Download Slides for Chapters 5, 6, 7, 8 and Review

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General Physics I
(aka PHYS 2013)
P ROF. VANCHURIN
( AKA V ITALY )
University of Minnesota, Duluth
(aka UMD)
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O UTLINE
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S ECTION 5.1: U SING N EWTON ’ S F IRST L AW
First Law. A body acted on by no net force, i.e.
X
~Fi = 0
i
has a constant velocity (which may be zero) and zero acceleration.
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S ECTION 5.1: U SING N EWTON ’ S F IRST L AW
Example 5.1. A gymnast with mass mG = 50 kg suspends herself
from the lower end of a hanging rope of negligible mass. The upper
end of the rope is attached to the gymnasium ceiling.
(a) What is the gymnasts’s weight?
(b) What force (magnitude and direction) does the rope exert on her?
(c) What is the tension at the top of the rope?
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S ECTION 5.1: U SING N EWTON ’ S F IRST L AW
Example 5.2. Find the tension at each end of rope in Example 5.1 if
the weight of the rope is 120 N.
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S ECTION 5.1: U SING N EWTON ’ S F IRST L AW
Example 5.4. A car of weight w rests on a slanted ramp attached to a
trailer. (See figure below. Angle α is given. ) Only a cable running
from the trailer to the car prevents the car from rolling off the ramp.
(The car brakes are off and its transmission is neutral.) Find the
tension in the cable and the force that the ramp exerts on the car’s
tires.
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S ECTION 5.2: U SING N EWTON ’ S S ECOND L AW
Second Law. If a net external force acts on a body, the body
accelerates. The direction of acceleration is the same as the direction of
the net force. The mass of the body times the acceleration vector of the
body equals to the net force vector, i.e.
X
~Fi = m~a
(1)
i
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S ECTION 5.2: U SING N EWTON ’ S S ECOND L AW
Example 5.6. An iceboat is at rest on a frictionless horizontal surface.
Due to the blowing wind, 4.0 s after the iceboat is released, it is
moving to the right at 6.0 m/s. What constant horizontal force FW
does the wind exert on the iceboat? The combined mass of iceboat and
rider is 200 kg.
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S ECTION 5.2: U SING N EWTON ’ S S ECOND L AW
Example 5.10. A toboggan loaded with physics students (total
weight w) slides down a snow-covered hill that slopes at a constant
angle α. The toboggan is well waxed, so there is virtually no friction.
(a) What is its acceleration? (b) What is normal force?
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S ECTION 5.3: F RICTION F ORCES
Friction force. There two types of contact forces between
macroscopic objects:
~n − normal force always perpendicular to the contact surface
~f − friction force is always parallel to the contact surface .
Both forces arise due to microscopic (electromagnetic) interaction between
molecules, but we shall only study their macroscopic properties.
Kinetic Friction. If there is a motion along the surface of contact, then
these two forces are related to each other by the so-called coefficient
of kinetic friction
fk
µk =
(2)
n
or
fk = µk n.
(3)
Note that the friction coefficients are dimensionless, i.e. have no units.
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S ECTION 5.3: F RICTION F ORCES
Static Friction If there is no motion along the surface of contact, then
the friction force is bounded from above
fs ≤ (fs )max = µs n.
(4)
It turns out that
µs > µk
and thus
fk < fsmax .
(5)
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S ECTION 5.3: F RICTION F ORCES
Table of Friction Coefficients Exact values of both coefficients
depend on the materials in constant, e.g.
Material
steel on steel
ice on steel
dry rubber on dry concrete
wet rubber on wet concrete
µs
0.74
0.03
1.0
0.3
µk
0.57
0.015
0.8
0.25
There is also rolling friction which is typically much smaller. For steel
wheels on steel rails it is ∼ 0.002 − 0.003.
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S ECTION 5.3: F RICTION F ORCES
Example 5.13. You want to move a 500 N crate across a level floor.
To start the crate moving, you have to pull with a 230 N horizontal
force. Once the crate starts to move, you can keep it moving at
constant velocity with only 200 N force. What are the coefficient of
static and kinetic friction?
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S ECTION 5.3: F RICTION F ORCES
Example 5.16. A toboggan loaded with physics students (from
Example 5.10) slides down a snow-covered hill. The wax has worn
off, so there is a nonzero coefficient of kinetic friction µk . The slope has
just the right angle α to make the toboggan slide with constant
velocity. Find the angle in terms of w and µk .
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S ECTION 5.3: F RICTION F ORCES
Fluid (air) resistance. The (magnitude of) force of fluid resistance
depends on the velocity,
(
kv
for “small” velocities
f =
(6)
2
Dv for “large” velocities
where the coefficients depend on: type of fluid, shape of object, etc.
Terminal velocity. For small velocities the second law implies
mg − (kv) = ma
(7)
This acceleration must stop when the terminal velocity is reached
mg
(8)
vt =
k
Similarly in the regimes of large velocities the second law implies
mg − (Dv2 ) = ma.
(9)
and thus the terminal velocity is
r
vt =
mg
.
D
(10)
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S ECTION 5.3: F RICTION F ORCES
Example 5.18. For a human body falling through air in a spread-eagle
position, the numerical value of the constant D in Eq. (5.6) is about 0.25
kg/m. Find the terminal speed for a 50 kg skydiver.
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S ECTION 5.4: D YNAMICS OF CIRCULAR MOTION
In Sec. 3.4 we derived the following eqs. for uniform circular motion
v2
R
(11)
4π 2 R
T2
(12)
a⊥ =
and
a⊥ =
where
a⊥
−
v −
R −
T
−
magnitude of acceleration
constant speed
radius of circular path
period of motion.
(13)
An object in such a motion experiences a constant (in magnitude)
acceleration and thus according to second law
Fnet = ma⊥ = m
v2
4π 2 R
=m 2 .
R
T
(14)
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S ECTION 5.4: D YNAMICS OF CIRCULAR MOTION
Example 5.19. A sled with a mass of 25.0 kg rests on a horizontal sheet of
essentially frictionless ice. It is attached by a 5.00 m rope to a post set in the
ice. Once given a push, the sled revolves uniformly in a circle around the
post . If the sled makes five complete revolutions every minute, find the force
F exerted on it by the rope.
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S ECTION 6.1: W ORK
Constant speed
Speed of an object is unchanged if
I
the sum of all forces is acting on it is zero.
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the sum of all forces is acting in the direction orthogonal to the
direction of motion.
An important observation is that in both cases any function of speed
would also not change, e.g.
f (v(t)) ∝ v(t)2 = const.
(15)
Changing speed
Speed of an object is changing on if the net force acting on the body
has a component parallel to the direction of motion. Then for example
f (v(t)) ∝ v(t)2 6= const.
(16)
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S ECTION 6.1: W ORK
Work
Let us define a physical quantity, we shall call work , which will
quantify whether there is a component of constant force ~F parallel to
a straight-line displacement, ~s,
W = ~F · ~s = Fs cos φ.
(17)
Although the work is done by each and every force acting on a body,
according to above equation it may be positive, negative or even zero.
Units of work
[Work] = [Force] × [Distance].
(18)
In SI units
1 J = 1 N×1 m.
(19)
Note that the dimensions of work can be also written as
[Work] = [Mass] × [Acceleration] × [Distance]
(20)
[Work] = [Mass] × [Velocity]2 .
(21)
or
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S ECTION 6.1: W ORK
Example 6.2. A farmer hitches her tractor to a sled loaded with
firewood and pulls it a distance of 20 m along ground. The total
weight of sled and load is 147000 N. The tractor exerts a constant
5000 N force at an angle of 36.9◦ above the horizontal. A 3500 N
friction force opposes the sled’s motion. Find the work done by each
force acting on the sled and the total work done by the forces.
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S ECTION 6.2: K INETIC E NERGY
Back in second chapter we argued that 1D motion with constant
acceleration is described by following equations
1
= x0 + v0x t + ax t2
2
v(t) = v0x + ax t
x(t)
a(t)
= ax .
(22)
Then we derived a useful relation
2ax (x(t) − x0 ) = vx (t)2 − v20x
(23)
which can be written as
max (x − x0 ) =
1 2 1 2
mv − mv .
2 x 2 0x
(24)
Note that this equation does not depend on time explicitly, but only through
time-dependence of x and vx .
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S ECTION 6.2: K INETIC E NERGY
Kinetic energy. In 3D the equation can be written as
m~a · ~s =
1 2 1 2
mv − mv0
2
2
(25)
where the left hand side takes the form of the equation for work.
What about the right hand side of (25)? If we define kinetic energy by
K(v) =
1 2
mv
2
(26)
then equation (25) tells us that the work acting on a system changes
its kinetic energy by the amount equal to work
W = K(t) − K(t0 ) = ∆K.
(27)
In fact what is relevant is the total work of all forces, i.e.
Wtot = ∆K.
(28)
This is known as the work-energy theorem or as we will see later an
equation representing conservation of energy.
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S ECTION 6.2: K INETIC E NERGY
Example 6.3. Let’s come beck to Example 6.2. What is the speed of the sled
after it moves 20 m? Suppose sled’s initial speed v1 = 2.0 m/s.
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S ECTION 6.2: K INETIC E NERGY
Example 6.4. The 200 − kg steel hammerhead of a pile driver is lifted
3.00 m above the top of a vertical I-beam being driven into the ground. The
hammerhead is then dropped, driving the I-beam 7.4 cm deeper into the
ground. The vertical guide rails exert a constant 60 − N friction force on the
hammerhead. Use the work-energy theorem to find (a) the speed of the
hammerhead just as it heats the I-beam and (b) the average force the
hammerhead exerts on the I-beam. Ignore effects of air.
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S ECTION 6.2: K INETIC E NERGY
Example 6.5. Two iceboats hold a race on a frictionless horizontal lake.
The two iceboats have masses m and 2m. The iceboats have identical sails, so
the wind exerts the same constant force ~F on each iceboat. They start from
rest and cross the finish line a distance s away. Which iceboat crosses the
finish line with greater kinetic energy.
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S ECTION 6.3: VARYING F ORCES
Up to now we were dealing with work done by constant forces
W = Fx xf − xi .
(29)
This can generalized in two different ways:
I
we can consider a motion along a curve and/or
I
we can consider forces which change with position.
In the latter case the motion can be broken into small segments
W1
W2
=
F1 ∆x
=
F2 ∆x
...
Wn
=
Fn ∆x
(30)
and by adding all these small pieces of work we get
n
X
i=1
Wi =
n
X
i=1
Fi ∆x
(31)
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S ECTION 6.3: VARYING F ORCES
In the limit of infinite n this sum reduces to integral
W ≡ lim
n→∞
n
X
Z
xf
Fx (x)dx
Fi ∆x =
(32)
xi
i=1
where Fx (x) is x-component of ~F(x) which is a function of position.
For example, force required to stretch an ideal spring is
Fx (x) = kx.
(33)
where k is the so-called spring constant measured
[k] =
[Force]
.
[Distance]
From (32) and (33) the work need to stretch an ideal spring is
Z
W=
0
X
1 2
Fx (x)dx =
kx
2
X
=
0
1 2
kX .
2
(34)
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S ECTION 6.3: VARYING F ORCES
Work done on an ideal spring to stretch it
X
Z X
1 2
1
W=
Fx (x)dx =
kx
= kX2 .
2
2
0
0
(35)
Note that the position for non-stretched spring was set to x = 0, but if
the origin is displaced then the equations would have to be modified
Fx (x)
W
= k(x − x0 )
1
2
=
k (X − x0 ) .
2
(36)
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S ECTION 6.3: VARYING F ORCES
Example 6.6. A woman weighing 600 N steps on a bathroom scale that
contains a stiff spring. In equilibrium, the spring is compressed 1.0 cm
under her weight. Find the force constant of the spring and the total work
done on it during compression.
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S ECTION 6.3: VARYING F ORCES
Example 6.7. An air-track glider of mass 0.100 kg is attached to the end of a
horizontal air track by a spring with force constant 20.0 N/m. Initially the
spring is unstretched and the glider is moving at 1.50 m/s to the right. Find
the maximum distance d that the glider moves to the right (a) is the air track
is turned on, so that there is no friction, and (b) if the air is turned off, so
that there is a kinetic friction coefficient µk = 0.47
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S ECTION 6.4: P OWER
I
So far we did not pay attention to time it takes to do the work.
I
The physical quantity which represents the rate at which the
work is done is called power.
I
This suggest that dimensions of power should be
[Power] =
I
(37)
In SI units power is measures in watts (W) which is defined as
1W =
I
[Work]
.
[Time]
1J
1s
(38)
Average power is defined from total work done in the same way
as average velocity was defined from total displacement
vavg
=
Pavg
=
∆x
∆t
∆W
.
∆t
(39)
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S ECTION 6.4: P OWER
Similarly the (instantaneous) power is defined as (instantaneous)
velocity by taking limit of interval to zero
v
≡
P ≡
∆x
dx
=
∆t
dt
dW
∆W
=
.
lim
∆t→0 ∆t
dt
lim
∆t→0
(40)
Since work is given by
W = ~F · ~s
the power can also be expressed as
~F · ~s
d
dW
d~F
d~s
P=
=
=
· ~s + ~F ·
dt
dt
dt
dt
(41)
(42)
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S ECTION 6.4: P OWER
But for constant (time-independent) forces acting on a particle
d~F
=0
dt
(43)
the power is
P=
d~F
d~s ~
· ~s + ~F ·
= F · ~v
dt
dt
(44)
d~s
.
dt
(45)
where
~v =
is the velocity of particle.
Example 6.9. Each of the four jet engines on an Airbus A380 airliner
develops a thrust (a forward force on the airliner) of 322000 N (72000 lb).
When the airplane is flying at 250 m/s (900 km/h, or roughly 560 mi/h) what
is horsepower does each engine develop.
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S ECTION 7.1: G RAVITATIONAL P OTENTIAL E NERGY
Potential Energy
I
So far the only form of energy that we defined was kinetic energy
which is a function of velocity
K(v) =
I
1 2
mv .
2
(46)
It turns out that it is also useful to define another form of energy
- potential energy - which is a function of position, e.g.
U(x) = mgx
or
U(x) =
1 2
kx .
2
(47)
I
The idea is that it might be possible to store energy in the form
of potential energy by placing an object at certain position.
I
The process of storing energy require work to be done on the
object, but later on the potential energy can be released.
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S ECTION 7.1: G RAVITATIONAL P OTENTIAL E NERGY
Gravitational Potential Energy
I
The most familiar example of potential energy is the
gravitational potential energy, when the work must be done on
an object to lift it up (or down).
I
For example, if on object has mass m and is lifted from height x1
to height x2 , then the work done by gravitational force is
Wgrav = −mg î (x2 − x1 ) î = mgx1 − mgx2
(48)
I
Then it is convenient to define a gravitational potential energy
Ugrav ≡ mgx
(49)
and then the work done by gravitational force can be written as
Wgrav = −∆Ugrav = Ugrav,1 − Ugrav,2 .
(50)
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S ECTION 7.1: G RAVITATIONAL P OTENTIAL E NERGY
Gravitational Potential Energy
I
Now if we assume that there are no other forces acting on a
body, then according to work-energy theorem
Wgrav = Kf − Ki
(51)
and by combining (50) and (51) we get
Ugrav,i − Ugrav,f = Kf − Ki
(52)
Ugrav,i + Ki = Ugrav,f + Kf .
(53)
or
I
The latter equation suggest that the total energy is unchanged
E ≡ K + Ugrav = constant.
I
More generally there might be other forces acting on a given
system and then the work energy theorem would imply
Wother = Ef − Ei
I
(54)
(55)
Conservation of energy is the first of many conservation laws.
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S ECTION 7.1: G RAVITATIONAL P OTENTIAL E NERGY
Example 7.1. You throw a 0.145 − kg baseball straight up, giving it an
initial velocity of magnitude 20.0 m/s. How high it goes?
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S ECTION 7.1: G RAVITATIONAL P OTENTIAL E NERGY
Example 7.2. In Example 7.1 suppose your hand moves upward by 0.50 m
while you are throwing the ball. The ball leaves your hand with an upward
velocity of 20 m/s. (a) Find the magnitude of the force (assumed constant)
that your hand exerts on the ball. (b) Find the speed of the ball at a point
15.0 m above the point where it leaves your hand. Ignore air resistance.
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S ECTION 7.1: G RAVITATIONAL P OTENTIAL E NERGY
Example 7.3. A batter hits two identical baseballs with the same initial
speed and from the same initial height but at different initial angles. Prove
that both balls have the same speed at any height h is air resistance can be
neglected.
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S ECTION 7.1: G RAVITATIONAL P OTENTIAL E NERGY
Example 7.6. We want to slide a 12 − kg crate up a 2.5 − m-long ramp
inclined at 30◦ angle. A worker, ignoring friction, calculates that he can do
this by giving it an initial speed of 5.0 m/s at the bottom and letting it go.
But, friction is not negligible; the crate slides only 1.6 m up the ramp, stops,
and slides back down. (a) Find the magnitude of friction acting on crate,
assuming that it is constant. (b) How fast is the crate moving when it
reaches the bottom of the ramp?
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S ECTION 7.2: E LASTIC P OTENTIAL E NERGY
I
Remember that the work done on a spring is given by
W=
1 2 1 2
kx − kxi
2 f
2
(56)
but so the work done by a spring is
Wel =
1 2 1 2
kx − kxf
2 i
2
(57)
Similarly to work done by gravity,
Wgrav = −∆Ugrav = mgyi − mgyf
(58)
the work done by spring can be expressed as
Wel = −∆Uel =
1 2 1 2
kx − kxf
2 i
2
(59)
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S ECTION 7.2: E LASTIC P OTENTIAL E NERGY
I
The main difference is that
Ugrav = mgy
(60)
but
1 2
kx
2
where the origin corresponds to unstretched spring.
Uel =
I
(61)
It is sometime useful to define the total potential energy
1
U = Ugrav + Uel = mgy + kx2
2
(62)
and the total mechanical energy as
E=U+K
I
(63)
We can also modify the work-energy theorem further
Wother = Ef − Ei .
(64)
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S ECTION 7.2: E LASTIC P OTENTIAL E NERGY
Example 7.7. A glider with mass m = 0.200 kg sits on a frictionless
horizontal air track, connected to a spring with constant k = 5.00 N/m. You
pull on the glider, stretching the spring 0.100 m, and releasing it from rest.
The glider moves back towards its equilibrium position (x = 0). What is its
x-velocity when x = 0.080 m?
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S ECTION 7.2: E LASTIC P OTENTIAL E NERGY
Example 7.9. A 2000 − kg (or 19600 − N) elevator with broken cables in a
test rig is falling at 4.00 m/s when it contacts a cushioning spring at the
bottom of the shaft. The spring is intended to stop the elevator, compressing
2.00 m as it does so. During the motion a safety clamp applies a constant
17000 − N frictional force to the elevator. What is the necessary force
constant k for the spring.
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S ECTION 7.3: C ONSERVATIVE AND
N ONCONSERVATIVE FORCES
I
We were able to define potential energy associated with work
done by gravitational and elastic forces. All such forces are
called conservative forces.
I
Work done by conservative forces:
I can always be expressed as difference between initial and
final values of a suitably defined potential energy
I it is reversible and is independent on the trajectory of the
body, but only on initial and final points
I
One might wonder if it is possible to do the same for all
macroscopic forces which would allow to rewrite the
work-energy theorem as a simple law of conservation of energy.
I
It turns out that there are other or non-conservative forces for
which it is not possible to define potential energy.
I
For example, frictional force or air resistance forces are
nonconservative.
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S ECTION 7.2: C ONSERVATIVE / N ONCONSERVATIVE
Example 7.11. In a region of space the force of an electron is ~F = C x ĵ,
where C is a positive constant. The electron moves around a square loop in
the xy-plane. Calculate the work done on the electron by the force ~F during a
counterclockwise trip around square.
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S ECTION 7.3: C ONSERVATIVE / N ONCONSERVATIVE
I
The non-conservative forces cannot be described in terms of
mechanical potential energies, but one can still associate with
the other energies such as internal energy.
I
For example the frictional force is non-conservative, but when
friction is applied to objects in contact the internal properties of
objects change. In particular friction leads to increase in
temperature or internal energy.
I
On the microscopic level it corresponds to the change of kinetic
energy of individual molecules. But from a macroscopic point of
view one can think of work done by all non-conservative forces
as a measure of change of internal energies
Wother = −∆Uint
(65)
and the work energy theorem implies the conservation law
∆K + ∆U + ∆Uint = 0.
(66)
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S ECTION 7.3: C ONSERVATIVE / N ONCONSERVATIVE
I
The main difference is that one cannot use internal energy to do
any useful work.
I
This follows from the so-called second law of thermodynamics
that we will see later in the course.
I
The conservation law of energy and the second law of
thermodynamics had passed a very large number of tests, but
this does not stop people from trying to build a perpetual
motion machine.
I
Perpetual motion machines:
I
I
first kind (do work without input of energy)
second kind (do work using internal energy).
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S ECTION 7.4: F ORCE AND POTENTIAL ENERGY
I
Work done on a particle when displacement is small
W = Fx (x)∆x
(67)
which can also be expressed as a change in potential energy
W = −∆U.
(68)
By equating (67) and (68) we get
Fx (x)∆x = −∆U
(69)
or in the limit of small infinitesimal displacement
Fx (x) = − lim
∆x→0
∆U
dU
=−
.
∆x
dx
(70)
For example
U=
1 2
kx
2
⇒
Fx (x) = −kx
(71)
Fx (x) = −mg.
(72)
or
U = mgx
⇒
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S ECTION 7.4: F ORCE AND POTENTIAL ENERGY
I
This can be easily generalized to 3D
~F = − ∂U(x, y, z) , ∂U(x, y, z) , ∂U(x, y, z) .
∂x
∂y
∂z
(73)
It is convenient to define a gradient operator (pronounced nabla)
~ ≡ ∂ , ∂ , ∂
(74)
∇
∂x ∂y ∂z
such that
~F = −∇U(x,
~
y, z).
(75)
For example the gravitational potential energy is
U = mgy
(76)
and so
∂
∂
∂
~F = −∇(mgy)
~
= − î
+ ĵ
+ k̂
(mgy) = −mgĵ.
∂x
∂y
∂z
(77)
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R EVIEW
S ECTION 7.4: F ORCE AND POTENTIAL ENERGY
Example 7.14. A puck with coordinates x and y slides on a level,
frictionless air-hockey table. It is acted on by a conservative force described
by the potential-energy function
U(x, y) =
1
k x2 + y2 .
2
Find a vector expression for the force acting on the puck, and find an
expression for the magnitude of the force.
(78)
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S ECTION 7.5: E NERGY D IAGRAM
I
Physical systems can be described in terms of energy diagrams.
I
For example harmonic oscillator whose potential energy is
U=
1 2
kx
2
(79)
and total energy is
E = K + U.
I
(80)
The main point is to study the possible solutions qualitatively
without having to solve equations of motion quantitatively.
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S ECTION 7.5: E NERGY D IAGRAM
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R EVIEW
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S ECTION 7.5: E NERGY D IAGRAM
Which statement correctly describes what happens to particle
when it is at the maximum (of red line) between x2 and x3 ?
1. The particle’s acceleration is zero.
2. The particle accelerates in the positive x-direction; the
magnitude of the acceleration is less than at any other
point between x2 and x3 .
3. The particle accelerates in the positive x-direction; the
magnitude of the acceleration is greater than at any other
point between x2 and x3 .
4. The particle accelerates in the negative x-direction; the
magnitude of the acceleration is less at any other point
between x2 and x3 .
5. The particle accelerates in the negative x-direction; the
magnitude of the acceleration is greater than at any other
point between x2 and x3 .
R EVIEW
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R EVIEW
S ECTION 7.5: E NERGY D IAGRAM
I
The total energy was defined as a sum
2
dx
1
dx
1
E x,
= m
+ kx2 .
dt
2
dt
2
(81)
I
However, a more fundamental quantity (called Lagrangian) is
2
1
dx
1
dx
= m
− kx.2
(82)
L x,
dt
2
dt
2
I
For any given trajectory x(t) Lagrangian L is a function of time
and thus can be integrated over some interval, i.e.
Z
ti
tf
dx
L x,
dt
dt.
(83)
I
Clearly, for any trajectory x(t) the integral would be some real
number and one might wonder what trajectory would produce
the smallest (or largest) number? Answer: Classical trajectories.
I
Note, however, that all other trajectories are also possible!
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S ECTION 8.1: M OMENTUM AND I MPULSE
I
Newton’s second law can be expressed as
X
~Fi = d~p
dt
(84)
i
where the new quantity (known as momentum) is defined as:
~p ≡ m~v = m
d~x
.
dt
(85)
I
In context of the so-called Hamiltonian mechanics it is defined as
∂E x, dx
dt
~p =
.
(86)
∂( dx
dt )
I
For example if
d~x
2
~
∂E
x
,
dt
d~x
1
d~x
1
d~x
E ~x,
= m
+ kx2 ⇒ ~p =
=m
x
dt
2
dt
2
dt
∂( d~
)
dt
(87)
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R EVIEW
S ECTION 8.1: M OMENTUM AND I MPULSE
I
The units of momentum are given by
[Momentum] = [Mass] × [Velocity]
(88)
and for example in SI units
1 N · 1 s = 1 kg · m/s.
I
Alternatively one can write units of momentum as
[Momentum]
I
(89)
=
[Mass] × [Velocity]
=
[Mass] × [Acceleration] × [Time]
=
[Force] × [Time]
The latter form suggests that when some force is applied to a
system for some time then, it may effect the momentum.
(90)
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S ECTION 8.1: M OMENTUM AND I MPULSE
I If we define impulse (of the time-independent net force) as
X
~J =
~Fi ∆t
(91)
i
then from (84) we get
~J = ~pf − ~pi .
I When the net force is time dependent the impulse is defined as
Z t X
f
~J =
~Fi dt
ti
(92)
(93)
i
and from (84) we get
~J =
Z
ti
tf
X
i
~Fi dt =
tf
Z
ti
d~p
dt = ~pf − ~pi
dt
I Note also that if the average force is defined as
R tf P
~
i Fi dt
~Favg ≡ ti
tf − ti
(94)
(95)
then the impulse is just
~J = ~Favg (tf − ti )
(96)
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S ECTION 8.1: M OMENTUM AND I MPULSE
I
Similarly to the energy conservation which is fundamentally
due to time-shift symmetry of physics laws, the momentum
conservation is due to space-shift symmetry.
I
Conservation of energy expresses changes in time
X
~Fi dt = ~pf − ~pi
~J =
(97)
i
I
Conservation of momentum expresses changes in space
X
~Fi · d~l = Kf − Ki .
W=
(98)
i
I
The fact that the two expressions look so much alike might be
surprising at first but this is what led people to eventually
discover a more fundamental and unified conserved quantity
the energy-momentum tensor as well as other conserved
quantities such as electric charge.
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R EVIEW
S ECTION 8.1: M OMENTUM AND I MPULSE
Example 8.1. We can now go back to the example 6.5 where we considered a
race of two iceboats on a frictionless frozen lake. The boats have masses m
and 2m, and the wind exerts the same constant horizontal force ~F on each
boat. The boats start from rest and cross the finish like a distance s away.
Which boat crosses the final line with greater momentum.
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R EVIEW
S ECTION 8.1: M OMENTUM AND I MPULSE
Example 8.2. You throw a ball with a mass of 0.40 kg against a brick wall.
It hits the wall moving horizontally to the left at 30 m/s and rebounds
horizontally to the right at 20 m/s. (a) Find the impulse of the net force on
the ball during its collision with the wall. (b) If the ball is in contact with the
wall for 0.010 s, find the average horizontal force that the wall exerts on the
ball during impact.
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S ECTION 8.2: C ONSERVATION OF M OMENTUM
I
Similarly to the energy conservation which is fundamentally
due to time-shift symmetry of physics laws, the momentum
conservation is due to space-shift symmetry.
I
Conservation of energy expresses changes in time
X
~Fi dt = ~pf − ~pi
~J =
(99)
i
I
Conservation of momentum expresses changes in space
X
~Fi · d~l = Kf − Ki .
W=
(100)
i
I
The fact that the two expressions look so much alike might be
surprising at first but this is what led people to eventually
discover a more fundamental and unified conserved quantity
the energy-momentum tensor as well as other conserved
quantities such as electric charge.
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R EVIEW
S ECTION 8.2: C ONSERVATION OF M OMENTUM
It is useful to distinguish two types of forces:
I
internal forces (forces exerted by objects inside the system)
I
external forces (forces exerted by objects outside of the system)
When there are no external forces the systems is said to be closed or
isolated. For isolated systems one can write down the
impulse-momentum theorem for each object separately
~FA on B
=
~FB on A
=
d~pB
dt
d~pA
dt
(101)
but because of the Newton’s third law
~FA on B = −~FB on A
(102)
d~pB
d~pA
=−
.
dt
dt
(103)
we get
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S ECTION 8.2: C ONSERVATION OF M OMENTUM
If we now define the total momentum of all particles as
~P = ~pA + ~pB
(104)
then we get the law of conservation of total momentum
d~P
= 0.
dt
This is true for an arbitrary collection of particles, i.e.
~P = ~pA + ~pB + ~pC + ... = mA~vA + mB~vB + mC~vC + ...
(105)
(106)
given that there are no external (but only internal) forces.
Note that since (105) is a vector equation we have three equatiions
dPx
dt
dPy
dt
dPz
dt
=
d(mA VAx +mB VBx +mC VCx +...)
dt
=0
=
d(mA VAy +mB VBy +mC VCy +...)
dt
=0
=
d(mA VAz +mB VBz +mC VCz +...)
dt
=0
(107)
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S ECTION 8.2: C ONSERVATION OF M OMENTUM
Example 8.4. A marksman holds a rifle of mass mR = 3.00 kg loosely, so it
can recoil freely. He fires a bullet of mass mB = 5.00 g horizontally with a
velocity relative to the ground of vBx = 300 m/s. What is the recoil velocity
vRx of the rifle? What are the final momentum and kinetic energy of the
bullet and rifle?
R EVIEW
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S ECTION 8.2: C ONSERVATION OF M OMENTUM
Example 8.6. Consider two batting robots on a frictionless surface. Robot
A, with mass 20 kg, initially moves at 2.0 m/s parallel to the x-axis. It
collides with robot B, which has mass 12 kg and is initially at rest. After the
collision, robot A moves at 1.0 m/s in a direction that makes an angle
α = 30◦ with its initial direction. What is the final velocity of robot B.
R EVIEW
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R EVIEW
S ECTION 8.3: I NELASTIC C OLLISIONS
In this section we will discuss instantaneous events - we call
collisions - which suddenly change the kinetic energies of objects.
I
If forces between colliding objects are conservative, then the
total kinetic energy right before and right after collision is the
same and the the collision is called elastic.
∆K = 0
I
(108)
If the forces are non-conservative, then the total kinetic energy is
not conserved and the collision is called inelastic (or completely
inelastic is objects stick together).
∆K 6= 0
(109)
The key point is that although the (kinetic) energy might not be
conserved, the momentum is still conserved for both elastic and
inelastic collisions:
∆~P = 0
(110)
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R EVIEW
S ECTION 8.3: I NELASTIC C OLLISIONS
Consider a completely inelastic two-bodies collision, i.e.
~vA2 = ~vB2 = ~v2
(111)
then conservation of momentum implies
mA~vA1 + mB~vB1 = (mA + mB ) ~v2
(112)
or
mA~vA1 + mB~vB1
.
(mA + mB )
For example if object B was originally at rest, then
mA
v2 =
vA1
(mA + mB )
~v2 =
(113)
(114)
and then after the (completely inelastic) collision
K2
mA
=
<1
K1
(mA + mB )
(115)
or the kinetic energy after the collision is lower than before collision
K2 < K1 .
(116)
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R EVIEW
S ECTION 8.3: I NELASTIC C OLLISIONS
Example 8.8. Consider a ballistic pendulum, a simple system of measuring
speed of a bullet. A bullet of mass mB makes a completely inelastic collision
with a block of wood of mass mW , which is suspended like a pendulum. After
impact, the block swings up to a maximum height y. In terms of y, mB , and
mW , what is the initial speed v1 of the bullet.
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R EVIEW
S ECTION 8.3: I NELASTIC C OLLISIONS
Example 8.9. A 1000 − kg car traveling north at 15 m/s collides with a
2000 − kg truck traveling east at 10 m/s. The occupants, wearing seat belts,
are uninjured, but the two vehicles move away from the impact point as one.
The insurance adjustor asks you to find the velocity of the wreckage just
after the impact. What is your answer?
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S ECTION 8.4: E LASTIC C OLLISIONS
In elastic collisions the total kinetic energy before and after collisions
is unchanged and so one can use both conservation of energy and
momenta
∆K
∆~P
= 0
= 0
(117)
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S ECTION 8.4: E LASTIC C OLLISIONS
Example 8.12. Consider an elastic collision of two pucks (masses
mA = 0.500 kg and mB = 0.300 kg) on a frictionless air-hockey table. Puck
A has an initial velocity of 4.00 m/s in the positive x-direction and a final
velocity of 2.00 m/s in an unknown direction α. Puck B is initially at rest.
Find the final speed vB2 of puck B and the angles α and β.
R EVIEW
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R EVIEW
S ECTION 8.5: C ENTER OF M ASS
The conservation of momentum can be restated in terms of center of
mass defined as the mass-weighted average position of particle
P
mi~ri
m1~r1 + m2~r2 + m3~r3 + ...
~rcm ≡
= Pi
.
m1 + m2 + m3 + ...
i mi
Then the total momentum can be written as
~P = (m1 + m2 + m3 + ...) d~rcm
dt
or simply
~P = M~vcm
where
M=
X
mi
i
is the total mass and
~vcm =
is the velocity of the center of mass.
d~rcm
dt
(118)
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S ECTION 8.5: C ENTER OF M ASS
Example 8.14. James (mJ = 90.0 kg) and Ramon (mR = 60.0 kg) are 20 m
apart on a frozen pond. Midway between them is a mug of their favorite
beverage. They pull on the ends of a light rope stretched between them.
When James has moved 6.0 m towards the mug, how far and in what
direction has Ramon moved?
R EVIEW
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S ECTION 8.5: C ENTER OF M ASS
If there are external forces acting on the objects, then the total
momentum changes as
X
~
~Fext = dP = d (M~vcm ) = M~acm .
dt
dt
Example. Will the center of mass continue on the same parabolic trajectory
even after one of the fragments hits the ground? Why or why not?
R EVIEW
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C HAPTER 0: M ATHEMATICS
I
Algebra
I
I
I
I
Trigonometry
I
I
I
I
I
Solving linear equations (numerically/algebraically)
Solving quadratic equations (numerically/algebraically)
Solving system of two equations with two unknowns
(numerically/algebraically)
Hypotenuse from two sides (numerically/algebraically)
Side from hypotenuse and another side
(numerically/algebraically)
Angle from hypothenuse and one side
(numerically/algebraically)
Angle from two sides, etc. (numerically/algebraically)
Calculus
I
I
I
I
Differentiation of standard functions
Product rule, quotient rule, chain rule.
Indefinite integrals (i.e. antiderivitives)
Definite integrals over interval
R EVIEW
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C HAPTER 5: A PPLYING N EWTON L AWS
I
I
I
I
I
I
Coordinate system
I Choosing 1D, 2D or 3D coordinate system
I Drawing a free-body diagram for each object
Using Newton’s First Law
I Identifying equilibrium or motion without acceleration
I Applying First Law to every object in equilibrium
Using Newton’s Second Law
I Identifying non-equilibrium or motion with acceleration
I Applying Second Law to every object out of equilibrium
Using Newton’s Third Law
I Identifying action-reaction pairs for objects in contact
I Applying Third Law to every action-reaction pair
Frictional Force
I Static and kinetic frictional forces and frictional coefficients
I Relation between normal force and frictional forces
Circular motion
I Relation between radius, velocity and acceleration
I Relation between radius, period and acceleration
R EVIEW
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C HAPTER 6: W ORK AND K INETIC E NERGY
I
Work
I
I
I
Kinetic Energy
I
I
I
Definition of kinetic energy
Applying the work-energy theorem to every object
Varying forces
I
I
I
I
Work done by constant force for straight-line displacement
Average acceleration and instantaneous acceleration
Work done by varying forces and along straight line
Work done by varying forces and along curved line
Applying the work-energy theorem to varying forces
Power
I
I
Definition of average and instantaneous power
Power from velocity of a particle and force acting on it
R EVIEW
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R EVIEW
C HAPTER 7: P OTENTIAL E NERGY
I
Potential Energy
I
I
I
I
I
Conservative and Nonconservative forces
I
I
I
I
Distinguishing conservative and non-conservative forces
Applying energy conservation for only conservative forces
Applying work-energy theorem with nonconservative force
Force and potential energy
I
I
I
Definition of gravitational potential energy
Definition of elastic potential energy
Understanding conservation of total mechanical energy
Applying the Work-energy theorem for every object
Expressing force as a derivative of potential energy in 1D
Expressing force as a gradient of potential energy in 3D
Energy diagrams
I
I
Drawing energy diagrams
Determining stable/unstable equilibriums
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C HAPTER 8: M OMENTUM , I MPULSE , C OLLISIONS
I
Momentum and Impulse
I Definition of momentum
I Definition of impulse
I Applying the impulse-momentum theorem to every object
I
Conservation of momentum
I Conservation of the total momentum of all objects
I Applying the conservation of total momentum
I
Inelastic collisions
I Distinguishing inelastic and elastic collisions
I Applying momentum conservation to inelastic collisions
I
Elastic collisions
I Applying momentum conservation to elastic collisions
I Applying energy conservation to elastic collisions
I
Center of mass
I Definition of the center of mass
I Applying second law for motion of center of mass
R EVIEW