse and Momentum
In addition to the equations of linear impulse and momentum considered in the previous lecture, there is a
parallel set of equations that relate the angular impulse and momentum.
Angular Momentum
We consider a particle of mass, m, with velocity v, moving under the influence of a force F. The angular
momentum about point O is defined as the “moment” of the particle’s linear momentum, L, about O.
Thus, the particle’s angular momentum is given by,
HO = r×mv = r×L . (1)
The units for the angular momentum are kg·m2/s in the SI system, and slug·ft2/s in the English system.
Using a a right handed cartesian coordinate system, the components of the angular momentum are calculated
as
HO = Hxi+Hyj +Hzk = m
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsingle
i j k
x y z
vx vy vz
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsingle
= m(vzy?vyz)i+m(vxz ?vzx)j +m(vyx?vxy)k .
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It is clear from its definition that the angular momentum is a vector which is perpendicular to the plane
defined by r and v. Thus, in some occasions it may be more convenient to determine the direction of HO
from the right hand rule, and its modulus directly from the definition of the vector product as
HO = mvrsinα ,
where α is the angle between r and v.
Rate of Change of Angular Momentum
We now want to examine how the angular momentum changes with time. Taking a time derivative of
expression (1), we have
˙HO = ˙r×mv +r×m˙v .
Here, we have assumed that m is constant. If O is a fixed point, then ˙r = v and ˙r×mv = 0. Thus, we end
up with,
˙HO = r×m˙v = r×ma .
Applying Newton’s second law to the right hand side of the above equation, we have that r×ma = r×F =
MO, where MO is the moment of the force F about point O. The equation expressing the rate of change
of angular momentum is then written as
MO = ˙HO . (2)
We note that this expression is valid whenever point O is fixed. The above equation is analogous to the
equation derived in the previous lecture expressing the rate of change of linear momentum. It states that
the rate of change of linear momentum about a fixed point O is equal to the moment about O due to the
resultant force acting on the particle. Since this is a vector equation, it must be satisfied for each component
independently. Thus, if the force acting on a particle is such that the component of its moment along a given
direction is zero, then the component of the angular momentum along this direction will remain constant.
Example Pendulum
Here, we revisit the simple pendulum problem introduced in lecture D6 and re-derive the pendulum equation
using equation (2).
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There are two forces acting on the suspended mass: the string tension and the weight. The string tension,
T, is parallel to the position vector r and therefore its moment about O is zero. On the other hand, the
weight creates a moment about O which is MO = ?lmgsinθk.
The angular momentum is given by
HO = r×mv = ler ×ml˙θeθ = ml2 ˙θ er ×eθ = ml2 ˙θk.
Therefore, the z component of equation (2) gives
ml2¨θ = ?lmgsinθ ,
or,
¨θ + g
l sinθ = 0 ,
which is precisely the same equation as the one derived in lecture D6.
Principle of Angular Impulse and Momentum
Equation (2) gives us the instantaneous relation between the moment and the time rate of change of angular
momentum. Imagine now that the force considered acts on a particle between time t1 and time t2. Equation
(2) can then be integrated in time to obtain
integraldisplay t2
t1
MO dt =
integraldisplay t2
t1
˙HO dt = (HO)2 ?(HO)1 = ?HO . (3)
Here, (HO)1 = HO(t1) and (HO)2 = HO(t2). The term
integraldisplay t2
t1
MO dt ,
is called the angular impulse. Thus, the angular impulse on a particle is equal to the angular momentum
change.
Equation (3) is particularly useful when we are dealing with impulsive forces. In such cases, it is often
possible to calculate the integrated effect of a force on a particle without knowing in detail the actual value
of the force as a function of time.
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Conservation of Angular Momentum
We see from equation (1) that if the moment of the resultant force on a particle is zero during an interval of
time, then its angular momentum HO must remain constant.
Consider now two particles ma and mb which interact during an interval of time. Assume that interaction
forces between them are the only unbalanced forces on the particles that have a non-zero moment about a
fixed point O. Let F be the interaction force that particle mb exerts on particle ma. Then, according to
Newton’s third law, the interaction force that particle ma exerts on particle mb will be ?F. Using expression
(3), we will have that ?(HO)a = ??(HO)b, or ?HO = ?(HO)a + ?(HO)b = 0. That is, the changes
in angular momentum of particles m1 and mb are equal in magnitude and of opposite sign, and the total
angular momentum change equals zero. Recall that this is true only if the unbalanced forces, those with
non-zero moment about O, are the interaction forces between the particles. The more general situation in
which external forces can be present will be considered in future lectures.
We note that the above argument is also valid in a componentwise sense. That is, when two particles interact
and there are no external unbalanced moments along a given direction, then the total angular momentum
change along that direction must be zero.
Example Ball on a cylinder
A particle of mass m is released on the smooth inside wall of an open cylindrical surface with a velocity v0
that makes an angle α with the horizontal tangent. The gravity acceleration is pointing downwards. We
want to obtain : i) an expression for the largest magnitude of v0 that will prevent the particle from leaving
the cylinder through the top end, and ii) an expression for the angle β that the velocity vector will form
with the horizontal tangent, as a function of b.
The only forces on the particle are gravity and the normal force from the cylinder surface. The moment of
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these forces about O (or, in fact, about any point on the axis of the cylinder) always has a zero component in
the z direction. That is, (MO)z = 0. To see that, we notice that for any point on the surface of the cylinder,
r and F are always contained in a vertical plane that contains the z axis. Therefore, the moment must be
normal to that plane. Since the moment has zero component in the z direction, (HO)z will be constant.
Thus, we have that
(HO)z = rmv0 cosα = constant .
For part i), we consider the trajectory for which the velocity is horizontal when z = a and let (v0)limit be
the initial velocity that corresponds to this trajectory. It is clear that for any trajectory for which v0 has a
larger magnitude than (v0)limit, the particle will leave the cylinder through the top end. Thus, for the limit
trajectory we have, from conservation of energy
1
2m(v0)
2
limit =
1
2mv
2
a +mga ,
and from conservation of angular momentum
(HO)z = rm(v0)limit cosα = rmva .
Here, va is the magnitude of the velocity for the limit trajectory when z = a. Eliminating va from these
equations we finally arrive at,
(v0)limit =
radicalbigg 2ga
1?cos2 α .
Therefore, for v0 ≤ (v0)limit the mass will not leave the cylinder through the top end.
For part ii), we also consider conservation of energy
1
2mv
2
0 =
1
2mv
2
b ?mgb
and conservation of angular momentum,
rmv0 cosα = rmvb cosβ .
Eliminating, vb from these two expressions we obtain,
β = cos?1
parenleftBigg
cosαradicalbig
1 + 2gb/v20
parenrightBigg
.
Example Spinning Mass
A small particle of mass m and its restraining cord are spinning with an angular velocity ω on the horizontal
surface of a smooth disk, shown in section. As the force Fs is slowly increased, r decreases and ω changes.
Initially, the mass is spinning with ω0 and r0. Determine : i) an expression for ω as a function of r, and ii)
the work done on the particle by Fs between r0 and an arbitrary r, and verify the principle of work and
energy.
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The component, along the spinning axis, of moment of the forces acting on the particle is zero. Therefore,
the vertical component of the angular momentum will be constant. For i), we have
mr0v0 = mrv, v0 = ω0r0, v = ωr → ω = r
20ω0
r2 .
For ii), we first calculate the force on the string
Fs = ?mv
2
r = ?m
r2ω2
r = ?m
r40ω0
r3 .
The work done by Fs, will be
W =
integraldisplay r
r0
Fsdr = ?mr40ω20
integraldisplay r
r0
dr
r3 = mr
4
0ω
2
0
1
2
parenleftbigg 1
r2 ?
1
r20
parenrightbigg
.
The energy balance implies that
T0 +W = T .
This expression can be directly verified since,
1
2m(ω0r0)
2
bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright
T0
+ 12m(ω0r0)2
parenleftbiggr2
0
r2 ?1
parenrightbigg
bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright
W
= 12m(ωr)2
bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright
T
.
Example Ballistic Pendulum
We consider a pendulum consisting of a mass, M, suspended by a rigid rod of length L. The pendulum is
initially at rest and the mass of the rod can be neglected. A bullet of mass m and velocity v0 impacts M
and stays embedded in it. We want to find out the angle θmax reached by the pendulum. The angle that
the velocity vector v0 forms with the horizontal is α.
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Because the rod is assumed to be rigid, we can expect that when the bullet impacts the mass, there will
be an impulsive reaction that the rod will exert on the bullet. If we use the principle of linear impulse and
momentum, it will be necessary to solve for this impulsive force. An alternative approach that simplifies the
problem considerably is to use the principle of angular impulse and momentum. We consider the angular
momentum about point O of the particles m and M just before and after the impact. The only external
forces acting on the two particles are gravity and the reaction from the rod. It turns out that gravity is not
an impulsive force and therefore its effect on the total angular impulse, over a very short time interval, can
be safely neglected (it turns out that in this case, the moment about O of the gravity forces at the time of
impact is also zero). On the other hand, we can expect the reaction from the rod to be large. However, the
moment about O of this reaction is zero, since it is directed in the direction of the rod. Therefore, we have
that during impact, the z component of the angular momentum is conserved.
The angular momentum before impact will be,
[(HO)z]1 = Lcosα mv0 .
After impact, the velocity v1 has to be horizontal. Thus, the angular momentum will be
[(HO)z]2 = L(M +m)v1 .
Equating these two expressions we get,
v1 = mM +mv0 cosα .
After impact, the system is conservative, and the maximum height can be easily obtained from conservation
of energy,
1
2(M +m)v
2
1 = (M +m)gL(1?cosθmax) .
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Thus,
θmax = cos?1
parenleftBigg
1?
parenleftbigg m
(M +m)
parenrightbigg2 v2
0 cos2 α
2gL
parenrightBigg
.
ADDITIONAL READING
J.L. Meriam and L.G. Kraige, Engineering Mechanics, DYNAMICS, 5th Edition
3/10
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