Download Slides for Chapters 9, 10, 11 and Review

General Physics I
(aka PHYS 2013)
P ROF. VANCHURIN
( AKA V ITALY )
University of Minnesota, Duluth
(aka UMD)
O UTLINE
C HAPTER 9: R OTATION OF RIGID BODIES
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Section 9.1: Angular velocity and acceleration
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Section 9.2: Rotation with constant angular acceleration
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Section 9.3: Relating linear and angular kinematics
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Section 9.4: Energy in rotational motion
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Section 9.5: Parallel-Axis Theorem
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Section 9.6: Moment-of-Inertia calculations
S ECTION 9.1: A NGULAR VELOCITY / ACCELERATION
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Angular position is describe in terms of radians defined by
2π rad ≡ 360◦ = 1 rev
or
1 rad ≡
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360◦
1
=
rev.
2π
2π
One can convert from one angular unit to another using,
2π rad
1 =
360◦
1rev
1 =
2π rad
360◦
1 =
1rev
(1)
(2)
(3)
S ECTION 9.1: A NGULAR VELOCITY / ACCELERATION
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Average linear (velocity and acceleration)
vavg ≡
I
∆x
∆t
∆x
dx
=
∆t→0 ∆t
dt
(4)
∆v
dv
d2 x
=
= 2
∆t→0 ∆t
dt
dt
a ≡ lim
(5)
Average angular (velocity and acceleration)
ωavg ≡
I
∆v
∆t
Instantaneous linear (velocity and acceleration)
v ≡ lim
I
aavg ≡
∆θ
∆t
αavg ≡
∆ω
∆t
(6)
Instantaneous angular (velocity and acceleration)
∆θ
dθ
=
∆t→0 ∆t
dt
ω ≡ lim
∆ω
dω
d2 θ
=
= 2.
∆t→0 ∆t
dt
dt
α ≡ lim
(7)
S ECTION 9.1: A NGULAR VELOCITY / ACCELERATION
Example 9.1-9.2. The angular position θ of a 0.36 m-diameter flywheel is
given by
θ(t) = 2.0 rad/s3 t3 .
(8)
a) Find θ, in radians and in degrees, at t1 = 2.0 s and t2 = 5.0 s.
b) Find the total distance (not displacement) that a particle on the flywheel
rim moves over the time interval from t1 = 2.0 s to t2 = 5.0 s.
c) Find the average angular velocity, in rad/s and in rev/min over that
interval.
d) Find the instantaneous angular velocity at t1 = 2.0 s and t2 = 5.0 s.
e) Find the average angular acceleration between t1 = 2.0 s and t2 = 5.0 s.
f) Find the instantaneous angular acceleration at t1 = 2.0 s and t2 = 5.0 s.
S ECTION 9.2: C ONSTANT ANGULAR ACCELERATION
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Angular velocity and acceleration are vectors pointing along the
axis of rotation and direction determined by right-hand rule.
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There is an equivalence of linear and angular quantities
x
→ θ
vx
→ ωz
ax
→ αz .
and so for motion with constant acceleration we have
a(t) = a
→
α(t)
= αz
v(t) = v0 + ax t
1
x = x0 + v0x t + ax t2
2
as well useful relations
→
ω(t)
=
→
θ(t)
1
= θ0 + ω0z t + αz t2
2
vx (t)2 = v20x + 2ax (x − x0 ) → ωz (t)2
1
x(t) − x0 = (v0x + vx (t)) t → θ(t) − θ0
2
ω0z + αz t
2
= ω0z
+ 2αz (θ(t) − θ0 )
1
=
(ω0z + ωz (t)) t.
2
S ECTION 9.2: C ONSTANT ANGULAR ACCELERATION
Example 9.3. You have finished watching a movie Blu-ray and the disk is
slowing to a stop. The disc’s angular velocity at t = 0 is 27.5 rad/s, and its
angular acceleration is a constant −10.0 rad/s2 . A line PQ on the disc’s
surface lies along the +x-axis at t = 0 s.
a) What is the disc’s angular velocity at t = 0.300 s?
b) What angle does the line PQ make with the +x-axis at this time?
S ECTION 9.2: C ONSTANT ANGULAR ACCELERATION
Example. Supposed the disc in Example 9.3 was initially spinning at twice
the rate (i.e. 55.0 rad/s) and slowed down at twice the rate (i.e.
−20.0 rad/s2 ).
a) Compared to the situation in Example 9.3 how long would it take the disc
to come to a stop?
(i) the same amount of time;
(ii) twice as much time;
(iii) four times as much time;
(iv) one half as much time;
(v) one fourth as much time
b) Compared to the situation in Example 9.3 through how many revolutions
would the disc rotate before coming to a stop?
(i) the same number of revolutions;
(ii) twice as many revolutions;
(iii) four times as many revolutions;
(iv) one half as many revolutions;
(v) one fourth as many revolutions.
S ECTION 9.3: R ELATING LINEAR AND ANGULAR
KINEMATICS
The linear displacement of a given point P on rotating body is
s = rθ,
where θ must be measured in radians.
(9)
S ECTION 9.3: R ELATING LINEAR AND ANGULAR
KINEMATICS
This suggests that the (magnitude of) linear velocity of the point P is
ds d(rθ) = r dθ v = = (10)
dt dt
dt
or
v = rω.
(11)
and the tangential (or parallel) component of linear acceleration
a|| =
dω
dv
=r
= rα.
dt
dt
(12)
and the centripetal (or perpendicular) component
a⊥ =
v2
= ω 2 r.
r
(13)
S ECTION 9.3: R ELATING LINEAR AND ANGULAR
KINEMATICS
Example 9.4. An athlete whirls a discus in a circle of radius 80.0 cm. At a
certain instant, the athlete is rotating at 10.0 rad/s and the angular speed is
increasing at 50.0 rad/s2 . At this instant, find the tangential and centripetal
components of the acceleration of the discus and the magnitude of the
acceleration.
S ECTION 9.3: R ELATING LINEAR AND ANGULAR
KINEMATICS
Example 9.5. You are designing an airplane propeller that is to turn at
2400 rpm (or revolutions per minute). The forward airspeed of the plane is
to be 75.0 m/s, and the speed of the tips of the propeller blades through the air
must not exceed 270 m/s. (This is about 80% of the speed of the sound in air.
If the speed of the propeller tips were greater than this, they would produce a
lot of noise.) (a) What is the maximum possible propeller radius? (b) With
this radius, what is the acceleration of the propeller tip?
S ECTION 9.3: R ELATING LINEAR AND ANGULAR
KINEMATICS
Example. Information is stored on a disc in a coded pattern of tiny pits. The
pits are arranged in a track that spirals outwards towards the rim of the disc.
As the disc spins inside a player, the track is scanned at a constant linear
speed. How must the rotation speed of the disc change as the player’s
scanning head moves over the track?
(i) The rotation speed must increase.
(ii) The rotation speed must decrease.
(iii) The rotation speed must stay the same.
S ECTION 9.4: E NERGY IN ROTATIONAL MOTION
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When a rigid body rotates around a fixed axis different parts of
it move with different linear velocities and so have different
kinetic energies:
1
1
Ki = mi v2i = mi r2i ω 2 .
(14)
2
2
The total (rotational) kinetic energy is then
X
X1
1 X
1
K=
Ki =
mi r2i ω 2 =
mi r2i ω 2 = Iω 2
(15)
2
2
2
i
where
I≡
X
mi r2i
(16)
is known as the moment of inertia.
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Note that the rotational kinetic energy looks very similar to the
translational kinetic energy
1 2
Iω
2
↔
1 2
mv .
2
(17)
S ECTION 9.4: E NERGY IN ROTATIONAL MOTION
Example 9.6. A machine part consists of three discs linked by light-weight
struts.
(a) What is this body’s moment of inertia about an axis through the center of
the disc A, perpendicular to the plane of the diagram?
(b) What is its moment of inertia about an axis through the centers of disks
B and C?
(c) What is the body’s kinetic energy if it rotates about the axis through A
with angular speed ω = 4.0 rad/s?
S ECTION 9.4: E NERGY IN ROTATIONAL MOTION
For more complicated objects the moment of inertia must be obtained
(as we shall see) by volume integration. Here we summarize some
useful results:
S ECTION 9.4: E NERGY IN ROTATIONAL MOTION
Example 9.7. We wrap a light, nonstretching cable around a solid cylinder
of mass 50 kg and diameter 0.120 m, which rotates in frictionless bearings
about a stationary horizontal axis. We pull the free end of the cable with a
constant 9.0 − N force for a distance of 2.0 m; it turns the cylinder as it
unwinds without slipping. The cylinder is initially at rest. Find its final
angular speed and the final speed of the cable.
S ECTION 9.4: E NERGY IN ROTATIONAL MOTION
Example 9.8. We wrap a light, non-stretching cable around a solid
cylinder with mass M and radius R. The cylinder rotates with
negligible friction about a stationary horizontal axis. We tie the free
and of the cable to a block of mass m and release the block from rest at
distance h above the floor. As the block falls, the cable unwinds
without stretching or slipping. Find expression for the speed of the
cylinder as the block strikes the floor.
S ECTION 9.5: PARALLEL -A XIS T HEOREM
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There are (infinitely) many axis passing through a given object,
and thus there are infinitely many moments of inertia.
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However, some of them are related to each other.
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Consider two axis of rotation:
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axis O: passing through the center of mass and
axis D: parallel to the first one, but displaced by distance d.
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Let’s choose z-axis to point along the direction of the first axis
and with origin at the center of mass.
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Then the O axis of rotation has coordinates
(x, y, z) = (0, 0, z)
(18)
and the D axis of rotation has coordinates
(x, y, z) = (a, b, z)
where
d=
p
a 2 + b2 .
(19)
(20)
S ECTION 9.5: PARALLEL -A XIS T HEOREM
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There are (infinitely) many axis passing through a given object,
and thus there are infinitely many moments of inertia.
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However, some of them are related to each other.
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Consider two axis of rotation:
I
I
axis O: passing through the center of mass and
axis D: parallel to the first one, but displaced by distance d.
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Let’s choose z-axis to point along the direction of the first axis
and with origin at the center of mass.
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Then the O axis of rotation has coordinates
(x, y, z) = (0, 0, z)
(21)
and the D axis of rotation has coordinates
(x, y, z) = (a, b, z)
where
d=
p
a 2 + b2 .
(22)
(23)
S ECTION 9.5: PARALLEL -A XIS T HEOREM
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Then the two moments of inertia are given by
X
IO =
mi x2i + y2i
(24)
i
and
ID =
X
2
2
mi (xi − a) + (yi − b) .
(25)
i
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The second equation can be expanded as
X
X
X
X
mi
mi x2i + y2i − 2a
mi xi − 2a
mi xi + a2 + b2
ID =
i
i
i
i
where there third and forth terms vanish.
P
I If we denote by total mass M =
i mi then the final expression is
ID = IO + Md2 .
also known as the parallel-axis theorem.
(26)
S ECTION 9.5: PARALLEL -A XIS T HEOREM
Example 9.9. A part of mechanical linkage has a mass of m = 3.6 kg. Its
moment of inertia IP about an axis l = 0.15 m from its center of mass is
IP = 0.132 kg · m2 . What is the moment of inertia Icm about a parallel axis
through the center of mass?
S ECTION 9.6: M OMENT- OF -I NERTIA CALCULATIONS
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In general moment-of-inertia is defined by replacing summation
with integration, i.e.
Z
X
I=
r2 mi
→
I = r2 dm.
(27)
i
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In the case when the density (or mass per unit volume)
dm
dV
is independent of position the moment of inertia
Z
I = r2 ρdV
ρ=
or
Z
I=
r2 ρ dx dy dz
where the limits of integration are set by the geometry of the
body.
(28)
(29)
(30)
S ECTION 9.6: M OMENT- OF -I NERTIA CALCULATIONS
Example 9.10. Consider a hollow cylinder of uniform mass density ρ with
length L, inner radius R1 , and outer radius R2 . Find its moment of inertia
about its axis of symmetry.
S ECTION 9.6: M OMENT- OF -I NERTIA CALCULATIONS
Example 9.11. Find the moment of inertia of a solid sphere of uniform mass
density ρ (like a billiard ball) about an axis through its center.
C HAPTER 10: D YNAMICS OF R OTATIONAL M OTION
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Section 10.1: Torque
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Section 10.2: Torque and Angular Acceleration for Rigid
Body
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Section 10.3: Rigid-body Rotation About a Moving Axis
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Section 10.4: Work and Power in Rotational Motion
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Section 10.5: Angular Momentum
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Section 10.6: Conservation of Angular Momentum
S ECTION 10.1: T ORQUE
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When force acts on an object it can change its translational as well
as rotational motion.
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The effect on the rotational motion depends not only on the
magnitude of the applied force, but also to which point the force
is applied.
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For example:
S ECTION 10.1: T ORQUE
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The relevant physical quantity which measures the twisting
effort of some force ~F applied to point A with respect to some
other point B is called torque.
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If we place origin of coordinates to point B then the torque can
be defined as
~τ ≡ ~r × ~F.
(31)
where
~F − force applied to object at point B
~r − position of B measured with origin at A.
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It is assume that both vectors ~F and ~r lie in the plane orthogonal
to the axis of rotation, then torque would point along the axis of
rotation determined by right-hand rule.
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It is a convention to use right hand coordinate system and so the
torque is positive if the force causes counterclockwise rotation
and negative if the force cause clockwise rotation.
S ECTION 10.1: T ORQUE
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Moreover the magnitude of torque is
τ = rF sin θ.
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(32)
First Interpretation: If we define lever-arm as distance between
line of action of force, i.e.
l ≡ r sin θ
(33)
where θ is the angle between vectors ~F and ~r then the torque is
product of force and lever-arm
τ = Fl.
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(34)
Secondly Interpretation: If we define tangential component of
the force as
F|| ≡ F sin θ
(35)
then torque is product of tangential force to radius of rotation
τ = F|| r.
(36)
S ECTION 10.1: T ORQUE
Example 10.1. To loosen a pipe fitting, a weekend plumber slips a piece of
scrap pipe (a “cheater”) over his wrench handle. He stands on the end of the
cheater, applying his full 900 N weight at a point 0.80 m from the center of
the fitting. the wrench and cheater make an angle of 19◦ with horizontal.
Find the magnitude and direction of the torque he applies about the center of
the fitting.
S ECTION 10.2: T ORQUE FOR R IGID B ODY
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Consider a single particle in a rigid body whose displacement
from axis of rotation is r1 , mass is m1 and force acting on it F1 .
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Since the only component of the force which contributes to
rotation is parallel, the second law for this force gives
F1,|| = m1 a1,|| .
(37)
But since the parallel acceleration is
a1,|| = r1 αz
(38)
τ1,z = F1,|| r1
(39)
τ1,z = m1 r21 αz .
(40)
and torque is
we get
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The same applies to every point-like particle in a rigid body, i.e.
X
X
τi,z =
mi r2i αz = Iαz .
(41)
i
i
S ECTION 10.2: T ORQUE FOR R IGID B ODY
Example 10.2. (setup as in Example 9.7) We wrap a light, nonstretching
cable around a solid cylinder of mass 50 kg and diameter 0.120 m, which
rotates in frictionless bearings about a stationary horizontal axis. We pull
the free end of the cable with a constant 9.0 − N force for a distance of 2.0 m;
it turns the cylinder as it unwinds without slipping. The cylinder is initially
at rest. What is the cable’s acceleration?
S ECTION 10.2: T ORQUE FOR R IGID B ODY
Example 10.3. (setup as in Example 9.8) We wrap a light, non-stretching
cable around a solid cylinder with mass M and radius R. The cylinder
rotates with negligible friction about a stationary horizontal axis. We tie the
free and of the cable to a block of mass m and release the block from rest at
distance h above the floor. As the block falls, the cable unwinds without
stretching or slipping. What are the acceleration of falling block and the
tension in the cable?
S ECTION 10.3: R OTATION A BOUT A M OVING A XIS
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More generally a given rigid body can have both rotational
motion (about some axis passing through center of mass) and
translation motion (of the center of mass).
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In this case the total kinetic energy is a sum of rotational and
translational kinetic energies, i.e.
K=
I
1
1
Mv2cm + Icm ω 2 .
2
2
(42)
This can be shown by imagining the rigid body made up of
many particles with masses mi moving with velocities ~vi so that
the total kinetic energy is
1X
K=
mi (~vi · ~vi ) .
(43)
2
i
If we now define a relative velocity of each particle with respect
to center of mass as
~v0i = ~vi − ~vcm
(44)
then (42) follows.
S ECTION 10.3: R OTATION A BOUT A M OVING A XIS
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An important example is rolling without slipping.
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Then the center of mass velocity is related to angular velocity
vcm = Rω
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Then the motion is purely rotational about point 1, or a
rotational around point 0 and translational motion of 0.
(45)
S ECTION 10.3: R OTATION A BOUT A M OVING A XIS
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To see that two views give the same kinetic energy we can use
the parallel-axis theorem, i.e.
K
=
=
=
=
I
1
I1 ω 2
2
1
Icm + MR2 ω 2
2
1
1
Icm ω 2 + MR2 ω 2
2
2
1
1
Icm ω 2 + Mv2cm .
2
2
(46)
If in addition the potential energy changes (as in the case of
gravitational potential energy), then the total mechanical energy
1
1
Mv2cm + Icm ω 2 + Mgycm
(47)
2
2
and thus one might think of potential energy coming from a
point particle with mass M located at the body’s center of mass.
K=
S ECTION 10.3: T ORQUE FOR R IGID B ODY
Example 10.4. You make a primitive yo-yo by wrapping a massless string
around a solid cylinder with mass M and radius R. You hold the free end of
the string stationary and release the cylinder from rest. The string unwinds
but does not slip or stretches the cylinder descends and rotates. Using
energy considerations, find the speed of the center of mass of the cylinder
after it has descended a distance h.
S ECTION 10.3: R OTATION A BOUT A M OVING A XIS
In a special case when the axis of rotation
I
is the axis of symmetry and
I
does not change direction,
the combined rotational and translational dynamics can be described
in context of second law, i.e.
X
~FI = M~acm
I
X
τzI
= Icm αz
(48)
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where the sum in both equations goes only external quantities (i.e.
forces and torques).
S ECTION 10.3: T ORQUE FOR R IGID B ODY
Example 10.6. You make a primitive yo-yo by wrapping a massless string
around a solid cylinder with mass M and radius R. You hold the free end of
the string stationary and release the cylinder from rest. The string unwinds
but does not slip or stretches the cylinder descends and rotates. Find
downward acceleration of the cylinder and the tension in the string.
S ECTION 10.4: W ORK AND P OWER IN R OTATIONAL
M OTION
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Similarly to how translational work is given by an integral
Z xf
W=
Fx (x)dx
(49)
xi
the rotational work is given by
Z
W=
Z
F|| ds =
Z
θf
F|| Rdθ =
τz (θ)dθ
(50)
θi
I
In the case of force independent of angle we have
W = τz (θf − θi ) = τz ∆θ.
(51)
S ECTION 10.4: W ORK AND P OWER IN R OTATIONAL
M OTION
I
Moreover, the power associated with rotational work is
P=
I
dW
dθ
= τz
= τz ωz .
dt
dt
(52)
And the integral of (50) can also be rewritten by applying the
(rotational) second law,
W=
1 2 1 2
Iω − Iωi
2 f
2
i.e expressing work as a change of rotational kinetic energy.
(53)
S ECTION 10.4: W ORK AND P OWER IN R OTATIONAL
M OTION
Example 10.8. An electric motor exerts a constant τ = 10 − N · m torque
on grindstone which has moment of inertia of I = 2.0 kg·m2 about its shaft.
The system starts from rest. Find the work W done by the motor in t = 8.0 s
and the grindstone kinetic energy K at this time. What average power Pav is
delivered by the motor?
S ECTION 10.5: A NGULAR M OMENTUM
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A rotation analog of linear momentum is angular momentum
~L ≡ ~r × ~p = ~r × (m~v) .
I
Then the rate of change change of angular momentum
d~L
d~r
d~v
=
× (m~v) +~r × m
dt
dt
dt
~
= ~r × (m~a) = ~r × F = ~τ
is nothing but torque of the net force acting on a particle.
(54)
(55)
S ECTION 10.5: A NGULAR M OMENTUM
I
For a rigid body the same is true for every particle and if we
choose the axis of rotation to point along z axis, then
X
X
X
Lz =
Lzi =
ri (mi vi ) =
ri (mi ri ωz ) = ωz .
(56)
i
I
i
i
If the axis of rotation is also a symmetry axis, then
~L = I~
ω.
I
(57)
If we apply (55) to every particle in a rigid body, then
X
I
~τI =
d~L
dt
where the sum on the left-hand-side is over all external torques.
I
Moreover if the rigid body rotates around a symmetry axis then
X
τz = Iαz .
S ECTION 10.5: A NGULAR M OMENTUM
Example 10.9. A turbine fan in a jet engine has a moment of inertia of
I = 2.5 kg·m2 about its axis of rotation. As the turbine starts up, its angular
velocity is given by
ωz = 40 rad/s3 t2.
(a) Find the fan’s angular momentum as a function of time and find its value
at t = 3.0s.
(b) Find the net torque on the fan as a function of time, and find its value at
t = 3.0 s.
S ECTION 10.6: C ONSERVATION OF A NGULAR
M OMENTUM
I
When the net external torque acting on a system is zero, the total
angular momentum of the system is constant.
~i
L~f = L
I
This does not mean that the change of angular momentum of
part of a given body must not changes.
S ECTION 10.5: A NGULAR M OMENTUM
Example 10.10. A physics professor stands at the center of a frictionless
turntable with arms outstretches and a m = 5.0 − kg dumbbell in each
hand. He is set rotating about the vertical axis, making one revolution in
T = 2.0 s. Find his final angular velocity is he pulls the dumbbells in to his
stomach. His moment of inertia (without the dumbbells) is Ii = 3.0 kg·m2
with arms outstretched and If = 2.2 kg · m2 with his hands at his stomach.
The dumbbells are ri = 1.0 m from the axis initially and rf = 0.20 m at the
end.
C HAPTER 11: E QUILIBRIUM AND E LASTICITY
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Section 11.1: Conditions for Equilibrium
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Section 11.2: Center of Gravity
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Section 11.3: Solving Rigid-Body Equilibrium Problems
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Section 11.4: Stress, Strain and Elastic Moduli
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Section 11.5: Elasticity and Plasticity
S ECTION 11.1: C ONDITIONS FOR E QUILIBRIUM
I
I
An object is said to be in equilibrium if two conditions are
satisfied:
I
First condition for equilibrium. The sum of all external
forces acting on the body is zero:
X
~F = 0.
(58)
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Second condition for equilibrium. The sum of all torques
(due to external forces about any point) on the body is zero:
X
~τ = 0.
(59)
Although we shall only apply these equilibrium conditions to
bodies at rest, the same conditions applies to bodies in uniform
translational motion (without rotation).
S ECTION 11.1: C ONDITIONS FOR E QUILIBRIUM
S ECTION 11.1: C ONDITIONS FOR E QUILIBRIUM
Example. Which situation satisfies both the first and second conditions for
equilibrium?
(i) a seagull gliding at a constant angle below the horizontal and at constant
speed;
(ii) an automobile crankshaft turning at an increasing angular speed in the
engine of a parked car;
(iii) a thrown baseball that does not rotate as it sails through the air.
S ECTION 11.2: C ENTER OF G RAVITY
I
Consider a collection of particles which make up a given object,
then their center of mass is then given by
P
mi~ri
m1~r1 + m2~r2 + m3~r3 + ...
~rcm =
= i
,
(60)
m1 + m2 + m3 + ...
M
where
X
M=
mi .
(61)
i
I
Now let us consider gravitational torque caused by gravitational
forces with axis of rotation in the origin
~τ
=
X
~τi
i
=
X
~ri × (mi~gi )
i
=
X ~ri mi i
M
× (M~gi )
(62)
S ECTION 11.2: C ENTER OF G RAVITY
I
In general the gravitational acceleration changes with attitude.
For example, the center of gravity of 452-m-tall Petronas Towers
in Malaysia is 2 cm below the center of mass.
I
Most of the time the gravitational acceleration can be assumed
to be constant and then the torque is given by
P
ri
i mi~
× (M~g)
~τ =
M
= ~rcm × (M~g) .
(63)
I
However. it turns out that even in a more general case when the
gravitational acceleration decreases with altitude
~τ = ~rcg × (M~g) .
where ~rcg is the center of gravity.
(64)
S ECTION 11.1: C ENTER OF G RAVITY
I
Finding the center of gravity (theoretically) is easy when the
object is symmetric, but what if the object has irregular shape?
I
To find the center of gravity experimentally one can suspend an
object from a given point.
.
I
By repeating the same experiment for two distinct points a pair
of such lines can be constructed and then the point where the
lines intersect should be the center of gravity.
S ECTION 11.2: C ENTER OF G RAVITY
Example 11.1. A uniform plank of length L = 6.0 m and mass M = 90 kg
rests on sawhorses separated by D = 1.5 m and equidistant from the center
of the plank. Cousin Throckmorton wants to stand on the right-hand end of
the plank. If the plank is to remain at rest, how massive can Throckmorton
be?
S ECTION 11.2: C ENTER OF G RAVITY
Example. A rock is attached to the left end of a uniform meter stick that has
the same mass as the rock. In order for the combination of rock in meter stick
to balance atop the triangular object (see figure below), how far from the left
end of the stick should the triangular object be placed?
S ECTION 11.3: R IGID -B ODY E QUILIBRIUM P ROBLEMS
For problem in two dimensions we can choose a coordinate system so
that the vectorial equilibrium conditions (58) and (59) are replaced by
three scalar conditions
X
Fi,x = 0
(65)
i
X
Fi,y
= 0
(66)
τi,z
= 0.
(67)
i
X
i
S ECTION 11.3: R IGID -B ODY E QUILIBRIUM P ROBLEMS
Example 11.2. An auto magazine reports that a certain sports car has 53%
of its weight on the front wheel. The distance between the axles is
D = 2.46 m. How far in front of the rear axle is the car’s center of gravity?
(Let’s denote is by Lcg )
S ECTION 11.3: R IGID -B ODY E QUILIBRIUM P ROBLEMS
Example 11.3. Sir Lancelot, who weights 800 N, is assaulting a castle by
climbing a uniform ladder that is 5.0 m long and weights 180 N. The bottom
of the ladder rests on a ledge and leans across the moat in equilibrium
against a frictionless, vertical castle wall. The ladder makes an angle of 53.1◦
with the horizontal. Lancelot pauses one-third of the way up the ladder.
(a) Find normal and friction forces on the base of the ladder.
(b) Find the minimum coefficient of static friction needed to prevent slipping
at the base.
(c) Find the magnitude and direction of the contact force on the base of the
ladder.
S ECTION 11.4: S TRESS , S TAIN , AND E LASTIC M ODULI
I
So far we considered (mostly) rigid bodies.
I
In reality all objects can deform when external forces are applied
to them, e.g. stretching, squeezing, twisting
I
For each type of deformation there are two relevant quantities:
I
I
stress which characterizes the cause of the deformation
strain which characterizes the effect of the deformation
I
Often the two quantities are directly proportional to each other.
I
Coefficient of proportionality is known as elastic modulus,
Stress
= Elastic modulus
Strain
I
(68)
We have already seen an example of this (or Hooke’s) law
F
= k.
x
(69)
S ECTION 11.4: S TRESS , S TAIN , AND E LASTIC M ODULI
Tensile/compressive stress. For stretching/squeezing deformations
the tensile/compressive stresses are defined as
σ=
F⊥
A
(70)
and the tensile/compressive strains are defined as
ε=
l − l0
∆l
=
.
l
l
The units of the stress is
[Stress] =
[Force]
[Distance]2
and in SI system a unit of stress is measures in Pascals,
1 Pa ≡ 1 N/m2 .
but the strain is dimensionless.
(71)
S ECTION 11.4: S TRESS , S TAIN , AND E LASTIC M ODULI
Young’s modulus.
I
In the case of tensile/compressive stress the elastic modulus is
known as Young’s modulus
Y=
I
F⊥ /A
F⊥ l0
σ
=
=
.
ε
∆l/l0
A∆l
(72)
This definition insures that Young’s modulus is positive if the
perpendicular component of force F⊥ is chosen to point
outwards:
Tensile: F⊥ > 0 and ∆l > 0
⇒
Y>0
Compressive: F⊥ < 0 and ∆l < 0
⇒
Y>0
(Note that since the strain was dimensionless, the Young’s
modulus has the same units as stress.)
I
For many materials the tensile and compressive Young’s
modulus are the same and the Hooke’s law works very well.
(73)
S ECTION 11.4: S TRESS , S TAIN , AND E LASTIC M ODULI
Here is a table of the modulus for some of the materials
However for composite materials (such as concrete or stone) the
compressive stress can be quite large, while the tensile stress easily
breaks the object and thus the Hooke’s law does not hold.
S ECTION 11.4: S TRESS , S TAIN , AND E LASTIC M ODULI
Example 11.5. A steel 2.0 m long has a cross-sectional area of 0.30 cm2 . It is
hung by one end from a support, and a 550 kg milling machine is hung from
its other end. Determine the stress on the rod and the resulting strain and
elongation.
S ECTION 11.4: S TRESS , S TAIN , AND E LASTIC M ODULI
Bulk stress.
I
There situations when the same tensile/compressive stress is
applied to all spatial directions simultaneously.
I
Such an isotropic (i.e. the same in all directions) stress is called
(sometimes bulk stress) or pressure
p=
F⊥
A
(74)
and an isotropic strain is called bulk or volume strain
εbulk =
∆V
.
V0
(75)
where ∆V = Vf − Vi .
I
Units of pressure
1 Pa ≡ 1 N/m2
1 atm ≈ 105 Pa
1 psi ≈ 6900 Pa.
(76)
S ECTION 11.4: S TRESS , S TAIN , AND E LASTIC M ODULI
I
Confusing Point #1: The pressure is usually positive which is in
conflict with our choice for perpendicular direction to point
outwards of an object. This can be fixed by either stating that
p=−
I
F⊥
.
A
(77)
Confusing Point #2: The bulk stress is defined not as pressure but
as a negative change in pressure as an object is moved from one
substance (e.g. air) to another (e.g. water), i.e.
σ bulk = ∆p.
(78)
where ∆p = pf − pi
I
Confusing Point #3: The corresponding elastic modulus (known
as bulk modulus) is defined with a minus sign
B=
I
σ bulk
−∆p · V0
=
.
εbulk
∆V
(79)
Two out of three confusion (i.e. #1 and #3) could have been
avoided if the pressure was defined with a minus sign as in (77).
S ECTION 11.4: S TRESS , S TAIN , AND E LASTIC M ODULI
Example 11.6. A hydraulic press contains V0 = 0.25 m3 (250 L) of oil. Find
the decrease in the volume ∆V of the oil when it is subjected to a pressure
increase ∆p = 1.6 × 107 Pa (about 160 atm). The bulk modulus of the oil is
B = 5.0 × 109 Pa (about 5.0 × 104 atm) and its compressibility is
k = 1/B = 20 × 10−6 atm−1 .
S ECTION 11.4: S TRESS , S TAIN , AND E LASTIC M ODULI
Shear stress. In addition to tensile/compressive deformations due to
perpendicular component of the applied force, a parallel component
of the applied force can cause a deformation.
For such deformations we defined shear stress as
σ shear =
F||
A
(80)
εshear =
x
h
(81)
shear strain as
and shear modulus as
S=
F|| /A
F|| h
σshear
=
=
.
εshear
x/h
Ax
(82)
If we imagine a shear deformation of rectangular parallelepiped of
dimensions x, y, z then
Fx y
S=
.
(83)
xz∆x
S ECTION 11.4: S TRESS , S TAIN , AND E LASTIC M ODULI
Example 11.7. Suppose a brass base plate of an outdoor sculpture
experiences shear force in an earthquake. The vertically oriented square plate
is x = 0.5 cm thick. Whats is the force exerted on each of its edges is the
resulting displacement is ∆x = 0.16 mm?
S ECTION 11.4: S TRESS , S TAIN , AND E LASTIC M ODULI
Example A copper rod of cross-sectional area A = 0.500 cm2 and length
1.00 m is elongated by 2.00 × 10−2 mm, and a steel rod of the same
cross-sectional area but 0.100 m in length is elongated by 2.00 × 10−3 mm.
a) Which rod has greater tensile strain?
I
the copper rod
I
the steel rod
I
the stain is the same
b) Which rod is under greater tensile stress?
I
the copper rod
I
the steel rod
I
the stain is the same
S ECTION 11.5: E LASTICITY AND P LASTICITY
For larger deformations the linear dependence of stress as functions
of stain (68) does not hold. A typical dependence for metals under
tension is illustrated in
and for rubber under tension is illustrated by
C HAPTER 9: R OTATION OF R IGID BODIES
I
Angular quantities
I
I
I
I
Constant angular acceleration
I
I
I
I
Angular velocity and position as functions of time
Angular velocity as function of angular position
Angular position as function of angular velocity
Relating angular and linear quantities
I
I
I
I
Angular position in terms of revs, degrees and radians
Average angular velocity and acceleration
Instantaneous angular velocity and acceleration
Relating linear velocity from angular velocity
Relating tangential acceleration and angular acceleration
Relating centripetal acceleration and angular velocity
Energy in rotational motion
I
I
I
Definition of rotational kinetic energy
Applications of parallel axis theorem
Calculation of moment of inertia
C HAPTER 10: D YNAMICS OF R OTATIONAL M OTION
I
Torque
I
I
I
I
Scalar and vectorial definitions of torque
Lever-arm interpretation of torque
Tangential interpretation of torque
Torque for rigid body
I
I
I
Rotational Newton’s second law
Rotational and translational kinetic energy
Relation of velocities for motion without slipping
I
Application of second law for rotations about a moving axis
I
Definitions and calculations of rotational work and power
I
Angular momentum
I
I
I
Definition of angular momentum
Conservation of angular momentum of closed systems
Change of angular momentum due to external torques
C HAPTER 11: E QUILIBRIUM AND E LASTICITY
I
Equilibrium conditions
I Translational equilibrium conditions
I Rotational equilibrium conditions
I Identification of of equilibrium conditions
I
Center of gravity
I Definition of center of mass
I Calculation of torque due to gravitational force
I Finding center of gravity (theoretical/experimental)
I
2D equilibrium problems
I Drawing free-body diagram for an equilibrium object
I Applying two translational equilibrium conditions
I Applying a rotational equilibrium condition
I
Stress/Stain/Modulus
I Tensile,compressive, bulk and shear stresses
I Tensile,compressive, bulk and shear strains
I Tensile,compressive, bulk and shear moduli