Physics 71 Final Exam Reviewer
guess that:
• Diandrew Lexter L. Dy •
∆energy , (1.1) ∆time which is the correct expression for average power. You can also guess the force and speed relation of average power. However, this method will not work when there are unitless constants involved. For example, the escape speed[should ] [ ] m3 be in ms , and must involve G, with units [G] = kg·s 2 . [ 2] Noting that [G · mass/distance] = ms2 , you can probably guess that: √ Gm vesc = . (1.2) r √ This formula however is off by a factor of 2, hence is wrong. You can still probably get away by guessing the formula because the exam is multiple choice, thus you can probably guess the factor. You can also guess the correct answer in multiple choice exams by the correct units. If the question requires work (pun intended), your answer should be in Joules, and you may discard the choices with different units. Remember, guessing is an art and a science, and is not just completely random. An amazing example of guessing, and my favorite problem so far, is in your third exam, the Launch problem (somewhere at the end). [ ] The problem states that you launch a projectile at 4 km from the ground, and you are asked of s its speed at a height 20km off the ground. The solution here is through energy conservation, then find speed after a (slightly) long algebra and calculator exercise (that is why it is someewhere at the end). However, you could have guessed the answer by simply looking at the choices. Looking at the choices, you can see that only one option has speed lesser than the initial speed. And intuitively, the speed should decrease at the top. So in short: guessing works sometimes, and can also be used as a check to your answers. Why am I babbling about this? Maybe because I realized that guessing is an important skill, and must be taught! Sorry, I should have discussed this before, never too late I hope. One of the advantages of SI units are the SI prefixes that are all based on powers of 10. In Physics 71, the important prefixes are: Power =
This is a very condensed reviewer for the Finals. This is essentially just writing all the needed equations, with some concepts. Exercises are suggested from our previous problem sets, quizzes, and long exmas. This summary in nature can be used as a guide for things you have to learn for the exam. If you cannot understand it, then reading Young and Freedman would be helpful.
1
Ch 1: Units, Physical Quantities and Vectors
The first chapter deals with preliminary topics for physics. This is not included in the exam, but the skills discussed here are prevalent all throughout the remaining chapters of Young and Freedman, as well as Physics in general. Let s briefly discuss the important skills:
1.1
What is Physics?
Such an important question that has various interpretations. In the words of my arrogrant physicist friend, ”Physics is everything!”. I however think Physics is Math in disguise, while Mathematicians think they are Philosophers, Philosophers want to be Psychologists, Psychologists goal is to explain Biology, Biologists envy Chemists, and Chemistry wants to be Physics (so circle!). Now what about the Engineers? Engineers want to be in Science, but they are too busy getting rich. Architects think they are gods.
1.2
Units
Units are important for experimentalist and engineers because they want to base the physical variables by some standard measurement. These standard scales however are based on convention, the most popular and useful of which are S.I. units. The S.I. units used for Physics are as follows: Variable length time mass Force Energy Power Pressure
Units meter [m] seconds [s] kilogram [kg] Newton [N] Joule [J] Watt [W] Pascal [Pa]
The fundamental units are the first three: length, time and mass. The rest of the variables has units that are expressible in terms of the fundamental units. The reason they introduce these units is probably convenience. Familiarization of units for the variables has important consequences in any Physics exam. First, you can guess the formula for the variable by looking [ ]at the units. For example, power is in Watts, which is Js . So you can probably Dy, D. L. Phys 71 1.2
Prefix kilo centi
power 103 10−2
syntax (in meter) km cm
The rest of the prefixes will be useful later in your Physics career. Hence conversion of units is a breeze! The time conversion however obeys the following rules: time 1 minute [min] 1 hour [hr] 1 year [yr]
in seconds 60 [s] 60 min =3600 [s] 525600 moments! 1
And also for days, months and other time quantities. This peculiarity of time is that it isn’t really S.I. because of the base 10 wasn’t converted. Maybe it’s too difficult to change the times. Theoretical physicists don’t care however, as sometimes they will just measure time in meters, hence the SI base 10 property still works!
1.3
Scalars and Vectors
Scalars are quantities without direction, while vectors have direction. Several vectors can be added using the parallelogram rule. Vectors can be broken down into components. We usually use Cartesian components, where the basis (unit vectors) are perpendicular to each other. Hence we can express any vector in three space dimensions as: b V = Vxbi + Vybj + Vz k
(1.3)
b are the basis unit vectors, while Vx , Vy , Vz are where bi, bj, k the components of V, which are scalar numbers (can be negative!). The magnitude of the vector can be solved using the ’Pythagorean theorem’: √ V = |V| = (Vx )2 + (Vy )2 + (Vz )2 . (1.4) The magnitude V is never negative! The direction can also be solved. For a vector F = Fxbi + Fybj, the usual angle is solved as: θ = Arctan
Fy Fx
(1.5)
For vectors, important operations include: inner product (e.g. W = F · s) or cross product (e.g. τ = r × F). Familiarization of these operations are important in the next chapters. That ends Chapter 1!
2
Ch 2: Motion along a straight line
2.1
Dictionary
Consider an object moving in a straight line. The motion of the object hence can be described by its position at all times, x(t). This position can be any real number (can be negative!). From x(t), we can define displacement, defined as: displacementt1
to t2
= x(t2 ) − x(t1 ) = ∆x.
(2.1)
Hence displacement is dependent only at the endpoints of the position, and the sign implies the direction (positive means its going to more positive position). This is different than our notion of distance, where we add magnitude of displacements even with a change of direction. Rule of thumb: Going in a full circle: zero displacement, distance > 0. We next define velocity vx as: velocity: vx =
dx . dt
(2.2)
This quantity is a vector (vx is actually the component, direction is bi) with the sign implying direction (can be negative). We define speed v (no subscript) as: speed: v = |vx | .
(2.3)
Now we can also add instantaneous as an adjective to the velocity and speed because it is evaluated at one time. Because derivatives are hard to imagine, we can define the average velocity and speed from time t1 to t2 as: x(t2 ) − x(t1 ) ∆x = . t2 − t1 ∆t = |vx,ave | . (2.4)
ave. velocity: vx,ave,(t1 ave. speed: vave
to t2 )
=
Note that in the limit that the interval t1 to t2 becomes infinitesimal t2 − t1 = δt, then the definition of average velocity becomes velocity! We finally define acceleration as: acceleration: ax =
dv d2 x = 2. dt dt
(2.5)
and the average counterpart: ave. acceleration: ax,ave,(t1
to t2 )
=
v(t2 ) − v(t1 ) ∆v = . t2 − t1 ∆t (2.6)
These formulas are important and must be memorized! Trivia: da dt is a jerk. Examples You can find several examples in Young (see the syllabus). In the first exam, understand Kinematics, Hindi Tamad and Stopping Distance. Dy, D. L. Phys 71 2.2
2
2.2
Plotting x − t, vx − t, ax − t diagrams
Examples
To picture the properties of moving objects, we can introduce graphs. These are the 2D graphs that we did with the horizontal axis as time, and the vertical axis can either be the position (x), velocity vx and acceleration ax . Important concepts include that the derivative of a function is related to the slope of the graph. For example, the velocity at t1 is the slope of the x − t graph at t1 .
We have done lots of examples for this (too many to mention). See PS1,2,3 and Q2,3 for example. In your exam, Stopping Distance, Skid marks, Catch, In free fall are important examples. That ends Ch 2!
Examples You can find several examples in Young (see the syllabus). In the first exam, understand X vs T, Plot I and Plot II. Also read about Problem Set 1, where most of the numbers are about diagrams.
2.3
Motion with constant acceleration
Let us first consider when vx is a constant. Then we obtain (by integration) the position at time t1 as: x(t1 ) = x(t0 ) + vx (t1 − t0 )
(2.7)
where t0 is the initial time, hence x(t0 ) is the initial position. This is hence relatively easy to memorize, but also easy to derive by integration. Next, we consider the case where ax is a constant (independent of time, hence is not a jerk!). We then obtain the equations of motion at time t1 : x1 = x0 + vx0 (t1 − t0 ) + 21 ax (t1 − t0 )2
(2.8)
vx1 = vx0 + ax (t1 − t0 )
(2.9)
(vx1 )2 = (vx0 )2 + 2ax (x1 − x0 )
(2.10)
where we introduce the notation: x0 = x(t0 ) as the initial position and vx0 = vx (t0 ) as the initial velocity, while x1 = x(t1 ), vx1 = vx (t1 ). Note that these equations should be familiar to you. I usually derive these equations by integration, but you can also memorize! In practice, I can survive constant acceleration problems with these three equations, and I just derive other things by algebra. I hate memorizing! Another possibly important relation is the average formula: vx1 + vx0 x1 − x0 = (t1 − t0 ), (2.11) 2 which can be derived from the previous three/ For freely falling bodies, we can approximate the motion as with acceleration constant ay = −g. Note that we refer the y-axis because it is associated with going up! Hence, we just replace ax and x in the above equations as: y1 = y0 + vy0 (t1 − t0 ) − 12 g(t1 − t0 )2
(2.12)
vy1 = vy0 − g(t1 − t0 )
(2.13)
(vy1 )2 = (vy0 )2 − 2g(y1 − y0 )
(2.14)
Note: These equations should be natural to you! breathing... Dy, D. L. Phys 71 2.3
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3
3
Ch 3: Motion in 2 or 3D
We now extend our discussion of the previous chapter to higher dimensions.
3.1
vy1 = vy0 − g(t1 − t0 )
(3.10)
(vy1 )2 = (vy0 )2 − 2g(y1 − y0 ),
(3.11)
Dictionary
Again we introduce the position, but this time in 3D as: b r = xbi + ybj + z k.
(3.1)
From hereon, if you want to consider 2D, then don’t include the third component, or set z = 0. We can also define displacement as: displacementt1
to t2
= r(t2 ) − r(t1 ) = ∆r.
(3.2)
We next define velocity v as: velocity: v =
dx dy dz b dr = bi + bj + k. dt dt dt dt
(3.3)
Hence we can define the components vx , vy , vz of velocity. We next define speed v as: √ speed: v = |v| = (vx )2 + (vy )2 + (vz )2 . (3.4) We can also define the average velocity and speed from time t1 to t2 as: ave. velocity: vave,(t1
to t2 )
r(t2 ) − r(t1 ) . t2 − t1
=
ave. speed: vave = |vx,ave | .
(3.5)
We finally define acceleration as: acceleration: a =
dv d2 r = 2. dt dt
(3.6)
3.3
and the average counterpart: ave. acceleration: aave,(t1
to t2 )
=
v(t2 ) − v(t1 ) . t2 − t1
where the x refers to the horizontal position, while y is the vertical position. From experience, the usual problems usually start with the object given an initial speed at a certain angle. From these initial conditions, you should be able to deduce the initial velocity components v0x , v0y . The projectile then follows a trajectory, and several properties of the projectile can be asked. For example, you may be asked the time of to reach maximum height, the time of flight, the range, the maximum height and the speed at certain points among others. You may be also asked from several scenarios, which can include on a flat ground or from a certain height. You should be able to solve them quickly. Here are my tips: 1. Determine the time immediately! Time is precious for these problems, and is usually the first thing to solve for these problems. 2. A sketch usually helps, especially with complicated problems. 3. Practice! And practice quickly.. Examples: You have to recall how to solve several problems about projectile. Know how to compute for maximum height, the time of flight, the range, the maximum height and the speed at certain points. I emphasize that you should never memorize these things, except for the equations of motion stated above. The reason is that, your memory can be brutally misleading, especially the range and maximum height. Focus your memorization powers elsewhere in the exam.
(3.7)
Motion in a circle
We now discuss motion in a circle. When an object is moving in a circle at constant speed v, then its net acceleration is always directed towards the center of the circle with magnitude:
These definitions are important and must be memorized! arad = Examples See young, Problem sets, in Quiz:Acceleration, Aave, Save
3.2
Projectile Motion
A projectile is defined as any body given an initial velocity and its trajectory is purely governed by gravity and possibly air resistance. The rule of thumb here is: solve horizontal and vertical components separately, the horizontal motion has constant velocity while the vertical motion is constant acceleration (−g). The equations are therefore: x(t1 ) = x(t0 ) + vx (t1 − t0 )
(3.8)
y1 = y0 + vy0 (t1 − t0 ) − 12 g(t1 − t0 )2
(3.9)
Dy, D. L. Phys 71 3.4
v2 r
(3.12)
where r is the circle’s radius. Now if the object is increasing speed along the circular path, then there must be a tangential acceleration atan along the direction of motion, such that the net acceleration is the vector sum arad + atan . If the motion is decreasing speed, then the tangential acceleration must be opposite to the movement. These concepts can be easier understood through the succeeding figure. Examples There are several problems that we did involving motion in a circle. Among which are in PS4-3, Q2-6, LE1-Circular Road, Toy Car. You may also be asked about the period of the motion, which we can get through T = 2πr/v. 4
4
Figure 1: Object moving in a circle with speed increasing. Notice the direction of the acceleration component vectors.
3.4
Relative motion
The object’s velocity is always relative to the one who is measuring it. In Amsterdam high way for example, they have a speed limit of 120kph (as opposed to Commonwealth Avenue’s 60kph limit). So even though the speed limit is fast, there is a tendency for the cars to have the same speed of about 120kph. So if you are in a car, then it seems that the car in front of you is not moving at all. However, with respect to the person standing along the road, the cars are moving very fast. Thus, when speed is mentioned, there must always be a ”relative to” next to it. In our discussion of speed so far, it is assumed that the observer is relative. We now generalize to those moving observers. We introduce the notation: vA|B = velocity of A relative to the observer B.
(3.13)
So in our car example, we have vcar1 |car2 = 0 while vcar1 |person = 120kph. We can now compare relative speed of object A for two observers B, C by the important relation: vA|B = vA|C + vC|B .
(3.14)
Notice the placement of the indices, which is a useful way of memorizing this equation. Example The raindrop problem in a moving car is my favorite example for this problem. Also, the boat in a moving stream is a common problem LE1-River. That ends Chapter 3!
Dy, D. L. Phys 71 4.1
Newton’s Laws of Motion
Isaac Newton tried to describe the motion of objects by describing a new physical variable: Force! Let us first define force as: The Force is an energy field created by all living things. It surrounds us and penetrates us. It binds the galaxy together. Life creates it, makes it grow. Its energy surrounds us and binds us. Luminous beings are we, not this crude matter. You must feel the Force around you; between you, me, the tree, the rock, everywhere. Yes, even between the land and the ship. A Jedi’s strength flows from the Force. But beware of the dark side. Anger, fear, aggression; the dark side of the Force are they. Easily they flow, quick to join you in a fight. If once you start down the dark path, forever will it dominate your destiny, consume you it will, as it did ObiWan’s apprentice! Seriously though, force is the push and pull of any two objects. In effect it defines how two objects interact and causes the other to move. We can naively give two types of foces depending if there’s physical contact (contact forces) and none (non-contact forces). Contact forces include forces due to pushing, or tensile forces. Non-contact forces include gravity, which does seem like there’s no physical contact. Newton decided to define clearly forces through the Newton’s laws of motion through: Law 1: Inertial reference frame: a = 0. Law 2: Fnet = manet . Law 3: Action-reaction: Fa on b = −Fb on a Here, we introduce the notation that Fa on b is the force of object a on another object b. The Newton’s laws of motion are only valid for inertial reference frames. Hence it should be obvious to you that a rotating reference frame is not inertial. For the second law, we emphasize the net subscript! This means if there are several forces acting on an object, the net force (or the sum of all the forces) will be related to the acceleration of the object. The net force is actually not a physical force, unless of course there’s only one force acting on the object. However, a common mistake of these problems is confusing between forces and the net force. The third law, action-reaction, must be given in order to be consistent with the conservation of momentum that will be define.
4.1
Weight, Normal force, Tension
We now define weight of the object as the force of earth on the object. For this section, we assume that the acceleration due to gravity is −g, hence the weight is W = mg directed downwards (by 2nd law).
Weight : W = mg(downwards)
(4.1)
From the above definition of weight, then a book on top of a table must have a weight towards the ground. However, the book is not accelerating at all towards the ground. This 5
is because there is a force due to the table on the book that exactly opposes the weight, such that the net force, hence the net acceleration is zero. This force from the table is called the normal force. The normal force is not necessarily equal to the weight! For example, an object on a frictionless inclined plane, the two forces are not exactly equal, hence the net acceleration is downward along the incline. The tension is the magnitude of the force due to a string or a rod. The direction of the force is along the length of the string or rod. Usually, we just consider the end of the string, where there is a force applied.
4.2
Free Body Diagrams(FBD)
Free body diagrams are an important tool to find the motion of the object. This is basically a sketch of an object, and drawing all the forces acting on the object. Care must be given on the direction of the forces, and all forces must be accounted for. After drawing all the forces, then by Newton’s 2nd law, the acceleration of the object can be solved. You must know how to draw FBD on several situations. The usual problems include several forces acting on an object, and find the acceleration of the object. Some problems can also include: given the acceleration, and several forces except an unknown force, find the unknown force. These concepts must be familiar to you. That ends Ch 4!
5 5.1
Ch5: Applying Newton’s Laws Several examples of Newton’s Laws
There are two main types of problems that you can encounter in applying Newton’s laws: equilibrium problems and dynamic problems. They are differentiated as follows: Equilibrium :Fnet = 0 Dynamic :Fnet = manet ̸= 0.
(5.1)
In solving problems, you must know which type are you solving. The key here is the net acceleration of the object considered. If it has zero acceleration (e.g. not moving, constant velocity), then that’s an equilibrium problem, and dynamic otherwise. After knowing which type, then you can draw the FBD of the object, then account all the forces, then use Newton’s 2nd law to solve for the needed variable. Examples Several examples have been discussed in the problem sets and quizzes. In general, the common problems include: 1. An object attached by a cord or a string. The object can be suspended or accelerating. You may be asked what is the acceleration, or the tension, or the weight, etc. See the pulley examples, or the Atwood machine examples. 2. Inclined plane problems, where objects are on an incline. You may be asked about the acceleration of the object, or the angle.
5.2
Friction
In a realistic world, the roughness of the surfaces causes an opposing force that causes a tendency of objects to slow down. This is called friction. There are two types of friction depending on the movement of the object. It could be static (not moving) and kinetic friction (already moving). You can better understand it with the following diagram
Figure 2: Friction plots and FBD. From the diagram, we give important relations to the kinetic friction fk as: fk = µk fn Dy, D. L. Phys 71 5.2
(5.2) 6
where µk is the coefficient of kinetic friction, and fn is the normal force. The direction of the frictional force is opposite to the motion. For the static friction fs , the relation is: fs ≤ µs fn
(5.3)
where µs is the coefficient of static friction. Notice the inequality, such that the static friction depends on the amount of force you push the object. To see this, imagine pushing a box on a rough surface. If you push with a small force, then the box won’t move because the static friction opposes your force equally. Increasing your force would also increase the static friction, until it reaches µs fn . After that, the box will move, with a constant kinetic friction. It is important that you understand these differences. Examples You must be ready with solving frictional problems. You must understand the concepts, as well as draw FBD like your LE1-FBD. Memorize the key definitions of frictional forces.
5.3
Banked curves
Banked curves are a cute application of frictional forces on road surfaces. The result is:
Figure 3: Banked curves and FBD.
tan β =
v2 gR
(5.4)
which you should be able to derive from the FBD. That ends Ch 5!
Dy, D. L. Phys 71 5.3
7
Physics 71 Final Exam Reviewer
6.2
• Diandrew Lexter L. Dy •
Kinetic energy (K) is a measure of how ’energetic’ an object is, in the layman sense. It is defined through an object’s mass and speed as:
6 6.1
Ch 6: Work and Kinetic Energy
Kinetic Energy
Work
Work to me is essentially a fancy word for effort. Precisely though, and in Physics context, it is defined as: ∫
r2
W =
F(r) · dr
(6.1)
Kinetic Energy: K =
where it shows that work is just a line integral, calculus speaking. In one dimension, this is just: ∫ x2 W = F(x)dx (6.2)
Work-Kinetic Energy Theorem
Work-Kinetic energy theorem is a staple and sometimes overused theorem for several problems encountered in Physics 71. It states that:
x1
For those who are not really good at integration, what this means is that work is the area under the curve of an F − x graph. From this, you shouuld be able to compute for the work given an F − t graph! Examples include: LE2 Jess1 and Jess2, plus the Quiz4-9. Be careful that by area under the graph, there can be negative area also. Understanding is the key! When force is a constant, such that F(r) = F , then we can express work (W) as: Work: W = F · s
(6.3)
where s is the object’s displacement. Note that work can be negative, zero and positive, depending on the direction of F and s. If they are along the same direction, then work is positive, and if in opposite then it is negative. If they are perpendicular, then work is zero. The work concept has been used several examples, including LE2 Work I, II and III, Motor Work among others. Read through Quiz 3 or PS4 for example to familiarize how to solve work problems.
(6.4)
Note that this definition will always be consistent independent on the force of interaction.
6.3
r1
1 mv 2 2
Wnet = ∆K =
1 1 mv 2 − mv 2 2 f 2 i
(6.5)
Emphasis is on the net work done on the object, and not just due to one force. Examples include: LE2 Jess II or PS 4-4, Q4-2. YOu should understand when to use this theorem, which is quite useful and maybe a bit overused.
6.4
Power
Power is the rate of work. A requirement for power is of course knowledge! Google ”Ernie Baron + knowledge + power”! Power is defined as: Power: P =
dW dt
(6.6)
When force is a constant, we can express this as: Power, constant force: P = F · v
(6.7)
In many instances where the force is a constant, you can use the above formula and hence integration is not needed. Sometimes though the speed is not given but the average speed. For this, you may use average power as: Power, average: Pav = F · vav
(6.8)
where vav is the average velocity defined earlier and you should memorize by now. Examples: Several examples include PS4-4, and several Young examples. Please keep in mind that these definitions must be memorized because some takes too much time to derive. That ends Ch 6!!
Dy, D. L. Phys 71 6.4
8
7 7.1
Potential Energy and Energy conserva- 7.4 Elastic Potential Energy tion The force due to a spring, aka elastic force, is also a conPotential Energy
The potential energy U (x) is the energy stored from the force. It is related by the force as: dU (x) (7.1) dx where we just consider one dimension for simplicity. Hence the force is negative to the slope of the U − x graph. From this expression, we can find the position where it’s in equilibrium, or when force is zero. This is when the slope is zero. We can however classify these equilibrium points through its concavity. These are: 1. Stable equilibrium: U-x graph is concaved up! d2 U/dx2 > 0. 2. Unstable equilibrium: U-x graph is concaved down! d2 U/dx2 < 0. 3. Neutral equilibrium: U-x graph has zero concavity: d2 U/dx2 = 0 Finally, we stress that we can express U in terms of the force as: ∫ x U = U0 − Fx′ dx′ . (7.2) Fx = −
This result is useful to find the expression of U , although it is probably beyond the coverage of Physics 71. Here U0 is some reference potential energy at some position.
7.2
Conservative forces
Let us first define conservative force by giving properties of the solved work. The work done by a conservative force always has four properties: 1. It can be expressed as the difference between the initial and final values of a potential-energy function. 2. It is reversible. 3. It is independent of the path of the body and depends only on the starting and ending points. 4. When the starting and ending points are the same, the total work is zero. Forces that contradict these properties are called nonconservative forces. You must remember these properties by heart. And by remember, try to understand them also.
7.3
Gravitational Potential Energy
The force due to gravity is a conservative force. So far, the force of an object to the gravity is defined as the [ weight ] W = mg ith g approximated as constant (g=9.81 sm2 ). From property 1 of the conservative force, we can express the potential (by integrating (7.2)) as: Ugrav = mgy
(7.3)
where y is the height from some reference point. For this systems, we can select any height to be the reference point, and you can choose the location where it is convenient. Dy, D. L. Phys 71 7.6
servative force. The force is given by F = −kx, where x is the displacement from the equilibrium and k is the spring’s constant. F here is a restoring force, something that should be familiar to you by now. From property 1 of the conservative force, we can express the potential (by integrating (7.2)) as: Uel =
7.5
1 2 kx . 2
(7.4)
Energy Conservation
Similar to the work-kinetic energy theorem, we can express work from a conservative force as: W x1
to x2
= U (x1 ) − U (x2 ) = −∆U
(7.5)
With Wnet = −∆K, then we can derive the conservation of energy for conservative forces: K2 + U2 = K1 + U1
(7.6)
where the subscript two refers to a state that could be different than state one (say a different position). Also, U refers to all potential energy form conserved forces. If gravity and spring are present, then U = Uel + Ugrav . This result is also called conservation of mechanical energy, because we define mechanical energy as K + U . When non-conservative forces are present, then the potential for these forces are not accounted for just by using (7.7). For these systems, we introduce Wothers as the work done by these nonconservative forces. We thus define the conservation of energy for these systems as: Ki + Ui + Wothers = Kf + Uf .
(7.7)
Here, we make a distinction between the initial state (before the nonconservative forces are doing the work) and after the nonconservatve work has been done. Note that the Wothers is in the initial side.
7.6
Examples
Conservation of energy is a very useful tool in solving physics problems. This is because energy is a scalar quantity, and directions are not involved (as opposed to forces). The usual questions answered through conservation of energy is speed, or position. I cannot stress enough how important this is. Try to solve several examples using conservation of energy. This is a bit overused, and hence should be familiar to you. If not, familiarize yourselves using the exam: June 21, Loop the loop, and so on. Quiz 4 is mainly about this topic, as well as PS4. That ends Ch 7!
9
8
Ch8: Momentum,Impulse and Collisions Inelastic collision
8.1
Impulse and Momentum Dictionary
We first define impulse J on a net force Fnet (t) at a time interval t1 to t2 as: ∫
t2
impulse: J =
Fnet (t)dt
(8.1)
t1
For constant force, we can express the impulse as: J = F(t2 − t1 ) = F∆t
(8.2)
We can also express the impulse in terms of average force as: J = Fav ∆t
(8.3)
Next, we define momentum as: momentum: p = mv.
(8.4)
This implies that we can express Newton’s second law for constant mass systems as Fnet =
dp . dt
(8.5)
This expression of the Newton’s second law is much more preferred to most physicist than the F = ma, which is uglier! Anyway, using the second law, and definition of impulse, we get the Momentum-impulse theorem:
Inelastic collision happens when the kinetic energy of the system after collision is lesser than before the collision. I stress that the momentum is still conserved! Recall that the kinetic energy is K = 12 mv 2 . So in order to show whether the collision is inelastic, then you have to compute for the sum of kinetic energies of each object before and after collision, then compare. When the colliding objects stick together after collision, we call it completely inelastic. Otherwise, it is just partially inelastic. In calculating inelastic collision problems, it is usually sufficient to use the conservation of momentum. For completely inelastic collisions of two objects with mass m1 and m2 , you have to bear in mind that the final object’s mass is the sum m1 + m2 . Several examples include LE2 Ballistic pendulum. Understanding how to solve these problems are important because they are pretty fundamental and easy to solve. Elastic collision Elastic collision happens when the kinetic energy of the system after collision is the same as that before the collision. Again I stress that the momentum is still conserved! We consider elastic collision of two objects with mass m1 and m2 . Now we specialize to the case where m2 is initially at rest, and m1 is approaching at speed v1i . Then just by energy conservation and momentum conservation, then you should be able to solve the final speeds of the two objects as:
(8.6)
v1f =
m1 − m2 v10 m1 + m2
(8.7)
This result has massive consequences in solving for the average force on objects hitting on a wall or the ground. These are definitions from the Physics dictionary, hence you have no choice but to memorize these expressions.
v2f =
2m1 v10 . m1 + m2
(8.8)
J = p2 − p1 = ∆p.
8.2
Conservation of Momentum
Momentum is an amazing physics variable because like energy, it is conserved for isolated systems. By isolated, we mean that the system is not given an external force (hence isolated). By the 2nd Law, then dp dt = 0, implying that p of the system is conserved. This important conservation law has massive consequences for lots and lots of problems, similar to that of conservation of energy. Among which are the collision problems. Collision problems are those where objects are colliding! Colliding should be obvious? Sorry.. it’s late, getting sleeepy... Anyway, on collision problems, it is usually assumed that the system is isolated, such that momentum is conserved. We can differentiate collision problems by its kinetic energy through: 1. Elastic collision: When Kaf ter = Kbef ore . 2. Inelastic collision: When Kaf ter < Kbef ore . Dy, D. L. Phys 71 8.2
Now deriving these equations is a bit tedious in algebra, hence it might be useful to just memorize them. An aide to your derivations or memorization is the ”stop ball” case, where when the two objects have the same mass, then the final speed of m1 should be zero. When both objects are initially moving, an important result is the relative velocity: v2f − v1f = −(v2i − v1i ).
(8.9)
You can also derive this result, or just memorize it. Examples We have encountered lots of elastic collision problems. Among which os LE2 Particles, Q4-7 and Q4-11. We stress that this is an important concept for this chapter.
10
8.3
Center of Mass
The center of mass of any object is the location where the weighted average of the mass for a system of particles. The weighted average formula is just: ∑ ri mi rCM = ∑i (8.10) i mi where mi are the masses of individual particles. The center of mass is not necessarily the geometric center of an object. However, for highly symmetric objects that are homogenous (uniform density), then these two different concepts are identical (e.g. homogenous sphere, cube). The center of mass doesn’t have to be at the object, e.g. a homogenous donut’s center of mass is at the hole in the middle. The center of mass for highly symmetric shapes can be ’estimated’ easily. You can also calculate it by the center of mass formula above, but it can be difficult. Familiarize yourselves with this concept through examples such as: LE2 Smiley. That ends Ch 8!
9
Ch 9: Rotation of Rigid Bodies
We now discuss rotation. This is different than the previous chapters, where the motion is translational. For these cases, we shall employ angular variables.
9.1
Dictionary of Angular variables
By angular variables, it means the movement of an object is along a certain angle, instead of being translational (along x-axis for example). We define the position of an object as s = rα
(9.1)
where s is the arclength of a circle of radius r over an angle α [in radians]. The displacement s can be thought of as a change of position in the same way we have: x − x0 = r(θ − θ0 )
(9.2)
We therefore can obtain an analogy for these two types of variables: translational ↔ rotational translational position: x velocity: vx = dx dt acceleration: ax = dv dt
↔ ↔ ↔ ↔
rotational angular position: θ angular velocity: ω = dθ dt angular acceleration: α = dω dt
mass: m Ktan = 12 mv 2
↔ ↔
Moment of Inertia: I Krot = 12 Iω 2
force: F = dp dt momentum: p
↔ torque: τ = dL dt = r × F ↔ angular momentum: L = r × p
Note that these variables will be discussed later in this chapter and the next.
9.2
Constant α and ω rotation
Similar to the result for constant v, we obtain the equations of motion for constant ω as: θ(t1 ) = θ(t0 ) + ωx (t1 − t0 )
(9.3)
where t0 is the initial time, hence θ(t0 ) is the initial position. Note that the resemblance to the translational analog should be obvious to you. Next, we consider the case where α is constant of which we have the equations of motion at time t1 : θ1 = θ0 + ω0 (t1 − t0 ) + 21 α(t1 − t0 )2
(9.4)
ω1 = ω0 + α(t1 − t0 )
(9.5)
(ω1 )2 = (ω0 )2 + 2α(θ1 − θ0 )
(9.6)
where we introduce the notation: θ0 = θ(t0 ) as the initial position and ω0 = ω(t0 ) as the initial velocity, while θ1 = θ(t1 ), ω1 = ω(t1 ). Again, the resemblance to the Dy, D. L. Phys 71 9.2
11
translational should be obvious to you. These equations should be easy to derive once you know the translational case: just change the translational variables to the rotational variables. Finally, we stress that from the analogous relation before: s = rα that the velocity and angular velocity are related by: v =rω arad =rα.
Examples Familiarize yourselves by looking at Young, or LE2 Sushi problems.
Rotational Kinetic energy
There is also the rotational counterpart for kinetic energy, defined as: Krot =
1 2 Iω 2
9.4
Parallel Axis Theorem
Sometimes you are only explicitly given the formula of the moment of inertia at a certain po . However you can solve for the moment of inertia about a parallel axis through:
(9.7)
Here, we emphasize arad because it is possible to have an atan depending on how ω is changing. A trick in solving these types of problem is units. When you are should [ asked ] about angular speed, then your [ munits ] be in rad s , while if its speed, it should be s . For that reason, in solving these types of problems, you must also solve for the units. This is helpful because sometimes you forget to multiply r, and that you will notice your answers is missing a meter in the units. So my tip here is units!
9.3
an object that is just rotating on an axis, the total kinetic energy is just the rotational. If it is also moving translationally, e.g. the center of mass is moving linearly at a distance, then you have to add the translational kinetic energy.
Ip = Icm + M d2
(9.9)
where M is the mass of the rigid object, and d is the distance of the parallel axis to that of the center of mass. You should try using parallel axis on several geometries, like solving the parallel axis at the end of a rod, or a sphere. Try reviewing most of PS5. This is an integral part!
9.5
Conservation of Energy
Conservation of energy also holds for these systems. You must be careful though of accounting all of the kinetic energy. A common mistake is just considering the translational or rotational kinetic energy to the energy, which leads to a mistake in the calculation. Please be careful about this! Examples: Try to solve PS5 again, also LE2: Axis, Rank the Cylinders and Disks and Rolling down, Ball 1 and Ball 2 problems are useful concepts that you have to know! That ends Ch 9!
(9.8)
where I is the moment of inertia, which is different depending on the geometry of the rotating rigid object. You can determine the moment of inertia through integration, but you can also refer to a table:
10
Ch 10: Dynamics of Rotational Motion!
This is just a continuation of Ch 9 for rotation.
10.1
Torque
Torque is the rotational analogue of force. Recall that F = ma, of which we obtain the analogue: torque: τ = Iα = r × F
(10.1)
Here, α has direction along the axis of rotation. In the second definition τ = r × F, r is the moment arm, which is the position vector from the axis of rotation to the point where the force F is acted upon. The direction of the torque is obtained by the right hand rule. The cross product can be computed either of the three relations: τ = rF sin α = r⊥ F = rF⊥ Figure 4: Moment of inertia for several geometries. You have to know how to calculate the moment of inertia given the formulas above. The total kinetic energy of an object is the sum of its rotational and translational i.e. Ktot = Ktrans + Krot . For Dy, D. L. Phys 71 10.1
(10.2)
In the first equality, α is the angle between r and F vectors. You must know how to find the angle! This approach can be difficult if you don’t know how to find the angle. The second and third approach are equivalent, that is find the component of any of r or F that is perpendicular to the other and just multiply them. This approach is sometimes easier and faster, and I highly recommend you to 12
study it! To find the direction, you just use the right-hand rule. Examples for torque problems can be found in Young or LE3 Torques. Static equilibrium problems in Chapter 11 also uses torque a lot!
where τz is the torque’s only component along the axis of rotation. There is also a work-kinetic energy counterpart:
10.2
which is very useful (LE2-Merry2). We can also define rotational power for constant torque analogously as:
Rolling without slipping
A circular object (cylinder or sphere) rolling in a surface can either slip (i need this!) or can be rolling without slipping (or in tagalog.. puyat! like what I am doing right now, thank you very much! :)) When the circular object of radius r is rolling without slipping, then the center of mass obeys:
W = ∆Krot = Krot,2 − Krot,1
P =τ ·ω
vcm =rωacm =
rα
(10.3)
The total kinetic energy is then Ktot = Ktrans + Krot ! You have to include the translational as well! When an object is rolling without slipping, there should be friction involved. This is a static friction, not kinetic friction! And since it is a static friction, the work done by this friction is zero (displacement is zero!). Hence the mechanical energy will not diminish and the object will roll forever (even though there’s friction!) Try to understand these concepts, or at least keep telling it to yourselves until these become true. When an object is rolling with slipping, friction can cause the object to slow down until friction is enough such that the rolling without slipping condition will hold. If there are no friction, then the object will slip forever (which i wish i could do.. but i’m not yet suicidal though.. it’s really late.. sorry.. almost done!).
10.3
Dynamics of rotational motion
In the same way that FBD + Newton’s second law are used in the translational counterpart, so will it work in the rotational, but note that the Newton’s second law counterpart has the form: τ net = Iαnet
(10.4)
Knowing this, you should be able to solve the yo-yo problem, objects rolling down an incline problem, pulley problems (LE2 Pulley 1,2). These concepts to be honest are difficult to understand. However, with practice comes familiarity. Familiarity breeds correctness. It may also lead to contempt, but you don’t have much choice do you?? That’s too sad. I don’t like rotation also.. nakakahilo e.
10.4
(10.7)
which resembles P = F · v.
10.5 scm =rθ
(10.6)
Angular momentum
Angular momentum is defined similarly as: L = r × p = Iω
(10.8)
Notice that it is similar to torque in the sense that the moment arm r is ’multiplied’ by its translational counterpart. The Newton’s second law in rotation can be expressed as: τ net =
dL dt
(10.9)
which should resemble something familiar. Thus when the net external torque of a system is zero, then L is a constant, and hence conserved. Solving through conservation of angular momentum should be similar to that of momentum, but with an added cross product. In many cases, r and p are perpendicular, and that you just multiply in magnitude. You must be careful in the direction though (use right-hand rule), as there might be a sign error otherwise. You may also use Iω definition, which is useful for rotating objects. This is probably the most difficult thing to comprehend in the exam, yet very easy to calculate algebraically. Try to read more about Young, and study LE2 stick. These problems are also important, and must be understood! That ends Chapter 5
Work and Power in Rotation
Again, we can calculate work in rotation analogous to the translational part. We do this through: ∫
θ2
W =
τz dθ
(10.5)
θ1
Dy, D. L. Phys 71 10.5
13
Physics 71 Final Exam Reviewer
Examples for Static Eq and CoG
• Diandrew Lexter L. Dy •
You have to familiarize yourselves with some possible problems that you can encounter for static equilibrium. Types of problems for this section include: 1. See-saw problems These are important problems because it’s pretty basic and unconvoluted. Examples include (PS6-1, PS6-2, PS6-3, PS6-5, PS6-6, PS6-13, PS6-16). You really have to know how to solve these problems before the exam, and while these problems might be repetitive in nature, they can be useful. Note that for many problems above, you can survive using only the condition τnet = 0. For other problems, you may have to use Fnet = 0. These problems include Q6-3, PS 6-14, PS 6-15. You don’t have to solve each problem though, just know how to solve them, and solve them quickly. 2. Hanging beam problems These problems are usually a beam attached to a hinge on a wall and supported by a string to keep it in static equilibrium. Problem examples include PS 6-11, PS 6-15, Q6-1. These questions most of the time requires trigo, because the string is usually at an angle. Determining the angle is at times the tricky part for these types of problem. A skill involved for these problems is finding the perpendicular component of the tension, so that you can just multiply it with the level arm r, hence the calculation is way easier. This skill is important if you wish to solve the problems quickly, and hence gives you enough time to review the answer. 3. Ladder problems These problems are quite convuluted in nature, because it requires you to solve a system of equations. Problem examples include: PS6-7 to 10, PS6-19.
This is a very condensed reviewer for the third Long exam. This is essentially just writing all the needed equations, with some concepts. Exercises are suggested from our previous problem sets and Quizzes. This summary in nature can be used as a guide for things you have to learn for the exam. If you cannot understand it, then reading the text would be helpful.
11
Ch 12: Static Equilibrium and Elasticity
This is the first Chapter for the Third long.
11.1
Static Equilibrium
For this section, you have to know what is static equilibrium. Basically, static equilibrium are situations when objects are not moving at all. Not even rotating, or moving at a constant speed. The two conditions of static quilibrium are: Fnet = 0; τ net = 0
[at all points]
(11.1) (11.2)
To differentiate this from equilibrium depends on how you define equilibrium. Young didn’t elaborate, hence we state that an object in equilibrium when it is moving translationally at constant speed. Don’t worry about this possible confusion, because I made sure that your exam, you only have to worry about static equilibrium, which has a precise definition.
11.2
Center of Gravity
A related topic is the center of gravity. This is related to the center of mass, but this time it’s the gravity that is being ’averaged’ over. More precisely, given a collection of n particles, then the center of gravity’s location is calculated using: ∑n ri Wi r = ∑i=1 (11.3) n i=1 Wi where for object i, ri is its position and Wi is its weight, which is just mi gi . When the acceleration due to gravity, gi , is constant for all particles, then the position of the center of gravity reduces to: ∑n ri mi r = ∑i=1 (11.4) n i=1 mi which is just the center of mass location. Hence the center of gravity corresponds to the center of mass when g is constant for all particles. This is true when the objects [ ]are near the surface of the earth, where the g = 9.81 sm2 is roughly constant. We can also find the center of gravity by noting that there is no net torque when the object is balanced at the center of gravity. This is what I usually use because I usually forget the average formula above. Dy, D. L. Phys 71 11.3
11.3
Elasticity
Objects are not rigid realistically, and they are at times stretched, squeezed or twisted in the presence of a force. For an object given a force F⊥ perpendicular to its cross sectional area A, we can define tensile stress as: Tensile stress =
F⊥ A
(11.5)
and we can also define tensile stress as: Tensile strain =
∆l l0
(11.6)
where l0 is the original length of the object, and ∆l is the amount of stretching and compression. We next define Young’s modulus Y as: Y =
Tensile stress . Tensile strain
(11.7)
In many objects, the Young’s modulus are relatively constant. We can similarly define bulk stress, bulk strain and bulk modulus when the force is all around the object, and 14
12
which we obtain the relations: Bulk stress =
F⊥ ; A
Bulk strain =
Bulk modulus =
∆V V0
Bulk stress . Bulk strain
We now review gravity. In our level, its classical, starting with the implications of Newton’s law of Gravitation. (11.8)
F⊥ ; A
12.1
Newton’s Law of Gravitation
Given two objects with masses m1 and m2 , the magnitude of the force of m1 on m2 is given by:
And similarly for shear stress: Shear stress =
Chapter 12: Gravity
Shear strain =
∆l l0
Shear stress Shear modulus = . Shear strain
F = (11.9)
In many objects, we can treat the modulus as constant. This is okay for small ∆l. For larger ∆l however, Y is not constant, and the behavior of stress vs strain is not linear. It has the shape as shown in Figure below
Gm1 m2 r2
(12.1)
where r is the distance between the two objects and G = 6.67 × 10−11 N m2 /kg 2 is the gravitational constant. This force is an attractive force, such that the direction of the force on m2 is towards m1 . Also, note that the Newton’s third law holds, because the reactive force has the same form as that above, that is: Fm1 onm2 = Fm2 onm1 =
Gm1 m2 r2
(12.2)
and the direction is opposite with each other (because its attractive). Examples Several examples include the pretty straightforward: PS78 or PS7-16. When there are many objects involved, then superposition is needed: Q6-7.
12.2
Figure 5: Stress vs strain of a metal You have to understand this figure. Point a is the point where the strain is still proportional to the strain. The region between a and b is a bit curved, but will not deform once released. Beyond b, plastic deformation follows, where there is permanent deformation once the object is relieved of stress. Finally, at point d, the object breaks. Examples Examples of this part involves mainly plugging the definition of stress or strain, or manipulating the equations to find the needed variable. Examples include PS7-1 to 5 and Q6-4. While some problems may require you to use previous knowledge or FBD like the elevator problem PS7-4. These examples are basic in nature, and does not require much derivations, but you have to at least know the definitions by heart. That ends Chapter 11
Dy, D. L. Phys 71 12.2
Acceleration due to gravity, mass and weight
When you consider one of the object as that of the earth, then we can calculate the force of earth on any object. The weight of the object is essentially the force by earth, and hence decreases as the object gets farther from earth. The acceleration at the surface of the earth on any mass m1 can be computed by equating to m1 g the force due to earth, hence m1 cancels and we obtain: g=
Gme re2
(12.3)
where me = 5.97 × 1024 [kg] is the mass of earth, and re = 6.38 × 106 [m] is the average radius of the earth. In any planet of mass mp and radius rp , the acceleration of gravity has a similar form: gp =
Gmp . rp2
(12.4)
Examples Examples include comparison of the surface gravity or weight at several heights: (PS7-6, PS7-22) and that of several planets (PS7-7, PS7-20). These problems are very basic and fundamental in nature, so it should be obvious that this is one of the many focus in any exam. 15
Note: A common mistake is not including the Earth’s radius for r. You have to be careful about this, if some height is given from the Earth’s surface, then r = h + re .
12.3
Circular Orbit
Any object (e.g. satellite) can orbit around earth (or any planet, just change me ) in a circular orbit if given the correct initial speed. The speed can be solved by equating the centripetal force mv 2 /r with that of the Newton’s gravitational force, which we get: √ Gme vc.o. = . (12.5) re You don’t have to memorize this equation though. Just know how to derive it!It’s even simpler, and you are sure about your result. Examples Examples can be pretty straightforward (PS7-10a). However, in many cases, the period is required. Recall that the period can be derived for circular orbits through T = 2π/ω where ω is the angular speed, which is related to the object’s speed by ω = v/r. These formulas should be natural to you by now; and if it isn’t, then you should try to memorize them as often as possible. Examples where orbits are needed include PS7-10, PS714, PS7-21. You should really solve these problems really quickly! Common problems include unfamiliarity with the calculator, and also non-inclusion of the earth’s radius. Be really careful. Try to solve these problems in under a minute. And also, be aware if your answer is correct and makes sense. Note that the answers to these problems are usually in km/s(for Earth). So answers that are very very far from these values magnitude-wise could be wrong.
12.4
Energy Methods
In many instances, it is often easier to calculate for speed using energy methods. Recall that energy is always conserved for these systems. Also, the kinetic energy still has the form K = 12 mv 2 . The potential energy though is different, and depends in your chosen location where potential is zero. We will use this point to be infinity, that is U (∞) = 0. Hence the potential energy has the form: U =−
Gm1 m2 . r
(12.6)
Note the negative sign, and that the denominator is just r, while the force is r2 . Energy methods can then be used by equating the initial energy with the final energy. For √ circular orbits, we can derive using the solved speed v = GM/r that E = U/2 and that K = −U/2.
because g is no longer constant. PS7-12 is an example of projectiles, and is an important and basic example. When the object from the surface is projected with enough speed such that it just escapes Earth, we call it escape velocity. Rule of thumb: an object reaches infinity at zero speed and hence the total energy at infinity is zero. Conservation of energy hence implies: Gme 1 mv 2 − = 0, 2 re
(12.7)
hence we obtain the escape speed: √ 2Gme . (12.8) vesc. = re √ Note the factor of 2 difference with the speed of circular orbits. Examples for escape speed problems include PS 7-9, PS7-15
12.5
Kepler’s laws
Briefly stated. Kepler law 1: Planets are moving in an ellipse with sun at the focus. Kepler law 2: Angular momentum about an axis perpendicular to the plane of orbit and through the sun is conserved. 4π 2 3 Kepler law 3: T 2 = GM a , where T is the period of s orbit and a is the semi-major axis of the planet’s orbit. Now, the important things to learn here is the second law and third law. Second law are used for problems when angular momentum can easily be calculated (when moment arm is perpendicular to the momentum, say when the planet is at the perihelion (nearest to sun) or aphelion (farthest). So learn to use it! The third law can be derived through circular orbits, then just change r into a. That is, assume it’s in circular orbit,√then calculate period as T = 2π/ω = 2πr/v with v = Gme /r solved previously for circular orbits. This can also be derived using integration methods, but its too complicated. Or you can memorize! Examples Straightforward application of 2nd law includes Q7-8. Sometimes, you might need conservation of energy, hence PS8-4. Third law has many examples, including: PS8-5 to 9. In my opinion, once you have the formula, then everything is just manipulating and calculator exercise. But of course, that’s my ’humble’ opinion.
Examples An object launched like a projectile, which has been covered in the previous chapters, is different in this chapter mainly Dy, D. L. Phys 71 12.5
16
13
Ch 14: Fluid Mechanics
Examples
Fluids is a nice chapter!
13.1
Familiarize yourselves with this concept through several examples: PS8-13 to 15, Q7-1, Q7-3, Q7-5. You must know how to solve these problems before exam!This is important.
Density
Density ρ is defined as: ρ=
mass m = . Volume V
13.4 (13.1)
Continuity equation
For incompressible fluids, the volume flow rate:
It is an intrinsic property: it is independent of the size of an object.
(13.6)
where Ais the cross sectional area and v is the fluid speed, This implies that:
Examples The calculation of density usually involves calculation of the volume. Now volume of regular objects like the sphere or cube should be familiar to you, and thus formulas will not be given on the exam. Problems include being careless (like forgetting the formula, or not taking the cube of a factor), or panicking because there is no time. Examples include PS8-10 to 11
13.2
dV = Av = constant dt
Pressure in a fluid
Pressure by a liquid is defined as: p=
F⊥ A
A1 v1 = A2 v2 This is the continuity equation. Examples
Examples include: PS 8-18, Q7-2, Q7-5, Q7-6, the shower head problem done in class. Familiarize yourselves with these examples as this will be important!
13.5 (13.2)
Bernoulli equation
The Bernoulli equation states that:
where A is the cross sectional area of the fluid, and F⊥ is the force perpendicular to the area, similar to our definition of stress. For liquids, pressure varies with depth through: p = p0 + ρgh
(13.3)
where h is the depth. We can select p0 to be exposed to the earth’s atmosphere (e.g. sea-level) such that p0 = patm = 101325 [Pa] = 1 [atm]. The pressure p above is also called absolute pressure. We also define gauge pressure as p − patm . When p0 = patm , then the gauge pressure (pg ) is: pg = ρgh
(13.7)
1 P + ρgh + ρv 2 = constant 2
(13.8)
While this is a long equation, you can think of this as the fluid counterpart for energy conservation, with the mass being the density. This has important concepts like as speed of fluid increases, the pressure decreases. Examples Examples include: PS 8-17, Q7-2, Q7-5. Another important concept!
(13.4)
Gauge pressure is hence easier to calculate. But be careful. The gauge doesn’t include the atmospheric pressure, so don’t include it in your calculations.
This ends Chapter 14!
Examples The examples include PS8-12,13. This concept is important, and you have to study how to solve these problems.
13.3
Buoyant force
When an object is floating or submerged in a fluid, then there is buoyant force brought about by the depth variation of the pressure. The buoyant force has the form F = ρgVdispbj.
(13.5)
Here Vdisp means the volume displaced by water, which is the volume of the object that is submerged in water. ρ here is the density of the fluid, not the object. Dy, D. L. Phys 71 13.5
17
14
Ch 13: Oscillation
Oscillations are by definition objects with back and forth motion.
14.1
Simple Harmonic Motion (SHM)
Simple harmonic motion has the property that you can express your equations of motion as: d2 x(t) = −ω 2 x(t) dt2
(14.1)
where ω is called the angular frequency. The solution for this equation is: x(t) = A cos(ωt + δ)
(14.2)
where δ is some phase angle that is determined at the initial position of the object. We can select this to be zero, which implies that x is initially at its amplitude:x(0) = A. The plot of x(t) is sinusoidal. From the graph, you should be able to find the period and the amplitude. The angular frequency can then be computed by T = 2π/ω. See Example: PS9-1 The object’s speed and acceleration can then be solved straightforwardly by differentiating the position with time. Hence: dx = −Aω sin(ωt + δ) dt
(14.3)
d2 x = −Aω 2 cos(ωt + δ) dt2
(14.4)
v(t) =
a(t) =
We therefore see that a = −ω 2 x at all times. Also, we note that when the position is maximum in magnitude (and hence also acceleration), then the speed is a minimum. And also vice versa. This is because when cos is max (in magnitude), then sin is minimum (in magnitude). The energy of SHM can be computed by noting that when the kinetic energy is zero, the x = ±A, hence E = U = 1 2 2 2 kA , where k = mω . This energy remains constant at all times. The kinetic energy is maximum when U = x = 0, or when the system is at equilibrium. Shadow The shadow of an object moving in uniform circular motion exhibits SHM. We can see this from the diagram below: The shadow then is the x-component of the position of the object. Hence x = r cos θ. Noting that the angular speed ωs = v/r is a constant, then we can express θ = ωs t + ω0 . Hence the shadow has the solution:
Figure 6: Shadow of an object in uniform circular motion exhibits SHM Spring-mass system k Spring-mass system exhibits SHM, with ω 2 = m , where k is the spring’s constant and m is the object’s mass. The period is hence:
T =
2π = 2π ω
√
m k
(14.6)
You must memorize these formulas, or derive it from FBD. The latter is however not advisable for the exam because it takes too much time. The spring-mass system has the same period if it is moving horizontally or vertically. Examples include: PS8-3, PS8-4. Simple pendulum For simple pendulum of small oscillations, we have ω 2 = gl , where l is the length of the spring. The period is hence: √ 2π T = = 2π ω
l g
(14.7)
You must memorize these formulas!!Very important! I think am more familiar with the period, because it is used very often. Then from that, I just calculate the frequency from T = 2π/ω. Examples include: PS8-6, PS8-8. You really must understand this part. Otherwise, you don’t have much hope for this part of the exam. Physical pendulum
x(t) = r cos(ωs t + θ0 ).
(14.5)
From the solution to SHM (14.2), we see that δ = θ0 and ω = ωs and the amplitude is r. Hence the shadow exhibits SHM. Dy, D. L. Phys 71 14.1
For physical pendulums, or for rigid bodies of small oscillations, we have ω 2 = mgd I , where m is the mass of the rigid object, d is the distance of the axis of rotation to the center of mass, and I is the moment of inertia about the axis. The 18
period is hence: √ T =
2π = 2π ω
I mgd
(14.8)
It doesn’t have to be said, but I will say it anyway.. this is important! Examples include: PS8-9, PS8-12. These examples are really important, so please understand them! Damped Oscillation
to zero. For overdamping, the rate of decay is slower than that of critical damping, but it also doesn’t oscillate but just decays. Forced Oscillation Oscillations can also be driven, that is you can simultaneously give a force on the object together with the restoring force. An interesting scenario occurs when the force is also sinusoidal, that is of the form F = F0 cos(ωD t), where ωD is the driving angular frequency. When this happens, then we obtain an amplitude plot with ωD as:
This part is more conceptual. For SHM, the amplitude does not decrease with time, which is not the case when friction is present. When the amplitude decays in time, then this is what we call damped oscillation. For several models, the frictional force has the form −bv such that the equations of motion has the form: d2 x dx = −ω 2 x − b dt2 dt
(14.9)
Note that b is the damping parameter. When b is large, then the systen is more damped, or high in friction. The solution to this equation is: x(t) = Ae−b/2m cos(ω ′ t + δ) where ω ′ has the form: ′
ω =
√
b2 k − . m 4m2
(14.10)
(14.11)
You don’t have to memorize these equations for the third exam, because it’s a bit convoluted I must say. But you have to understand the following cases: Case b = 0: then it is SHM. Case b < 2m, then ω ′ is real and greater than zero. This is called underdamping. Case b = 2m, then ω ′ is zero. This is called critical damping. Case b > 2m, then ω ′ is imaginary. This is called overdamping. Also, you have to understand the plots for these cases as shown below.
Figure 8: Amplitude vs driving frequency for different damping parameters We therefore see that as ωD approach the natural angular frequency (angular frequency without damping or driving force, or that solved using SHM), then the amplitude peaks! This peaking of the amplitude is called resonance, hence the resonance frequency is also the natural frequency. Note that: frequency is ”f” and angular frequency is ω. This is important! Do not be confused about this. They are related by ω = 2πf , such that T = 1/f . That ends Chapter 13!
Figure 7: Oscillation for different damping What this means is that underdamping oscillates but with amplitude decaying to zero. The critical damping however does not oscillate and the amplitude immediately goes Dy, D. L. Phys 71 14.1
19
15
Ch 15: Mechanical Waves
15.2
Mechanical waves are waves that travel in a medium. They have the form f (x − vt); f (x + vt);
wave propagating to the positive x wave propagating to the negative x (15.1)
For sinusoidal mechanical waves, it has the form: y(x, t) = A cos(kx − ωt + δ) or A cos(kx + ωt + δ) (15.2) depending on the propagation of the wave. We call ω still as the angular frequency, while k is the wave number. We define λ = 2π/k as the wavelength of the wave, which is similar to the period but with the space instead of time. The speed of propagation of the wave is defined as v = ω/k = f λ. Let us list the important formulas: 2π λ 1 2π T = = f ω k=
ω = 2πf v=
ω = f λ. k
Pav =
1√ µFT ω 2 A2 2
(15.4)
is a transverse wave, because the particles of the medium is moving in the y-direction. Longitudinal waves are however, when the movement is on the same direction of the direction of propagation. Note about transverse waves: The propagation speed is the speed of the movement of the ’pulse’. The medium did not move in the propagation direction, but just in the y direction. But there is a seeming motion of the pulse, because the crest seems to move. We can also define transverse speed as the speed of the particles of the medium. This is just vy = dy/dt. This is not the same as the propagation speed!
(15.6)
where ω is the angular frequency of the wave, and A is its amplitude. There is also the maximum power Pmax = 2Pav . Don’t bother too much with Pmax , just Pav . The loudness of the sound of the string is described by the intensity, defined by: Pav 4πr2
(15.7)
where r is the distance of the listener to the source of the sound. These formulas are important and must be memorized, because it will be too involved to derive them. Examples Several examples that require some or all of these formulas include PS10-4,5 PS 10-14,15. You have to know these problems, because they are important!
15.3
Types of Mechanical waves: Transverse waves and Longitudinal waves. Transverse waves are those where the particles in the medium move in a transverse direction (perpendicular to the direction of propagation). Hence,
Dy, D. L. Phys 71 15.3
where µ = mass/length = m/L is the mass density of the string. Here FT is the tension of the string. The average power due to the string is given by:
I=
Types of Mechanical Waves
y(x, t) = A cos(kx − ωt + δ)
When a string with tension is given a transvers speed vy , then there is a speed of the transverse wave, or propagation speed, v given by: √ FT v= (15.5) µ
(15.3)
These are equations that you have to learn by heart! When given a plot of y vs x and y vs t, you should be able to determine easily the amplitude, the period and the wavelength. After this, you should be able to compute f, k, ω, v. Do this quickly!Remember, if it takes you a minute to find any of the variables, then you are too slow, and hence needs more practice. Examples: See PS10-1, PS10-3 for example. Do this quickly! Solve each in under two minutes, and you should be fine. Remember, practice makes cliches!!
15.1
Transverse wave on a string
Standing waves
When two waves ’collide’, then they superpose, or you just add them! There are some cases though that after two waves are added, the resulting wave doesn’t seem to propagate, that is a crest of the wave doesn’t move. These cases are called standing waves. Two waves that have the same frequency and wave number, and same amplitude but moving in opposite propagation direction can be a standing wave. We define the node of a standing wave as the points which does not move. The antinode are the minimum and maximum values of the wave, which
Figure 9: Node and antinode 20
Note that you must know how to draw a wave with a certain number of nodes or antinodes. For example: draw a string of length L with 10 nodes. You should also be able to show that the frequency of the same string with 10 antinodes is higher than that of 10 nodes, or 5 antinodes or 5 nodes.
15.4
Harmonic frequencies of a string
For a string with both ends fixed, you can obtain several natural frequencies as shown in the diagram below.
The fundamental wavelength is the largest wavelength that you can achieve. This happens when λ1 = 2L. With v = f λ, then we obtain the fundamental frequency as: v 2L
(15.8)
For other harmonics, we can calculate the wavelength and the frequencies as: λn =
2L λ1 = ; n n
fn = nf1 =
nv ; 2L
n = 1, 2, 3...
16
Ch 16.8: Doppler Effect
When a source emitting a frequency is moving, then the frequency heard by the listener will be different than the frequency from the source. This is Doppler effect. These are the important concepts you have to understand for Doppler effect: When the source and/or listener is approaching each other, then the frequency heard by the listener will increase. If its going away, then its the opposite. That is: fL > fs :
approaching
fL < fs :
moving away
(16.1)
Next.. you have to remember this factor: v ± vL fL = fs (16.2) v ± vs where v is the speed of sound in air. Note that we can choose either plus or minus in both numerator and denominator, depending on whether the direction of movement of the listener and the source. We just have to remember that if it is approaching, then the frequency of the listener has to increase, and when moving away, it has to decrease. Then we adjust the signs accordingly. For example, suppose that the listener is going to the −x direction with speed vL and the source is going to the +x direction with speed vs . Then both movement should cause an increase in fL because both are approaching. Hence to increase, the numerator should add, and denominator should be negative. Hence the equation is: v + vL fL = fs (16.3) v − vs
Figure 10: Harmonic frequencies
f1 =
That ends Chapter 15!
(15.9)
Now suppose that the listener is going to the x direction with speed vL and the source is also going to the +x direction with speed vs . Then the listener is going away to the source, hence causing a decrease of the frequency, so that we take the minus sign.Hence the equation is: v − vL fL = fs (16.4) v − vs Try it for other cases just to familiarize yourselves. While these equations look complicated, it is actually not as convuluted as it seems. You just have to know that: fL > fs :
approaching
The fundamental frequency has the highest amplitude, and is usually heard most prominently (I ∝ A2 ). You have to memorize these formulas, or just draw the largest wavelength you can achieve when both ends are fixed. However, it might be easier to derive them.
Examples
Examples
Examples include: PS10-10 and 10-11. Try to read more examples in the book for practice. This is very important!
Familiarize yourselves with some important examples about strings. For example: PS 10-2, PS 10-6, PS 10-9, PS 10-13. Dy, D. L. Phys 71 16.0
fL < fs :
moving away v ± vL fs fL = v ± vs
(16.5)
That ends Ch 16.8!
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