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Chapter 1 Mathematical Prelude
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Calculus-Based Physics I
by Jeffrey W. Schnick Copyright 2005-2008, Jeffrey W. Schnick, Creative Commons Attribution Share-Alike License 3.0. You can copy, modify, and re- release this work under the same license provided you give attribution to the author. See http://creativecommons.org/
- 1 Mathematical Prelude.............................................................................................................. This book is dedicated to Marie, Sara, and Natalie.
- 2 Conservation of Mechanical Energy I: Kinetic Energy & Gravitational Potential Energy ......
- 3 Conservation of Mechanical Energy II: Springs, Rotational Kinetic Energy ..........................
- 4 Conservation of Momentum..................................................................................................
- 5 Conservation of Angular Momentum ....................................................................................
- 6 One-Dimensional Motion (Motion Along a Line): Definitions and Mathematics ...................
- 7 One-Dimensional Motion: The Constant Acceleration Equations ..........................................
- 8 One-Dimensional Motion: Collision Type II .........................................................................
- 9 One-Dimensional Motion Graphs..........................................................................................
- 10 Constant Acceleration Problems in Two Dimensions ..........................................................
- 11 Relative Velocity.................................................................................................................
- 12 Gravitational Force Near the Surface of the Earth, First Brush with Newton’s 2nd Law........
- 13 Freefall, a.k.a. Projectile Motion .........................................................................................
- 14 Newton’s Laws #1: Using Free Body Diagrams ..................................................................
- 15 Newton’s Laws #2: Kinds of Forces, Creating Free Body Diagrams....................................
- 16 Newton’s Laws #3: Components, Friction, Ramps, Pulleys, and Strings..............................
- 17 The Universal Law of Gravitation .....................................................................................
- 18 Circular Motion: Centripetal Acceleration.........................................................................
- 19 Rotational Motion Variables, Tangential Acceleration, Constant Angular Acceleration.....
- 20 Torque & Circular Motion.................................................................................................
- 21 Vectors: The Cross Product & Torque...............................................................................
- 22 Center of Mass, Moment of Inertia ....................................................................................
- 23 Statics ...............................................................................................................................
- 24 Work and Energy ..............................................................................................................
- 25 Potential Energy, Conservation of Energy, Power .............................................................
- 26 Impulse and Momentum....................................................................................................
- 27 Oscillations: Introduction, Mass on a Spring .....................................................................
- 28 Oscillations: The Simple Pendulum, Energy in Simple Harmonic Motion .........................
- 29 Waves: Characteristics, Types, Energy..............................................................................
- 30 Wave Function, Interference, Standing Waves ..................................................................
- 31 Strings, Air Columns.........................................................................................................
- 32 Beats, The Doppler Effect .................................................................................................
- 33 Fluids: Pressure, Density, Archimedes’ Principle ..............................................................
- 34 Pascal’s Principle, the Continuity Equation, and Bernoulli’s Principle...............................
- 35 Temperature, Internal Energy, Heat, and Specific Heat Capacity.......................................
- 36 Heat: Phase Changes .........................................................................................................
- 37 The First Law of Thermodynamics....................................................................................
Chapter 1 Mathematical Prelude
1 Mathematical Prelude
Just below the title of each chapter is a tip on what I perceive to be the most common mistake made by students in applying material from the chapter. I include these tips so that you can avoid making the mistakes. Here’s the first one:
The reciprocal of x y
- (^) is not x + y. Try it in the case of some simple numbers.
Suppose x=2 and y=4. Then 4
x y
and the reciprocal
of 4
is 3
which is clearly not 6 (which is what you obtain if you take the
reciprocal of 4
- to be 2+4). So what is the reciprocal of x y
+? The
reciprocal of x y
x y
This book is a physics book, not a mathematics book. One of your goals in taking a physics course is to become more proficient at solving physics problems, both conceptual problems involving little to no math, and problems involving some mathematics. In a typical physics problem you are given a description about something that is taking place in the universe and you are supposed to figure out and write something very specific about what happens as a result of what is taking place. More importantly, you are supposed to communicate clearly, completely, and effectively, how, based on the description and basic principles of physics, you arrived at your conclusion. To solve a typical physics problem you have to: (1) form a picture based on the given description, quite often a moving picture, in your mind, (2) concoct an appropriate mathematical problem based on the picture, (3) solve the mathematical problem, and (4) interpret the solution of the mathematical problem. The physics occurs in steps 1, 2, and 4. The mathematics occurs in step 3. It only represents about 25% of the solution to a typical physics problem.
You might well wonder why we start off a physics book with a chapter on mathematics. The thing is, the mathematics covered in this chapter is mathematics you are supposed to already know. The problem is that you might be a little bit rusty with it. We don’t want that rust to get in the way of your learning of the physics. So, we try to knock the rust off of the mathematics that you are supposed to already know, so that you can concentrate on the physics.
As much as we emphasize that this is a physics course rather than a mathematics course, there is no doubt that you will advance your mathematical knowledge if you take this course seriously. You will use mathematics as a tool, and as with any tool, the more you use it the better you get at using it. Some of the mathematics in this book is expected to be new to you. The mathematics that is expected to be new to you will be introduced in recitation on an as-needed basis. It is anticipated that you will learn and use some calculus in this course before you ever see it in a mathematics course. (This book is addressed most specifically to students who have never had a
Chapter 1 Mathematical Prelude
original
new original Change
While it’s certainly okay to memorize this by accident because of familiarity with it, you should concentrate on being able to derive it using common sense (rather than working at memorizing it).
Substituting the given values for the case at hand we obtain
s
m
s
m s
m
Change .
% Change==== 3. 5 %
Problems Involving Right Triangles
Example 1-1: The length of the shorter side of a right triangle is x and the length of the
hypotenuse is r. Find the length of the longer side and find both of the angles, aside
from the right angle, in the triangle.
Draw the triangle such that it is obvious which side is the shorter side
Pythagorean Theorem r 2 =x^2 +y^2 Subtract x 2 from both sides of the equation (^) r 2 −x^2 =y^2 Swap sides (^) y 2 =r^2 −x^2 Take the square root of both sides of the equation y = r^2 −x^2
By definition, the sine of θ is the side
opposite θ divided by the hypotenuse r
x
sin θ =
Take the arcsine of both sides of the
equation in order to get θ by itself r
1 x
θ = sin −
By definition, the cosine of ϕ is the side
adjacent to ϕ divided by the hypotenuse r
x
cos ϕ =
Take the arccosine of both sides of the
equation in order to get ϕ by itself r
1 x
ϕ = cos−
r
x
y
Solution:
Chapter 1 Mathematical Prelude
To solve a problem like the one above, you need to memorize the relations between the sides and the angles of a right triangle. A convenient mnemonic^2 for doing so is “SOHCAHTOA” pronounced as a single word.
Referring to the diagram above right:
SOH reminds us that: Hypotenuse
Opposite
sin θ = (1-3)
CAH reminds us that: Hypotenuse
Adjacent
cos θ = (1-4)
TOA reminds us that: Adjacent
Opposite
tan θ = (1-5)
Points to remember:
1. The angle θ is never the 90 degree angle.
- The words “opposite” and “adjacent” designate sides relative to the angle. For instance,
the cosine of θ is the length of the side adjacent to θ divided by the length of the
hypotenuse.
You also need to know about the arcsine and the arccosine functions to solve the example problem above. The arcsine function is the inverse of the sine function. The answer to the question, “What is the arcsine of 0.44?” is, “that angle whose sine is 0.44 .” There is an arcsine button on your calculator. It is typically labeled sin-1, to be read, “arcsine.” To use it you probably have to hit the inverse button or the second function button on your calculator first.
The inverse function of a function undoes what the function does. Thus:
sin −^1 sinθ = θ (1-6)
Furthermore, the sine function is the inverse function to the arcsine function and the cosine function is the inverse function to the arccosine function. For the former, this means that:
sin ( sin −^1 x ) =x (1-7)
(^2) A mnemonic is something easy to remember that helps you remember something that is harder to remember.
θ Adjacent
Opposite
Hypotenuse
Chapter 1 Mathematical Prelude
equation. The second multiplicand in the expression ( a x++++ b)x==== 0 is x itself, so
x = 0
is a solution to the equation. Setting the first term equal to zero gives:
ax++++ b==== 0
a x====−−−− b
a
b x =−
Now suppose the b in the quadratic equation a x^2 ++++ bx++++c==== 0 , equation 1-8, is zero. In that case,
the quadratic equation reduces to:
ax^2 ++++ c==== 0
which can easily be solved without the quadratic formula as follows:
a x^2 ====−−−− c
a
c x 2 =−
a
c x =± −
where we have emphasized the fact that there are two square roots to every value by placing a plus-or-minus sign in front of the radical.
Now, if upon arranging the given equation in the form of the quadratic equation (equation 1-8):
ax^2 ++++ bx++++c==== 0
you find that a, b, and c are all non-zero, then you should use the quadratic formula. Here we present an example of a problem whose solution involves the quadratic formula:
Chapter 1 Mathematical Prelude
Example 1-2: Quadratic Formula Example Problem
Given
x
x (1-10)
find x.
At first glance, this one doesn’t look like a quadratic equation, but as we begin isolating x, as we always strive to do in solving for x, (hey, once we have x all by itself on the left side of the equation, with no x on the right side of the equation, we have indeed solved for x—that’s what it means to solve for x) we quickly find that it is a quadratic equation.
Whenever we have the unknown in the denominator of a fraction, the first step in isolating that unknown is to multiply both sides of the equation by the denominator. In the case at hand, this yields ( x+ 1 )( 3 +x)= 24
Multiplying through on the left we find
3 x+ 3 +x^2 +x= 24
At this point it is pretty clear that we are dealing with a quadratic equation so our goal becomes getting it into the standard form of the quadratic equation, the form of equation 1-8, namely:
ax 2 + bx+c= 0. Combining the terms involving x on the left and rearranging we obtain
x^2 + 4 x+ 3 = 24
Subtracting 24 from both sides yields
x^2 + 4 x− 21 = 0
which is indeed in the standard quadratic equation form. Now we just have to use inspection to identify which values in our given equation are the a, b, and c that appear in the standard
quadratic equation (equation 1-8) ax 2 + bx+c= 0. Although it is not written, the constant multiplying the x^2 , in the case at hand, is just 1. So we have a = 1, b = 4, and c = −21.
Substituting these values into the quadratic formula (equation 1-9):
a
b b ac x 2
− ±^2 − 4
yields
x=
which results in x = 3 , x=− 7
Chapter 2 Conservation of Mechanical Energy I: Kinetic Energy & Gravitational Potential Energy
2 Conservation of Mechanical Energy I: Kinetic Energy &
Gravitational Potential Energy
Physics professors often assign conservation of energy problems that, in terms of mathematical complexity, are very easy, to make sure that students can demonstrate that they know what is going on and can reason through the problem in a correct manner, without having to spend much time on the mathematics. A good before-and-after-picture correctly depicting the configuration and state of motion at each of two well-chosen instants in time is crucial in showing the appropriate understanding. A presentation of the remainder of the conceptual- plus-mathematical solution of the problem starting with a statement in equation form that the energy in the before picture is equal to the energy in the after picture, continuing through to an analytical solution and, if numerical values are provided, only after the analytical solution has been arrived at, substituting values with units, evaluating, and recording the result is almost as important as the picture. The problem is that, at this stage of the course, students often think that it is the final answer that matters rather than the communication of the reasoning that leads to the answer. Furthermore, the chosen problems are often so easy that students can arrive at the correct final answer without fully understanding or communicating the reasoning that leads to it. Students are unpleasantly surprised to find that correct final answers earn little to no credit in the absence of a good correct before-and- after picture and a well-written remainder of the solution that starts from first principles, is consistent with the before and after picture, and leads logically, with no steps omitted, to the correct answer. Note that students who focus on correctly communicating the entire solution, on their own, on every homework problem they do, stand a much better chance of successfully doing so on a test than those that “just try to get the right numerical answer” on homework problems.
Mechanical Energy
Energy is a transferable physical quantity that an object can be said to have. If one transfers energy to a material particle that is initially at rest, the speed of that particle changes to a value which is an indicator of how much energy was transferred. Energy has units of joules, abbreviated J. Energy can’t be measured directly but when energy is transferred to or from an object, some measurable characteristic (or characteristics) of that object changes (change) such that, measured values of that characteristic or those characteristics (in combination with one or more characteristics such as mass that do not change by any measurable amount) can be used to determine how much energy was transferred. Energy is often categorized according to which measurable characteristic changes when energy is transferred. In other words, we categorize energy in accord with the way it reveals itself to us. For instance, when the measurable characteristic is temperature, we call the energy thermal energy; when the measurable quantity is speed, we call the energy kinetic energy. While it can be argued that there is only one form or kind of energy, in the jargon of physics we call the energy that reveals itself one way one kind or form of energy (such as thermal energy) and the energy that reveals itself another way another kind or form of energy (such as kinetic energy). In physical processes it often occurs that the
Chapter 2 Conservation of Mechanical Energy I: Kinetic Energy & Gravitational Potential Energy
way in which energy is revealing itself changes. When that happens we say that energy is transformed from one kind of energy to another.
Kinetic Energy is energy of motion. An object at rest has no motion; hence, it has no kinetic energy. The kinetic energy K of a non-rotating rigid object in motion depends on the mass m and
speed v of the object as follows^1 :
2 2
K = 1 m v (2-1)
The mass m of an object is a measure of the object’s inertia, the object’s inherent tendency to maintain a constant velocity. The inertia of an object is what makes it hard to get that object moving. The words “mass” and “inertia” both mean the same thing. Physicists typically use the word “inertia” when talking about the property in general conceptual terms, and the word “mass” when they are assigning a value to it, or using it in an equation. Mass has units of kilograms,
abbreviated kg. The speed v has units of meters per second, abbreviated m/s. Check out the
units in equation 2-1: 2 2
K =^1 m v
On the left we have the kinetic energy which has units of joules. On the right we have the
product of a mass and the square of a velocity. Thus the units on the right are (^2)
2
s
m kg and we
can deduce that a joule is a (^2)
2
s
m kg.
Potential Energy is energy that depends on the arrangement of matter. Here, we consider one type of potential energy:
The Gravitational Potential Energy of an object^2 near the surface of the earth is the energy (relative to the gravitational potential energy that the object has when it is at the reference level about to me mentioned) that the object has because it is "up high" above a reference level such as the ground, the floor, or a table top. In characterizing the relative gravitational potential energy of an object it is important to specify what you are using for a reference level. In using the concept of near-earth gravitational potential energy to solve a physics problem, although you are free to choose whatever you want to as a reference level, it is important to stick with one and the same reference level throughout the problem. The relative gravitational potential energy U g of
(^1) In classical physics we deal with speeds much smaller than the speed of light c = 3. 00 × 108 m/s. The classical
physics expression K = 21 m v^2 is an approximation (a fantastic approximation at speeds much smaller than the speed
of light—the smaller the better) to the relativistic expression K =( 1 / 1 − v 2 /c^2 − 1 )mc^2 which is valid for all speeds. (^2) We call the potential energy discussed here the gravitational potential energy “of the object.” Actually, it is the
gravitational potential energy of the object-plus-earth system taken as a whole. It would be more accurate to ascribe the potential energy to the gravitational field of the object and the gravitational field of the earth. In lifting an object, it is as if you are stretching a weird invisible spring—weird in that it doesn’t pull harder the more you stretch it as an ordinary spring does—and the energy is being stored in that invisible spring. For energy accounting purposes however, it is easier to ascribe the gravitational potential energy of an object near the surface of the earth, to the object, and that is what we do in this book. This is similar to calling the gravitational force exerted on an object by the earth’s gravitational field the “weight of the object” as if it were a property of the object, rather than what it really is, an external influence acting on the object.
Chapter 2 Conservation of Mechanical Energy I: Kinetic Energy & Gravitational Potential Energy
(1) No energy is transferred to or from the surroundings. (2) No energy is converted to or from other forms of energy (such as thermal energy).
Consider a couple of processes in which the total mechanical energy of a system does not remain the same:
Case #
A rock is dropped from shoulder height. It hits the ground and comes to a complete stop.
The "system of objects" in this case is just the rock. As the rock falls, the gravitational potential energy is continually decreasing. As such, the kinetic energy of the rock must be continually increasing in order for the total energy to be staying the same. On the collision with the ground, some of the kinetic energy gained by the rock as it falls through space is transferred to the ground and the rest is converted to thermal energy and the energy associated with sound. Neither condition (no transfer and no transformation of energy) required for the total mechanical energy of the system to remain the same is met; hence, it would be incorrect to write an equation setting the initial mechanical energy of the rock (upon release) equal to the final mechanical energy of the rock (after landing).
Can the idea of an unchanging total amount of mechanical energy be used in the case of a falling object? The answer is yes. The difficulties associated with the previous process occurred upon collision with the ground. You can use the idea of an unchanging total amount of mechanical energy to say something about the rock if you end your consideration of the rock before it hits the ground. For instance, given the height from which it is dropped, you can use the idea of an unchanging total amount of mechanical energy to determine the speed of the rock at the last instant before it strikes the ground. The "last instant before" it hits the ground corresponds to the situation in which the rock has not yet touched the ground but will touch the ground in an amount of time that is too small to measure and hence can be neglected. It is so close to the ground that the distance between it and the ground is too small to measure and hence can be neglected. It is so close to the ground that the additional speed that it would pick up in continuing to fall to the ground is too small to be measured and hence can be neglected. The total amount of mechanical energy does not change during this process. It would be correct to write an equation setting the initial mechanical energy of the rock (upon release) equal to the final mechanical energy of the rock (at the last instant before collision).
Case #
A block, in contact with nothing but a sidewalk, slides across the sidewalk.
The total amount of mechanical energy does not remain the same because there is friction between the block and the sidewalk. In any case involving friction, mechanical energy is converted into thermal energy; hence, the total amount of mechanical energy after the sliding, is not equal to the total amount of mechanical energy prior to the sliding.
Chapter 2 Conservation of Mechanical Energy I: Kinetic Energy & Gravitational Potential Energy
Applying the Principle of the Conservation of Energy for the Special Case
in which the Mechanical Energy of a System does not Change
In applying the principle of conservation of mechanical energy for the special case in which the mechanical energy of a system does not change, you write an equation which sets the total mechanical energy of an object or system objects at one instant in time equal to the total mechanical energy at another instant in time. Success hangs on the appropriate choice of the two instants. The principal applies to all pairs of instants of the time interval during which energy is neither transferred into or out of the system nor transformed into non-mechanical forms. You characterize the conditions at the first instant by means of a "Before Picture" and the conditions at the second instant by means of an "After Picture.” In applying the principle of conservation of mechanical energy for the special case in which the mechanical energy of a system does not change, you write an equation which sets the total mechanical energy in the Before Picture equal to the total mechanical energy in the After Picture. (In both cases, the “total” mechanical energy in question is the amount the system has relative to the mechanical energy it would have if all objects were at rest at the reference level.) To do so effectively, it is necessary to sketch a Before Picture and a separate After Picture. After doing so, the first line in one's solution to a problem involving an unchanging total of mechanical energy always reads
Energy Before = Energy After (2-3)
We can write this first line more symbolically in several different manners:
E 1 = E 2 or E (^) i = Ef or E = E′ (2-4)
The first two versions use subscripts to distinguish between "before picture" and "after picture" energies and are to be read "E-sub-one equals E-sub-two" and "E-sub-i equals E-sub-f." In the latter case the symbols i and f stand for initial and final. In the final version, the prime symbol is added to the E to distinguish "after picture" energy from "before picture" energy. The last equation is to be read "E equals E-prime." (The prime symbol is sometimes used in mathematics to distinguish one variable from another and it is sometimes used in mathematics to signify the derivative with respect to x. It is never used it to signify the derivative in this book.) The unprimed/prime notation is the notation that will be used in the following example:
Chapter 2 Conservation of Mechanical Energy I: Kinetic Energy & Gravitational Potential Energy
The solution presented in the example provides you with an example of what is required of students in solving physics problems. In cases where student work is evaluated, it is the solution which is evaluated, not just the final answer. In the following list, general requirements for solutions are discussed, with reference to the solution of the example problem:
- Sketch (the before and after pictures in the example). Start each solution with a sketch or sketches appropriate to the problem at hand. Use the sketch to define symbols and, as appropriate, to assign values to symbols. The sketch aids you in solving the problem and is important in communicating your solution to the reader. Note that each sketch depicts a configuration at a particular instant in time rather than a process which extends over a time interval.
- Write the "Concept Equation" ( E = E′in the example).
- Replace quantities in the "Concept Equation" with more specific representations of the same quantities. Repeat as appropriate.
In the example given, the symbol E representing total mechanical energy in the before picture is replaced with "what it is,” namely, the sum of the kinetic energy and the potential energy K + U of the rock in the before picture. On the same line E′^ has been replaced with what it is, namely, the sum of the kinetic energy and the potential energy K ′^ +U′in the after picture. Quantities that are obviously zero have slashes drawn through them and are omitted from subsequent steps.
This step is repeated in the next line ( m gy = 21 m v ′^2 ) in which the gravitational potential
energy in the before picture, U, has been replaced with what it is, namely m gy, and on the
right, the kinetic energy in the after picture has been replaced with what it is, namely, 21 m v ′^2. The symbol m that appears in this step is defined in the diagram.
- Solve the problem algebraically. The student is required to solve the problem by algebraically manipulating the symbols rather than substituting values and simultaneously evaluating and manipulating them.
The reasons that physics teachers require students taking college level physics courses to solve the problems algebraically in terms of the symbols rather than working with the numbers are:
(a) College physics teachers are expected to provide the student with experience in "the next level" in abstract reasoning beyond working with the numbers. To gain this experience, the students must solve the problems algebraically in terms of symbols.
(b) Students are expected to be able to solve the more general problem in which, whereas certain quantities are to be treated as if they are known, no actual values are given. Solutions to such problems are often used in computer programs which enable the user to obtain results for many different values of the "known quantities.” Actual values are
Chapter 2 Conservation of Mechanical Energy I: Kinetic Energy & Gravitational Potential Energy
assigned to the known quantities only after the user of the program provides them to the program as input—long after the algebraic problem is solved.
(c) Many problems more complicated than the given example can more easily be solved algebraically in terms of the symbols. Experience has shown that students accustomed to substituting numerical values for symbols at the earliest possible stage in a problem are unable to solve the more complicated problems.
In the example, the algebraic solution begins with the line m gy ==== 21 m v ′′′′^2. The m's appearing
on both sides of the equation have been canceled out (this is the algebraic step) in the solution provided. Note that in the example, had the m's not canceled out, a numerical answer to the problem could not have been determined since no value for m was given. The next two lines represent the additional steps necessary in solving algebraically for the final speed v ′^. The
final line in the algebraic solution ( v ′′′′==== 2 gy in the example) always has the quantity being
solved for all by itself on the left side of the equation being set equal to an expression involving only known quantities on the right side of the equation. The algebraic solution is not complete if unknown quantities (especially the quantity sought) appear in the expression on the right hand side. Writing the final line of the algebraic solution in the reverse order, e.g.
2 gy ==== v ′′′′, is unconventional and hence unacceptable. If your algebraic solution naturally
leads to that, you should write one more line with the algebraic answer written in the correct order.
- Replace symbols with numerical values with units, ) 16 m s
2 ( 980 m
v ′′′′====. 2. in the example; the
units are the units of measurement: (^2) s
m and m in the example).
No computations should be carried out at this stage. Just copy down the algebraic solution but with symbols representing known quantities replaced with numerical values with units. Use parentheses and brackets as necessary for clarity.
- Write the final answer with units ( s
v ′′′′==== 5. 6 m in the example).
Numerical evaluations are to be carried out directly on the calculator and/or on scratch paper. It is unacceptable to clutter the solution with arithmetic and intermediate numerical answers between the previous step and this step. Units should be worked out and provided with the final answer. It is good to show some steps in working out the units but for simple cases units (not algebraic solutions) may be worked out in your head. In the example provided, it is easy to see
that upon taking the square root of the product of (^2) s
m and m, one obtains s
m (^) hence no additional
steps were depicted.
Chapter 3 Conservation of Mechanical Energy II: Springs, Rotational Kinetic Energy
into 2π parts and call each part a radian. Thus a radian is
of a rotation. The fact that an
angle is a fraction of a rotation means that an angle is really a pure number and the word “radian” abbreviated rad, is a reminder about how many parts the rotation has been divided up into, rather than a true unit. In working out the units in cases involving radians, one can simply erase the word radian. This is not the case for actual units such as meters or joules.
The moment of inertia I has units of kg ⋅ m^2. The units of the right hand side of equation 3-2,
2 2
K =^1 I w , thus work out to be
2
2 2 s
rad kg ⋅ m. Taking advantage of the fact that a radian is not a
true unit, we can simply erase the units rad 2 leaving us with units of (^2)
2
s
m kg ⋅ , a combination
that we recognize as a joule which it must be since the quantity on the left side of the equation 2 2
K =^1 I w (equation 3-2) is an energy.
Energy of Rolling
An object which is rolling is both moving through space and spinning so it has both kinds of
kinetic energy, the 21 m v 2 and the 21 I w^2. The movement of an object through space is called
translation. To contrast it with rotational kinetic energy, the ordinary kinetic energy K = 21 m v^2
is referred to as translational kinetic energy. So, the total kinetic energy of an object that is rolling can be expressed as
K (^) Rolling = KTranslation+K Rotation (3-3)
2 2 2
KRolling = mv + I w (3-4)
Now you probably recognize that an object that is rolling without slipping is spinning at a rate
that depends on how fast it is going forward. That is to say that the value of w depends on the
value of v. Let’s see how. When an object that is rolling without slipping completes one
rotation, it moves a distance equal to its circumference which is 2π times the radius of that part of the object on which the object is rolling.
Distance traveledinonerotation==== 2 π r (3-5)
Now if we divide both sides of this equation by the amount of time that it takes for the object to complete one rotation we obtain on the left, the speed of the object and, on the right, we can interpret the 2π as 2π radians and, since 2π radians is one rotation the 2π radians divided by the time it takes for the object to complete one rotation is just the magnitude of the angular velocity
w. Hence we arrive at
v =w r
which is typically written:
v = r w (3-6)
Chapter 4 Conservation of Momentum
4 Conservation of Momentum
A common mistake involving conservation of momentum crops up in the case of totally inelastic collisions of two objects, the kind of collision in which the two colliding objects stick together and move off as one. The mistake is to use conservation of mechanical energy rather than conservation of momentum. One way to recognize that some mechanical energy is converted to other forms is to imagine a spring to be in between the two colliding objects such that the objects compress the spring. Then imagine that, just when the spring is at maximum compression, the two objects become latched together. The two objects move off together as one as in the case of a typical totally inelastic collision. After the collision, there is energy stored in the compressed spring so it is clear that the total kinetic energy of the latched pair is less than the total kinetic energy of the pair prior to the collision. There is no spring in a typical inelastic collision. The mechanical energy that would be stored in the spring, if there was one, results in permanent deformation and a temperature increase of the objects involved in the collision.
The momentum of an object is a measure of how hard it is to stop that object. The momentum of an object depends on both its mass and its velocity. Consider two objects of the same mass, e.g. two baseballs. One of them is coming at you at 10 mph, and the other at 100 mph. Which one has the greater momentum? Answer: The faster baseball is, of course, harder to stop, so it has the greater momentum. Now consider two objects of different mass with the same velocity, e.g. a Ping-Pong ball and a cannon ball, both coming at you at 25 mph. Which one has the greater momentum? The cannon ball is, of course, harder to stop, so it has the greater momentum.
The momentum p of an object is equal to the product^1 of the object’s mass m and velocity v :
p = m v (4-1)
Momentum has direction. Its direction is the same as that of the velocity. In this chapter we will limit ourselves to motion along a line (motion in one dimension). Then there are only two directions, forward and backward. An object moving forward has a positive velocity/momentum and one moving backward has a negative velocity/momentum. In solving physics problems, the decision as to which way is forward is typically left to the problem solver. Once the problem solver decides which direction is the positive direction, she must state what her choice is (this statement, often made by means of notation in a sketch, is an important part of the solution), and stick with it throughout the problem.
The concept of momentum is important in physics because the total momentum of any system remains constant unless there is a net transfer of momentum to that system, and if there is an ongoing momentum transfer, the rate of change of the momentum of the system is equal to the rate at which momentum is being transferred into the system. As in the case of energy, this means that one can make predictions regarding the outcome of physical processes by means of
(^1) This classical physics expression is valid for speeds small compared to the speed of light c = 3. 00 × 108 m/s. The
relativistic expression for momentum is p= mv 1 − v^2 /c^2. At speeds that are very small compared to the speed of light, the classical physics expression p = m v is a fantastic approximation to the relativistic expression.
