Download REVIEW: Rotational Equilibrium (Chapter 11)

Lecture 15: Elasticity (Chapter 11) and Universal Gravity (Chapter 12)
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REVIEW: Rotational Equilibrium (Chapter 11)
With the use of torques one can solve problems in rotational equilibrium.
Rotational equilibrium is the rigid body equivalent of the equilibrium of a particle
for which the net external force is 0.
Rotational equilibrium of a rigid body is summarized by the following statement:
The net clockwise torque equals the net counterclockwise torque
Statement of Ladder Problem, pages 360–361 (somewhat modified)
Sir Lancelot, a knight of the Round Table who weighs 800 N, is standing 1/3 the
way up on a ladder which is 5 meters long. The ladder is leaning against a wall
of the Camelot Castle, at an angle of 53.1o above the horizontal. The ladder,
the wall, and the length along the ground constitute a 3–4–5 right triangle.
The ladder itself weighs 180 N. The wall is assumed to be frictionless (much
harder to climb up the wall using ropes that way), but there is static friction
between the ground and the feet on the ladder. Exactly why Lancelot is ascending the ladder is not known, but it is rumored that Queen Guinevere’s chambers
are on this side of the castle, and King Arthur is off chasing foxes.
Answer the following purely physics questions.
1) What is the normal force of the castle wall on the ladder?
2) What is the force of the ground on the feet of the ladder?
3) What is the force of kinetic friction exerted by the ground on the feet of
the ladder?
4) Suppose the coefficient of static friction between the ladder and the ground
is µs = 0.35. Can Lancelot climb all the way to the top of the ladder, where
there is a window into the castle?
Lecture 15: Elasticity (Chapter 11) and Universal Gravity (Chapter 12)
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Worked Ladder Problem in Rotational Equilibrium
Method of Solution
Again this is an equilibrium problem. Nothing is being accelerated. So the net
force must be zero in the vertical and the horizontal directions. However, there
is more than that. There is no rotation either. So the clockwise torque must be
equal to the counter clockwise torque. Essentially we have three equations: one
for the vertical forces, one for the horizontal forces, and one for the torques. We
have three unknowns:
1) The horizontal force of the ground on the ladder. This force comes about
because of static friction. If there was ice on the ground, such that the
static friction force was very small, then it would be foolish for Lancelot to
try to climb the ladder.
2) The vertical force of the ground on the ladder. This is a normal force since
it is perpendicular to the ground.
3) The horizontal force of the wall on the ladder. This is also a normal force
since it is perpendicular to the wall. There is no friction between the wall
and the ladder, meaning that there can be no vertical force exerted on the
ladder by the wall.
Lecture 15: Elasticity (Chapter 11) and Universal Gravity (Chapter 12)
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Worked Ladder Problem in Rotational Equilibrium
Solution
We have three equations to work with: 1) zero net force in the horizontal direction, 2) zero net force in the vertical direction, and 3) the sum of the torques
about any convenient axis is zero.
X
Fx = 0 = R cos θ − F = fs − F =⇒ fs = F
The angle between the force from the ground and the horizontal is θ. The
horizontal component of that force is the force of static friction: fs = R cos θ.
From the above we see that the normal force of the wall F is equal in magnitude
to the force of the static friction. Even better, the force of static friction must
be equal to the horizontal force of the wall. For the vertical force equality:
X
Fy = 0 = R sin θ−WL −WK = N −WL −WK =⇒ N = WL +WK = 980 Newtons
We use WL for the weight of the ladder, and WK for the weight of Knight
Lancelot. The vertical component of the force from the ground is the normal
force of the ground on the ladder. The normal force N determines the the
maximum of the static friction force, and is just the sum of the two weights.
Next, we can compute the torques about the bottom of the ladder 1 .
X
5
τ = F ∗5 sin(53.1)−WL ∗2.5 cos(53.1)−WK ∗( ) cos(53.1) =⇒ F = 268 Newtons
3
=⇒ fs = F = 268 Newtons
We now have the vertical component of R (=980 N) and the horizontal component of R (=268 N). From these two components we can work out that the
magnitude R = 1020 N, and that the angle θR = 75o above the horizontal. We
can also verify that the static friction force required (=268 N) is less than the
maximum possible static friction force
fsmax = µs N = 0.35 ∗ 980 = 343 Newtons
Lastly, if Lancelot goes to the top of the ladder, the first two equations are the
same but the torque equation changes to have 5 meters instead of (5/3) meters
as the lever arm for Lancelot’s weight. That makes F = 667 N, which is bigger
than 343 N. The ladder will slip before Lancelot gets to the top. This is why
climbing ladders can be dangerous. Just because it is safe climbing part way up
the ladder doesn’t mean it is safe all the way up the ladder.
1
The torque equation below has been corrected from the mistaken version show in class which left out the
trig factors.
Lecture 15: Elasticity (Chapter 11) and Universal Gravity (Chapter 12)
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Chapter 11: Deformations of Non-Rigid Bodies
Real World Behavior of Objects Subjected to Outside Forces
Real world objects are not perfectly rigid. If force is applied to these objects,
then there will usually be some small change in the dimensions of the objects.
Just as in the spring case, if the force is small then the change in dimension will
reverse itself when the force is removed. The change in dimension will also scale
linearly with the size of the applied force
Linear Stretching of a Real Object
The simplest change is dimension is an increase in length of an object, say a
string or a rod, when the object has a tension force applied at both ends. Lest
us symbolize the initial length of the object when there is no applied force as l 0 .
Also, assume that the object has a cross-sectional area A. Then a tension T is
applied at both ends of the object. The object increases its length to l 0 + ∆l.
We can then define two physical quantities which are related by one equation.
Tensile Stress ≡
T
( in units of N/m2 )
A
Since the units ratio N/m2 occurs often in physics (and chemistry) we give it
a special name: 1 N/m2 ≡ 1 Pascal, and Pa is the two-letter abbreviation for
Pascal2 . The second new definition is
Tensile Strain ≡
∆l
( dimensionless ratio)
l0
Make sure that you do not confuse these two quantities which used similarly
in everyday language but are very different in physics language. As long as
the strain is not too large, the ratio of stress to strain for a given material is
a constant called the elastic modulus of the material. For a linear stretching
the name of that elastic modulus is the Young’s Modulus symbolized by the
letter Y .
Stress
(in units of Pa)
Y =
Strain
A table of Young’s Modulus numbers for different materials is given on page 365
in the text. For most materials, the Young’s Modulus number for compression
is the same as the Young’s Modulus number for stretching.
2
Blaise Pascal was a noted mathematician and physicist in the 17th century France who contributed to the
science of hydrodynamics which we will study in two weeks.
Lecture 15: Elasticity (Chapter 11) and Universal Gravity (Chapter 12)
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Chapter 11: Deformations of Solid Volumes and Liquids
Compression of a Volume
Objects can also be compressed if the object is subjected to an increase in
external pressure. You are all aware that the Earth’s atmosphere is exerting
pressure. Air pressure is simply the weight of a column of air on a unit area, the
column going from the Earth’s surface to an arbitrarily high altitude where the
atmosphere has essentially disappeared.
Since weight is measured in Newtons, and area in m2 , then pressure is measured
in N/m2 or Pascals. Normal air pressure at the surface of the earth has a value
of 1.013 × 105 Pa, which is also called the one atmosphere unit of pressure.
Pressure is exerted uniformly on all surfaces. Pressure is not a vector. Pressure
has a magnitude and units, but no direction. You can get the force exerted on
a given area by a certain pressure by multiplying the pressure value by the size
of the area.
When an object of initial volume V0 is subjected to a uniform increase in external
pressure ∆p, then the object will be compressed, meaning that its volume will
be changed by an amount ∆V . The ∆V will be negative. We can define the
bulk strain and bulk stress as
Bulk strain ≡
∆V
(dimensionless)
V0
Bulk stress ≡ ∆p (in units of Pa)
As with stretching, the ratio of bulk stress to bulk strain will be a constant for
a specific material. That constant is called the Bulk Compression Modulus and
is given the symbol B
B ≡ ∆p/
∆V
(in units of Pa)
V0
Again, values of B for specific solid materials are given in the table on page 365.
For liquids, it is conventional to use the inverse of the Bulk Compression Modulus
which is called the Compressibility. The compressibility numbers for common
liquids are given in the separate table on page 367
Lecture 15: Elasticity (Chapter 11) and Universal Gravity (Chapter 12)
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Chapter 11: Shear Deformations
Shear of a Volume
Solid volumes, but not liquid volumes, can also be subjected to shear. A shear
is provided by opposite directly force acting to twist or rotate an object out of
its original shape. You have probability heard of the term wind shear which can
occur in violent storms.
A shear stress is best illustrated by a figure such as Fig. 11.17 on page 367,
or at the class web site. Just as with tensile stress and tensile strain, we can
define shear stress and shear strain in terms of the applied forces and the relative
deformation. Again, for small amounts of strain, the ratio of stress to strain will
be a constant for a given material. That constant is called the Shear Modulus
with symbol S
In terms of the given figures
(F/A)
S≡
(∆x/h)
Values for S for common materials are also given in the table on page 365.
Lecture 15: Elasticity (Chapter 11) and Universal Gravity (Chapter 12)
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Chapter 12: Newton’s Law of Gravity
Newton deduced his Law of Universal Gravity based on the astronomical observations made in the previous two centuries.
Law of Universal Gravitation
m1 m2
G = 6.674 ± 0.001 × 10−11 N–m2 /kg2
2
r
This force has infinite range, and also an inverse square dependence.
FG = G
Gravity acceleration as a function of altitude
The g constant of gravitational acceleration has a value of 9.8 m/s2 on the Earth’s
surface. However, g will get smaller with altitude. Suppose we are at a distance
h above the Earth’s surface. What will the value of g be then?
We first solve for g at the Earth’s surface. In that case the r in Newton’s Law
of Universal Gravitation is RE . We take the weight force of a mass m. On the
Earth’s surface we have
w(r = RE ) = mg(r = RE ) = G
mME
2
RE
ME
2
RE
Now consider an altitude h such that r = RE + h. We now have
g(r = RE ) = G
w(r = RE + h) = mg(r = RE + h) = G
mME
(RE + h)2
2
ME
RE
g(r = RE + h) = G
=g
(RE + h)2
(RE + h)2
So if we put in a value h = 2RE , an altitude of twice the Earth’s radius we get
2
RE
g
g(r = 3RE ) = g
=
(3RE )2
9
So an object weighing 270 N on the Earth’s surface will weigh only 30 N when
placed at an altitude h = 2RE twice the Earth’s radius.
Lecture 15: Elasticity (Chapter 11) and Universal Gravity (Chapter 12)
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Chapter 12: Newton’s Law of Gravity
We showed in the previous page that the acceleration due to gravity varies as
the distance above the Earth’s surface
2
RE
g(r = RE + h) = g
(RE + h)2
We can now apply this formula to the acceleration that the Moon feels because
of the Earth’s gravity. According to the above formula we need the EarthMoon distance to get h, and that is some 3.84 × 108 m. The Earth’s radius is
RE = 6.37 × 106 m. So the value of g at the Moon’s distance is
(6.37 × 106 )2
2
−3
g(r = 3.84 × 10 ) = g
=
2.75
×
10
m/s
(3.84 × 108 )2
8
What Newton did was to compare this with the centripetal acceleration which
was being felt by the Moon. This centripetal acceleration could be calculated
independently, knowing that the Moon had an orbital period T of 27.32 days.
This time T established the speed vM of the Moon
2πrM
T
established the centripetal acceleration:
vM =
In turn vM
2
4π 2 rm
vM
= 2.72 × 10−3 m/s2
aM =
=
2
rM
T
These two values, aM and g(r = rM ) were just about equal, and this was the
first great proof of Newton’s Law of Universal Gravity.
Lecture 15: Elasticity (Chapter 11) and Universal Gravity (Chapter 12)
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Kepler’s Three Laws
Before Newton formulated his law of universal gravitation, the astronomer Kepler had worked out three laws of planetary motion. As far as Kepler was
concerned, these were three independent laws. However, Newton showed that
all three of Kepler’s Laws could be derived from Newton’s one law of gravity. So
Newton accomplished the first “unification” in physics.
Kepler’s First Law
Each planet moves in an elliptical orbit with the Sun at one focus
Actually, all the planets except Pluto, have very near circular orbits. The elliptical orbit can be shown to be the general solution to Newton’s gravity law with
an inverse square dependence.
Kepler’s Second Law
The radius vector from the Sun to a planet sweeps out equal areas in equal
amounts of time.
This is know as the “equal areas” law. In fact, it is simply a consequence of the
fact that the gravity force is a central force. It exerts no torque on the planet.
Then it can be shown that the areas swept out are just proportional to the
angular momentum which is conserved. The equal areas law does not depend
on the inverse square dependence.
Kepler’s Third Law
The square of the period T of a planet’s orbit is proportional to the cube of its
semi-major axis.
This was originally thought to be some magical property of the planets (“harmony of the spheres” was the phrase used in Kepler’s time) but actually it’s an
extremely simple consequence of the inverse square force of gravity law.
Lecture 15: Elasticity (Chapter 11) and Universal Gravity (Chapter 12)
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Kepler’s Third Law Proved
It is very easy to prove Kepler’s Third Law for circular orbits
The square of the period T of a planet’s orbit is proportional to the cube of its
radius R.
We take the mass of the Sun to be MS and the mass of the planet to be mp .
The orbital radius is taken as R. Newton’s gravity law states
MS mp
R2
However, this gravity force is also the centripetal force, just as we saw for the
moon. And centripetal forces have simple expressions in terms of the speeds:
F =G
mp v 2
F =
R
Now the speed v is just the circumference of the orbit divided by the time of the
orbit
2πR
T
So equate everything concerning the force:
v=
MS mp
mp v 2
mp 4π 2 R2
F =G
=
=
R2
R
T2
The mp term cancels out and we are left with
4π 2 R2
MS
G 2 =
R
T2
4π 2 R3
T2 =
= KS R 3
GMS
4π 2
KS ≡
GMS
For planetary orbits the cube of the Radius
Is proportional to the square of the Period
The square of the time is proportional to the cube of the radius. Moreover, once
you have found out what G value is, then you can determine from KS what the
mass of the Sun is.
Lecture 15: Elasticity (Chapter 11) and Universal Gravity (Chapter 12)
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Energy in Planetary Orbits
All planets have negative mechanical energy?
Believe it or not, all planets have negative total mechanical energy !
How can this be? It surely can’t be because of the Kinetic Energy:
1
K = mp v 2
2
which has to be positive because all planets have a non-zero speed.
Thus it must the the “fault” of the potential energy. For the Universal Gravity
force, we can derive from the integral definition that the potential energy is given
by the formula
U ≡−
Z
F (r)dr =⇒ UG (R) = −G
mp MS
R
Recall that the absolute value, or zero reference, of potential energy is arbitrary.
Conventionally, we say the gravity potential energy is zero when two masses are
infinitely far apart. In the above formula, mp is the mass of the planet, R is the
orbital radius of the planet (distance from the Sun), and MS is the mass of the
Sun.
Lecture 15: Elasticity (Chapter 11) and Universal Gravity (Chapter 12)
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Energy in Planetary Orbits
The Escape Speed from the Earth’s Surface
For a given initial speed vi at the Earth’s surface where ri = RE we have the
total energy conserved at any later distance r:
1
mME
mME
1
= mv 2 − G
= E( arbitrary distance r)
E = Ei = mvi2 − G
2
RE
2
r
To get the maximum distance rmax we set the “final” speed equal to 0. This
then gives us the expression for rmax
1
1
1
mME
( mvi2 − G
)
rmax
GmME 2
RE
Now suppose we want the mass to travel infinitely far away for the Earth, so
then 1/rmax = 0. Then the expression above in parenthesis must also be zero:
=
v
u
u 2GME
t
1
mME
( mvi2 − G
) = 0 =⇒ vi (rmax = ∞) ≡ vesc =
2
RE
RE
Objects with a speed vesc at the Earth’s surface will travel to infinity and never
return. Obviously, you can substitute for any other planet or even the Sun to
get their escape speed. The concept of escape speed is critical to understand
why the moon has no atmosphere, why the Earth has no hydrogen or helium
in its atmosphere, and why the outer heavy planets all contain hydrogen and
helium as their most abundant elements.
Lecture 15: Elasticity (Chapter 11) and Universal Gravity (Chapter 12)
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The Ultimate in Escape Speed: The BLACK HOLE
According to relativity, the fastest speed anything can have is the speed of
light c. Nothing, not even light, can go faster then c. So what happens when
the escape speed becomes bigger than the speed of light? What you have then is
something called a Black Hole ! This is an object so massive, and so compact,
that nothing can escape from it not even light itself. So you can’t see it, but it’s
there. And if you get to close to it, then you will never escape yourself.
Do Black Holes exists? Most physicists think so. It is believe that Stars with
a mass several times that of the Sun will eventually explode into a ”Super-nova”
and leave behind an incredibly dense core. For example, the core might be the
mass of the Earth and the size of a dime. The only evidence for a Black Hole
would be if another object is too close, like a binary star companion, and one
sees the companion gradually consumed by the Black Hole.