vectors and their rules - illinois state

PHY110 LECTURE NOTES BY DR. SU

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©19© Vectors and their rules

When a particle does not move along a line, a sign is not sufficient (we need 3 signed

numbers) to represent its direction of motion, it must be represented by a vector.

Special properties of vectors (1) easy to write physical laws (2) new algebra

A scalar quantity has only magnitude and no direction in space (eg. mass, distance, time,

energy, temperature ...), and follows ordinary rules of algebra.

A vector quantity has both magnitude and direction in space (eg. displacement, velocity,

acceleration ...) and the combination of vector quantities follow new rules. (ordinary algebra in

3 directions and a new rule for combing them or resolving them.)

To see the vector rules which work for any vector, let us examine a special kind of vector,

displacement, a vector that represents a change in position for a point. If a particle changes its

position by moving from A to B, we say that it undergoes a displacement from A to B, we

represent this with an arrow pointing from A to B (the arrow specifies the vector graphically).

ex: car moved from Bloomington to Indianapolis (200 mi=magnitude, due East=direction)

Draw on paper length proportional to the magnitude of the vector. (eg. A and 2A)

The symbol for vector: (1) printed boldface Α, handwriting symbols (2) r A or (3) A

The magnitude: A or | A | (always ≥ 0) both signify a scalar.

ex: A is the displacement vector from Bloomington to Indy

ex: B is the displacement vector from Bloomington to Milwaukee (about 200 mi, also)

good: use A, B (instead of 200 mi due East or 200 mi due North)

bad: be careful | A | = | B |, but A ≠ B

fortunately it means no more than one city is 200 mi due East and the other one is 200 mi due

North


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Many quantities already have directional information, so they may naturally be represented by

a vector. Velocity has a direction, but speed does not. When you are near a cliff you care not

only your speedometer reading, but also in which direction you are driving. And we know the

change of velocity vector (not speed) corresponds to a force vector.

Adding vectors: graphical method

We know how to add 2 numbers algebraically How do we add two vectors?

ex: car moves from Bloomington->Urbana (A) then later Urbana->Chicago (B)

We can represent the overall displacement with two successive displacement vectors, A and B,

the net effect of these two displacements is a single displacement from B to C (S). We call S the

vector sum (or resultant) of the vectors A and B. (not the usual algebraic sum)

A

BS

Written as: S = A + B

“+” or “sum” or “add” have different meanings for vectors than they do in the usual algebraic

operations (somehow the horizontal displacement caused by the vector A due east is cancelled by

that from B)

Graphic procedure: lay out B with its tail at the head of A, construct S by drawing an arrow

from the tail of A to the head of B

Properties:

(1) the order of addition does not matter: A + B = B + A (commutative law)

Note: vectors in the brackets are the same as un-bracketed ones


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A

BS

(A)

(S)

(2)When dealing with more than two vectors, grouping doesn’t matter: (A+B) + C = A + (B+C)

(associative law)

A

BS

C

(-B) has same magnitude but opposite direction as B

B + (-B) = 0 (adding -B has the effect of subtracting B)

D ≡ A - B = A + (-B) (by adding -B to the A)

note: We can still only add vectors of the same kind.

ex: if 2 vectors' tails are together, how to draw the sum and the difference of the two?

©20© Vectors and their components

Adding vectors graphically can be tedious, a nice and easier technique involves algebra, but it

requires that the vectors be placed on a rectangular coordinate system. We ignore z for now and

add it in later.

Drop ⊥ lines from ends of A to x-, y- axes, Ax and Ay these are called the components of A in


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the x and y directions. The procedure of forming them is called resolving the vector.

Note: A = Ax + Ay but components are either horizontal or vertical

A

Ay

Ax

The net effect of A is indistinguishable from the sum of vector components Ax and Ay

Note:

Unit vectors: A unit vector has a magnitude of 1 and points in a particular direction. It has no

dimension or unit. Its sole purpose is to point, basically telling direction.

Ax is a signed number, the scalar component Ax = A cosθ

Here: θ is the angle that vector A makes with +x, and A = | A |

Ax < 0 makes the x-component points in the - x direction

Ax = Ax i

Here i is an unit vector in the +x (due east) direction.

Similarly,

Ay =Ay j with j due North

and Ay, (scalar) components Ay = A sinθ, Ax = A cosθ

Ay < 0 makes the y-component point in the - y direction

A is equivalent to ( |A|, θ ) or (Ax, Ay)

A =(Ay2 + Ax2) and tanθ = Ay / Ax , the new resolving / combining rules.

Adding vectors revisited (by components)

Using the graphic method is tedious, the accuracy is limited, and it is challenging in 3d. So,

combine components, axis by axis

(i+j) + (2i+j) move 1E, 1N then 2E, 1N

=(i+2i) + (j+j) = 3i + 2j = ( √(32+22), tan-1(2/3) )


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S = A + B is equivalent to: Sx = Ax + Bx , Sy = Ay + By , Sz = Az + Bz

component of the sum is the sum of the components

- simpler looking - earlier to do (scalars)

(1) resolve the vectors into scalar components

(2) axis by axis, combine scalar components to get the components of the sum S

(3) combine to get S (unit vector notation or magnitude / direction)

other vector rules (multiplications) later

ex: (2 i + j) + 2 (i - j) = (2 i + j) + 2 i - 2 j = 4 i - j

magnitude 17= 4.123, direction θ = tan-1(-1/4)= 14 ° , south of east

ex: Find force C such that 3 forces balance A+B+C=0, for (1) A = B =1 (2) A = 1, B = 2.

Assume that A and B are 60° above and below x.

A

B

C

x

y�

Note: (1) by symmetry C = 1 and it is along -x. (not obvious but check it)

(2) Cx = - (Ax+Bx) = -(cos60°+2 cos-60°) = -1.5

Cy = - (Ay+By) = -(sin60°+2 sin-60°) = 0.866

magnitude of C is (-1.5)2+(0.866)2 = 1.73

direction θ = tan-1(0.866/-1.5)= 180° - 30° = 150°, north of west

3D extension: A = Ax i + Ay j + Az k This is the unit vector notation of A


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Unit vectors in + x, +y and +z axes are labeled i, j, and k.

A

Ax

Ay

Az

x

y

z

S = A + B, Sx = Ax + Bx, Sy = Ay + By, Sz = Az + Bz

eg. 3 vectors S = A + B + C

ex: A car moves from a to b. Draw position and displacement vectors.

A

B

B-A

distance between end points: | A-B |

©21© The characterization of 2D motions

Things don’t always move in straight lines. We will gain experience in a 3D motion.

Examples:

(1) A ball rolling on a table top

(2) A ball rolling down an inclined plane

(3) A baseball flying off McGuire’s bat

(4) A car moving in a racing track

(5) You falling off the Watterson tower

(6) The Moon around the Earth, Earth around the Sun, ...

Keep track of the position (location) of the object as a function of time. Are you a Chicago or

St. Louis fan?


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Example: Birthday present: Hot dogs and beer in a major league ball park. a

b

o

(1) Ken Griffy, Jr.’s base-run: create a drawing for mom (bird-eye view) showing where

home-base (point a) 1st-base (b) and you (o) are. He went straight from a to

b.

(2) Call your brother (with your cellular) and describe the situation: a

b

o

40

2045

25

45-20

40-2525

15

29.15

Ken was 20m East/40m North of you when started, and 45m East/25m North when he

was finished. His net Eastward movement was 25m, and his Southward movement was 15m;

therefore, the net (south of east) movement was 29.15m, according to the Pythagorean

theorem. Note 29.15=√[(25)2+(-15)2].

Ken finished in 5 seconds so his average speed was 29.1m/5s = 5.8 m/s.

The direction of his run was θ = tan-1(-15/25) = - 31° (31° South of East)

To summarize:

Ken’s speed (for a uniform motion) may be defined as

v = d/ε (over an amount of time ε)

Since he ran straight from a to b, Ken’s movement has a direction from a to b

Since vector d carries both the magnitude of d and the direction of the movement required,

define V = d/ε (divide by ε has no effect on the direction, V points along d)

V has both magnitude (speed) and the direction (of motion).


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The velocity defined is actually the average velocity, however, for Ken’s uniform velocity, it

coincides with his instantaneous velocity. If his velocity changes, (either non-uniform speed or

changing direction) an instantaneous velocity needs to be defined at every point along the path.

(Instantaneous) velocity is defined as the limit when the time increment is small (very close to

zero).

V = lim!"0d/ε

Direction is tangent to the path according to this definition.

r(t)

r(t+!t)

d=!ra

b

o

Position vectors r(t) and r(t+Δt) locate points a and b, it is not difficult to find that

d = r(t+Δt)-r(t) =Δr (or displacement) while time taken ε = Δt.

Thus in new notation, V = Δr/Δt and V = lim!t"0 Δr/Δt

To calculate, let’s break things into x- and y- components:

a

b

o xb

yb

xa xb xa-

!x!yya

yb ya-

d = !( )

!x2

!y 2+

V = [r(t+Δt)-r(t)] /Δt = [(xb i + yb j) -(xa i + ya j)]/Δt = [(xb-xa) i + (yb-ya) j]/Δt

= (Δx/Δt) i + (Δy/Δt) j

= V x i + V y j (sum of average velocities in x-, y- directions)

and V = lim!t"0 Δr/Δt =

lim!t"0 (Δx/Δt) i +

lim!t"0 (Δy/Δt) j

= (dx/dt) i + (dy/dt) j (small dx, dy, dt)

= Vx i + Vy j (horizontal, vertical V’s)


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sum of velocities in x-, y- directions (Vy < 0 means points down)

Ah-Ha: 2D motion is 2-1D motions combined!

Recipe: Get Vx and Vy, combine them to get V.

speed V = | dr / dt | = √[(dx)2+(dy)2]/dt =√[(dx/dt)2+(dy/dt)2] = √[(Vx)2+(Vy)2]

direction θV = atan (dy / dx) = atan[ (dy/dt) / (dx/dt) ]=atan[ Vy /Vx ]

©22© 2D Acceleration

(3) Ken’s Eastward and Northward position x(t) and y(t) are, t ≤ 5s:

x(t) = 20 + 5 t y(t) = 40 - 3 t (runner.f: data t-x-y, graph x-y)

Velocities:

Eastbound Vx = dx/dt = d/dt (20 + 5 t) = 5 m/s (uniform motion)

Northbound Vy = dy/dt = d/dt (40 - 3 t) = -3 m/s (uniform, and Southbound really)

Speed V = √(Vx2+Vy2) = √ [ (5)2 + (-3)2 ] = 5.8 m/s

V-direction θ = tan-1(Vy /Vx) = tan-1(-3 /5) = - 31° (31° South of East)

Next example: Shoot for the bulls-eye:

You project the bullet straight at the target center but always miss by a bit.

Check the high speed video analyzer: bulls_eye.f

t x(t) y(t) 0.00 0 0.00000 0.01 10 -0.00049 0.02 20 -0.00196 0.03 30 -0.00441 0.04 40 -0.00784 0.05 50 -0.01225 0.06 60 -0.01764 0.07 70 -0.02401 0.08 80 -0.03136 0.09 90 -0.03969 0.10 100 -0.04900


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Fit (kaleidagraph): x(t) = 1000 t (m) y(t) = -0.5*9.8*t2 (m)

Velocities: Vx(t) = dx / dt = d/dt (1000 t ) = 1000 m/s (uniform)

Vy(t) = dy / dt = d/dt (-0.5*9.8*t2) = -9.8 t m/s (increasing, downward)

Vx is independent of time

Vy increases as time goes by

V is diagonal of parallelogram Vx

Vy

v

Accelerations: ax(t) = dVx / dt = d/dt (1000) = 0 m/s2 (no gravity, affects no x(t))

ay(t) = dVy / dt = d/dt (-9.8*t) = -9.8 m/s2 (gravity act only vertically)

Ah-Ha: 2D projectile is the combination of uniform horizontal motion and vertical free fall!

Position is characterized by: r(t) = x(t) i + y(t) j = (1000 t)m i + (-0.5*9.8*t2)m j

Velocity: V(t) = Vx(t) i + Vy(t) j = (1000)m/s i + (-9.8*t)m/s j

Acceleration: a(t) = ax(t) i + ay(t) j = -9.8m/s2 j

By magnitude and direction

Position: r(t) = √[ x(t)2 + y(t)2 ] = √[ 106 t2 + 24 t4 ] m (distance to origin)

θr = tan-1[ y(t) / x(t) ] = tan-1[-0.0049 t] =-tan-1[0.0049 t] Velocity: v(t) = √[ Vx(t)2 + Vy(t)2 ] = √[ 106 + 96 t2 ] m/s (speed, changes)

θV = tan-1[Vy(t)/Vx(t)]= tan-1[-0.0098 t] =-tan-1[0.0098 t] (if hit something)

Acceleration: a(t) = √[ ax(t)2 + ay(t)2 ] = 9.8 m/s2


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θa = tan-1[ -9.8 / 0 ] = Error (Oops!) θa = -90° (downward)

Demo: Data t, x, y, Vx, Vy with FORTRAN, plot x-y, demonstrate angles, when Vy change sign.

Summary:

Position : r(t) = x(t) i + y(t) j

Velocity: V(t) = Vx(t) i + Vy(t) j = (dx/dt) i+(dy/dt) j (i, j: consts, move inside d/dt)

= d/dt (x i+y j)

so V(t) = dr(t)/dt

Acceleration: a(t) = ax(t) i+ay(t) j = (dVx(t)/dt) i+(dVy(t)/dt) j (i, j: consts, move inside d/dt)

= d/dt (Vx(t) i + Vy j)

so a(t) = dV(t)/dt

Average V and a: V(t) = Δr(t) / Δt (refer to Griffy’s B-A)

a(t) = ΔV(t) / Δt

©23© 2D-Projectile Motion

Appl: If a ball follows x(t) = 10 t (m) y(t) = 2t2 - 4t3 (m)

(1) What’s the position when y reaches the maximum

At the maximum what are the (2) velocity and (3) acceleration?

Fortran plot x-y ball_surface.f

Sol: (1) At the max dy(t)/dt = Vy(t) = d/dt (2t2 - 4t3) = 4t - 12t2 = 0

t = 0 (drop) or t =1/3 s

xmax = x(t=1/3s) = 10*1/3 (m) = 3.33 m (limit taking)

ymax = y(t=1/3s) = 2*(1/3)2-4*(1/3)3 (m) = 0.074 m

(2) Vx(t=1/3s) =dx/dt (t=1/3s) =d/dt(10t) (t=1/3s) =10 m/s (limit after derivative)

Vy(t=1/3s) =dy/dt (t=1/3s) =d/dt(2t2 - 4t3) = 0 m/s


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(3) ax(t=1/3s) =dVx/dt (t=1/3s) =d/dt(10) (t=1/3s) = [4t - 12t2] (t=1/3s) = 0 m/s2

ay(t=1/3s) =Vx/dt (t=1/3s) =d/dt(4t - 12t2 ) = [4-24t] (t=1/3s) = -4 m/s2

Fortran plot x(t), Vx(t), ax(t) on one, y(t), Vy(t), ay(t) on another ball_surface2.f

display

Appl: If a ball moved according to r = 2t i + 5t2 j (m)

find (1) v from 1s to 2 s (2) V(t) and (3) aav (from 1s to 2 s)?

Sol: (1) v = Δr/Δt = [r(2s)−r(1s)]/(2s-1s)=[(4i + 20j)m-(2i + 5j)m]/1s =2i + 15j m/s

| v | = √(22+152) = 15.1 m/s θ = tan-1(15/2) =82.4°

(2) V = dr/dt = d[2t i + 5t2 j]/dt (move i, j out of d/dt with consts.)

=2i dt/dt+ 5j d(t2 )/dt=[ 2 i +10 t j ] (m/s)

(3) a = ΔV/Δt = [V(2s)−V(1s)]/(2s-1s)=[(2i + 20j)m-(2i + 10j)m]/1s =10j m/s

| a | = 10 m/s θ = tan-1(10/0) = ∞ [Oops!] 90° , or upward

projectile vs. freefall:

vertical drops will be the same

vertical motion is independent of horizontal motion

Demo: Game time: drop a bullet at the same time as fire a bullet horizontally,

take another look at shoot the bulls-eye, y = - 0.5 g t2, same equation as free fall

which bullet hits the ground first? (start at the same height)

Same time, because they fall at the same rate (horizontal motion has no effect on vertical motion)

Draw vertical drops after same elapsed times.

Note the two examples differ in horizontal motions

Ex: Drop at shoulder height of 4 ft.

The distance fallen is d = 0.5 g*t2 or 4 ft =0.5*32ft/s2*t2

So t = √(1s2/4) = 0.5 s

Same downward acceleration and velocity: ay, Vy.


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If horizontal speed is 1000 ft/s, it goes 500 ft in 0.5 s

Demo: Fortran: draw x-y for two bullets on one plot

Now let’s look at the example of Monkey hunting.

Shoot the money hanging on the tree and horizontally to the right of the hunter, on gun fire the

monkey drops, Q. Will the monkey get shot?

Hint: Compare this with the previous example, Draw the vertical drop after equal times.

What happens if the gun is fired upward at an angle?

in absence of gravity, bullet follows straight line

with gravity it falls away from that straight line

Another thought-exp: Monkey & hunter (down with animal exp if you are related to monkeys)

Hunter aims his gun directly at a monkey hanging from a tree, the monkey cleverly

releases his grip at the exact moment the hunter fires the gun, what happens?

Vertical falling is independent of forward motion, (should have learned physics for

monkeys)

in the case when the bullet goes fast the bullet and monkey fall only a little way

if bullet is slow, they fall farther, but still both d=0.5gt2 from the same straight

line

Demo: draw x=1000t y=0.5*32t2 with x=500 y = same over 0.5s

Demo: real apparatus

yb = Vy0 t -1/2*gt2 ym = H = 1/2*gt2

convince yourself that when a bullet passes over the vertical line of monkey Vy0 t = H


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©24©

Range and Maximum Height for a projectile

y

x

H

R

Vy=0

Mark McGuire hits 550 ft at 300 ft max height, what’s ball’s speed off the bat and angle?

x = Vx0 t = (V0 cosθ0) t y = Vy0 t - 0.5gt2 = (V0 sinθ0) t - 0.5gt2

at maximum height: Vy = dy/dt = 0 = V0 sinθ0 - gt so t 1/2 = V0 sinθ0 / g

H = y(t 1/2)=(V0 sinθ0)(V0 sinθ0/g) - 0.5g(V0 sinθ0/g)*(V0 sinθ0/g)=0.5V02 sin2θ0 /g

There are still 2 unknowns

ball hits the ground when y(t1) = 0 = (V0 sinθ0) t - 0.5gt2

get t = 0 (initially at zero level, drop this solution)

t1 = 2V0 sinθ0 / g = 2 t 1/2

R = x(t1) =(V0 cosθ0) t1 =(V0 cosθ0) 2V0 sinθ0 / g = V02 sin2θ0 / g

300 ft = 0.5V02 sin2θ0 /32 (1)

550 ft = V02 sin(2θ0)/32 (2)

(1)/(2) 300 / 550 =[ 0.5V02 sin2θ0 /32 ]/[ V02 sin2θ0 /32 ] / = 0.5 sin2θ0/sin2θ0=0.25tanθ0

we get θ0 = tan-1[4*300/550] = 65.4°


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into (1) 300 = 0.5 V02 sin265.4°/32

we get V0 = √[2*300*32/sin224.6°]=√23233=152.4 ft/s

Quizzes

Case: Which ball hits 1st, both with 152 ft/s, one at 65.4° the other at 45° [t1=2V0 sinθ0/g]

[45° | 65.4° | same]

Case: Which hits farther at 65.4° or at 45°, both with 152.4 ft/s [R=V02 sin2θ0/g]

sin[ 2*65.4° ] = 0.757 sin[ 2*45° ] = 1 [45° | 65.4° | same]

R65.4° = 550 ft (as should be) R45°= 726 ft (Mark can do this in the vacuum)

Case: What about 24.6°? R24.6° = 550 ft = R65.4° generally Rθ = R90°-θ [same |45° | 65.4°]

Case: Which hits higher: 65.4° or at 45°, both with 152.4 ft/s [H=0.5V02 sin2θ0 /g]

H65.4° = 300 ft and H45° = 181 ft [ 65.4° |45° | same]

note H24.6° = 63 ft ≠ H65.4°

Case: same θ0 if V0 doubled

R, H increase 4 times each [ 4 | 3 | 2 | 1 ]

t increase 2 times [ 2 | 3 | 4 | 1/2 | 1/3 | 1/4 | none ]

Case: Send Mark to the Moon, how does he do under same θ0 and V0?

Gravity on the Moon = (1/6) Gravity on the Earth

R, H, t increase 6 times each [ 6 | 3 | 2 | 1| none ]

Demo: project them x-y plot for: 65.4°, 55.4°, 45.4°, 35.4°, 25.4° f9 mark.f plot

©25©


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2D uniform circular motion: x, v, a

Ex: Turning a car on flat surface whirl a ball tight to a string

Drive a car over a bump Drive a car out of a dip

It helps to draw

r

v!

!x

y

Vx

Vy

s

period = time to travel circumference of circle, speed could stay the same when velocity changes

position: x(t) = r cos θ = r cos(s/r) (θ in rads ; s=rθ)

y(t) = r sin θ = r sin (s/r)

r = r cosθ i + r sinθ j

magnitude: r

direction: θ

velocity: From the figure Vx(t) =-v sinθ Vy(t) = v cosθ

check: Vx(t) = dx/dt =d/dt [r cos(s/r )] = d/dt[ r cos(θ) ]=r d/dθ[cos(θ)] d(s/r )/dt (by chain

rule)

= -ds/dt sinθ = -v sinθ

xxx --Homework-- xxx Vy(t) = dy/dt =d/dt [r sin(s/r)] = d/dt[ r sin(θ) ]

=r d/dθ[sin(θ)] d(s/r)/dt = ds/dt cosθ = v cosθ

V = -v sinθ i + v cosθ j


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magnitude: v

direction: ⊥ to r and θV = 90°+θr

acceleration:

ax(t) = dVx/dt =d/dt [-v sin(s/r)] = d/dt[-v sin(θ) ]=-v d/dθ[sin(θ)] d(s/r)/dt

= -v v /r cosθ = -v2/r cosθ

xxx similarly, ay(t) = dVy/dt =d/dt [v cos(s/r)] = d/dt[v cos(θ) ]= v d/dθ[cos(θ)]

d(s/r)/dt

= -v v /r sinθ= -v2/r sinθ

What do these formulas mean? Center pointing a, try draw it

!

a

-a cos!

-a sin !

magnitude: a = v2/r = ar

xxx --Homework-- xxx a = √[ax2+ay2]=√[(-v2/r cosθ)2+(-v2/r sinθ)2]

= v2/r√[cos2θ+sin2θ] = v2/r

direction of a: opposite of r =θ+180° (toward center)

a is usually referred to as radial / centripetal acceleration

Alternatively, strictly from geometry. A force toward the center at the top


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sx

2r-s

xx

Because of similar triangles of the shaded area

x / s = (2r-s) / x ≈ 2r / x

so s = x2 / (2r)

but x = v t so s = [v2/(2r)] t2

v = ds/dt = 2[v2/(2r)] t = [v2/r] t

a = d v/ dt = v2/r

©26© Applications of Uniform Circular Motion

Appl: Calculate the Moon’s acceleration around the Earth.

[Sol] T =30∞24 hr = 30∞24∞3600s = 2358720 s

r = 3.84 ∞ 108 m

c = 2π r = 2 ∞ 3.14 ∞ 3.84 ∞ 108 m = 2.41∞ 109 m

v = c / T = 2.41∞ 109 m/86400 s=1022 m/s

ar = v2 /r = (1022 m/s)2 / 3.84 ∞ 108 m = 2.72∞10-3 m/s2 ≈ 1/3600 = 1/602

Appl: Acceleration of Indy car (c = 2.5 mi, v = 200 mi/h)

r = c/2π = 398 m, v = 89.4 m/s, so ar = (89.4 m/s)2 / 398 = 20 m/s2

Appl: Calculate an electron’s acceleration around the hydrogen nucleus, if its orbiting speed

is 1/137 the speed of light in vacuum and orbiting radius of 0.5∞ 10-10 m.

[Sol] v = 3 ∞ 108 m/s /137 = 2.2∞ 106 m/s

r = 0.5∞ 10-10 m


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ar = v2 /r = (2.2∞ 106 m/s)2 / 0.5∞ 10-10 m = 9.7 ∞ 1022 m/s2

2D non-uniform circular motion: x, v, a

Roller Coaster: radial and tangential acceleration

position: magnitude: stays const r, θ: not changing at a const rate

x(t) = r cos θ = r cos(s/r)

y(t) = r sin θ = r sin(s/r)

r = r cosθ i + r sinθ j

velocity: magnitude: v not a const, θ: not changing at a const rate

Vx(t) = -v sinθ

Vy(t) = v cosθ

V = -v sinθ i + v cosθ j

acceleration:

ax(t) = dVx/dt =d/dt [-v(t) sin (θ(t))] = d/dt[-v] [sin(θ)] + [-v] d/dt[sin(θ)]

= -dv/dt sinθ -v d/dθ[sin(θ)] d(s/r)/dt = -dv/dt sinθ -v2/r cosθ

xxx Similarly ay(t) = dVy/dt =d/dt [v(t) cos(θ(t))] = d/dt[v] [cos(θ)] + [v] d/dt[cos(θ)]

= dv/dt cosθ +v d/dθ[cos(θ)] d(s/r)/dt = dv/dt cosθ -v2/r sinθ

What do these mean?

!

!ar

at

ax(t) = -at sinθ - arcosθ ay(t) = at cosθ - ar sinθ

When comparing with the above statements, we have the identification: ar = v2/r at =


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dv/dt

direction: θr=θ+180° (toward center as before)

θt=θ+90° (at ⊥ to ar)

total acceleration is a = ar + at

magnitude: a = √[ (v2/r)2+(dv/dt)2 ]

direction: θa = tan-1(ay / ax)

xxxx check magnitude: a = √[ax2+ay2] = √[(-dv/dt sinθ -v2/r cosθ)2+(dv/dt cosθ-v2/r sinθ)2]

= √[ (v2/r)2+(dv/dt)2 ]

Appl: A stunt pilot follows a vertical circular path (clockwise) of radius 500 m. At the top

of this path, the pilot has a speed of 100 m/s which is decreasing at the rate of 5 m/s2. What is

the (a) magnitude and the (b) direction of the acceleration of the pilot at this maximum height?

(c) Express acceleration in unit vector notation.

[Sol] Given: r = 500 m, v = 100 m/s, dv/dt = -5 m/s2

ar = v2/r = (100 m/s)2/(500m) = 20 m/s2

at = dv/dt = -5 m/s2

(a) a = √(ar2 + at2)=√((20 m/s2)2 + (-5 m/s2)2)=20.6 m/s2

(b) θ = tan-1(-5/-20)=14°+180°=194°

(c) a = -5 m/s2 i -20 m/s2 j

©27© Comment on non-uniform non-circular motion

Ex: Driving over a bump, air plane looping


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At any instance find the tightest circle of rotation. We may want to consider circular, non-

uniform motion from the point on.

Comment on 3D motions

position: x(t), y(t), z(t), and r(t) = x(t) i + y(t) j + z(t) k

magnitude: r = √[x2+y2+z2]

directions: θx = cos-1(x/r), θy = cos-1(y/r), θz = cos-1(z/r)

velocity: Vx(t)=dx(t)/dt, Vy(t)=dy(t)/dt, Vz(t)=dz(t)/dt,

and V(t) = Vx(t) i + Vy(t) j + Vz(t) k

mag: V = √[Vx2+Vy2+Vz2]

dirs: θVx = cos-1(Vx/V), θVy = cos-1(Vy/V), θVz = cos-1(Vz/V)

acceleration: ax(t)=dVx(t)/dt, ay(t)=dVy(t)/dt, az(t)=dVz(t)/dt,

and a(t) = ax(t) i + ay(t) j + az(t) k

mag: a = √[ax2+ay2+az2]

dirs: θax = cos-1(ax/a), θay = cos-1(ay/a), θaz = cos-1(az/a)

Newton’s Laws of motion

Introduction

Newton’s 3 laws and the universal law of gravitation constitute the backbones of

classical (as oppose to modern) mechanics. Classical mechanics is also known to as Newtonian

mechanics.

With his three laws, Newton summarized all the experimental laws to his time:

Galileo’s experiment and Kepler’s theoretical laws of planetary motion that stemmed from

astronomical data. His theories are the ones that best fit all experiments for objects not moving

extremely fast or that have a very small mass. He could now begin to explain why do apples fall,

why the moon orbits, or why toys move, but what is the deepest secret?


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The only limitation to his theory was the fact that they require significant

modifications when: (1) speed of the motion is approaching the speed of light, and (2) if the

masses in question are on the atomic scale. Theoretical extensions have been developed for

those cases, which must recover “Newtonian” limit out of those ranges.

The idea of linking the cause and the motion (via equation of motion), with complete

predictability of the future from the present is very profound in physics. This brings the

universality to problem solving in physics. All mechanical problems may be solved through

more or less the same procedure, by solving the Newton’s equation with terms in it that are

specific to the system at hand. We know how to do this, at least in principle, and it has become

truer with ever more powerful computers. Refer to the success of the Moon-landing project.

Many “versions” of the Newton’s laws for different “branches” of physics may be

deduced from the original version, to name a few: fluid mechanics, plasma mechanics,

kinematics of gas, etc.

Within Newtonian mechanics, other laws may be deduced from Newton’s laws. We

will illustrate this as we go along. Mechanics is Newton's laws plus some definitions.

So let the fun begin. Let’s learn the laws.

Statement of Newton’s 1st Law (principle of inertia):

If an object is left alone, is not disturbed, it continues to move with a constant velocity in a

straight line if it was originally moving, or it continues to stand still if it was just standing still.

Reference frames in which the 1st law is true is referred to as inertial frames.

Statement of Newton’s 2nd Law

Acceleration of an object is directly proportional to the net force experienced and inversely

proportional to the mass of the object. a = F / m or F = m a

Mass is different from weight, though they are proportional on the Earth surface. Mass is


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measured in kg. Acceleration is a vector, and its direction is the same as the net force. It is

measured in m/s2 . Force is a push or a pull, and is measured in Newtons,

1N = 1kg*1m/s2 = 1kg m/s2

The 1st law is the case when F=0, so a = 0 = dv/dt, so v = constant = unchanged

Force in the 2nd law should be the net result of all of the forces experienced, meaning the vector

sum of them.

Approximation comes in for a dropping apple. If we wanted to be completely accurate, we

would have to consider the fact that it experiences forces from stars...

Important forces: Weight (earth attraction), normal force, friction, tension, elastic force in

springs, gravitational force.

Statement of Newton’s 3rd Law

Whenever an object exerts a force on a second object, the second object exerts an equal and

opposite force on the first object.

Statement of the Universal Law of Gravitation

The gravitational attraction between any two objects with masses m1 and m2 and by a distance r

apart is F = G m1 m2 / r2, with G as a constant.

Object may be earthly or heavenly. The forces we are considering are directed toward each

other.

©28© Newton’s First Law

A little history about the 1st law

For centuries physicist slept in the shadow of Aristotle (484-322 BC), who believed

"Natural" motion of celestial objects (moon, stars) was circular


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Terrestrial objects (apple, rocks, you) tend "naturally" to fall

Thus

The moon's motion therefore, has no need to invent gravity (no Force for Celestial

motion)

Earthly objects, after falling, come to rest, unless forces push them sideways.

(force is necessary to maintain motion, as in un-engined cars) (need F to keep V)

Galileo (1564-1642) claimed: No force is needed to keep an object in uniform, straight-line

motion. (no F to keep V)

Forces are needed to change the motion. Left alone, objects will stay moving, it’s the

force of friction that slows them down. (F to change V, The force of his idea stopped

Aristotle.

But now we know there are 3 basic types of forces on the microscopic scale

1) weak nuclear and electromagnetic 2) strong nuclear 3) gravitational )

(eg. friction changes V)

For a frictionless rubber mat, the ball likes to roll to the same height on the other side, if there

were no other side, it would roll on forever

Isaac Newton (1642-1727) summarized Galileo's idea as Newton's 1st law

An object at rest tends to stay at rest, an object in motion tends to continue in motion at constant

speed in a straight line. (where there are no forces, objects move with constant V).

eg. coin over the cup

eg. skipping car

Inertia vs. Mass


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Inertia: is the property that makes them "tend" to obey the 1st law,

or resistance to changes in motion. m proportional to 1/a

Mass: measures the amount of inertia a body has

(massive things have a lot of inertia)

x Forces are needed to overcome a body's inertia that is in motion.

x F = d/dt (m V) is a form of Newton's second law.

x for const m: F = m (dV/dt) = m a (force overcomes inertia and produce

acceleration)

x more F->more a F = m a (fix m)

x more m->more it resists acceleration, a = F/m (fix F)

x of course one can find the mass by m = F/a (fix F)

x Look at the falling apple and the Moon.

x Earth pulls on the apple to cause downward acceleration

x (if not for gravity, you would hang in mid air)

x Earth pulls on the moon with the same force, causing it to fall away from its straight

path

x ar = v2 /r = 2.72∞10-3 m/s2 ≈ 1/3600 g= 1/602 g Earth’s gravity decreases with

1/r2.

x the force is universal, objects are attracted to each other, so the laws are universal

We will go directly to analyzing the implications of the second law.

Free-Fall Motion revisited

W = m g (W is gravitational force due to the earth g ∝ Me / r2) m, W different


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Neglect all other forces on the apple, due to you, Monica, sun, stars

Mass (the quantity of matter in an object) measures

how much gravity it exerts on other objects

how much it resists acceleration, how much inertia it has

W depends on where you are, on the moon 1/6th that of earth, in outer-space may be 0, but "m"

stays unchanged

2nd law suggests F = mg = m ay ( -mg = m ay , if up is positive )

the equation of motion for the falling apple system is the same as g = ay (free fall

vert. down)

develop a procedure (1) force analysis (free body diagram)

(2) apply Newton’s law in vertical direction

we may apply 2nd law in horizontal direction 0 = m ax , same as ax = 0

The solutions of the equation of motion

g = ay goes to

g = dVy /dt = d2 y / dt2 (a differential equation)

claim: solution y(t) = y0 + v0 t + 1/2 g t2 (w/ consts y0 and v0 )

check dy(t)/dt = d/dt (y0 + v0 t + 1/2 g t2)= d/dt (y0)+d/dt (v0 t)+d/dt (1/2 g t2)

= 0+v0d/dt (t)+1/2g d/dt (t2)=v0+1/2g 2t=v0+gt

RHS = d2y/dt2 = d/dt(dy(t)/dt ) = d/dt(v0+gt)=d/dt (v0)+d/dt

(gt)=0+gdt/dt=g=LHS

What are the constants y0 and v0 ?

dy(t)/dt = Vy(t)=v0+gt Vy(0)=v0+(g)(0)=v0 (the initial vertical velocity)

y(t) = y0 + v0 t + 1/2 g t2 y(0) = y0 + (v0)(0) + 1/2g(0)2 =y0 (init. vert. position)

usually taken both zero from free fall y(t) = 1/2 g t2 = 4.9 t2 (4.9m of drop per sec)


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©29© Spring force

If V0 and y0 are given, can we predict v(t) and y(t) for all "future" times?

note w/ y points up y(t) = -4.9 t2 and -4.9m per sec, since up is positive, it also means drop

x note if not free fall but with initial velocity or height, solution should take v0, y0 terms.

x (equation of motion generates a family of solutions, v0, y0 picks one solution out)

eg. with y(t)=4.9 t2 the apple's future is determined 100% before it hits ground

future height y(t=1.23sec)=7.4m drop from its original height, prediction

future velocity Vy(t=1.23sec)=12m/s

Q: This is a simple example which may be solved analytically to the end. Here is a more

complicated force.

Motion by a Spring Force

x

equilibriumposition

Direction of displacement is opposite of directional force. Force magnitude grows with the

magnitude of the displacement. What equation does the spring obey according to the "law"?

Equation of motion: F = -kx (Hooke’s law), k is a constant that depends on

spring


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-k x = m (dVx/dt) (2nd law)

same as -x = dVx/dt (if constants k/m = 1)

together with Vx = dx/dt determines the dynamics of the mass

Goal: find x(t) and v (t) (knowing initially Vx (0) = 0, and x(0) = 1.00)

Q: one step predictor: given x(t) and Vx(t), can we predict x(t+ε) and Vx(t+ε)? for small

ε

Vx (t)= dx/dt is approximately Vx (t)= [x(t+ε)-x(t)]/[t+ε-t]=[x(t+ε)-x(t)]/ε

so x(t+ε)-x(t)=ε∗Vx (t) or x(t+ε)=x(t)+ε∗Vx (t)

similarly -x = dVx/dt is approximately -x (t)= [Vx(t+ε)-Vx(t)]/ε

or Vx(t+ε)=Vx(t)-ε*x(t)

Predict: Solution "ε" sec into the future (LHS) from knowledge about the present (RHS)

If repeated, we can find solutions even farther into the future. The prediction gets better with

a smaller ε and more rounds of repetition.

Numerical solution of the equations spring

assume ε = 0.100 sec from Vx (0) = 0, and x(0) = 1.00, spring.f

x(t+ε)=x(t) +ε ∗Vx (t) Vx(t+ε)=Vx(t) -ε *x(t)

x( 0) = 1.00 Vx( 0)= = 0

x(0.1) = 1.00 + 0.100 ∗ 0 = 1.00 Vx(0.1)=0 -0.100*1.00=-0.100

x(0.2) = 1.00 + 0.100 ∗(−0.100) = 0.99 Vx(0.2)=-0.100 - 0.100*1.00=-0.200

x(0.3) = 0.99 + 0.100 ∗(−0.200) = 0.97 Vx(0.3)=-0.200 - 0.100*0.99=-0.299

x(0.4) = 0.97 + 0.100 ∗(−0.299) = 0.94 Vx(0.4)=-0.299 - 0.100*0.97=-0.396

...


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The x(t) fits well with a cos(t), we will come back to this later.

generalize this to: F = m dv/dt, v = dx/dt, v = v + F/m*Δt x = x + v*Δt

Numerical accuracy for free fall objects, is found by changing a minimum in the program.

free_fall.f. Computer demonstration about spring motion with different values of ε:

t x(ε=.1) x(ε=.01) x(ε="0")

.0000 .0000 0000 .0000

.1000 .0000 .0441 .0490 .2000 .0980 .1862 .1960 .3000 .2940 .4263 .4410

.4000 .5880 .7644 .7840 .5000 .9800 1.2005 1.2250 .6000 1.4700 1.7346 1.7640

.7000 2.0580 2.3667 2.4010 .8000 2.7440 3.0968 3.1360 .9000 3.5280 3.9249 3.9690

1.0000 4.4100 4.8510 4.9000

Accuracy of the Euler method as a function of ε.

ε 4.9-x(t=1s) Nt

.1000E+00 .4900E+00 10 .4642E-01 .4662E+00 21

.2154E-01 .1920E+00 46 .1000E-01 .4900E-01 100 .4642E-02 .4285E-01 215

.2154E-02 .1391E-01 464


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.1000E-02 .4916E-02 1000 .4642E-03 .4299E-02 2154

.2154E-03 .2173E-02 4641 .1000E-03 .7014E-03 10000 .4642E-04 .1268E-03 21544

©30© Newton's 3rd Law

Action = Reaction

Forces come in pairs, there is always action on 2 bodies.

Create, grow, decrease, and disappear at same time. Love and hate or when considering

relationships: (Shakespeare)

Love? You kiss your lover, you get kissed back.

Hate? You Slap someone on the face, you get a force in return. The harder you slap the

harder force you get.

Push against the wall, the force you get increases/decreases with the wall’s deformation/damage.

Why avoid a lot of splash off a diving board? Splash is an indication of the force the water

received, indirectly, what you received as well.

eg. When considering the free fall apple, there is one force on the apple, while the other force is

acting on the Earth. This latter force is as large as the net force on the apple during free fall, it is

non-zero. If we decide to look at apple+Earth as a system, the sum of the two forces will add

to zero. (They fall toward each other in a way the net force for the system of two bodies is zero.

From Newton’s 2nd law, acceleration must be zero? The point called the center of mass, it does

not accelerate)

The application of 2nd law on the apple, then again the 2nd law applied on the Earth will give

the same information about the Earth-apple attractive motion.


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e.g. Study a book sitting on the table, what is the action-reaction pair of the weight of the book.

not support force from the table?

"If the Earth attracts the book, the origin of its weight, then the book attracts the Earth"

The table pushes the book upward (F), thus the net force on the book is the vector sum of that

and the weight. Usually, the forces on the book are balanced, W=F. But if support is not enough,

F< W, net force is downward and it will break the table and fall.

Horse pulls a cart, if the cart pulls back with an equal force, how does it ever move ahead?

look at the cart: if pull of horse > friction on the cart, cart move ahead (2nd law)

look at the horse if push of ground > pull of cart, horse move ahead (2nd law)

pull of cart = pull of horse (3rd law) so push of ground > friction of cart to move

push of horse = push of ground (3rd law) so push of horse > friction of cart to move

More about forces: Applications of Newton's Laws

Newton's Laws:

(1) Without any forces, objects maintain constant velocity.

(2) A force produces an acceleration proportional to the force (and inversely prop to the

mass.)

(3) Objects exert equal but opposite forces on each other.

Forces are vectors (magnitude & direction), if forces F1, F2, F3 don't act in the same direction,

the acceleration is pointing in the direction of the net force.

If balanced: F1 + F2 + F3 = 0 (ma, a =0), Resolve the forces to get 3 equations.

F1x + F2x + F3x = 0, F1y + F2y + F3y = 0, F1z + F2z + F3z = 0

Similarly the forces on a skier (constant speed) must add to zero, too.

Generally ∑i Fix = max = m d2x/dt2

∑i Fiy = may = m d2y/dt2


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∑i Fiz = maz = m d2z/dt2

are required for the solution of x(t), y(t), z(t)

This is the most general form of the Newton’s 2nd law for single particle system.

Ex: A skier (of 200kg) slides down hill of 30°, how much distance is covered after 5 sec?

Assume no friction.

x

y

W

N

x

y

W

N

Wcos30

Wsin30

x

yN

Wcos30

Wsin30

∑i Fix = 200*9.8*sin30° = 200 ax = 200 d2x/dt2

∑i Fiy = N-200*9.8*cos30° = 200 (0) = 200 d2y/dt2 (ay =d2y/dt2 =0)

so ax = 9.8*sin30° = 4.9 m/s2

N = 200*9.8*cos30° = 1960 N

4.9 m/s2 = d2x/dt2 x 5 = 0.5*4.9*52 = 61.25 m

HW: Apparent weight in the elevator, N = mg+ma

HW: A skier (of 200kg) slides down hill of 30° maintains a constant velocity, what is the

friction?

©31©

Tension

Tug of war, each team pulls with a force of 980 N. What is the tension of the rope?

Tension: the value a spring scale would read if the rope were cut and the scale inserted.


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Compare with 100 kg mass with a spring scale hanging from the ceiling. W=980N=T

Mass pulls down with 980N, scale pulls up on the mass with the same F. The scale also pulls

down on the ceiling, and the ceiling pulls back with 980 N. (Mass less spring and scale)

This is a version of tug of war, oriented vertically. The rope transmits the force from one team

(mass) to the other. (ceiling)

Ex: m1 m2 F

Tension in the string between

masses=?

Neglect friction on the surface. Any prediction before hand?

Force analysis (free body diagram)

F

N

m g

2

2

TT

N 1

m g1

x

y

for m1: T = m1 a (1) N1-m1g = 0 (1’)

for m2: F-T = m2 a (2) N2-m2g = 0 (2’)

givens: F, m1, m2 unknowns: T, a

(1) + (2) F = (m1+m2) a so a = F/(m1+m2) back in (1)

T = m1/(m1+m2) * F so T < F

case1: m2 = 0 T = F

case2: m1 = m2 T = 0.5 F

case3: m2 =100m1 T =1/101 F ≈ 0.01 F

Ex: find out the acceleration and the tension in the string. (No friction) a =W2/m1 wrong!


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m g2

T

T

N 1

m g1

x

y

y

a

a

note the coordinate system does not have to been identical for m1 and m2,

but a, T must be the same because linked by string, pulley to change the direction.

for m1: T = m1 a (1) N1-m1g = 0 (1’)

for m2: m2 g - T = m2 a (2)

givens: m1, m2 unknowns: T, a

(1) + (2) m2 g = (m1+m2) a so a = m2/(m1+m2) g back in (1)

T = m1 m2/(m1+m2) g

note: a =W2/(m1+m2) not a =W2/m1 as guessed.

note: a < g always a ≈ g for m1<<m2 a << g for m1>>m2

T < min(W1,W2) T≈W1 T≈W2

case1: m1=m2 a = 0.5 g T = 0.5 W

case2: m2=100m1 a = 100/101 g T=100/101W1

case3: m1=100m2 a = 1/101 g T=100/101W2

since a is known constant, one can again find out x(t)

HW: Atwood's Masses with pulleys. a=(m2-m1)/(m2+m1)*g, T=2m1m2/(m2+m1)*g

©32© Friction: Parallel component of the contact force

Ex: Stopping a car on the road drop a ball in the honey sky dive

Friction is always against the tendency of motion

before sliding: grows with other force up to a max value (static friction)


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fs, max = µs N

sliding: taken over by kinetic friction

fk = µk N

µs, µk are independent of the area and force, and velocity depend only on material, µs>µk.

µs, µk are called the static or kinetic coefficient of friction.

Ex: find out the acceleration and the tension in the string. (µk=0.3)

m g2

T

T

N 1

m g1

x

y

y

a

a

f

for m1: T - f = m1 a (1) N1-m1g = 0 (2)

f = µk N1 (3)

for m2: m2 g - T = m2 a (4)

givens: m1, m2, µk unknowns: T, a

(1) + (2) m2 g -f = (m1+m2) a so a = (m2 g-f)/(m1+m2) (5)

(2) in (3) f = µk m1 g back in (5) a = (m2 g-µk m1 g)/(m1+m2) (5’)

back in (4) T = m2 (g-a) = m2 [g-(m2 g-µk m1 g)/(m1+m2)] = m1m2(1+µk)/(m1+m2)*g

reduce a, increase T

case1: m1=m2 a = 0.35 g T = 0.65 W

HW(5.61) - Kinetic friction A block is placed on a plane inclined at 35° relative to the

horizontal. If the block slides down the plane with an acceleration of magnitude g=g/3, determine

the coefficient of kinetic friction between the block and the plane.


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x

y

W

N

Wcos30

Wsin30

f

∑i Fix = mg*sinθ-f = max =mg/3 --> f=mg*sinθ−mg/3

∑i Fiy = N-mg*cosθ = 0 --> N=mg*cosθ

µ = f/N = (mg*sinθ−mg/3)/(mg*cosθ)

= (sinθ−1/3)/cosθ=(0.574-0.333)/0.819=0.29

©33© Friction: Drag forces (in fluids or air)

Drag force is proportional to speed, opposite direction of velocity.

low speed: f =-k v high speed: f =-k2 v2

k is constant

Ex: While diving, determine x(t), assume low speed

mg -k v = m dv/dt (assume k/m = 0.01 = κ)

g - κ v = d v / dt and v = d x / dt

Numerical solution:

x(t+dt) = x(t) + v(t)*dt and v(t+dt)=v(t)+[g-κv(t)]*dt

Ex: Free Fall with drag dive.f dt=0.1, nt=1000 t x vx ax .00000E+00 .00000E+00 .00000E+00 .98000E+01


-76-

.10000E+02 .35871E+03 .62129E+02 .35871E+01 .20000E+02 .11113E+04 .84870E+02 .13130E+01 .30000E+02 .20081E+04 .93194E+02 .48060E+00 .40000E+02 .29576E+04 .96241E+02 .17591E+00 .50000E+02 .39264E+04 .97356E+02 .64390E-01 .60000E+02 .49024E+04 .97764E+02 .23568E-01 .70000E+02 .58809E+04 .97914E+02 .86251E-02 .80000E+02 .68603E+04 .97968E+02 .31563E-02 .90000E+02 .78401E+04 .97988E+02 .11544E-02 .10000E+03 .88201E+04 .97996E+02 .42195E-03

Terminal velocity, at which there is no net acceleration due to a balance of forces: g -κ vt = 0 thus

vt = g/κ=98m/s

Newton’s laws with central forces

Set up equations for the following

Ex: car curving an unbanked road

f = m ac f = µs N =µs m g = m v2/r vmax =√(µs gr)

Ex: car curving a banked road w/o friction

N sinθ = m ac and N cosθ = mg so g tanθ = v2/r vmax =√(gr tanθ)

Ex: car over a bump. what is the support force from the bump

m g - N =m ac = m v2/r so N = m g - m v2/r < mg


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to get zero apparent weight, ie N = 0 v = √(gr)

Case 6.14 (a) v =9m/s, r =1m, mg =600N

N =600N-(600N/9.8m/s2)*(9m/s)2/(1m)=149N

(b) N = 0 v=√(1m*9.8m/s2) =10.4 m/s

Ex: car in a dip

N - m g =m ac = m v2/r so N = m g + m v2/r > mg

to get double weight, ie N =2W v = √(gr)

Ex: pendulum ball at an angle θ

mg sinθ = mat = m dv/dt (1)

and T-mg cosθ =mac = mv2/r (2)

but v = ds/dt = d(rθ)/dt = r dθ/dt (3)

put (3) into (1) we obtain g/r *sinθ = d2θ/dt2 use this equation to get θ then get v from (3)

T =mg cosθ+mv2/r

Ex: Circular motion with static friction: Coin on a turntable before slipping

N

mg

f

yx

∑i Fiy = N-mg = 0 --> N=mg

∑i Fix = f =m v2/r --> µs=f/N=m v2/r /(mg)=v2/(rg)=(50cm/s)2/(30cm 980

cm/s2)

Four (or three) basic forces


-78-

gravitational (electromagnetic nuclear weak) nuclear strong

From which the net force is entered into Newton’s 2nd law. ©34©

Work and Kinetic Energy

Doing work on a something adds energy to it. Work taken from a system causes its energy to

decrease. Energy is the capacity to do work.

Push box across with a constant force, it moves (d is the displacement)

calculate F d = m a d = m a 0.5 a t2 = 0.5 m (at)2 = 0.5 m v2

work = ΔKE

work = increase in KE (or decrease if < 0)

for more general initial condition:

F = m a a = F/m (constant) recall v2(t)-v2(0) = 2 a [x(t)-x(0)]

quick check on this v(t) =v(0) + a t thus t = [v(t)-v(0)]/a into

x(t) = x(0) +v(0) t + 1/2 a t2

x(t)-x(0) = v(0)*[v(t)-v(0)]/a + 1/2 a [v(t)-v(0)]2/a2

2 a [x(t)-x(0)] =2v(0)*v(t)-2v2(0)+v2(t)+v2(0)-2v(0)*v(t)=v2(t)-v2(0) √

use a = F / m 2 F / m*[x(t)-x(0)] = v2(t)-v2(0) we get

F*[x(t)-x(0)] = 1/2 m v2(t) -1/2 m v2(0) = Δ (1/2 m v2)

work = F*d = Δ(1/2 m v2)=change in kinetic energy (KE=1/2 m v2)

This is the work-energy theorem.

SI units is Nm or Joule, BE units is foot pound

(tell them about J, I am going to try pounding

feet...)


-79-

Kinetic energy definition: KE = 1/2 m v2 = 1/2 m v2x+1/2 m v2y + 1/2 m v2z

Work definition: 1d, const F: W = F*d = F*[x(t)-x(0)] (negative if F oppose to Δx,)

1d, varying F(x) W = ∑i Fi Δx --> ∫ F(x) dx

Ex: gravity Wg = (-mg) Δy (Δy is the drop)

Ex: spring Ws = ∫ (-kx) dx = =-k ∫ x dx = -k x2/2

only // component counts if F ⊥ d W = 0 (waitress’ tray, ball on a string, Moon around Earth)

generally W = F// d = F cos(θF,d) d ≡ F•d

Products of two vectors

scalar product A•B ≡ A B cos(θA,B) θA,B = angle between vectors

vector product |A∞B| ≡ A B sin(θA,B), ⊥ to the plane formed by A,B, follow RHR

unit vector calculation of scalar product

note A•B=B•A (c1A)•(c2B) =c1c2 (A•B)

A•B=(Ax i +Ay j + Az k)•(Bx i +By j + Bz k)

= AxBx (i•i)+AxBy (i•j)+AxBz (i•k)

+AyBx (j•i)+AyBy (j•j)+AyBz (j•k)

+AkBx (k•i)+AkBy (k•j)+AkBz (k•k)

A•B = AxBx + AyBy + AzBz

note i•i=j•j=k•k=(1)(1)cos0°=1 and i•j=i•k=j•i=j•k=k•i=k•j=(1)(1)cos90°=0

V•V = V2 = Vx2+ Vy2 + Vz2

A

Ax

Ay

Az

x

y

z

!Ax2 2Ay+

ex: (KE = 1/2 m V2 = 1/2 m V•V)


-80-

xxx for vector product note A×B=-B×A (c1A)×(c2B) =c1c2 (A×B)

xxx note i×i = j×j = k×k =0 and i×j=k i×k=-j j×i=-k

xxx j×k=i k×i=j k×j=-i

xxx A×B=(Ax i +Ay j + Az k)×(Bx i +By j + Bz k)

xxx = AxBx (i×i)+AxBy (i×j)+AxBz (i×k)

xxx + AyBx (j×i)+AyBy (j×j)+AyBz (j×k)

xxx + AzBx (k×i)+AzBy (k×j)+AzBz (k×k)

xxx = AxBy k -AxBz j -AyBx k +AyBz i+AzBx j-AzBy i

xxx A∞B = (AyBz-AzBy) i +(AzBx-AxBz) j+ (AxBy-AyBx) k

W = ∫ Fx dx + ∫ Fy dy + ∫ Fz dz = ∫ F•dr

©35© Power definition: P = W/t P(t) = dW / dt = F•dr/dt =F•V

Ex: power delivered by an elevator motor

mg

f

T

T-f-mg=ma=0 T=f+mg=4∞103N+(1.8∞103kg)(9.8m/s2)=2.16∞104N

P = TV=(2.16∞104N)(3m/s)=6.49∞104W=64.9kW=6.49∞104W*hp/746W=87hp

HW7.37 A 40-kg box initially at rest is pushed 5m along a rough, horizontal floor with a

constant applied horizontal force of 130N. If the coefficient of friction between box and floor is

0.3, find (a) the work done by the applied force (b) the energy lost due to friction (c) the change

in kinetic energy of the box (d) the final speed of the box


-81-

(a) WF=130N*5m=650J

(b) Wf=-µmgd =-0.3*40*9.8*5=-588J

(c) Wnet = WF+Wf =650-588 =62J =Δk

(d) Δk=mvf2/2-mvi2/2=62J-0 so vf =√(2*62/40) =1.76 m/s

Potential energy and energy conservation for gravitational field

Lift you up in the air, I do work W h = mgh, you at the end have certain capacity to do work if

released.

Potential energy definition: Ug=PEg= mgy (potential zero chosen arbitrarily)

from 2nd law we have y(t) = h - 0.5*g*t2 v(t) = g*t

now calculate KE= 0.5*mv2(t)=0.5*mg2t2

PE=mgy(t)=mgh-0.5*mg2t2=mgh-KE

so E(t) = KE(t) + PE(t) = mgh = conserved (constant, not a function of time)

For F=F(x) one may always define a potential energy PE = -∫ F(x) dx

Ex: Potential energy in a spring PEs= ∫ kx dx = 0.5*k x2

also have E(t) = KE(t) +PE(t) = conserved

x challenge: check from 2nd law

x F(x) = m dv /dt

x F(x) dx / dt = m v dv/dt = d(0.5mv2)/dt = d(KE)/dt

x d(KE) = F(x) dx = -d(PE) define PE = -∫ F(x) dx

x KE = -PE +const or KE + PE = const

Conservative Forces

g-force and s-force are examples of conservative forces, since mechanical energy is conserved.


-82-

path-2

cat

path-1

A

BC Q: which path does more work?

path-1 W1 = WAC+WCB = mgAC +0 = mgAC

path-2 W2 = WAB = mgABcosA = mg AC = W1

statement1: Work by conservative force is independent of the path taken.

Toss the cat, or loop the loop W<0 on the way-up W>0 on the way-down

A

B

W<0 W>0

statement2: total W by con f along a closed path must 0.

These two statements are equivalent

For cons. F one can always find a potential energy function U = -∫ F(x) dx or F = -dU/dx

Ex: Energy conservation is usually easier to solve a problem then going through F=ma.

Loop the loop mgh = 0.5mv2 v = √(2gh)

Ex8-11 A bead slides w/o friction around a loop-the-loop. If the bead is released from a

height h=3.5 R, what is its speed at point A (R=3m, m=123kg) A

Ki + Ui =Kf + Uf 0 + mgh = .5*mvf2 +mg(2R)

vf = √[2g(h-2R)] =√[2g(3.5R-2R)] = √[3gR] = √[3*9.8*3] =9 m/s

Demo: loop-the-loop


-83-

Demo: how high does the pendulum swing up? peg

h=?

Ex8-9 the spring loaded popgun

given compression x = 0.12 m, hmax = 20 m m =35 g

Ki + Ui =Kf + Uf 0+.5*kx2 = 0+mgh k = 2mgh/x2 =953 N/m

©36©

non-conservative forces

What if there is non-conservative force present Theorem extended to Wnc = ΔE

since Wext-nc = ∫ Fext-nc dx = ∫ (F-Fnc) dx =∫ m dv /dt*dx +ΔPE=∫ d(0.5mv2) + ΔPE

=ΔKE+ΔPE = Δ(KE+PE) = ΔE

Other forms of energy relations: Wnc = ΔE = ΔK + ΔU

Wnc = Ef -Ei or Wnc + Ei = Ef or -Wnc + Ki + Ui = Kf + Uf

Note:

(1) For work done by a conservative force (gravity, spring) exerted on a particle moving

through any closed path is zero, i.e. ΔE = ΔK + ΔU = 0

(2) A nonconservative force (e.g. friction) causes loss in mechanical energy and a net amount of work is done, i.e. Wnc = ΔK + ΔU = ΔE < 0

Ex: push box up incline with friction. What is the stopping distance = ?

Ex8-11 given m1=1kg, m2=2kg, k=25N/m, h=0.1m


-84-

m1

unstretchedinitially

m2

V = 0 find µ = ?

Wnc = f h cos180° = -fh = -µ(m1g)h ΔK = 0 ΔUg =-m2gh ΔUs =.5*kh2

-µ(m1g)h = -m2gh + .5*kh2 µ = [ m2g - .5*k h ]/(m1g)=(2*9.8-0.5*25*.1)/(9.8)=1.9

Ex8.31

x=0 vi=4m/s

xm=0 v=0

x=0 vf=3m/s

(a) energy loss =Ef-Ei=mvf2/2-mvi2/2=.5*8*(32-42)=-28J

(b) total d=2xm Wf =-f d =−µ mg 2 xm xm =28/(2*.4*8*9.8)=0.446m

General energy conservation principle: Energy may not be created nor destroyed,

and

it may transfer from one form to another.

Everything may usually be related to atoms/molecules, chemical energy, to solar energy, to the

big bang.

©37© Momentum and collision

Q: Which is harder to stop? A high flying bullet or a slowly moving ship.

Momentum is not just velocity nor just mass; it is a combination of the two, P = m V


-85-

Vector points along the velocity direction.

To change the momentum we need to supply force, they are related via 2nd law F = dP/dt

Note: F=dP/dt=d(mV)/dt = m dV/dt=m a

To stop a moving object amounts to decreasing it momentum down to zero by a force against P.

Ex: Water Balloon catching egg catching parachute landing:

F = ΔP / Δt, for small F, Δt should be long

Ex: Breaking a board large F, short Δt

ΔP = F Δt if F not constant Pf - Pi = ∫ F dt ≡ I (impulse)

change in momentum is the impulse (force over time) for large I maximize

product

Impulse is a vector so ΔPx = Ix ΔPy = Iy ΔPz = Iz or

ΔP = Pf - Pi = I

baseball hit

Pf

Pi

I

Collision

Now consider collision (coming together and flying apart)

ex: car crash, bomb sets off,...

during a collision a pair of forces induced with equal strength and opposite orientation, each

body thus receive an equal and opposite impulse during the collision, since impulse is the change

of P

ΔP1 = - ΔP2 the increase of one momentum must be the decrease of the other

P1f - P1i = - (P2f - P2i) = P2i - P2f or P1i + P2i = P1f + P2f


-86-

P1 + P2 is conserved (momentum conservation: no change in total

mom)

Ex: Push off on ice m1 v1 + m2 v2 = 0 (started at rest) v2 = - m1 v1/ m2

Momentum conservation works for direction in which no external force is exerted.

Ex: Bullet-Sandbox

v

V

mom. cons. (energy not) energy cons. (mom. not)

m v + 0 = (M+m) V so V = m v / (M+m)

check energy has been lost during the collision

ΔK = 1/2*(M+m)V2 -1/2*mv2 = 1/2*(M+m)*m2 v2 / (M+m)2 -1/2*mv2

= 1/2*mv2 [m/(M+m)-1]=1/2*mv2[m/(M+m)-(M+m)/(M+m)]=-1/2*m*M/(M+m)*v2<0

!energy is lost

after collision swing: no mom. cons. since g-force is not ⊥ to V

but g-force is conservative force, so mechanical energy is conserved

0.5*(M+m)*V2 = (M+m)*g*h V = √(2gh)

m v / (M+m) = √(2gh) we get v = (M+m)/m*√(2gh)

Inelastic: Collision in which energy is lost (other forms like heat) (ex any entanglement)

Elastic: Collision in which energy is conserved. (stretchy, spring like)


-87-

wet tissue wood steel rubber

Rubber: is totally elastic since we know it bounces back to original height and retains all KE.

Others: are inelastic since energy is lost, for wet tissue the degree is higher because more KE

is lost.

©38©

Ex: collision of loaded freight car

100k-kg traveling 3m/s toward a 50k-kg at rest,

what is the v for the coupled cars?

100,000 kg * 3 m/s = 150,000 kg* v v = 2 m/s

2D Collision momentum conservation in each of the two directions

Ex: tackled football player

100kg

80kg

2m/s

3m/s

v=?

Px = 100kg*2m/s = 200 kgm/s

Py = 80kg*3m/s = 240 kgm/s

they should be conserved separately, so

Vx = Px / (m1+m2) = 200 kgm/s / (100kg + 80kg) =1.1 m/s

Vy = Py / (m1+m2) = 240 kgm/s / (100kg + 80kg) =1.33 m/s

V = √((1.1)2+(1.33)2) = 1.7 m/s


-88-

θ = tan-1(1.33 / 1.1) = 50°

Elastic collision in 1D: a demo

???

All the balls are identical. What happens if one ball is dropped?

why not 2 balls? with half the speed each?

m v = m (1/2v) + m (1/2v) (momentum is conserved)

KEi = 1/2 m v2

KEf = 1/2 m (1/2v)2 + 1/2 m (1/2v)2 = 1/4 m v2 =1/2 KEi < KEi (energy loss)

The fact that one flies off and that it reaches the same height, shows that it elastic.

Elastic collision in 1D: quantitative V1i V2i

V1f V2f

given v1i, v2i, looking for v1f and v2f

(1) m1 v1i + m2 v2i = m1 v1f + m2 v2f (momentum conservation)

(2) .5*m1v1i2 + .5*m2*v2i2 = .5*m1v1f2 + .5*m2*v2f2 (energy conservation)

from (1) v1f = ( m1 v1i + m2 v2i - m2 v2f ) / m1 into (2)

and solve for v2f and back to get v1f (define M = m1+m2)

v1f = (m1-m2) /M v1i + (2m2) /M v2i

v2f = (2m1) /M v1i + (m2-m1) /M v2i

with the help of demo

case1: same weight m1 = m2 we get v1f = v2i and v2f = v1i (speed exchanged)


-89-

head on collision of billiard balls

case2: one at rest v2i = 0 we get v1f = (m1-m2)/M v1i v2f = (2m1)/M v1i

m1>>m2 heavy hits light v1f ≈ v1i v2f ≈ 2v1i

m1<<m2 light hits heavy v1f ≈-v1i v2f ≈ 0

©39©

Center of Mass

If system of particles, 1 point behaves like a single particle with the total amount of mass on it.

significance: drop cat (cm obeys free fall), drivers projectile

a cm = F / M (also P tot = M Vcm)

When looking at the motion of a very odd shaped object, we can simply look at the point

that is the center of mass and treat the system as the motion of a point particle.

Ex: baseball bat, sweet spot (least shock)

Ex: bend down while against the wall, not fall over

Definition: Balance two balls (through the center of mass)

m1 = m2 xcm = x2 / 2

m1 = 1/2*m2 xcm = 2x2 /3 (m2/M x2) and if m1 is at x1 not x=0

definition: average, weighted position of mass

xcm = (m1/M)x1+(m2/M)x2=(m1x1+m2x2)/M

Vcm=d/dt [xcm]= (m1/M) x1'+(m2/M)x2' = (m1V1+m2V2) / M = Ptot / M

acm = d/dt [Vcm] = (dPtot/dt) / M = F / M (F is the net force on the system)

CM moves as an imaginary single particle of total mass M

Ex: Determine the center of mass of a T-ruler


-90-

5

4

(1/2, 2)

(2, 4 1/2)

y

x

xcm = ( 1/2 m + 2 m ) / ( 2m ) =1 1/4

ycm = ( 2m + 4 1/2 m ) / ( 2m ) =3 1/4

more than 2 particles xcm = (m1x1+m2x2+...) / M =( ∑i=1N mi xi ) / M

ycm = ( ∑i=1N mi yi ) / M zcm = ( ∑i=1N mi zi ) / M

is not discrete xcm = [ ∫ x dm ] / M ycm = [ ∫ y dm ] / M zcm = [ ∫ z dm ] / M

Ex: If rod is non-uniform and λ = α x (Here λ is the mass length density, α is a constant.)

x dx

L

M = ∫ λ dx = ∫α x dx =[αx2/2] 0 L = αL2/2

∫ x dm = ∫ x λ dx = ∫α x2 dx =[αx3/3] 0 L = αL3/3

xcm= [ ∫ x dm ] / M =( αL3/3 ) / ( αL2/2 ) = 2L/3

©40©

Chap 10 Rotation of a rigid object about a fixed axis

Clock, fan, wheel, skater

Rigid object: non-deformable, separations between all pairs of particles remain constant

Ex: rotation about z-axis (fixed)

Any particle on the object undergoes circular motion in polar coordinates (r, q) arc length s (from

x-axis, q=0) angular position q

S=r q or q=s/r

Artificial unit for q: radian (rad)

One radian =the angle subtended by an arc length equal to the radius of the arc


-91-

360 degrees corresponds to 2pr/r rad=2p rad

or one revolution

1 rad = 360/2p=57.3 degrees =180/p

generally q(rad)=p/180 q (deg)

p/180 * 60 =p/3 rad

p/180 *45 =p/4 rad

rotation from q1 to q2, angular dispacement dq=q2-q1

time interval dt=t2-t1

average angular speed wav=(q2-q1)/(t2-t1)=dq/dt

instantaneous angular speed w= lim dt->0 dq/dt=dq/dt

Simple Harmonic Motion: Motion by Spring Force

x

equilibriumposition

may be used as a clock, break into a watch and you will find a spring

-k x = m (dV/dt) (2nd law)

same as -x = dV/dt (if constants k/m ≡ ω2 = 1)

together with V = dx/dt determines the dynamics of the mass

Find: x(t) and V(t) numerically x(t+ε)=x(t)+ε∗V(t) V(t+ε)=V(t)-ε*x(t)

guess solution: x1 = cos(t)

check V1=dx1/dt=d/dt [cos(t)] = -sin(t)

x1=-dV1/dt=-d/dt [-sin(t)]=d/dt[sin(t)]=cos(t)=x1 √


-92-

guess solution: x2 = sin(t)

check V2=dx2/dt=d/dt [sin(t)] = cos(t)

x2=-dV2/dt=-d/dt [cos(t)]= sin(t)=x2 √

Similarity of x1 and x2: both oscillatory, both repeat itself after t = 2π, same period

x1(0)= cos(0)=1, V1(0)=-sin(0)=0

x2(0)= sin(0)=0, V2(0)= cos(0)=1

x =10

V =00

x =00

V =10

different phase of the same oscillation, only what we define as the starting point of time.

Q: Can you start with x(0)=1 and V(0)=1? Numerically one can do it.

x(t) = x1(t)+x2(t) =cos(t)+sin(t)

check: x(0)=cos(0)+sin(0)=1

V(t) =x’(t)=-sin(t)+cos(t) V(0)=-sin(0)+cos(0)=1

check that it satisfies the Newton’s 2nd law

dV/dt = (-sin(t)+cos(t))’=-cos(t)-sin(t)=-x = F/m √

Numerical demo, compare with analytical solution.

Similarly solution x(t) = Acos(t)+Bsin(t) has x0=A and v0=B is a general solution

x(t) = x0cos(t)+v0sin(t) may also be written as x(t) = C sin(t+φ)

since x(t) = Acos(t)+Bsin(t)=√(A2+B2) [ A/√(A2+B2)*cos(t) +B/√(A2+B2)*sin(t) ]

For any A, B we can construct a right triangle

A

B

!(A +

B )2

2

!

such that A/√(A2+B2) = sinφ B/√(A2+B2)= cosφ


-93-

with definitions φ = atan(A/B) and C =√(A2+B2)

Using equation (1), we can write:

x(t) = C[ sinφ*cos(t) + cosφ*sin(t) ] = C sin(t+φ)

This is an alternative form of a simple harmonic motion

with amplitude C = √(A2+B2)=√(x02+v02) and initial phase angle φ

Either using (x0, v0) and x(t) = x0cos(t)+v0sin(t) or

[C=√(x02+v02), φ = atan(x0/v0)] and x(t) = C sin(t+φ)

as a starting point has the same effect, both describe the exact same solution.

Demo: numerically compare x(t)=x0cos(t)+v0sin(t) with x(t) = C sin(t+φ)

we discussed k/m ≡ ω2 =1 , if ω≠1 all the discussions are similar but

x(t)=x0cos(ωt)+(v0/ω)sin(ωt) or

x(t)= C sin(ωt+φ) with C=√(x02+(v0/ω)2), φ = atan(ωx0/v0)

x0

slope=v0

C

!="# t*

T

t

x

©41©

Period: T = 2π/ω = 2π(m/k)

Frequency: f = 1/T =ω / 2π

Phase: Φ=ωt+φ (in radians)

Amplitude: C=√(x02+(v0/ω)2)

Initial phase: φ = atan(ωx0/v0)


-94-

Geometric interpretation

!

"#t

y=Csin(")

C

note: T, f, ω independent of x0 and v0, because F is linear in x, not independent for m, k.

It may be used as a clock when we start with any amplitude, and adjust m to change T

C, φ depend on initial conditions

V=d/dt[C sin(ωt+φ)] = Cωcos[Φ]

KE = 1/2mv2 = 1/2m(Cωcos[Φ])2=1/2mC2ω2cos2[Φ]

PE = 1/2kx2 = 1/2(mω2)*(Csin[Φ])2=1/2mC2ω2 sin2[Φ]

E = KE + PE =1/2mC2ω2 [ cos2[Φ]+sin2[Φ] ] =1/2mC2ω2 (doesn’t vary with t!)

we just checked the conservation of mechanical energy (spring force is conservative)

Demo: FORTRAN check energy conservation

Simple Harmonic Motion: Simple pendulum

!

"

!mg

mg sinθ =mg sin(s/L)= m at = m dv/dt = -m d2s/dt2

for small s/L sin(s/L) ≈ s/L recall sin(x)/x -> 1 for small x

-(g/L) s = -d2s/dt2 compare with spring -(k/m) x = -d2x/dt2

define ω = √(g/L) the solution is simple harmonic as well

the solution is s(t) = C sin (ωt+φ)

period T = 2π/ω = 2π√(L/g) again doesn’t depend on initial condition but L

adjust L to change the period of the pendulum - grandfather clock


-95-

Other applications: chemical bonds-vibrations,

SHO is of 2 models of bound systems (that is soluble in physics)

linear medium has SHO solution, locally, propagate in space time --

wave

ex of waves: sound, transverse wave in string, water, light,... (later)

Periodic motion: simply repeat itself after period T, in general this depends on

amplitude

Simple harmonic motion: special periodic motion that takes the sinusoidal form. T is

independent of A

Chaotic: if the force is non-linear slightly diff initial conditions produce big diff future

motion

©42© Gravitation: Special 2D Dynamics

Newton’s law of gravitation states the attraction between any two masses M and m, r distance

apart

F = G M m/r2

The Earth attraction on the Moon is of same nature as

Earth attraction on the apple only weaker by 1/r2

note that rEM = 3.84 x 108 m rEA = 6.37 x 106 m

we knew that in 1 sec Moon falls 1.36∞10-3 m

apple falls 4.9 m

the ratio 4.9 m / 1.36∞10-3 m = 3.6∞103 and

(rEM/rEA )2=(3.84 ∞ 108 m/6.37 ∞ 106 m)2 = (60.3)2=3.6∞103


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G: is universal: determined for two known masses at known distance with known attraction.

Cavendish (1798) determined: G = 6.67 ∞ 10-11 Nm2/kg2

With G, the mass of apple, and radius of earth, one may estimate the mass of the earth.

So Cavendish was known to be the first person to have weighed the earth.

The reason for tides (twice daily rise and fall of ocean water) is because gravity gets weaker with

distance (the inverse square law, like the moon is pulled less as apple)

Moon this water is pulledmore strongly thanthe earth

the earth is pulledmore strongly thanthis water

Earth

Weight: is the gravitational force acting on you

m is smaller (you went on a diet)

M is smaller (the earth had less mass or you are on the moon)

r is larger (you were farther from the earth, on the roof)

jump off the roof, what’s the acceleration? note two ways to express the g-force on you

from Newton’s 2nd law F = mg

from universal gravitation: F = G M m/r2

setting these equal, we find mg = G M m/r2 so g = GM/r2= GM/(R+h)2

on surface h=0, g = 6.67x10-11*5.98x1024/(6.37x106)2 = 9.83 m/s2 !!!

notice g is independent of m, so same for any object, g only depends on M and slightly on

h

W = mg shows the distinction between weight and mass

m: -is the quantity of matter in an object,

-measures how much gravity it exerts on other objects,


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-how much an object resists acceleration (how much inertia it has)

W: is the amount of gravitational pull on the object, and it varies according to where you are

in deep space h -->∞ W -->0 but m stays the same

units: m (kg) W (N) 50kg person is 490 N

technically incorrect to say someone “weighs” 50kg,

in BE force unit is pound, while unit of mass is slug

a person weighing 160 pounds has a mass m=W/g = 160 lb / 32ft/s2 = 5 slugs

©43© A closer look at the Moon’s motion: numerical solutions

Now let’s talk about the solution to the Moon’s motion (set Earth at fixed center)

Earth-Moon form a line, pick the plane containing the initial velocity of the Moon, since there is

no force out of that plane it will stay in that plane. (call it x-y plane)

!

!

F(x, y)

sin! = y/!(x + y )22

cos! = x/!(x + y )2 2

Fx = - F cosθ =-G M m /(x2+y2)*x / √(x2+y2) =-GMm x / (x2+y2)3/2

Fy = - F sinθ =-G M m /(x2+y2)*y / √(x2+y2) =-GMm y / (x2+y2)3/2

Newton’s 2nd law

m dVx/dt = -GMm x / (x2+y2)3/2 and dx/dt = Vx

m dVy/dt = -GMm y / (x2+y2)3/2 and dy/dt = Vy

cancel m and let GM = 1 for simplicity

dVx/dt = - x / (x2+y2)3/2 (1)


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dx/dt = Vx (3)

dVy/dt = - y / (x2+y2)3/2 (2)

dy/dt = Vy (4)

the x and y motion are coupled together, solve 4 equations together

from above, it is not difficult to show that: (challenge) [via (1)/(2) + (3)/(4)] 1/2 (Vx2 +Vy2) -1/(x2+y2)1/2 = constant which is special form of

1/2 m (Vx2 +Vy2) -GMm/(x2+y2)1/2 = E conservation of Moon’s mechanical energy

PE = -GMm/(x2+y2)1/2 KE = 1/2 m (Vx2 +Vy2)

the orbit is in general elliptical. If the initial speed is small E<0 , and can be circular

for larger initial speed E = 0 the trajectory becomes parabola

numerically, we can check to see what initial speed was needed for the Moon to

escape from Earth

for even larger initial speed E>0, trajectory is hyperbolic

estimate Moon’s period if its orbit were circular.

GM/r2 = v2/r = (2πr/T)2/r = (2π/T)2r so (2π/T)2 = GM/r3

T = 2π /(GM)1/2 r3/2 = 2π /(6.67x10-11x5.98x1024)1/2 (3.84x108)3/2 =2.37x106s=27.4 days

The analytical solutions of x(t) and y(t) are possible (1 of 2 known solvable examples in

physics)

In general the exact solution of the Moon with other interaction included (Sun), needs to be

numerical, and in the same spirit as we have been demonstrated. kep.f

vectors and their rules - illinois state

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