Download BLACKBOARD COURSE PHYSICS 1.2. PHYS 1433

BLACKBOARD COURSE
PHYSICS 1.2. PHYS 1433
4. DYNAMICS: NEWTON’S LAWS OF MOTION.
We are ready to make next step now: to transfer from Kinematics to
Dynamics. Dynamics is the branch of Mechanics that relates the geometrical
characteristics of motion with the causes of motion. The causes of the
motion are forces. The Force in Mechanics is considered as any kind of a
push or a pull on an object. It is a vector quantity. We designate it by the
capital letter F. Several forces may simultaneously act on the same object.
Using the vector addition, we can find total or net force.
Fnet = Σ F = F1+ F2 + F3 + ………..
(4.1)
Where Greek letter Σ means sum.
4.1 Newton’s First Law of Motion.
The Newton’s Laws are the core of dynamics. The First Law is about
situation when no net force acting on an object. From ancient times, it was
supposed that object eventually comes to rest, so the rest is the natural state
of each object. Galileo revised this concept. He introduced the concept of the
frictional force and stated that when there is no this force, the motion with
the constant velocity will be also natural state for each object. This concept
incorporated into
First Newton’s Law of Motion: A body acted on by no net force moves
with constant velocity (which may be zero).
Constant velocity means zero acceleration. Finally the 1st Newton’s Law can
be analytically written as following:
………………If Σ F = 0, then v = const, a = 0……………………(4.2)
It is convenient to write first equation in components:
Σ Fx = 0
Σ Fy = 0
Σ Fz = 0
(4.2a)
The tendency of a body to keep moving once it is in motion (or to be at rest
if it was initially at rest) results from a property that is called inertia.
Because of this the 1st Newton’s Law of motion sometimes is called the Law
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of Inertia. The reference frames in which the law of inertia is valid is called
an Inertial Reference Frames.
Along with a force, a very important in dynamics is the physical quantity
called Mass. Newton treated mass as quantity of matter. Mass actually can
be considered as the measure of the inertia of an object. The greater mass an
object has, the greater force is needed to set this object in motion or to stop
it. For example, a huge truck much more difficult to stop than a small
compact car. Mass is scalar, the unit to measure mass in the SI system is
kilogram.
4.2 Newton’s Second Law of Motion.
We can perform the simple experiments applying to object different forces.
We will see that greater the force, greater the acceleration of an object. If we
apply the same force to the objects with different masses, we will deduce
that greater the mass of an object, smaller the acceleration. Based on the
experiment observation The Newton’s Second Law of Motion can be
formulated as following.
The acceleration of an object is directly proportional to the net force
acting on it, and is inversely proportional to its mass. The direction of
the acceleration is in the direction of the net force acting on the object.
The 2nd Law can be written as following:
a = Σ F/m
(4.3)
or in more wide used form:
Σ F = ma
(4.4)
From the left side of the Eq. (4.4) we have a vector addition. To find it we
use the method of components (see Chapter 3); therefore it is convenient to
write Eq. (4.4) in components. For 3D case we will have:
Σ Fx = max
Σ Fy = may
Σ Fz = maz
(4.5)
Using dimensional analysis we can determine from Eq. (4.4) the SI system
unit for force and introduce physical quantity Force to full extent. It vector
quantity. SI unit of force is called after Isaac Newton newton:
1 newton = 1 N = 1 (kg m)/(s²).
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4.3. Newton’s Third Law of Motion.
Forces are causes of motion, but where do forces come from? They come
from the interaction between objects. Forces always come in pairs. Force
acting on an object is always the result of its interaction with another object.
Newton supposed that two interacting objects must be treated on an equal
basis and formulates his
Newton’s Third Law of motion as following.
The forces with two objects interact with each other are equal in magnitudes
and opposite in directions. These two forces act on different objects.
If object 1 exerts a force on object 2 (an “action” F1→2), then object 2 exerts
a force on object 1 (a “reaction” F2→1). In this case the 3rd Law can be
written as
F1→2 = -- F2→1
(4.6)
Sometimes this law is interpreted as “to every action there is an equal and
opposite reaction”. But we should remember that two forces described in
this law act on different objects.
4.4. Forces in Mechanics.
Analysis of the Newton’s Laws of motion show, that if we know forces
acting on an object and its mass (dynamical physical quantities), we, in
principle, can solve corresponding equations and find acceleration of the
object (kinematical characteristic). Therefore Newton’s Laws of motion
relate causes of motion and resulting motion. But, before we try to use this
approach, we need to specify forces that usually can be applied to the objects
in mechanics.
Force of Gravity.
When object is in free fall (undergoes force of gravity) near the surface of
the Earth, it is moving with the constant acceleration – acceleration due to
gravity g. Using the 2nd Newton’s Law of Motion, we can write for the
magnitude of the gravitational force following expression:
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FG = mg
(4.7)
Direction of the force of gravity is always down. The force of gravity
often is called Weight. In every day life the concepts of weight and mass
are often to get confused. For example, there is serious problem in many
countries, which is called “how to lose weight”. In physics, solution of the
problem “how to lose weight” very simple. Take a look on Eq. (4.7); person
can lose the weight by coming to the places where g is smaller. In New York
City g = 9.803 m/s², but in the vicinity of the equator g = 9.780 m/s², on the
Moon surface g six times smaller than on the Earth surface. Coming to the
places with smaller g, person really will lose weight. But there will be no
satisfaction. Therefore the right title of this complicated problem should be
“how to lose mass”. You should remember, that weight is the force of
gravity, vector quantity directed in space and measured in newtons. Mass,
from other hand, is scalar, amount of matter in an object measured in
kilograms. Considering forces acting on an object, remember that if object
has mass, there is the gravitational force acting on it at the earth surface.
Normal Force.
When can observe a lot of objects at rest that are situated on some surfaces.
We know that gravitational force always acting on the objects. Why these
are objects are at rest but not in the free fall? We can deduce that some other
forces balance their weights. These forces result from reaction forces of
surfaces in contact. The normal (perpendicular to the surface) component of
these contact forces is called normal force FN. There is no formula that
allows us to calculate normal force. It could be found by solving whole
problem. Direction of the normal force is always perpendicular to the
surface on which an object is situated. It is directed up only if the surface is
horizontal (it is not so, for example, on the incline).
Tension Force.
When a cable attached to the object pulls object, we should take into account
the Tension Force FT. Direction of this force -- along the cable. There is no
special formula for the tension force. It could be found by solving whole
problem. The cables in the problems suppose to be with zero mass and
unstretchable. It means that we do not need additional force to accelerate the
cable or to stretch it. The cable just transfers force along itself. We will also
suppose that when a cable connects objects they are moving with the same
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magnitude of acceleration. If there are pulleys in a problem, we will suppose
that they only change the direction of the force, but not its magnitude. When
two objects are attached to each other by cable we should take into account
Newton’s 3rd Law of Motion. The forces with which these objects interact
through cable are equal in magnitude and directed in opposite directions
along the cable.
Friction Force.
When object is moving along the solid surface of other object, the atoms of
these two surfaces come close enough and interatomic interaction between
two surfaces appears. The tangent component of these contact forces is
called Friction Force Ffr. This force could be much smaller than other
forces in the problem in question. In this case, you will be informed by
conditions of problem that surface is frictionless. In many problems, surface
could not be treated as frictionless. The magnitude of the friction force
directly proportional to the normal force and depends on the material and
quality of a surface that characterized by material constant – the coefficient
of friction. If an object is moving, the friction is called kinetic friction. In
this case, the magnitude of the friction force can be found from the
following equation:
Ffr = µ k FN
(4.8)
Where µk is the coefficient of kinetic friction. When object is at rest, the
friction is called the static friction. The magnitude of the friction force in
this case can be estimated from the relationship
Ffr ≤ µ s FN
(4.9)
Where µs is the coefficient of static friction. Usually µ s > µ k. Both of them
are dimensionless because actually they can be treated as a ratio of two
forces.. The direction of the friction force –always opposite with the
respect to direction of motion (in the case of the kinetic friction) or with
respect of the possible motion (in the case of the static friction).
4.5. Solving Problems Related to the Newton’s Laws of Motion.
1. Draw a sketch of the situation as a whole.
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Take a look on the formulas of the Newton’s Laws of Motion (Eq. (4.2) –
Eq. (4.6)). We deal with forces. They are vectors directed in space. We need
to perform vector addition of vectors directed in different directions of the
space. Because of this, it is very important to draw a diagram showing all the
forces acting on each object involved (to get right side of the 1st or 2nd
Newton’s laws). This diagram is the specific feature of the solving process
and it has its own title. It is called
Free—body Diagram or Force Diagram. In this diagram, object should be
shown free from its environment. This is why the diagram is called freebody diagram.
2. Draw a free-body diagram for only one object (at a time). Show all forces
acting on this body. Do not show any forces that this body exerts on other
objects. Why? Take a look on the left side of the formulas of the 1st and 2nd
Laws. Only forces acting on it determine the motion of an object. There is
no any mention of the forces that object in question applies to other objects.
3. If several objects are involved, draw a free—body diagram for each
object.
4. To find a net force, we need to perform a vector addition. We will use the
method of components (see the Chapter 3). Therefore we need to introduce
coordinate system (again for each object separately). You are absolutely free
to choose the direction of a coordinate axes. Usually the most convenient
choice is the choice in which one of the axes is directed along of the
direction of an object’s motion.
5. Resolve all forces acting on an object into components using formulas
(3.2) – (3.3).
6. Try to figure out from the problem’s conditions, what Newton’s Law of
Motion should you use. If it is written in the conditions that object is moving
with constant velocity or it is at rest, you should apply the 1st Law (use
formulas (4.2a)). If you are given acceleration or you are asked to find it,
you should apply 2nd Law (use formulas (4.5)).
7. Write corresponding Newton’s Law equations in components for an
object in question. Specify left side of equations (forces acting on the object
with their signs – positive if the component is in the positive direction of
coordinate axis, otherwise – negative). Any material object experiences the
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gravity. Formula for the force of gravity is (4.7), direction – always down. If
an object situated on the surface, there is normal force, direction –
perpendicular to the surface. If there is a cable (string, rope and so on) by
which object is pulled, there is tension force directed along the cable. If an
object is moving along the surface and there is no mention in conditions that
surface is frictionless, if you are given coefficient of friction or I are asked to
find it, there is the friction force acting upon an object. Formula for the
friction force is given by Eq. (4.8) (if the object at rest you can use for
estimation Eq. (4.9)).
8. Writing left side of equations, remember that if there is no motion along
considered coordinate axis, the component of acceleration along this axis is
zero.
9. Because friction force is determined by the normal force (see Eq. (4.8)). It
is useful to diminish the number of unknowns as following. If the X-axis is
chosen as a direction of motion and right side of Y-component of equation
of motion, we can use this component to express normal forces through
other forces and substitute this expression into definition of friction force
(Eq. (4.8)) and then put it into X-component of equation of motion. As a
result we will diminish the number of equations describing the motion of an
object in question.
10. If there are several bodies in the problem, repeat all steps described
above for each object separately. If cables connect these objects, apply 3rd
Newton’s Law for tension forces acting on the connected objects.
Remember, that if objects connected by cables, they are moving with the
same magnitude of acceleration.
11. Bring all equations for all objects together and solve this system of
equations simultaneously for unknowns.
Actually in this course we will consider usually one or two objects in the
system. Some examples of typical problems are given below.
EXAMPLE 4.1. Apparent weight of an object in an elevator.
The real weight Wr of an object is the gravitational force that acts on it Wr =
mg. The apparent weight of the object Wa is the force object exerts on
whatever it rests on. Think about apparent weight as the reading on a scale
an object is placed on (see Fig. 4-1 on which the object is shown situated on
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the scale and both of them are inside elevator). It is supposed that scale is
situated on the horizontal surface. In this case, Wa = FN. The object’s mass is
m = 100 kg. Find the apparent weight of this object when: (a) elevator is
moving up with constant velocity v = 1 m/s.; (b) elevator is moving up with
the acceleration a = 1 m/s²; (c) elevator is moving down with the
acceleration a = 1 m/s²; (d) elevator is moving down with the acceleration a
= g = 9.8 m/s².
There are only two forces acting on the object: gravity force directed down
and normal force numerically equal to the apparent weight directed up. They
are shown on the Free-Body Diagrams for the cases (a) – (d) in Fig. 4.1. The
object is moving along vertical direction. Therefore it is convenient to
choose Y-axis of coordinate system directed up. There are no forces directed
along X-axis. Because of this we can analyze only Y-component of the
Newton’s second law of motion. In each case we need to find normal force
FN that numerically equal to the apparent weight. It could be equal to the real
weight (case (a)), greater than the real weight (case (b)), or smaller than real
weight (case(c)). In the case (d), apparent weight is zero and object is in the
state of weightless. This situation is used to train astronauts. The plane is
moving along the special trajectory for which all objects inside this plane
experience the state of weightless for some small interval of time.
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Fig.4-1. Example 4.1. Apparent weight of an object in an elevator.
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EXAMPLE 4.2. A 5.0-kg block is at rest on a horizontal surface. The
coefficient of friction between the block and the surface is µk = 0.30.
Constant horizontal force F then acts upon this block and increases its
velocity to 10.0 m/s in 5.0 s. (a) Draw the free-body diagram showing all
forces acting on the block. (b) What is the acceleration of the block? (c)
What is the magnitude of the force? (d) What distance covered by block
during 5.0 s of motion?
In this problem, we have again forces acting along vertical axis, that we
again choose as Y-axis: normal force and gravity force. But there are now
forces acting along horizontal that is convenient to choose as X-axis:
external force acting in positive direction of X-axis and frictional force
always acting in the direction opposite in the direction of a motion. FreeBody Diagram is shown in the Fig.4.2.
External force acting on the object is constant; therefore the acceleration of
the object must be also constant. Acceleration of the object now can be
found by using equation for the motion of an object with constant
acceleration and information given in conditions.
We will use now both X- and Y-components of the Newton’s second law of
motion. It looks as we have two equations and three unknowns: F, Ff, and FN.
But Ff, and FN are related by definition of the frictional force, which we can add
to the system of equations. From Y-component of the second law we can
find FN and substitute it into definition of frictional force. Substituting it into
X-component of Newton’s second law we will get one equation with one
unknown. This allowed us to determine unknown external force F.
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Fig. 4-2. Example 4.2.
EXAMPLE 4.3. Consider the mechanical system shown in Figure, where the
block m1 = 5.00 kg sliding on a flat frictionless surface is connected by a
string over a pulley to the hanging block m2 = 8.00 kg. The system is
released from the rest. (a) Draw the free-body diagram for each block. (b)
Calculate the acceleration of the system. (c) Find the tension in the string.
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There are two objects: object # 1 is situated at the surface; object # 2 is
hanging. They are connected through the cable. Besides gravity forces and
normal force, the tension force FT appears in this case. Important new point
is that we need to prepare now separately two free-body diagrams. Each
object should be considered separately (see Fig. 4.3). It is convenient to
choose positive directions of the coordinate axes along directions of
motions: X-axis -- to the right for object # 1; Y-axis -- down for the object #
2. There is no motion along Y-axis for the object # 1, and along X-axis for
the object # 2. Correspondingly the components of the acceleration vectors
along these axes equal zero for object # 1 and object # 2 respectively. There
is no influence on motion of the object # 1 from forces acting on it along Yaxis. Therefore there is no necessity to consider Y-component of the
Newton’s second law for object # 1. Similar situation is for the Xcomponent of the second law for object # 2. We can reduce number of
unknowns taking into account that because of cable connection, two objects
are moving with the same acceleration: a1 =a2 = a, and because of the
Newton’s third law of motion FT1 = FT2 = FT. Equating FT1 to FT2, we will get
one equation with one unknown. Solving it, we will find the acceleration of
the system of objects a, and then the tension force FT.
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Fig. 4.3. Example 4.3.
EXAMPLE 4.4. Suppose now that, in an Example 4.3, the coefficient of
friction between the block m1 and the surface is µk = 0.400. (a) Draw the
free-body diagram for each block. (b) Calculate the acceleration of the
system. (c) Find the tension in the string.
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Besides gravity forces, normal force, and the tension forces, frictional force
Ff appears in this example. Frictional force depends on the normal force FN.
To decrease numbers of unknown, we can to determine normal force from
the Y-component of the Newton’s second Law of motion for the object # 1
and put it into definition of the frictional force. Then the succession of
actions is the same as in the Example 4.3. It is useful to analyze solution by
comparing it with the Example 4.3. The comparison show that introduction
of the frictional force decreases the acceleration of the system. It looks
reasonable.
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Fig. 4.4. Example 4.4.
5. STATIC EQUILIBRIUM.
We studied two basic branches of the mechanics – the kinematics and
dynamics. Now we will begin to study the third basic branch of the
mechanics that is called Statics. Statics is the determination of the forces
within a structure at equilibrium. A body at rest (or in an uniform motion) is
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said to be in equilibrium. These branch of the Mechanics is very important
for applications in civil engineering, architecture, medicine and so on.
5.1. First Condition of equilibrium.
The object will be in equilibrium with respect of the translational motion
when net force acting on it is zero. Actually at rest velocity of object is zero,
so the acceleration is also zero. Using the second Newton’s Law of motion,
we can get the First Condition of Equilibrium.
ΣF=0
(5-1)
We will use (5-1) in components. In 2 dimensional problems, the first
condition of equilibrium can be written as following:
Σ Fx = 0
(5-1a)
Σ Fy = 0
(5-1b)
These relationships are necessary condition for an object to be in
equilibrium, however they are not always a sufficient condition. Really,
consider two forces with equal magnitudes and opposite directions, but
directed not along the same straight line. The first condition of equilibrium
will be satisfied in this case. Nevertheless the object will not be at rest and
will have a tendency to rotate. Therefore we need additional condition of
equilibrium with respect to a rotation. To introduce additional condition we
need to introduce concept of torque.
5.2. Torque.
It is understandable that to make object rotating about an axis of rotation we
need a force. But it is also important not only the magnitude of this force but
also the perpendicular distance from the axis of rotation to the line along
which a force is acting (line of action of a force). This perpendicular is
called lever arm. The rotating effect of a force also depends on the angle
between the lever arm and a force. We will take all these circumstances into
account if we introduced new physical quantity Torque that plays the same
role for rotational motion as force plays for translational motion. The
magnitude of the torque can be determined as following
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τ= r F sinθ
(5-2)
Where r is the distance from the axis of rotation to the point where force F
applied; F is the magnitude of the applied force; θ is the angle between the
line of action of the force and a line connecting axis of rotation and the point
at which the force is applied (see Fig. 5.1). Usually we will use the following
sign convention: a positive sign is assigned to torques that act to rotate the
object counterclockwise, and a negative sign to torques that act to rotate the
object clockwise. From relationship (5-2) we can deduce that unit of torque
in SI system is m N. There is no special name for this unit.
5.3. The second condition of equilibrium.
Now we can formulate the Second Condition of Equilibrium: The sum of
the torques acting on a body must be zero. In the 2D problems, we usually
suppose that axis of rotation is directed along Z-axis and all forces are lying
in the plane XOY. The all set of equations describing conditions of
equilibrium can be written for this case as following:
Σ Fx = 0
(5.3a)
Σ Fy = 0
(5.3b)
Στ=0
(5.3c)
Sum of X—components and sum of Y-component of all forces acting on an
object and sum of torques of all these forces applied in the XOY plane must
be equal zero. There are some specifics related to the Second condition of
equilibrium. Torque of the force can be calculated with respect to different
axis of rotation (real or virtual). So, with respect to which axis of rotation net
torque must be zero? The answer is: in equilibrium, net torque must be zero
with respect to any axis of rotation. Therefore, solving problems, we can
choose any axis of rotation that makes our calculation easier. For example,
we can choose as axis of rotation point at which unknown force is applied.
In this case, the torque of this force will be zero. The number of unknown in
the set of equations will be decreased by this way. It does not matter is the
chosen axis real or virtual.
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6. CIRCULAR MOTION; GRAVITATION.
Now we will apply the Newton’s law of motion to the consideration of the
circular motion of the objects. The circular motion is widely represented in
the technological base of our civilization. It occurs also in the nature. Very
important example is the motion of planets, including our own planet Earth,
and satellites motion around the Earth and other planets. All these types of
motion could be treated as the simple type of circular motion – the uniform
circular motion.
6.1. Kinematics of the Uniform Circular Motion.
The Uniform Circular Motion is the motion of an object in a circle with
constant speed. Suppose that the radius of the circle is r and the speed of an
object is v, then we can introduced some useful physical quantities that
characterized the uniform circular motion. We introduce Period of rotation
T. Period is the time of one rotation. It can be determined according to its
definition. During time interval T, object, moving with the speed v, cover
distance equals to circumference length 2 π r. Therefore the period can be
determined from the following equation:
T = (2 π r)/v
(6.1)
Unit in which we measure period is second (s). We also introduce physical
quantity Frequency f. Frequency is the number of rotations per unit time. In
SI system the unit of frequency is called hertz (after great German physicist
H. Hertz). 1 hertz = 1 Hz = 1/s. The period and frequency are related by the
equation:
f = 1/T
(6.2)
Now we will consider the basic physical quantity of the uniform circular
motion starting with velocity. We know that the speed of this motion is
constant, but what about velocity? Speed is only the magnitude of velocity.
Speed is the scalar, but velocity is vector quantity, it has the magnitude
(speed) and direction. Magnitude is always constant but direction of velocity
all time changing. Change in velocity causes appearance of the acceleration.
It does not matter what is changing during the motion magnitude of the
velocity (speed) or direction of the velocity or both of them. Anyway the
acceleration during uniform circular motion will not be zero. It could be
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shown that acceleration during the uniform circular motion always is
directed to the center of the rotation. This acceleration is called Centripetal
Acceleration are (from seeking the center in Latin). The formula for the
magnitude of the centripetal acceleration can be written as following:
aR = v²/r
(6.3)
It is not surprising that the aR depends on v and r. The greater the speed v, the
faster the velocity changes direction; the larger the radius , the slowly the
velocity changes direction.
6.2. Dynamics of the Uniform Circular Motion.
Now we will apply the Newton’s Laws of motion for the consideration of
the uniform circular motion. According to the 2nd law, when nonzero net
force is applied appears an acceleration directed in the same direction as the
net force. From this statement, we can deduce that if there is acceleration,
there is a nonzero net force that causes this acceleration appearance. This
force is directed in the same direction as the acceleration. Therefore if there
is the centripetal acceleration directed to the center of rotation, the net force
causes this acceleration directed to the center of rotation. This force is called
centripetal force. According to the 2nd Newton’s law of motion we can write:
ΣFR = m aR
(6.4)
Combining (6.4) and (6.3), we finally get expression of the 2nd Newton’s
law of motion for the Uniform Circular Motion.
ΣFR = m v²/r
(6.5)
Centripetal force is not the new type force of nature. Actually different
forces can that are directed to the center of rotation and keep an object on the
circular path (the orbit) during the circular motion can play the role of the
centripetal force. It could be the tension force, frictional force, component of
the normal force, gravity force. Below we will consider examples of
different forces that keep the object on its path around the center and
participate in the appearance of the centripetal force. Solving corresponding
problems, you should remember that problems are the problems in which
Newton’s laws of motion should be solved. Therefore, we will use the same
strategy that we used before in the Chapter 4: creation of the free body
diagram for an object; writing the corresponding Newton’s laws of motion
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(in the case of the uniform circular motion this is the equation (6.5));
solution of these equations and so on.
EXAMPLE 6.1. Revolving ball (tension force participates in the
appearance of the centripetal force).
A ball with the mass m = 0.300 kg on the end of a string is revolved at
constant speed v = 4.00 m/s in a vertical circle with the radius R = 0.720 m.
Calculate the tension in the string when the ball is (a) at the bottom of its
path, (b) at the top of its path. (c) Determine the minimum speed the ball
must have at the top of its arc so that the ball continues it’s moving in a
circle.
m = 0.300 kg
v = 4.00 m/s
r = 0.720 m
------------------(a) FTb -- ?
(b) FTt --?
(c) vmin--?
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Fig. 6.1 Example 6.1.
(a). Free-body diagram for the object at the bottom of its path is shown in
the bottom of the Fig. 6.1. There two forces acting on the object: gravity
force mg directed down and tension force FRB directed up. We will choose
the Y-axis directed up as positive. Then we will write the Newton’s 2nd
law of motion for this case:
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Σ FR = m are FRB – mg = m v²/r FTb = m (v²/r + g) = 9.61 N
(b). Free-body diagram for the object at the top of its path is shown in the
top of the Fig. 6.1. There two forces acting on the object: gravity force
mg directed down and tension force FTt directed down. We will choose
the Y-axis directed down as positive. Then we will write the Newton’s
2nd law of motion for this case:
Σ FR = m aR FTt + mg = m v²/r FTt = m ( v²/r -- g)
FTt = 3.73 N
(6.6)
(c). The ball continues it’s moving in a circle until the string will remain
taut. It will remain taut as long as there is tension force in it that is until FTt >
0. But if the tension disappears (because v is too small) the ball will fall out
of its circular path. Thus, the minimum speed will occur if FTt = 0.
Substituting FTt the in relationship, we will get
0 = m ( v²min/r -- g) vmin =
gr = 2.66 m/s
EXAMPLE 6.2. Car rounds curved part of the highway (frictional force
as centripetal force). The car (m = 1000 kg) rounds the curved part of a
highway with the radius 50.0 m. What should be the maximum speed at
which frictional force keeps the car at the curved path (a) at the dry
conditions (µs = 0.60) and (b) at the icy condition (µs = 0.25). (c). Is the result
independent of the mass of the car?
m = 1000 kg
r = 50.0 m
(a) µs = 0.60, vmaxdry ?
(b) µs = 0.25, vmaxicy ?
(c) Is the result independent of the m?
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Fig. 6.2. Example 6.2.
The free body diagram for this example is shown in Fig. 6.2. The following
forces acting on the car are the force of gravity mg is directed down, the
normal force FN exerted by the road directed up and there is horizontally
directed static friction force. Car is moving along the road but the static
frictional force Ffs keep the car on the track. This force is directed to the
center of rotation and there is no motion in this direction. We can choose
vertical direction as the positive direction of Y-axis. The direction to the
center of rotation can be chosen as the positive. The Newton’s 2nd low of
motion in components can be written as followings.
Σ Fx = m a
Σ FR = m aR -
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Ffs = m v²max/r
(6-6)
Σ Fy = 0 FN -- mg = 0 FN = mg
(6-7)
To diminish the number of unknowns, we can add the definition of the static
frictional force and get the expression for this force
Ffs = µs FN = µs m g
(6-8)
Substituting (6-8) into (6-6) we will get
m v²/r = mg. Mass is canceled
and we will finally get expression for the maximal speed.
_____
vmax = √ µs r g
(6.9)
But what actually vmax means? If the velocity of a car will be v ≤ vmax, then
the car will move along the highway. If the velocity of the car will be v >
vmax, then sufficient friction force cannot be applied and the car will skid. out
of circular path into a more nearly straight path. Substituting correspondent
values from the condition, we will get:
(a) vmaxdry = 17.1 m/s.
(b) vmaxicy = 11.1 m/s
(c) Mass is absent in the final expression (6.9), so the vmax is the same for
all cars disregarding their mass.
Drive car carefully, pay attention to traffic signs specifying the maximal
speed at the curved parts of the highways and take onto account weather
conditions.
6.3. Gravitation.
The force of gravity attracts all objects with mass on the surface of the Earth.
This force causes all objects to fall with the same acceleration – acceleration
due to gravity g – at the same location on the Earth. We have seen that this
motion (free fall motion) was analyzed by the Galileo. He had no real
explanation of this fact. Newton was also thinking about the problem of
gravity and used in the analysis his laws of motion. Since falling object is
moving with acceleration, there must be the force acting on it (2nd Newton’s
Law). This is gravity force. This force is caused by the interaction with
another object (3rd Law). The force is directed in the same direction as the
acceleration of the free fall – to the center of the Earth. Newton concluded
that it must be the Earth exerts the gravitational force on objects on its
surface. The next step made by Newton was described in the
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Newton’s Apple Story.
This is the real event happened with Newton. He told his friend about what
happened. Once upon a time, Isaac Newton was in a garden and noticed an
apple drop from a tree. As a result of this observation, Newton has been
struck with a sudden inspiration. He made an extremely important
revolutionary conclusion. If gravity acts at the tops of trees, and even at the
tops of mountains, then perhaps it acts all the way to the Moon.
Construction of the Solar System. Geocentric and Heliocentric Models.
Here, we should stress, that all scientist before Newton starting from
Aristotle thought that terrestrial and celestial objects are governed by
different laws of nature. Aristotle stated that Sun and planets are rotating
around the Earth (geocentric model). The celestial objects are moving by
rotation of crystal spheres. The motion of celestial objects is absolutely
perfect, orbits are circular and speed of motion is constant. In this time, there
was other idea in the ancient world. Aristarchus thought that the Earth is
moving around the Sun. But this idea contradicted the observations of
ancient scientists. If the Earth is moving around the Sun, The positions of the
stars measured from the different points of orbit (for example, in the
Summer and in the Winter) must be different. This phenomenon is called the
Stellar Parallax. But, the measurements of star positions made by ancient
scientist show no any measurable difference in the positions of the stars
during different seasons. Stellar parallax does occur, but it is too small to
detect with the unaided eye. (Actually the possibility to measure stellar
parallax was achieved astronomers only in XIX century). For this reason,
and because of the dominant influence of Aristotle ideas, Aristarchus model
was not accepted. The ancient Greek scientist developed further Aristotle
geocentric model (we can mention Hipparchus and especially Ptolemy). The
basic aspects of a geocentric model explained most of the motions of the
celestial objects by various geometric devices. The model was strong enough
to survive for a long time – more than a thousand years. In new times,
Copernicus introduced his heliocentric model in which Earth and other
planets are rotating around the Sun. He supposed that orbits of the planets
are circles and that the planets are moving along these circles with constant
speed. We know now that heliocentric model is right model, however
Copernicus predictions came out no better than those based on the Ptolemaic
geocentric model. Why?
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The explanations were found when German scientist Johannes Kepler
processed experimental date about planetary motion of planets (especially
the planet Mars) collected by Danish astronomer Tycho Brahe. Johannes
Kepler derived from these data three empirical Laws.
Kepler’s Laws of Planetary Motion.
1st Law. The path of each planet about the Sun is an ellipse.
2nd Law. Each planets moves so that an imaginary line drawn from the Sun
to the planet sweeps out equal areas in equal periods of time.
3rd Law. The ratio of the square of the period of planet rotation around the
Sun to the cube of its mean distance from the Sun is the same for all planets.
We can only imagine the titanic work done by Kepler to derive these results
from the raw experimental data fixing positions of the planets.
It is clearly seen from these laws that orbits of planets are not circle, but
ellipses (1st Law), and that planets speed of motion along their orbits is not
constant (2nd Law). This explained why the prediction of Copernicus
heliocentric model was not precise. Nobody could explain the Kepler’s Law
of Planetary Motion. This work was done by Newton.
6.4. Newton’s Law of Universal Gravitation.
When Newton came to the revolutionary idea that the motion of the celestial
object – the Moon is governed by the same law s as the terrestrial objects, he
decided to apply to the Moon motion his laws of motion and compare the
gravitational force acting on the Moon with the gravitational force acting on
the object at the surface of the Earth. According to the 2nd law the force is
proportional to the acceleration. Using astronomical data, Newton calculated
the centripetal acceleration of the Moon. It turned out that acceleration of the
Moon is 3600 = 60² times smaller than acceleration due to gravity at the
Earth surface. But the Moon is 60 times further from the center of the Earth
than the object at the Earth surface. But, according to the 2nd Newton’s Law
of motion, the force is proportional to the acceleration. Therefore Newton
concluded that the gravitational force exerted by the Earth on any object
decreases with the square of its distance r from the Earth’s center. It is
understandable that the force of gravity depends not only on distance but
also on the object’s mass. According to Newton’s third law, when the Earth
exerts its gravitational force on the Moon, the Moon exerts an equal and
opposite force on the Earth. Because of this symmetry, the magnitude of the
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force of gravity must be proportional to both of the masses. Thus, Newton
proposed the Newton’s Law of Universal Gravitation:
Every particle in the universe attracts every other particle with a force
that is proportional to the product of their masses and inversely
proportional to the square of the distance between them. This force acts
along the line joining the two particles.
F = G (m1 m2) / r²
(6-10)
Where m1 and m2 are the masses of the two particles, r is the distance between
them, and G is a universal constant unknown in the time of Newton. It must
be measured experimentally and has the same numerical value for all objects
in the Universe. Strictly speaking the formula (6-10) could be applied to a
point-like objects. When extended objects are small compared to the
distance between them (as for the Earth – Sun system), little inaccuracy
results from considering them as point-like particles. But how we can apply
formula (6-10) to the case when objects could not be treated as point-like
objects (for example the object at the surface of the Earth)? Newton showed
for two uniform spheres, relationship (6-10) gives the correct force where r
is the distance between their centers.
Force of gravity is comparatively small. For example, the electric force of
interaction between two electron is 10^(42) times smaller than force of force
of the gravitational interaction between them. In 1798, over 100 years after
Newton published his law, English physicist Henry Cavendish performed
fine experiment that is treated as the landmark in the experimental Physics
and confirmed The Newton’s law of universal gravitation. Cavendish
measured F, m1, m2, and r for two lead balls and determine the numerical
value of the Universal Gravitational Constant. Its accepted value today is
G = 6.67 10^(-11) (N m²)/(kg ²)
(6-11)
These results allowed to Cavendish to estimate the mass of the Earth. Let us
consider the gravity force acting on the object with mass m at the surface of
the Earth. We know that it could be written as Fg = mg. But, according to the
Newton’s Law of Universal gravitation (6-10), the same force can be written
as following FG = F = G (m mE) / (RE)², where mE is the mass of the Earth and
RE is the radius of the Earth. Because Fg = FG, we can write
27
mg = G (m mE) / (RE)²
(6-12)
Mass of the object is gone, and we can derive from relationship (6-12)
expression for the mass of the Earth:
ME = g (RE)²/G
(6-13)
Cavendish used this expression, value of G and calculated the mass of the
Earth. He published his results in the article with the title “I weighed the
Earth”. It allows to calculate the average density of the Earth ρE = mE/VE,
where VE It turned out that ρE is much greater than the density of the rock. It
was the hint that there is heavy iron core at the center of the Earth.
From equation (6-12) we can derive other important relationship:
g = G mE / (RE)²
(6-12)
We can understand now the mystery faced during study of Free Fall motion.
All objects are falling with the same acceleration (acceleration due to
gravity) at the same point of the Earth surface. Acceleration due to gravity g
on the surface of the earth has a little different value at different locations
because the Earth is not a perfect sphere. The value of G can vary locally
because of the presents of irregularities and rocks of different densities.
Geophysicists use the precise measurement of g as part of their
investigations into the structure of the Earth’s crust, and in mineral and oil
exploration.
The equation (6-12) can be used to determine acceleration due to gravity on
different celestial objects like planets and their moons. For example, if we
put in (6-12) astronomical data for the Moon, we will see that the
acceleration to the gravity at the Moon surface 6 times smaller than on the
Earth. Now we can understand why American astronauts were able to make
so high jumps at the surface of the Moon.
The Satellite Motion.
We apply the Newton’s Law of Universal gravitation to the satellite motion.
It could be natural satellites (for example, Earth’s natural satellite the Moon)
or artificial satellite. The possibility to launch artificial satellites was
discussed by Newton. If it will be launch with high speed it can orbit the
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Earth. But what about gravity force and Free Fall? In fact, a satellite is
falling (accelerating toward the Earth), but its high tangential speed keeps it
on the orbit. To describe the satellite motion we will use Newton’s 2nd Law
of motion. The force at the left side of the equation is the gravitation force
(6-10) acting on the satellite from the Earth. For simplicity, we consider the
satellite that is moving along the circular orbit with constant speed (uniform
circular motion).
Σ FR = m aR G (m mE) / (r)² = m v²sat/r
(6-13))
Where m is the mass of the satellite and vsat is its speed, r is the radius of the
orbit. From equation (6-13) we can derive equation for the satellite speed:
______
vsat = √ G mE/r
(6-14)
Analysis of the equation (6-14) shows that that we can not choose the radius
r of the orbit and the speed of the satellite vsat independently. If we launch the
satellite with the speed vsat, then the radius of the orbit r is determined. The
motion of the satellite does not depend on its mass, because it does not
appear in the equation (6-14).
EXAMPLE 6.3. The Hubble space telescope is orbiting at a height h of 596
km above the Earth surface and has a mass 11 000 kg. Determine: (a) the
speed of the Hubble space telescope; (b) the period of its revolution around
the Earth. We need some astronomical data: the mass of the Earth is mE =
5.98 10^(24) kg; the radius of the Earth is RE = 6.38 10^6 m.
h = 596 km = 5.96 10^5 m
mE = 5.98 10^(24) kg
RE = 6.38 10^6 m
G = 6.67 10^(-11) (N m²)/(kg ²)
___________________________
(a) vsat ?
(b) T?
(a) Before using expression (6-14) we should take into account that the
radius of the orbit, the height of the satellite, and the radius of the
Earth are related:
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r = h + RE
Then, substituting r into (6.14) we will get
___________
vsat = √ G mE/(RE + h) = 7.56 10^3 m/s
(6-15)
(6-16)
(b) We can use relationship (6-14) for determination the rotation period
(time of one rotation) of the satellite combining (6-14) with (6.1).
_____
T = (2 π r)/v = (2 π r³′²)/√ G mE = 5800 s = 97 min = 1.61 h
(6-17)
Gravity and Construction of the Universe.
The introduction of the Law of gravitation (6.10) was not the explanation of
the planetary motion yet. The Robert Hook the rival of Isaac Newton stated
that he knew inverse square law for force of gravity before Newton. But
Newton not only introduced the law of gravitation (6.10). He showed that
this Law allows explaining Kepler’s Laws of planetary motion. To proof this
he invented calculus. Therefore he was not only great physicist but also the
great mathematician. (Many scientists believe that Isaac Newton was the
greatest scientist ever). We could not demonstrated here how to get 1st
Kepler’s Law from the Law of universal gravitation, but we will do this for
3rd Kepler’s Law. We will modify expression (6.17) for the case of the solar
system. The period of rotation of the planet (natural satellite of the Sun) Tp
can be found from the relationship
_____
Tp = = (2 π r³′²)/√ G mS
(6-18)
Where mS is the mass of the Sun.
T² / r³ = (4 π²) / (G mE)
(6.18)
At the right side of the equation (6.18) are the same for all the planets of the
solar system as was predicted by 3rd Kepler’s Law of planetary motion.
The Newton’s Law of Universal Gravitation explained all known at those
time astronomical phenomena data. Mankind understood the construction of
the universe. It was a great achievement of science. Later some discrepancy
was found. The orbit of more distant from the Sun planet Uranus deviated
from orbit predicted by the Newton’s Law of Universal Gravitation. The
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English astronomer Adams and then French astronomer Leverrier explained
these discrepancies by gravitational influence of unknown planet and
predicted the position of this plant. German astronomer Galle really found
this planet. It was called Neptune. A huge blue planet was discovered as said
at the tip of the pen. It was a triumph of Newtonian mechanics. A little later,
in 1865, the new astronomical data was collected that needed explanation:
the orbit of the Mercury, the planet closest to the Sun, changes its position in
space with time. It is extremely small but measurable change. But, there was
any explanation of this phenomenon until 1916 when Albert Einstein
introduced its General Theory of Relativity. In this theory, a new concept of
gravity was suggested. According to Einstein, matter bends or curves space
and thus controls the behavior of nearby bodies. Now we could not
understand new phenomena discovered in the universe like black holes,
neutron stars and so on without the General Relativity. This theory is both
conceptually and mathematically difficult and far beyond the level of this
course. However, it should be noted that in weak gravitational fields,
Einstein’s theory reduces to Newton’s, so that everything that we have done
so far, remains correct. It is only in strong gravitational fields, such as close
to the Sun or other stars, that the differences become important.
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