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Note 11
Rotational Motion II
Sections Covered in the Text: Chapter 13
We continue here with the study of the dynamics of
rotational motion. Towards the end of this note we
shall review some of the physics you will encounter in
the experiment “The Compound Pendulum” in the
laboratory. We consider some simple problems of
statics and continue with rotational energy and
angular momentum.
Rotational Dynamics
Consider the hypothetical case of a small rocket of
mass m attached to the end of a rigid massless rod
pivoted from an immovable point (Figure 11-1). We
consider the rocket as a single point particle. The
rocket is pointing in an arbitrary direction φ with
respect to the rod and the thrust of the rocket engine
produces a force on the end of the rod as shown. You
should be able to see that the rocket and rod will spin
as one system counterclockwise at a steadily
increasing speed.
rFt = mr 2α .
…[11-2]
The left hand side of eq[11-2] is just the torque
produced by the force about the pivot. Thus we have
finally €
τ = mr 2α .
…[11-3]
We shall see in the next section that we shall find it
convenient to define the quantity mr 2 as the moment of
inertia of€the mass m about the pivot. In so doing we
shall arrive at the rotational analogue of Newton’s
Second Law.
Moment of Inertia and
Newton’s Second Law
We now move from a single point particle to a solid
rigid body. It is convenient for the moment to think of
the body as a collection of point particles. The
collection is constrained by an axis of rotation (Figure
11-2). Three arbitrary point particles are shown in the
body along with the force each receives. The whole
collection rotates with the same angular acceleration α
about the rotation axis.
Figure 11-1. The hypothetical case of a rocket attached to the
end of a rigid massless rod.
r
The force of the engine Fthrust is resolvable into
transverse and radial components F t and F r. F t gives
the system a tangential acceleration at, i.e., causes the
rocket to change speed.
The resultant of the tension
€
force T and F r causes the rocket to change direction.
Thus we have
Ft = mat = mrα ,
…[11-1]
where α is the angular acceleration of the system
(Note 10). Multiplying both sides of eq[11-1] by r we
get
€
Figure 11-2. Three forces applied to arbitrary positions in a
rigid body.
The net torque produced by the forces is the sum of
torques
11-1
Note 11


τ net = ∑ τ i = ∑ ( mi ri2α ) =  ∑ mi ri2 α .…[11-4]
 i

i
i
€
Example Problem 11-1
Moment of Inertia of a Rod about a Pivot at one End
where i runs over all the particles in the collection. We
now define the sum within brackets as the moment of
inertia of the collection. It is conventionally denoted I:
Calculate the moment of inertia of a thin, uniform rod
of length L and mass M that rotates about a pivot at
one end.
I = m1r12 + m2 r22 + ... = ∑ mi ri2 .
Solution:
We take the line of the rod as the x-axis and put the
end of the rod at x = 0. The rod is thin so we can
assume y ≈ 0 for all points on the rod. We imagine the
rod divided into an infinite number of mass elements
dm each of length dx (Figure 11-3). According to the
definition eq[11-8]:
…[11-5]
i
The units of moment of inertia are kg.m2. A body’s
moment of inertia, like torque, depends on the axis of
rotation. Once the axis is specified, so that the ri can be
€
determined, then the moment of inertia about that
axis can be calculated from eq[11-5].
Eqs[11-4] and [11-5] allow us to write the angular
acceleration as
α=
τ net
.
I
…[11-6]
This is just the rotational equivalent of Newton’s
Second Law (a = F/m).
Most bodies
€ of interest are not collections of point
particles but rather are continuous distributions of
mass. We can depict this by replacing the individual
“particles” with cells 1,2,3… each of mass Δm. This
amounts to converting the moment of inertia
summation to an integration
I = ∑ ri2Δm = lim
(∑ r Δm) = ∫ r dm .
2
i
Now
…[11-7]
€
where r is the distance from the rotation axis. If we let
the rotation axis be the z-axis then we can write the
moment of inertia as
I=
∫ (x
2
+ y ) dm .
2
…[11-8]
The trick in successfully applying eq[11-8] to a specific
body and a specific axis is to replace d m by the
appropriate
expression in dx and/or dy. Much
€
€
practice in doing this is given in a first year course in
calculus. Let us consider an example.
11-2
€
2
∫ x dm .
…[11-9]
M
dm =   dx ,
L
…[11-10]
I=
2
i
Δm →0
Figure 11-3. Finding the moment of inertia about one end of
a long thin rod.
and the rod extends from x = 0 to x = L. Thus eq[11-9]
becomes
Iend
€
M
=
L
L
∫
0
x 2 dx =
M x2L 1
2
  = ML . …[11-11]
L  3 0 3
The answer is seen to be simple in form.
Very often if you need an expression for the moment
of inertia of a body of simple shape, you need not
calculate the expression from from first principles, but
simply look it up in a table. Figure 11-4 illustrates a
number of bodies of simple shape and their moments
of inertia about major axes. In a typical exam or test
question you will be given the necessary formula; you
will NOT be required to memorize these formulas.
Note 11
The Parallel Axis Theorem
The moment of inertia of a body about an axis of
symmetry or other axis of convenience is often given
in a table. But sometimes you need to know the
moment of inertia of the body about some other,
parallel, axis (Figure 11-5). In this case you needn’t
calculate the moment of inertia from first principles.
All you need to do is use the so-called Parallel Axis
Theorem. The parallel axis theorem states that the
moment of inertia I of a body about an axis that is
parallel to the axis through the centre of mass a
distance d away is
Figure 11-5. Rotation about an off-centre axis.
I = Icm + Md 2 .
…[11-12]
where Icm is the moment of inertia of the body about
an axis through the centre of mass.
We can
€ easily prove eq[11-12] for a one dimensional
body pivoted about an axis O (Figure 11-6).
Figure 11-6. Rotation about an off-centre axis.
Figure 11-4. Some examples of the moments of inertia about
major axes of bodies of simple shape. These examples are
given for purposes of comparison. You will NOT be
expected to memorize these formulas.
We define two coordinate systems x and x’, where the
pivot is at x = 0 and the centre of mass is at x’ = 0. The
centre of mass is located a distance d from the pivot
point. Thus we have x’ = x – d. By definition the
11-3
Note 11
moment of inertia about O is
I=
2
∫ x dm = ∫ ( x'+d) dm = ∫ ( x'
2
=
€
∫ ( x')
2
2
+2dx'+d 2 ) dm
dm + 2d ∫ x' dm + d 2 ∫ dm
= Icm + 2d ∫ x' dm + Md 2
€
= Icm + Md 2
directions. The torques these forces produce about O
are equal in magnitude and opposite in direction, and
therefore add to zero. The net torque is zero as is the
net force. This body is in a state of translational and
rotational equilibrium.
Figure 11-7b shows two non-concurrent forces acting on a body. These forces are of unequal magnitude
and act in opposite directions. The magnitude of the
resultant torque about O produced by these forces is
seen to be
…[11-13]
€ because the middle term contains the definition of the
position of the centre of mass in the x’ system which is
€ zero. This result will prove useful in the theory for the
experiment “The Compound Pendulum”.
A Rigid Body in Equilibrium
We saw in Note 04 that a body is in a state of
translational equilibrium if the sum of the external
forces acting on it is zero, that is, if
∑F =0.
…[11-14a]
Now that we have studied rotational motion (and in
particular Example Problem 10-5) we can add that a
body is in a state of rotational equilibrium if the sum
of external torques on it is zero:
∑τ = 0 .
…[11-14b]
A body is in a state of both translational and rotational equilibrium if both conditions, eqs[11-14] are
satisfied.1
You must remember that eqs[11-14] are vector equations. Thus in 3D space eq[11-14a] yields the three
component equations:
∑F
x
=0
∑F
y
=0
∑F
z
=0.
Concurrent and Non-Concurrent Forces
Two forces are said to be concurrent if they act along
the same line. Two or more forces can act on a body
either concurrently or non-concurrently. Figure 11-7a
shows two concurrent forces acting in opposite
1
It is commonly understood that a body is in a state of equilibrium if both conditions, eqs[11-14], are satisfied. If in addition, the
body is at rest, then the body is said to be in the special state of
static equilibrium.
11-4
Figure 11-7. Two forces act on a body, concurrent forces (a)
and non-concurrent forces (b).
∑ τ = 2rF – 2rF = 0 .
This body is therefore in a state of rotational equilibrium. However, the net force on the body is not zero:
∑ F = 2F – F = F .
This means that though the body is in a state of rotational equilibrium, it is NOT in a state of translational
equilibrium. The body will have a translational
acceleration to the right.
In physics, a whole class of problems can be
described as static problems, meaning they involve a
body at rest. In such problems the body is described
by the conditions of equilibrium. Using these conditions we are able to solve for unknown quantities.
Note 11
We consider two classic examples, a horizontal beam
in equilibrium and the leaning ladder.
Example Problem 11-2
A Horizontal Beam in Equilibrium
A uniform horizontal beam of length 8.00 m and
weight 200 N is attached to a wall at its near end by a
pin connection. The far end of the beam is supported
by a cable that makes an angle of 53.0˚ with the horizontal (Figure 11-8a). If a 600 N man stands 2.00 m
from the wall, find the tension in the cable and the
force exerted by the wall on the beam.
Figure 11-8b. The free-body diagram of the beam.
∑F
x
∑F
y
= Rcosθ – T cos53.0 o = 0
= Rsinθ + T sin53.0 o – 600N − 200N = 0 .
Since we have three unknowns, R , T and θ, we need
an additional equation. We therefore redraw Figure
11-8b to show the components of R and T (Figure 118c) so we can apply the torque condition.
Figure 11-8a. A man stands on a beam supported on its left
(near) end by a pin fixed to a wall and on its right (far) end
by a cable.
Solution:
The system comprising man and beam is in a state of
static equilibrium. The free-body diagram of the beam
is drawn in Figure 11-8b. The pin exerts a reaction
force R on the beam at an angle (unknown) of, say, θ
with respect to the beam. We assume that the weight
of the beam acts downwards through the center of
mass of the beam, which in the case of a uniform
beam is its geometric center.
The beam is in a state of static equilibrium so the
sum of x-components of forces and the sum of ycomponents of forces must both add to zero:
Figure 11-8c. The figure of 11-8b with x- and y-components resolved.
We deliberately pick the point of the pin on the wall
as the point about which to calculate torque. Doing
this eliminates R x and R y from the equation since they
both pass through that point. Thus
∑ τ = (T sin53.0
o
)(8.00m) – (600N)(2.00m)
–(200N)(4.00m) = 0 .
Solving yields
T = 313N .
11-5
Note 11
Substituting T into the two equations above we get
Rcos θ = 188N
Rsin θ = 550N .
and
respectively. Applying the force condition, eq[11-14a]
gives
Fx = f – P = 0 ,
∑
∑F
and
y
= n – mg = 0 .
Dividing the second equation by the first we get
tan θ =
and therefore
Finally,
R=
550N
= 2.93
188N
θ = 71.1o
188N
188N
=
= 581N .
cosθ cos71.1o
The tension in the rope is 313 N. The reaction force
exerted by the wall on the beam is 581 N at an angle
of 71.1˚ with respect to the beam.
Figure 11-9b. The free-body diagram of the ladder.
Example Problem 11-3
The Leaning Ladder Problem
A uniform ladder of length l and mass m rests against
a smooth, vertical wall (Figure 11-9a). If the coefficient of static friction between the ladder and ground is
µs = 0.40, find the minimum angle θmin at which the
ladder will not slip.
Here f is the force of static friction the ground exerts
on the ladder. R is the vector resultant of f and the
normal force n . P is the normal force the wall exerts
on the ladder (the wall is smooth so the force of static
friction between wall and ladder at that point is
assumed to be negligible). Thus we have immediately
n = mg. We apply the torque condition, eq[11-14b],
about O because that is the point through which most
of the forces pass. Thus
∑τ
0
l
2
= Pl sin θ – mg cosθ = 0 .
This expression applies at just the stage when the
ladder is about to slip (or rotate about O), that is, at θ
= θmin . Thus, rearranging and substituting values we
get
tan θ min =
Figure 11-9a. The leaning ladder problem.
mg
n
n
1
=
=
=
2P 2 f s.max 2( µsn) 2(0.40)
= 1.25 ,
Solution:
The ladder is not moving in any way and is therefore
in a state of static equilibrium. The free-body diagram
of the ladder is drawn in Figure 11-9b. We take x- and
y-axes in the horizontal and vertical directions,
11-6
and finally
θ min = 51o .
This means that the ladder will not slip so long as the
angle of inclination exceeds 51˚. Notice that the
Note 11
1
K mech = K rot + U g = Iω 2 + Mgy cm
2
minimum angle is independent of the mass of the
ladder; it depends only on µ s .
is conserved. Let us consider an example.
Rotational Energy
A rotating body possesses rotational kinetic energy by
definition because the particles making up the body
also rotate. Consider three arbitrary particles in a
rotating body (Figure 11-10). Assume the body is
rotating about a fixed axis (axle).
€
Example Problem 11-4
Finding the speed of a Rotating Rod
A rod of length 1.0 m and mass 200.0 g is hinged at
one end and connected to a wall (Figure 11-11). It is
held out horizontally, then released. What is the speed
of the tip of the rod as it hits the wall?
Figure 11-10. Three arbitrary particles are shown in a
rotating body.
All particles in the body rotate with the same angular
speed ω. A particle i of mass m i moves in a circle of
radius r i with speed vi = ωri. Thus the total rotational
kinetic energy of the body is the sum of the kinetic
energies of the particles:
1
1
1
1
K rot = m1v12 + m2v 22 + ... = m1r12ω 2 + m2 r22ω 2 + ...
2
2
2
2

1
1
= ∑ mi ri2 ω 2 = Iω 2 ,
2 i
2

€
€
Figure 11-11. A before- and-after representation of the rod.
Solution:
Mechanical energy is conserved so we equate the
rod’s final mechanical energy (at position 1) to its
initial mechanical energy (at position 0):
1 2
1
Iω1 + Mgy cm1 = Iω 02 + Mgy cm 0 .
2
2
…[11-15]
where I is the moment of inertia of the body about the
axle chosen.
In the event that the rotation axis does not pass
through the centre of mass then rotation of the body
may cause the centre of mass to move up or down. In
that case, the body’s gravitational potential energy Ug
= Mgycm will change. If the system is isolated (i.e., no
energy is lost to dissipative forces and no work is
done on the body from outside) then the body’s total
mechanical energy
The initial conditions are ω 0 = 0 and y cm0 = 0. The
centre of mass moves to ycm1 = –L/2 as the rod hits the
€wall. Using I = (1/3)ML2 for a rod rotating about one
end (Figure 11-4) we have
1 2
1
1
Iω1 + Mgy cm1 = ML2ω12 − MgL = 0
2
6
2
Solving for the rod’s angular velocity we get
€
11-7
Note 11
ω1 =
3g
.
L
The tip of the rod is moving in a circle of radius r = L.
Thus the final speed of the tip is
€
v tip = ω1L = 3gL = 5.42 m.s−1 .
Angular Momentum of a Particle
We now define the rotational equivalent of linear
momentum which we shall call (not surprisingly)
angular momentum.
In Figure 11-13a is shown a particle moving along a
trajectory. At this instant, the particle’s linear
r
momentum vector p makes an angle
β with the
r
position vector r . The particle’s angular momentum
relative to the origin is defined as the vector
€
€
€
Torque on a Particle
We have already seen how torque can be defined for a
solid body with respect to some axis. But torque can
also be defined for a single
particle. Consider a
r
particle subject to a force F in the xy-plane as shown
in Figure 11-12a.
€
Figure 11-13. The angular momentum vector L.
where
r r r
L=r×p
…[11-16a]
r
L = mrv sin β .
…[11-16b]
€
Figure 11-12. The torque vector is perpendicular to the plane
of r and F.
The units of angular momentum are kg.m2.s–1.
A relationship exists between torque and angular
momentum.
To see this we take the time derivative of
€
eq[11-16a]:
r
r
r
dL d r r dr r r dp
= ( r × p) =
× p+ r ×
dt dt
dt
dt
r r r r
= v × p + r × Fnet
The torque produced by the force wrt to the origin is,
as before,
r r r
τ = r × F.
Thus you can see that in this particular example the
torque vector is along the positive z-axis.
€
11-8
r r
r
= r × Fnet = τ net
€
€
€
Note 11
r
dL r
= τ net
dt
Thus
…[11-17]
r
r
Eq[11-7] follows because v and p are parallel and
therefore their cross product is zero. Eq[11-17] states
in words that
€ a net torque causes a system’s angular
momentum to change.
It is the rotational equivalent of
€
Newton’s Second Law. €
Angular Momentum of a Rigid Body
We can easily extend Eq[11-17] to apply to a rigid
body consisting of a collection of particles. The total
angular momentum of such a system is just the vector
sum:
r r r r
r
L = L1 + L2 + L3 + ... = Li . …[11-18]
r
momentum Li . Please note that if the angular
momentum is conserved, then both the magnitude
and the direction of the angular momentum vector are
unchanged.
€
Angular Momentum and Angular Velocity
Angular momentum and angular velocity are related
by
v
r
L = Iω
…[11-20]
and as shown in Figure 11-14.
€
∑
Combining eqs[11-17] and [11-18] we get
r
r
r r
dL
dLi
=∑
= ∑ τ i = τ net
dt
i dt
i
€
…[11-19]
The only forces that contribute to the net torque are
the forces exerted on the system by the environment
(external
forces). The forces producing torques
€
internal to the system cancel out (Newton’s Third
Law).
Figure 11-14. The angular momentum vector of a rigid body
rotating about an axis of symmetry.
Conservation Law of Angular Momentum
Recall that for an isolated system the total linear
momentum is conserved. By the same token, for an
r
isolated system (one for which τ net = 0) the angular
momentum is conserved.
Or in other words, the final
r
angular momentum L f is equal to the initial angular
€
€
11-9
Note 11
Addendum
Calculation of the Moment of Inertia of a Rectangular Flat Plate
The pendulum used here is not exactly a rod in the
strictest sense of the word because its width is not
zero. A steel ruler is in fact a flat rectangular plate
—albeit an elongated one. This means that its
moment of inertia about its end is given only
approximately by eq[3-8]. Let us calculate it properly now to see how good an approximation of
eq[3-8] it is.
Consider a flat plate (ruler) of length l and width
w (Figure 3-4).
w l
2 2
ICM =
∫ ∫ σ(x
w l
− −
2 2
2
+ y 2 )dxdy .
…[A-2]
Now σ is a constant with the value M/lw. Taking σ
outside the integral in eq[A-2] and integrating first
€
over y we get
ICM =
M
lw
l
2
w
2

∫ x
−
2
l
2
y+
1 3
y
dx ,
3 
−
l
w
2
dx
dy
y
O
w
x
CM
=
€
M
lw
l
2

2
1
3

∫ x w + 12 w dx ,
−
l
2
l
Figure 3-4. An element of area dA in a rectangular plate.
€
M 1
w 3 2
=  x 3w + x  .
lw  3
12 − l
2
This object is flat, with a thickness that is definitely
negligible in comparison to its other dimensions.
The basic definition of the moment of inertia I of a
flat object is
I=
2
2
∫∫ σr dA = ∫ ∫ σr dxdy
y
€
Substituting the limits and simplifying the final
result we get
€
ICM =
…[A-1]
1
M (l 2 + w 2 ) .
12
x
where σ is the object’s surface mass density and r
locates an element of area dA relative to the axis of
interest.
Let us first calculate the moment of inertia of this
object about an axis perpendicular to the object
and passing through its center—its center of mass
(CM). Later we can apply the parallel axis theorem
to get the moment of inertia about the point O at
its end. Substituting r 2 in terms of x and y and
inserting the upper and lower limits of x and y into
eq[10] we get the double integral
According to the parallel axis theorem the moment of
inertia about the endpoint O is given by
€
I0 = ICM + Md 2 ,
…[A-4]
where d is the distance that O is located relative to
CM. This distance here is l/2. Thus substituting this
€
fact and eq[A-3] into eq[A-4] we get
I0 =
1
1
M ( l 2 + w 2 ) + Ml 2 ,
12
4
which, when simplified, is
€
11-10
…[A-3]
Note 11
I0 =
1
M ( 4l 2 + w 2 ) .
12
…[A3-5]
ible in comparison to twice the length, then eq[A35] reduces to eq[3-8].
Clearly, if w2 << 4l2, that is, if the width is neglig-
€
11-11
Note 11
To Be Mastered
•
•
Definitions: angular displacement, radian, average angular velocity, average angular speed, instantaneous angular
velocity, instantaneous angular acceleration, average angular acceleration
equations of rotational kinematics (to be memorized):
ω f = ω i + αt
•
1
θ f = θi + ω it + αt
2
2
1
θ f = θi + (ω i + ω f )t
2
2
2
ω f = ω i + 2α (θ f – θi )
relations between the magnitudes of translational and rotational quantities
x = θr
a = αr = ω 2r
v = ωr
Typical Quiz/Test/Exam Questions
1.
(a) Define torque
(b) What are the units of torque?
2.
State the following as being a vector or a scalar. Give the units of each.
(a) coefficient of friction
(b) potential energy
(c) linear momentum
(d) angular velocity
3.
A body is subject to non-concurrent forces. Write down the conditions for the body to be in a state of
mechanical equilibrium. Explain the meaning of each quantity in your expressions.
4.
A uniform plank of length 2.00 m and weight 100.0 N is to be balanced on a fulcrum or support point (see the
figure). A 500.0 N weight is suspended from the right end of the plank and a 200.0 N weight suspended from
the left end. Answer the following questions.
right
left
fulcrum
(a) Describe the state of the plank when it is balanced.
(b) What are the conditions for this state?
(c) How far from the left end of the plank must the fulcrum be placed?
(d) What is the reaction force exerted by the fulcrum on the plank?
11-12