Introductory Physics I - Lecture notes - 1 - 32 PDF

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BRIEF COURSE INTRODUCTION Full course information, schedule, exam policy etc is at: http://www.pa.msu.edu/courses/2005fall/PHY231/ Some things to note immediately: - Weekly LON-CAPA homeworks are due Tuesdays at 11pm. The first homework is due Tuesday September 6. Your MSU account should give you access to LON-CAPA - REGISTER YOUR CLICKER using LON-CAPA - Get to know the exam formula sheet OUTLINE - Mechanics, Thermal physics and Waves Mechanics Introduction to physics, units, scales and accuracy (1 lecture - Chapter 1) Motion in one dimension (2 lectures - Chapter 2) Motion in two dimensions (3 lectures - Chapter 3) Newton’s laws of motion (3 lectures - Chapter 4) - First Midterm Work, energy and power (3 lectures - Chapter 5) Momentum, collisions (2 lectures - Chapter 6.1-3) Rotational motion and gravity (3 lectures - Chapter 7) Forces and rotational motion (2 lectures - Chapter 8.1-5 Energy and momentum in rotational motion (1 lecture - Chapter 8.1-8.7) - Second Midterm Thermal Physics Solids and fluids (3 lectures - Chapter 9) Thermal physics (2 lectures - Chapter 10) Energy in thermal processes (3 lectures - Chapter 11,12.1) Laws of thermodynamics (1 lecture - Chapter 12.2-3) - Third Midterm Laws of thermodynamics (1 lecture - Chapter 12.4-5) Waves Vibrations and waves (3 lectures - Chapter 13) Sound and sound waves (3 lectures - Chapter 14) - Final

Lecture 1 - Units, Accuracy and Estimates 1

UNITS, DIMENSIONAL ANALYSIS Units The base SI (syst`eme internationale) units used in mechanics are: meter (m), kilogram (kg), second (s). This set of units is also called the mks system. Two other systems in common use are the cgs (centimeter, gram, second) and the US system (foot, pound, second). Besides the mks units used in mechanics, there are four other SI base units, which we will use later in the course, and for completeness it is worth noting them now: ampere (A), Kelvin (K), mole (mol), candela (cd). Dimensional analysis When using equations all terms which are added or subtracted in the equation must have the same units. Testing whether this is true is called dimensional analysis and is essential to any correct equation. To illustrate dimensional analysis consider the following quantities which describe motion in one dimension: time, t - [t] = T position x or displacement ∆x - units [x] = [∆x] = L speed or velocity v - units [v] = L/T acceleration a - units [a] = L/T 2 Now consider a dimensional analysis of the following equations, if ∆t = v0 + 2a∆x; then T =

L L + 2 L (INCORRECT EQU AT ION ) (1) T T

and if,

L L L (2) v 2 = v02 + 2a∆x; then ( )2 = ( )2 + 2 L T T T Equation (1) is dimensionally incorrect so it cannot be a valid law of physics. Equation (2) is dimensionally correct so it may be a valid law of physics. In many simple problems the simplest combination of the variables which is dimensionally correct is actually the correct answer. It is a very good problem solving strategy to try this if you have no better ideas! If you write down or use equations that are dimensionally incorrect on the other hand, examiners tend to be rather hard on you.

2

Changing units Given the fact that 1meter = 3.281f eet and 1mile = 5280f eet, convert 50miles/hour to meters/second 50miles/hour = 50

5280f t 1m = 22m/s 3600s 3.281f t

(3)

ACCURACY It is really important to know how accurate numbers are. Yet in everyday life the accuracy of statistical results are often incorrect or in many cases not quoted at all. We will not get into the details of putting the correct error bars on measurements, though you will do this in the PHY251 laboratory. However we also need to know how to calculate the accuracy when we combine results in equations. For example if we have an equation like v = v0 + at, and we have known accuracy in v0 , a and t, what is the accuracy of v? You need to know two different ways of treating the accuracy of a number: The number of significant figures Some examples: 0.0012 has two significant figures 1.2 × 10−3 has two significant figures 15000 has five significant figures 1.5 × 104 has two significant figures 1.5000 × 104 has five significant figures Rule for multiplication and division When dividing or multiplying two quantities A, B, the result has its number of significant figures equal to the smallest number of significant figures of the two quantities A, B . In our example v = v0 + at, the product at yields a value with its number of significant figures equal to the smallest number of significant figures of the two quantities a and t. The number of decimal places Some examples: 0.0012 has four decimal places 1.2 × 10−3 has four decimal places 15000 has no decimal places 3

1.5 × 104 has no decimal places 1.5000 × 104 has no decimal places Rule for addition and subtraction When adding or subtracting two quantities A, B, the result has its number of decimal places equal to the smallest number of decimal places of the two quantities A, B . In our example v = v0 + at, the product at yields a value with a number of decimal places. The result v has its number of decimal places equal to the smallest of the number of decimal places of the quantity v0 and the quantity at. Numerical example Consider v0 = 1.25m/s, a = 9.8m/s2 , t = 0.1s. The product at = 1m/s (one significant figure). The sum v = v0 + at = 2m/s (no decimal places) SCIENTIFIC NOTATION AND ABBREVIATIONS In science and technology there is an enormous range of values for many of the quantities of interest. For example the power used by a bacteria to swim is about 0.00000000000000001watts, while the power produced by a nuclear power plant is about 1, 000, 000, 000watts. It is clumsy to write all of these zeros so instead we introduce different notation. Though the terms million and billion are used most frequently in the news, they are not typical in science. One reason is that the billion has a different meaning in the US compare to in Europe. Instead we use scientific notation and abbreviations, which are universaly accepted. Scientific notation Examples (keeping the full number of significant figures) 0.012 = 1.2 × 10−2 0.000, 000, 000, 000, 000, 01watts = 10−17W 1, 000.000 = 1.000000 × 103 Abbreviations tera (T) = 1012 , giga (G) = 109 , mega (M) = 106 , kilo (k) = 103 , milli(m) = 10−3 , micro (µ) = 10−6, nano (n) = 10−9 pico(p) = 10−12, femto (f) = 10−15

4

Examples 0.012W = 1.2 × 10−2 W = 12mW 0.000, 000, 000, 000, 000, 01watts = 1 × 10−17W = 0.01f W 1, 000.000 = 1.000000 × 103 = 1.000000kW 1, 000, 000, 000watts = 1GW (ignoring the implied 10 significant figures) ORDER OF MAGNITUDE ESTIMATES Find the order of magnitude of the gravitational force of attraction between two students at opposite ends of the lecture hall. F =

GMm ; r2

G = 6.67 × 10−11

N m2 kg 2

(4)

this leads to the estimate 2

F ≈

6.67 × 10−11 Nkgm2 100kg 100kg (30m)2

≈ 10−10N

(5)

Is this a big force? For comparison, the force of gravity on each of the two students is mg ≈ 9.8m/s2 × 100kg ≈ 1000N . Note that though this is small, gravitational detectors are sensitive enough to notice it. Gravitational detection is used for mining exploration, particularly in the search for heavy metals such as gold. It is important to be able to estimate so that you can quickly determine whether a hypothesis is likely to be true. For example is it likely that the electric field from power lines affects human health? An analogous question would be: Does living near a large gold repository (eg. fort Knox) affect human health?

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Lecture 2 - Motion in one dimension Covers Sections 2.1 - 2.4 Mechanics is the study of motion (kinematics) and of the forces which produce motion (dynamics). We shall first study kinematics in one dimension. To describe kinematics in one dimension, we will introduce several different quantities. We all have a colloquial understanding of terms like speed, velocity, acceleration, distance. The physics usage of these terms is similar to the colloquial usage, but there are important differences. These differences will become very pronounced when we study motion in two dimensions. The quantities we need to understand and to calculate with are: (i) Position, x, displacement ∆xif = xf − xi , distance d (ii) Average velocity v, instantaneous velocity v, average speed, instantaneous speed. (iii) Average acceleration a and instantaneous acceleration a. A scalar quantity is just a number and has no direction. A vector quantity has a magnitude and a direction. Distance and speed are scalars, while position, velocity and acceleration are vectors. Position, velocity and acceleration To define a position we have to first decide on our reference point or origin. In general we have to define a reference frame, as we will see in the case of two dimensional motion. The position of point P is an arrow which points from the origin to P. This arrow has a magnitude, and a direction, which is either positive or negative in this one dimensional case. The displacement between two positions P and Q is an arrow which points from P to Q, and we define ∆xP Q = xQ − xP . For example if xP = 10 and xQ = 12, ∆xP Q = xQ − xP = 2. Notice that if xP = 10 and xQ = 8, ∆xP Q = −2, indicating that the arrow points in the negative direction. In kinematics, we are interested in motion and a useful way of looking at motion is through motion diagrams or motion equations, which show position (and/or velocity and acceleration) as a function of time. For example consider motion described by, x = −3 + 4t − t2 (1) where x is in meters and t is in seconds. Note that to ensure that this equation is dimensionally correct, the units of the number -3 are meters, the units of the number 4 are m/s and the units of the -1 prefactor of the t2 term 1

are m/s2 . When motion equations are written down these units are usually omited, but they are implied and sometimes you will be asked to figure out what the units must be to ensure that the equation is dimensionally correct. We can draw many different graphs of x vs t. Are all of them physically reasonable? One restriction is that at any given time, the position is single valued because we can’t be in two places at the same time. However we can return to the same position at a later time, as occurs in the motion described by Eq. (1). From motion diagrams or motion equations, we can find the displacement between two times, ti and tf , so that ∆xif = x(tf ) − x(ti ). Now we are ready to talk about velocity. Velocity is a vector so it can also be thought of as an arrow which has a magnitude and a direction. The average velocity is the rate of change of position with time. In mathematical terms this is, x(tf ) − x(ti ) ∆xif = v= (2) tf − ti ∆tif where we have defined ∆tif = tf −ti . The instantaneous velocity is the special case where ∆tif becomes very small. In fact in the limit where this quantity goes to zero is the limit in which the ratio ∆xif /∆tif becomes a derivative and we enter the world of calculus. Though we shall not use calculus, we are carrying out similar calculations when we calculate the instantaneous velocity. Lets look at the motion diagram corresponding to Eq. (1). Some points to note: (i) The average velocity in going from B to D is zero! (ii) The average velocity in going from A to B is 3m/s (iii) The average velocity in going from C to D is −1m/s (iv) The instantaneous velocity at any position is the slope of the graph at that position. The acceleration is the rate of change of velocity and it is also a vector so it can be thought of as an arrow with a magnitude and a direction. In mathematical terms the average acceleration is then given by, a=

v (tf ) − v (ti ) ∆vif = tf − ti ∆tif

(3)

where v(t) is the instantaneous velocity at time t and we use ∆tif = tf − ti , as in Eq. (2). The instantaneous acceleration is the special case where ∆tif becomes very small. Looking at the motion curve corresponding to Eq. (1), note the following: 2

(i) The acceleration is the change in the slope of the motion curve. (ii) The change in slope is negative for the whole motion curve, which means that Eq. (1) describes a system which is accelerating in the negative x direction. (iii) Draw a graph and/or write an equation for the case where the change in slope is positive? Distance and speed As remarked above, the distance and speed are scalars. To be frank, there is quite of bit of confusion about the definition of these quantities, even in introductory textbooks. The case where confusion arises is illustrated by the example Quiz 2.1 in the text (page 27). The total displacement is zero, however the total distance is 200 yards. In our calculations, we will take the distance to be the total distance travelled. Note that the magnitude of the displacement is the net distance which in this example is zero. In a similar way we shall define the average speed as follows, average speed =

total distance total time

(4)

Note the following about distance and speed (i) The distance is taken to be the total distance travelled (ii) Distance and speed are always positive (iii) The instantaneous speed is the modulus or magnitude of the instantaneous velocity. (iv) The average speed is always greater than or equal to the modulus of the average velocity, over the same time interval. Example Considering running from the origin (ie. xi = 0) directly to a position x = 25m and then reversing direction and running back to the final position xf = 11 in a total time of ∆tif = 3s. The displacment ∆xif = 11, so the average velocity v if = 3.67m/s, but the total distance travelled is 39m so the average speed is 39m/3s = 13m/s.

3

Lecture 3 - Motion at constant acceleration - an important special case An example of motion at constant acceleration is motion near the earth’s surface due to gravity. To restrict this motion to one dimension we consider just up and down motion. Later when we study motion in two dimensions, we will analyse projectile motion (eg shoot the monkey). It is possible to write down formulas for displacement and velocity when the acceleration is constant. To do this, first notice that if the acceleration, a, is constant, then the velocity increases at a constant rate. This means that if we plot velocity as a function of time it is a straight line with slope a. We thus deduce that the velocity as a function of time has the form, v = v0 + at

(1)

Now the average velocity over a time period, t, is given by, v=

∆x 1 (v0 + v) = t 2

(2)

Here we used Eq. (2) of Lecture 2, but used t instead of ∆t. We also dropped the subscipts used in Lecture 2. This makes the formula more compact, but we have to keep in mind the meaning of each of the variables. Using Eq. (1) for v in Eq. (2), we find, ∆x 1 , (v0 + v0 + at) = t 2

(3)

and solving for ∆x, leads to, 1 ∆x = v0 t + at2 . 2

(4)

We may also eliminate t from this equation in favor of v using Eq. (1), after some algebra this yields, (5) v 2 = v02 + 2a∆x. Eqs. (1), (2), (4) and (5) are the key constant acceleration formula and they appear in many problems due to their broad importance. These formula will also be the basis of understanding many types of motion in two dimensions.

1

An application - Whooping it up An excited cowboy fires his handgun(at head height) vertically upward with an initial speed v0 = 311m/s. Ignoring air drag on the bullet, how long before the bullet lands on the cowboy’s head? How high does the bullet go? Take the gravitational acceleration to be −9.81m/s2 . (Note that this acceleration is the same regardless of whether the motion is positive (up) or negative (down)) To find the time taken for the bullet to reach its highest altitude note that at the highest point of its motion, v = 0 so we must have(using Eq. (1)), 0 = 311m/s − 9.81m/s2 t (6) which gives t = 31.7s. The time until the bullet hits the cowboy’s head is then 2t = 63.4s. The displacement at the top of its trajectory is found from Eq. (4), which yields ∆x = 4930m (about 3 miles!). What is the velocity of the bullet when it hits the cowboy? Answer 311m/s! How would these results be affected by air drag?

2

Lecture 4 : Motion in two dimensions I: Co-ordinate systems, vectors and projectiles Co-ordinate systems (fixing landmarks) Anyone who has adventured into the wilderness or has driven in an unfamiliar neighborhood without a map is well aware of the need to carefully keep track of where they are. A simple way to do this is to identify a set of landmarks which can be used to retrace their steps. Global positioning technology (GPS - Garmin or Magellan) has made this obsolete as there are now a set of satellites which have positions that are precisely known and can be used as landmarks to determine any location on earth to an accuracy of a couple of meters or better. Any positioning system uses co-ordinate or reference systems in which distances are measured. In everyday life we almost always treat a landmark on the earth’s surface as our reference location. A position on the earth surface is determined by a distance east or west and a distance north or south from this origin, for example we may consider Sparty as the origin of our co-ordinate system and measure distances east or west and north or south from Sparty. In the case of an airplane we also have to specify an altitude. Our reference altitude is usually ”sea level” so that an altitude of 35, 000f t means 35, 000f t above sea level. We shall first work with two dimensional co-ordinate systems, as this already allows us to treat a wide range of problems, such as projectile motion (eg. throwing a ball) and instead of east, west, north and south we shall use a distance along the x-direction and a distance along the y-axis. In fact instead of a distance which is always positive we use an “x-co-ordinate”, x, which may be positive or negative and a “y-co-ordinate”, y, which may also be positive or negative. Any position, ~r P , in our two dimensional co-ordinate system is then specified by two numbers (x, y). This position vector is defined with respect to (x, y) = (0, 0), which is the origin of our co-ordinate system. This co-ordinate system is called the Cartesian co-ordinate system and was invented by Ren´e Descartes in 1637. Polar co-ordinates Although it is natural to define position using the x and y co-ordinates, that does not tell us the distance of P from the origin. To find the distance we need to use the Pythagoras law, r 2 = x2 + y 2 1

(1)

The distance r tells us the distance from the origin to point P, and is the magnitude of the vector ~r but it does not tell us the direction. The correct direction is set by defining an angle θ, which is defined as the angle from the x − axis in a counterclockwise direction. The vector ~r is a vector with magnitude r and direction determined by θ. This way of describing the position of point P is called polar co-ordinates. Using trigonometry we write equations relating the cartesian co-ordinates (x, y) to the polar co-ordinates (r, θ). That is, x = rcos(θ) ; y = rsin(θ) (2) The velocity and acceleration are also vectors and may be written in either cartesian co-ordinates or polar co-ordinates. For example the acceleration at the earth surface may be written in cartesian co-ordinates as, ~a = (ax , ay ) = (0, −9.81m/s2 )

(3)

and in polar co-ordinates as, ~a = (a, θ) = (9.81m/s2 , 270o) = (9.81m/s2 , −90o )

(4)

Vector operations To describe motion in two dimensions, we need to add, subtract and carry out a basic multiplication operation with vectors. For example lets consider the final position of a student who starts at Sparty and walks 10m east, then turns and walks 7.07m south west. How far (what distance) is the student from Sparty. This is a problem in vector addition, which we can write as, ~rf = ~r1 + ~r2 = (x1 , y1 ) + (x2 , y2) = (x1 + x2 , y1 + y2 )