General Physics I (aka PHYS 2013) › ~vvanchur › 2015PHYS2013 › Slides.pdf · CHAPTER 1...

CHAPTER 1 CHAPTER 2 CHAPTER 3 CHAPTER 4 REVIEW

General Physics I(aka PHYS 2013)

PROF. VANCHURIN(AKA VITALY)

University of Minnesota, Duluth(aka UMD)


OUTLINE

CHAPTER 1

CHAPTER 2

CHAPTER 3

CHAPTER 4


SECTION 1.1: THE NATURE OF PHYSICS

I MathematicsI is the language of science: physics, astronomy, chemistry,

engineering, geology, etc.I very abstract math ideas find their way into science (e.g.

complex numbers, differential geometry)I It was conjectured that all of the mathematics has a

physical realization somewhere (e.g. multiverse theories)I Physics

I serves as a bridge (or foundation) to other sciences.I one can never prove anything in physics, but math is also

not as pure as seems (Godel’s theorems)I it is however remarkable how successful the language of

math is in describing the world around us.I Think of physics as a toolbox of ideas which can be used in

building scientific models in chemistry, biology, geology,astronomy, engineering, etc.


SECTION 1.2: SOLVING PHYSICS PROBLEMS

I ConceptsI in most disciplines the more material you can memorized

the better your final grade will be. Not the case in physics.I concepts is what you have to understand (actually there is

only a single concept and almost everything follows)I learning how to learn physics concepts will be your first

and perhaps most difficult task.I Problems

I the only way to evaluate if you really understand conceptsis to solve problems. understanding solutions not enough.

I you might have hard time solving problems at first, but it isessential to learn how to solve problems on your own.

I keep track of how many problems you solved by yourselfin each chapter (without anyone helping or googling)


SECTION 1.3: STANDARDS AND UNITSI Units

I Physics is an experimental science and experimentsinvolves measurements (e.g. time, length, mass).

I To express the results of measurements we use units. Unitsof length, units of time units of mass.

I There are some standard units just because historically wedecided so (e.g. seconds, meters, kilograms).

I One can always convert from one system of units toanother given a conversion dictionary.

I Time: measured in seconds, milliseconds, microseconds,...I What is time and arrow of time is a deep philosophical question. We might discuss it a bit in the

last day of classes in context of the second law of thermodynamics.

I Length: measured in meters, millimeters, micrometers, ....I Despite of the fact that length and time appear to us very differently, there is a very deep

connection (symmetry) between them. We might discuss is briefly when we discuss gravitation.

I Mass: measured in units of gram, milligram, microgram, ...I Mass is also something very familiar to us in everyday life, but also has very deep properties

connecting it so length and time. We might mention it briefly in connection to black-holes.

I Other units can be formed from seconds, meters and kilograms


SECTION 1.4: CONVERTING UNITS

I Dimension.I Any physical quantity expressed in units of TIME, LENGTH or MASS is said to have dimensions of

time, length or mass respectively. More generally one can have physical quantities which havemixed dimensions. For example if d has units of LENGTH and t has units of TIME, then quantity

v =d

t(1)

has units of LENGTH/TIME.I Evidently Eq. (1) has quantities with the same dimension (i.e. LENGTH/TIME) on both sides of the

equation. This must be true for any equation that you write. Checking that the quantities on bothsides of equation have the same dimension is a quick, but very important test that you could dowhenever you setup a new equation. If the dimension is not the same than you are doingsomething wrong.

Conversion.I Sometimes you will need to convert from one system of units to another. This can always be done

with the help of conversion dictionary. For the case of conversion from standard system of units toBritish system of units the dictionary is:

1 in = 2.54 cm (2)

1 pound = 4.448221615260 Newtons (3)

I If we are given a quantity in units of speed, then we can convert it from one system of units toanother.


SECTION 1.5: UNCERTAINTY AND SIGNIFICANT

FIGURES

Experimental Measurements.I Measurements are always uncertain, but it was always hoped

that by designing a better and better experiment we can improvethe uncertainty without limits. It turned out not to be the case.

I There is a famous uncertainty principle of quantum mechanics,but you will only learn it next year in (PHYS 2021) if you decideto take it.

I From our point of view uncertainty is nothing but uncertainty inmeasurements. This (as well as significant figures) will bediscussed in your lab course.


SECTION 1.6: ESTIMATES AND ORDER OF

MAGNITUDE

Theoretical Estimates.I Similarly to uncertainties in experimental measurements,

theoretical predictions are never exact. We always makesimplifying assumption and thus the best we can hope for is anestimate for the physical quantities to be measured.

I A useful tool in such estimates is known as order-of-magnitudeestimate (also know as outcome of “back-of-the-envelopecalculations”).

I Such estimates are often done using the so-called dimensionalanalysis - i.e. just use the known quantities to form a quantitywith the dimension of the quantity that you are looking for.


SECTION 1.7: VECTORS AND VECTOR ADDITIONVectors.

I Some physical quantities are describe by a single real number. We call thesequantities - scalar quantities or scalars.

I Other quantities also have a direction associated with them and thus are describeby three real numbers.

~A = (Ax,Ay,Az). (4)

We call these quantities - vector quantities or vectors. There are also tensors etc.I When dealing with vectors it is often useful to draw a picture:

I Vectors are nothing but straight arrows drawn from one point to another.Zero vector is just a vector of zero length - a point.

I Length of vectors is the magnitude of vectors. The longer the arrow thebigger the magnitude.

I It is assumed that vectors can be parallel transported around. If you attachbeginning of vector ~B to end of another vector ~A then the vector ~A + ~B is astraight arrow from begging of vector ~A to end of vector ~B.

Coordinates.I The space around us does not have axis and labels, but we can imagine that

these x, y and z axis or the coordinate system to be there.I This makes it possible to talk about position of, for example, point particles

using their coordinates - real numbers.I Since one needs three real numbers to specify position it is a vector. Similarly,

velocity, acceleration and force are all vectors.


SECTION 1.8: COMPONENTS OF VECTORSI Symmetry. You might complain that there is arbitrariness in how one chooses

coordinate system or what components of the vector are and you would be right.It turns out that the physically observable quantities do not depend on thechoice of coordinate systems and thus one can choose it to be whatever is moreconvenient. Moreover, this symmetry is an extremely deep property which givesrise to conservation laws that we will learn in this course.

I Magnitude. The length of vector or magnitude is a scalar quantity

~A = | ~A| = A (5)

or in components

(Ax,Ay,Az) =√

A2x + A2

y + A2z . (6)

I Direction. One can also find direction of vector using trigonometric identities.I Addition. Two vectors can be added together to get a new vector

~C = ~A + ~B (7)

an in component form

(Cx,Cy,Cz) = (Ax,Ay,Az) + (Bx,By,Bz) = (Ax + Bx,Ay + By,Az + Bz). (8)


SECTION 1.9: UNIT VECTORSI Unit vectors is a vector (denoted with a hat) that has magnitude one.

|u| =√

u2x + u2

y + u2z = 1. (9)

There are three unit vectors are so important that there are special lettersreserved to denote these vectors

i = (1, 0, 0)

j = (0, 1, 0)

k = (0, 0, 1). (10)

I Multiplication / division by scalar. Any vector can be multiplied by a scalar toobtain another vector,

~A = C~B. (11)

In components from

(Ax,Ay,Az) = C(Bx,By,Bz) = (CBx,CBy,CBz). (12)

I Components. Any vector can be written in components in two ways:

~A = (Ax,Ay,Az) = Ax i + Ay j + Azk (13)


SECTION 1.10: PRODUCT OF VECTORS

I Scalar (or dot) product. Dot product is a multiplication between two vectorswhich produces a scalar:

~A · ~B = ~B · ~A = C (14)

In components

~A · ~B = (Ax,Ay,Az) · (Bx,By,Bz) = AxBx + AyBy + AzBz = AB cosφ. (15)

One can also derive multiplication table for unit vectors.I Vector (or cross) product. Cross product is a multiplication between two vectors

which produces a vector:

~A× ~B = −~B× ~A = ~C (16)

In components

~A×~B = (Ax,Ay,Az)×(Bx,By,Bz) ≡ (AyBz−AzBy,AzBx−AxBz,AxBy−AyBx) = ~C.(17)

One can also derive multiplication table for unit vectors.


SECTION 2.1: DISPLACEMENT, TIME, AVG. VELOCITYI Point particles.

I Objects (car, ball, stone) often modeled as point particles.I Position is described by a vector in some coordinate systemI 1D coordinate system for motion along a straight line.

I Position.I As time progresses position vector changes with time

~r(t) = (x(t), y(t), z(t)). (18)

I For motion in 1D only one component is relevant: x(t).I Average velocity.

I In 1D average velocity is defined as

vavg ≡∆x∆t

=x(t2)− x(t1)

t2 − t1. (19)

where ∆x ≡ x(t2)− x(t1) and ∆t ≡ t2 − t1.I Question: Can traveled distance be larger than displacement?

Can it be smaller? Can it be the same?


SECTION 2.2: INSTANTANEOUS VELOCITY

I Instantaneous velocity.I Instantaneous velocity (or just velocity) is defined as

v(t) ≡ lim∆t→0

vavg = lim∆t→0

∆x(t)∆t

=dx(t)

dt. (20)

I Given x as a function of t one can always find instantaneousvelocity at every moment of time by simple differentiation.

I Graphical representation.I average velocity is a slope of a line joining the coordinates

of initial (x1, t1) and final (x2, t2) points points on x(t) graphI instantaneous velocity if a slope of the tangent line to x(t) at

a given time.

I Question: Can instantaneous velocity be larger than average?Can it be smaller? Can it be the same?


SECTION 2.3: ACCELERATIONI Average acceleration.

I Given instantaneous velocity as a function of time v(t), onecan calculate average acceleration:

aavg ≡∆v∆t

=v(t2)− v(t1)

t2 − t1. (21)

I This is analogous to how the average velocity was defined.I Instantaneous acceleration.

I Instantaneous acceleration (i.e. an acceleration at a givenmoment of time) is defined by taking a limit,

a(t) ≡ lim∆t→0

aavg = lim∆t→0

∆v∆t

=dv(t)

dt. (22)

I Graphical representation.I avg. velocity is slope of line joining (v(t1), t1) and (v(t2), t2)I instantaneous velocity if a slope of the tangent line to v(t) .

I Question: Can instantaneous acceleration be larger thanaverage? Can it be smaller? Can it be the same?


SECTION 2.4: MOTION WITH CONST. ACCELERATIONConstant Acceleration.

I Motion with constant acceleration is a motion for whichinstantaneous acceleration is a constant function

a(t) = ax. (23)

I Velocity for such motion changes as

vx(t) = v0x + axt (24)

I Position changes as

x(t) = x0 + v0xt +12

axt2. (25)

I Note that ax, v0x, x0 are some fixed numbers representingacceleration, velocity and position at time t = 0, but t is avariable which can take any non-negative value.


SECTION 2.4: MOTION WITH CONST. ACCELERATION

Useful relations.I One can obtain other useful relations from Eqs. (24) and

(25) by expressing t in one equation and plugging it intoanother

vx(t)2 = v20x + 2ax (x(t)− x0) . (26)

I Another useful relation is obtained from Eqs. (24) and (25)by expressing ax in one equation and plugging it intoanother

x(t)− x0 =12

(vx(t) + v0x) t. (27)

I Question: What would change in equations (23), (24), (25), (26)and (27) if the initial time is at t0 6= 0?


SECTION 2.4: MOTION WITH CONST. ACCELERATION

Example 2.4. A motorcyclist heading east through a small townaccelerates at a constant acceleration 4.0 m/s2 after he leaves citylimits. At time t = 0 he is 5.0 m east of the city-limits while he moveseast at 15 m/s. (a) Find position and velocity at t = 2.0 s. (b) Whereis he when his speed is 25 m/s?


SECTION 2.5: FREE FALLING BODIES

Free falling.I Free fall is a very deep concept in physics. All it means

that in 4-dimensional space-time free falling object movealong straight lines (geodesics).

I For now we are only interested in motion very close tosurface of the Earth and such motions can beapproximated as motions with constant acceleration

ax = g = 9.80 m/s2. (28)

I Thus all of the concepts and equations considered in theprevious section apply.

I Note that g is taken to be positive and thus it makes senseto choose x-axis to point vertically and downward. If youchoose the x-axis to point upward then ax = −9.80 m/s2.


SECTION 2.5: FREE FALLING BODIES

Example 2.6. One-euro coin is dropped from the Leaning Tower ofPisa and falls free from rest. What are its position and velocity after1.0 s?


SECTION 2.5: FREE FALLING BODIESExample 2.7. You throw a ball vertically upward from the roof of a tall building. The ballleaves your hand at a point even with the roof railing with an upward speed of 15.0 m/s; theball is then in free fall. On its way back, it just misses the railing. Find (a) the ball’s positionand velocity 1.00 s and 4.00 s after leaving your hand; (b) the ball’s velocity when it is 5.00 mabove the railing; (c) the maximum height reached; (d) the ball’s acceleration when it is at itsmaximum height.


SECTION 2.6: VELOCITY/POSITION BY INTEGRATION.Velocity by Integration. From definition of acceleration

ax(t) =dvx(t)

dt(29)

and thus in differential fromdvx = ax(t)dt (30)

or in integral form ∫ vx(t2)

vx(t1)dvx =

∫ t2

t1

ax(t)dt. (31)

Therefore for t1 = 0 and t2 = T we have

vx(T) = vx(0) +

∫ T

0ax(t)dt. (32)

If we replaceT → t t→ τ vx(0)→ v0x (33)

then we get

vx(t) = v0x +

∫ t

0ax(τ)dτ (34)

and in the case of constant acceleration

vx(t) = v0x + axt. (35)


SECTION 2.6: VELOCITY/POSITION BY INTEGRATION.Position by Integration. From definition of velocity

vx(t) =dx(t)

dt(36)

and thus in differential fromdx = vx(t)dt (37)

or in integral form ∫ x(t2)

x(t1)dx =

∫ t2

t1

vx(t)dt. (38)

Therefore for t1 = 0 and t2 = T we have

x(T) = x(0) +

∫ T

0vx(t)dt. (39)

If we replaceT → t t→ τ x(0)→ x0 (40)

then we get

x(t) = x0 +

∫ t

0vx(τ)dτ (41)

and in the case of constant acceleration

x(t) = x0 +

∫ t

0(v0x + axτ)dτ or x(t) = x0 + v0xt +

12

axt2. (42)


SECTION 2.6: VELOCITY/POSITION BY INTEGRATION

Example 2.9. Sally is driving along a straight highway in her 1965 Mustang. At t = 0, whenshe is moving at 10 m/s in the positive x-direction, she passes a signpost at x = 50 m. Herx-acceleration as a function of time is

ax(t) = 2.0 m/s2 −(

0.10 m/s3)

t. (43)

(a) Find her x-velocity vx(t) and x(t) as functions of time. (b) When is her x-velocity greatest?(c) What is that maximum x-velocity? (d) Where is the car when it reaches that maximumx-velocity?


SECTION 3.1: POSITION AND VELOCITY VECTORS

Extra dimensionsI We now generalize motion to 2D and 3D, but one can extend it furtherI Extra dim. (six of them) are also needed to formulate string theoryI These extra dim. are usually very small and can often be neglected

PositionI Vector~r specifies position of an object in three dimensions

~r = (x, y, z) = xi + yj + zk (44)

I If the object is in motion, then the position changes with time

~r(t) = (x(t), y(t), z(t)) = x(t)i + y(t)j + z(t)k. (45)


SECTION 3.1: POSITION AND VELOCITY VECTORSVelocity

I Average velocity vector defined as

~vavg ≡∆~r∆t

=~r(t2)−~r(t1)

t2 − t1, (46)

I Instantaneous velocity vector defined as

~v(t) ≡ lim∆t→0

~vavg = lim∆t→0

∆~r∆t

=d~r(t)

dt. (47)

or in components as

~v =d~r(t)

dt=

(dx(t)

dt,

dy(t)dt

,dz(t)

dt

)=

dx(t)dt

i +dx(t)

dtj +

dx(t)dt

k (48)

and if we denote ~v ≡(vx, vy, vz

)then

vx =dx(t)

dt

vy =dy(t)

dt

vz =dz(t)

dt. (49)


SECTION 3.1: POSITION AND VELOCITY VECTORS

Example 3.1. A robotic vehicle, or rover, is exploring the surface of Mars. The stationary Marslander is the origin of coordinates, and the surrounding Martian surface lies in the xy-plane.The rover, which we represent as a point, has x- and y-coordinates that vary with time:

x(t) = 2.0 m−(

0.25 m/s2)

t2

y(t) = (1.0 m/s) t +(

0.025 m/s3)

t3

z(t) = 0 m (50)

(a) Find the rover’s coordinates and distance from the lander at t = 2.0 s. (b) Find the rover’sdisplacement and average velocity vectors for the interval t = 0.0 s to t = 2.0 s. (c) Find ageneral expression for the rover’s instantaneous velocity vector ~v. Express ~v at t = 2.0 s incomponents from and in terms of magnitude and direction.


SECTION 3.2: THE ACCELERATION VECTORAcceleration

I Average accelaration vector defined as

~aavg ≡∆~v∆t

=~v(t2)− ~v(t1)

t2 − t1, (51)

I Instantaneous velocity vector defined as

~a(t) ≡ lim∆t→0

~aavg = lim∆t→0

∆~v∆t

=d~v(t)

dt. (52)

or in components as

~a =d~v(t)

dt=

(dvx(t)

dt,

dvy(t)dt

,dvz(t)

dt

)=

dvx(t)dt

i +dvy(t)

dtj +

dvz(t)dt

k (53)

and if we denote~a ≡(ax, ay, az

)then

ax =dvx(t)

dt=

d2x(t)dt2

ay =dvy(t)

dt=

d2y(t)dt2

az =dvz(t)

dt=

d2z(t)dt2

. (54)


SECTION 3.2: THE ACCELERATION VECTOR

Example 3.2. Let’s return to motion of the Mars from the previoussection. (a) Find the components of the average acceleration for theinterval t = 0.0 s to t = 2.0 s. (b) Find the instantaneousacceleration at t = 2.0 s.


SECTION 3.2: THE ACCELERATION VECTORParallel and Perpendicular Components.I A useful example of a moving coordinate system in 2D is when one of the axis

point in the direction of the velocity vector ~v.I Then we can decompose the acceleration vector into component parallel~a|| and

perpendicular~a⊥ to the direction of motion.I Magnitude of velocity vector changes as

dv(t)dt

= v(t) ·~a||(t) (55)

I Direction of the velocity vector changes as

dv(t)dt

=~a⊥(t)v(t)

. (56)

I Parallel component is responsible for changes in magnitude of the velocityvector (i.e. speed), but not the direction

~a|| =

+

a||v ~v speed is increasing

0 speed is not changing−

a||v ~v speed is decreasing.

(57)

I Perpendicular component is responsible for changes in direction of velocity,butnot in magnitude.

I If there is only a perpendicular component then the object is in circular motion.


SECTION 3.3: PROJECTILE MOTION

Projectile motionI Projectile motion is completely determined by gravitationally acceleration and

air resistance starting from initial condition determined by position and velocity.I The fact that you have to specify an even number of initial data is a consequence of the

fact that (differential) equations of motion are of the second order in time.I Projectile motion is always confined to a 2D plane determined by two vectors:

gravitational acceleration vector and initial velocity vector.I It is convenient to choose a coordinate system so that one of the axis is vertical,

the other one is horizontal (and there is no motion in z-direction.)I For y-axis pointing upward and x-axis horizontally

~a = (0,−g) = −gj (58)

and thus

~v(t) = v0x i + (v0y − gt)j

~r(t) = (x0 + v0xt) i +

(y0 + v0yt−

12

gt2)

j (59)


SECTION 3.3: PROJECTILE MOTION

Example 3.6. A motorcycle stunt rider rides off the edge of a cliff.Just at the edge his velocity is horizontal with magnitude 9.0 m/s.Find the motorcycle’s position, distance from the edge of the cliff, andvelocity 0.50 s after it leaves the edge of the cliff.


SECTION 3.4: MOTION IN A CIRCLEUniform motionI Uniform motion is when the direction of velocity vector changes, but magnitude

(or speed) does not change, i.e.

dvdt

= 0 anddvdt6= 0, (60)

I This implies that

~v ·d~vdt

= 0 (61)

but if the magnitude of acceleration also does not change,

a(t) = a⊥(t) = const. (62)

then

0 =ddt

a2 =ddt

(~a ·~a) =ddt

(d~vdt·

d~vdt

)= 2

d2~vdt2·

d~vdt

(63)

I By combining Eqs. (61) and (63) in 2D we must conclude that

~v ∝d2~vdt2

. (64)

I If the proportionality constant is negative, then solutions of the above equationare sines and cosines (you will see this differential equation over and over inphysics courses).


SECTION 3.4: MOTION IN A CIRCLE

Uniform motionI We have shown that

~v(t) ∝d2~v(t)

dt2. (65)

I With little more of calculus (that we shall skip) one can show the such motiongives rise to circular motion described by equation

R =

√(x(t)− X)2 + (y(t)− Y)2 (66)

where (X,Y) is at the center of the circle and R is its radius.I By choosing the origin of coordinates at the center of circle, i.e. making

(X,Y) = (0, 0) (67)

we can simplify Eq. (66) to get

r =√

x(t)2 + y(t)2 = R (68)


SECTION 3.4: MOTION IN A CIRCLERadial (or centripetal) acceleration.By differentiating Eq. (66) with respect to time once we get

0 =dx(t)

dtx(t) +

dy(t)dt

y(t) (69)

or~r · ~v = 0 (70)

and by differentiating it twice we get

0 =

(dx(t)

dt

)2+

(d2x(t)

dt2

)x(t) +

(dy(t)

dt

)2+

(d2y(t)

dt2

)y(t) (71)

orv2 = −~a ·~r = −~a⊥ ·~r. (72)

But since the two vectors~a⊥ and~r point in opposite directions we have

v2 = a⊥R (73)

or

a⊥ =v2

R. (74)


SECTION 3.4: MOTION IN A CIRCLEPeriodic motion.

I Since the motion is circular the object must come to were it started at some finitetime. This time is called period, T, and the motion is called also periodic.

I It is useful to write an exact solution for periodic motion around origin

~r(t) =

(R sin

(2π

tT

+ φ

),R cos

(2π

tT

+ φ

))(75)

where R, T and φ are some constants.I Note that as time goes from t to t + T the arguments of the sin and cos functions

change by 2πand thus the position vector does not change

~r(t) =~r(t + T). (76)

I Evidently, an object in circular motion is traveling a distance 2πR with a constantspeed v and thus the period must be

T =2πR

vand v =

2πRT. (77)

I By combining it with

a⊥ =v2

R(78)

we get

a⊥ =4π2R

T2. (79)


SECTION 3.4: MOTION IN A CIRCLE

Example 3.12. Passengers on a carnival ride move at a constant speed in a horizontal circle ofradius 5.0 m, making a complete circle in 4.0 s. What is their acceleration?Step 1: What is the coordinate system? Let choose a (moving) coordinate system with ||direction in the direction of motion and ⊥ direction in the radial direction.Step 2: What is given? Period

T = 4.0 s

and radius

R = 5.0 m

Step 3: What do we have to find? Then

a⊥ =4π2(5.0 m)

(4.0 s)2= 12 m/s2 = 1.3 g

anda|| = 0.


SECTION 3.5: REFERENCE FRAME.Reference frame.

I We have already mentioned how the choice of coordinate system is importantfor calculations, but should not matter for physically observable quantities.

I We have also discussed (in context of 2D motions) how it is sometime useful towork with respect to a moving coordinate system.

I More generally, let object A move with respect to a coordinate system (orreference frame) of object B, described by position vector

~rA/B(t) (80)

and let object B move with respect to a coordinate system (or reference frame) ofobject C, described by position vector

~rB/C(t) (81)

then we say that to describe object A with respect to a coordinate system (orreference frame) of object C, we use the following rule

~rA/C(t) =~rA/B(t) +~rB/C(t). (82)

I By taking a time derivative of Eq. (82) we get a rule for adding velocities

d~rA/C(t)

dt=

d~rA/B(t)

dt+

d~rB/C(t)

dt~vA/C(t) = ~vA/B(t) + ~vB/C(t). (83)


SECTION 3.5: REFERENCE FRAME.Example 3.14-15. There is a 100 km/h wind from west to east. (a) If an airplane’s compassindicates that it is headed due north, and its airspeed indicator shows that it is moving throughthe air at 240 km/h, what is the velocity of the airplane relative to earth? (b) What directionshould the pilot head with speed 240 km/h to travel due north?Step 1: Choose coordinate system. Let x-axis to point east and y-axis to point north.Step 2: What is given?

~vA/E = 100 km/h i

Step 3: What do we have to find? (a) The airspeed indicator tells us that

~vP/A = 240 km/h j

and thus~vP/E = ~vP/A + ~vA/E = 100 km/h i + 240 km/h j

(b) In general velocity of airplane relative to air is

~vP/A = x i + y j with vP/A =√

x2 + y2 = 240 km/h

and thus

~vP/E = ~vP/A + ~vA/E =(

x i + y j)

+ 100 km/h i = (x + 100 km/h)i + y j = y j.

Thus we have two equations with two unknowns with solution

~vP/A = x i + y j = −100 km/h i + 218 km/h j.


SECTION 4.1: FORCES AND INTERACTIONSFundamental forces.

I There are four types of fundamental forces:1) electromagnetic,2) weak,3) strong and4) gravitational.

I The electromagnetic and weak (and to some extend strong)forces had been successfully unified into electroweak (and tosome extend GUT) theory.

I The situation is much worse with regards to gravitational forcewhich is manifests itself not through exchange of participles likeother forces, but through curvature of space-time.

I The string theory does describe a way of how to think aboutgravity (perturbatively), but it is too naive to expect that we willknow the final answer any time soon.

I In our everyday experience the (microscopic) forces manifestthemselves trough (macroscopic) forces.


SECTION 4.1: FORCES AND INTERACTIONS

Superposition of forces.There might be a number of different forces action on a given object,and the total force (or net force) is given by

~R =∑

i

~Fi = ~F1 + ~F2 + ~F3 + ... (84)

where the usual vector addition is used, i.e.

(Rx,Ry,Rz) =

(∑i

Fix,∑

i

Fiy,∑

i

Fiz

). (85)

Strength of the net force is then

R =√

R2x + R2

y + R2z =

√√√√(∑i

Fix

)2

+

(∑i

Fix

)2

+

(∑i

Fix

)2

. (86)


SECTION 4.1: FORCES AND INTERACTIONS

Example 4.1. Three professional wrestlers are fighting over a champion’sbelt. The forces are on horizontal plane and have magnitudes and directions:

F1 = 250 N and θ1 = 127◦

F2 = 50 N and θ2 = 0◦

F3 = 120 N and θ3 = 270◦. (87)

Find the components of the net force on the belt and it magnitude anddirection.


SECTION 4.2: NEWTON’S FIRST LAW

First Law.I A body acted on by no net force, i.e.∑

i

~Fi = 0 (88)

has a constant velocity (which may be zero) and zero acceleration.

I (This tendency for a body to continue its motion is know asinertia and is extremely important concept in theory of generalrelativity.)



Example 4.2. In the classic 1950 science-fiction film Rocketship X-M,a space-ship is moving in the vacuum of the outerspace, far from anystar or planet, when it engine dies. As a result, the spaceship slowsdown and stops. What does Newton’s first law say about this scene?



Example 4.3. You are driving a Maserati GranTurismo S on astraight testing track at a constant speed of 250 km/h. You pass a1971 Volkswagen Beetle doing a constant speed 75 km/h. On whichcar is the net force greater?



Inertial frames.I It is important to note that the Newton’s first law is not obeyed

in all reference frames (e.g. inside of an accelerating train).

I Those frames of references where it is obeyed is called inertialframe of reference (e.g. inside a train moving with constantvelocity).

I Surface of Earth is not exactly an inertial reference frame (why?)but it is pretty close to being inertial.

I For inertial reference frames one can easily go from one frame toanother using

~rA/C(t) =~rA/B(t) +~rB/C(t). (89)

I and~vA/C(t) = ~vA/B(t) + ~vB/C(t). (90)

which makes such frames particularly useful.



Examples. In which of the following situations is there zero net force on thebody?

I An airplane flying due north as a steady 120 m/s and at aconstant altitude?

I A car driving straight up a hill with a 3o slope at a s constant 90km/h

I A hawk circling at a a constant 20 km/h at a constant height of15 m above an open field?

I A box with slick, firctionless surface in the back truck as thetruck accelerates on a a level road at 5 m/s2.


SECTION 4.2: NEWTON’S SECOND LAWSecond Law.

I If a net external force acts on a body, the body accelerates. The directionof acceleration is the same as the direction of the net force. The mass ofthe body times the acceleration vector of the body equals to the net forcevector, i.e. ∑

i

~Fi = m~a (91)

or

~a =

∑i~Fi

m. (92)

I As a vector equation in 3D it is equivalent to three equations∑i

Fix = max∑i

Fiy = may∑i

Fiz = maz. (93)


SECTION 4.2: NEWTON’S SECOND LAW

Units.I With the help of second law we can now relate the units of

force to units of mass and acceleration.I Since each equation must have the same units on both

sides we see thatN = kg ·m/s2. (94)

I If a 1 kg object moves with acceleration 1 m/s2 then theremust be a net fore of 1 N applied to it.

I In British system of units

1 lb = 4.448 N. (95)


SECTION 4.2: NEWTON’S SECOND LAW

Example 4.4. A worker applies a constant horizontal force withmagnitude 20 N to a box with mass m = 40 kg resting on a level floorwith negligible friction. What is the acceleration of the box?


SECTION 4.3: MASS AND WEIGHT

Weight.I Is a gravitational force acting on an object close to the surface of

the Earth (to be precisely at the see level)

~w = m~g. (96)

I It is a vector, but one often refers to the magnitude of the weightforce as weight

w = mg (97)

or even to mass itself just because one, i.e. m, is simply related tothe other, i.e. w.

I One should however be careful when weight is measured aboveor below the sea level (e.g. on the airplane), as the weight forcecan vary.


SECTION 4.3: MASS AND WEIGHT

Example 4.7. A 2.49× 104 N Rolls-Royce Phantom traveling in the+x direction makes an emergency stop; the x-component of the netforce acting on it is −1.83× 104 N. What is its acceleration?


SECTION 4.4: NEWTON’S THIRD LAW

Third Law. If a body A exerts a force on body B (an “action”), thenbody B exerts a force on body A (a “reaction”). These two forces havethe same magnitude, but are opposite in direction. These two forcesact on different bodies.

~FA on B = −~FB on A. (98)



Example 4.8. After your sport car breaks down, you start to push itto the nearest repair shop. While the car is starting to move, how doesthe force exert on the car compare to the force the car exerts on you?How do these forces compare when you are pushing the car along at aconstant speed?



Example 4.9. An apple sits at rest on a table, in equilibrium (i.e.static). What forces act on the apple? What is the reaction force toeach of the forces acting on the apple? What are the action-reactionpairs?



Example 4.10. A stonemason drags a marble block across a floor bypulling on a (weightless) rope attached to the block. The block is notnecessarily in equilibrium. How are the various forces related? Whatare the action-reaction pairs?



Example. You are driving a car on a country road when a mosquitosplatters on the windshield. Which has the greater magnitude: theforce that the car exerts on the mosquito or the force that the mosquitoexerted on the car? Or are the magnitudes the same? If they are thedifferent, how can you reconcile this fact with Newton’s third law? Ifthey are equal, why is mosquito splattered and damaged while the caris undamaged?


SECTION 4.5: FREE-BODY DIAGRAM

Free-body diagram. A useful tool in solving problems on Newton’slaws is to draw free-body diagram, where a chosen body appears byitself without of its surroundings, and with vectors drawn to show themagnitude and directions of all the forces that act on the body.The tricky part here is to include all of the forces that act on achosen body, but not to include any other forces that act onother bodies.


SECTION 4.5: FREE-BODY DIAGRAMFree-body diagram.


CHAPTER 0: MATHEMATICS

I AlgebraI Solving linear equationsI Solving quadratic equationsI Solving system of two equations with two unknowns

I TrigonometryI Hypotenuse from two sidesI Side from hypotenuse and another sideI Angle from hypothenuse and one sideI Angle from two sides, etc.

I CalculusI Differentiation of standard functionsI Product rule, quotient rule, chain rule.I Indefinite integrals (i.e. antiderivitives)I Definite integrals over interval


CHAPTER 1: UNITS, PHYSICAL QUANTITIES AND

VECTORSI Units

I SI system of unitsI British system of unitsI Conversion of units

I Physical quantitiesI Scalars, Vectors, etc.

I Vector algebraI Vector representations (graph., comp., unit vec.)I Magnitude and direction of a 2D vector from componentsI Components of a 2D vector from magnitude and directionI Vector addition/subtraction (graph., comp., unit vec.)I Vector multiplication by scalar (graph., comp., unit vec.)I Dot product between 2 vectors (graph., comp., unit vec.).I Finding angle between 2 vectors from dot productI Cross product between 2 vectors (graph., comp., unit vec.)I Determine direction of cross product using right hand rule.


CHAPTER 2: MOTION ALONG A STRAIGHT LINE

I Definition of relevant physical quantitiesI Position and displacementI Average velocity and instantaneous velocityI Average acceleration and instantaneous acceleration

I Graphing of physical quantities vs. timeI Graphs a-vs-t, v-vs-t and x-vs-t (obtain one from another)I Calculate quantities from graphs of a-vs-t, v-vs-t or x-vs-t

I Differentiation and integrationI Calculate functions a(t), v(t) or x(t) (calc. one from another)I Calculate quantities from functions of a(t), v(t) or x(t)

I 1D kinematic problem (e.g. free fall)I Step 1: Choose 1D coordinate system (x-axis, origin)I Step 2: Determine initial conditions (e.g. position, velocity)I Step 3: Determine final conditions (e.g. position, velocity)I With many stages of motion consider each stage separatelyI Final cond. for one stage are initial cond. for next stage


CHAPTER 3: MOTION IN TWO OR THREE DIMENSIONS

I 2D kinematic problem (e.g. projectile motion)I Step 1: Choose 2D coordinate system (x-axis, y-axis, origin)I Step 2: Determine initial conditions (e.g. position, velocity)I Step 3: Determine final conditions (e.g. position, velocity)I With many phases solve for each phase separately.I Final cond. for one phase are initial cond. for next phase

I HintsI Intermediate answers can be algebraic, not numeric.I Every equation must have the same units on both sides.I Solve system if there are as many equations as unknowns.I Remember that quadratic equation may have two solutions.

I Uniform circular motionI Acceleration is perp. to velocity and points towards centerI Relations between: acceleration, velocity, radius, period.

I Relative motionI Find relative quantities by vector addition/subtraction


CHAPTER 4: NEWTON’S LAWS OF MOTION

I Newton’s First LawI Distinguishing inertial and non-inertial reference frames.I Apply the first law to object moving with zero acceleration

(which may or may not have zero velocity)I Newton’s Second Law

I Determine magnitudes of all forces applied to object, thenet force and acceleration along a given direction.

I Apply the second law to object moving with non-zeroacceleration (which may or may not be have const. speed).

I Newton’s Third LawI Determine all action-reaction pairs of objects.I Apply the third law to every pair of objects.

I Forces problemsI Step 1: Choose a coordinate system (1D or 2D).I Step 2: Draw a free-body diagram for each object.I Step 3: Apply Newton’s Laws (1st, 2nd and/or 3rd)

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