CH 1

Dimensional Analysis (units are important)
Use algebraic rules for units
ex: Area: l*w=10m*10m=10*10*m*m=100m²
cost:$120/gal* 5gal=$6
Radius of the earth: Front of the book: 6.4*10⁶ meters
area of sphere=4/3 r³ look up relationship is APPENDIX B 5.15m²
5.15m² =1.99E+08mi²

#2 a) Density of saturn: m/v=5.62*10²⁶kg/(4/3 (6.00*10⁷m³)=623kg/m³
b) would the planet float? Density of water: 1000kg/m³
#15: volume = At³+B/t where t=time V=m³
v=m³ At³=K*m³ A=K m³/sec³ B/t=Km³ B=K m³s K=constant
#20: class=40x20x12 m³ volume of air:V=9600m³ =3.39E+05ft³
weight of air: density=1.29kg/m³ * V(9600m³)=12384kg*2.2lb/kg=27245lb

Chapter one
motion in one dimension
Average velocity == distance traveled / time taken
symbolic representation:V=X/t
units: m/s

Instantaneous velocity==the limiting value of the ratio x/t
V==lim_t->0 x/t V=dx/dt

---

Average acceleration==change in velocity/change in time
=v/t=a
units: m/s/s=m/s²


one dimensional motion with constant acceleration
a=dv/dt at=dv/dt*dt (since a is defined as a constant, we can take it out)
at=dv=v (taking the integral of the derivative is inverse)
at+C=v if t=0, C=v so C=v_o
at+v_o=v velocity at a given time no position t=v-v_o/a v=dx/dt
v dt = dx/dt*dt = x
(v_o+at)dt=x v_odt+at dt=x v_ot+a +atdt=x v_ot+1/2(at²)+x_o=x
Constant of integration (C) becomes x_o when t=0
v_ot+1/2(at²)=(x-x_o) x-x_o=v_o((v-v_o)/a) + 1/2a((v-v_o)/a)² No final velocity

x-x_o=v_ov/a + v_o²/a + (1/2)(v²/a) - (v_o)/a + 1/2(v_o/a)²
v_o² + 2a(x-x_o)=v² getting rid of time to have v with respect to x
x-x_o=[(v+v_o)/2]t acceleration replaced by time No accelleration
v=dx/dt
a=dv/dt if a=constant

Free falling bodies: An object thrown upward (or downward) will experience constant acceleration
Acceleration due to gravity: 9.8m/s² down = g =9.8 (NOT -9.8) remember
to put in negative if the direction is down

What is V of a falling ball as a function of T
v=v_o+at v=v_o gt: + if g is in your positive direction / - if g is opposite to your negative direction

x-x_o=v_ot+1/2 at² --> x-x_o=v_ot+1/2 gt² in the case of freefall

v_o² + 2a(x-x_o)=v² --> v_o² + 2g(x-x_o)=v²

what is accelleration on a slope: no final velocity
x-x_o=v_ot+1/2 at²
2.4m=0+1/2 a4²sec a=.3m/s² = 30cm/s²
1.35m=[(v+0)/2]3s .9m/s

#15
0=60m/s+a15s a=-4m/s² negative means acceleration is in the opposite direction to the movement

#25
x-x_o=v_ot+1/2 at²
3-5=12cm/s *2s + 1/2 a(2s)² -2cm-24cm/s²

#34
CRT 2*10⁴m/s to 6*10⁶m/s over a distance of 1.5cm
a) how long does it take?
x-x_o=[(v_o+v)/2]t
1.5cm=[(2*10⁴+6*10⁶)/2]t
.015m(2/(2*10²+6*10²))=t=4.98*10^-9sec =4.98ns
b) what is the accelleration?
(v²-V_o²)/(2(x-x_o))=a=1.2*10¹⁵m/s²

#48
hot air balloon moving up at constant speed of 5m/s
21m above ground: a package is dropped
a) how long is the package in the air?
21m=-5m/sT+1/2(9.8m/s²)t²
4.9t²-5t-21=0 5 (25+4(4.9)(21))]/9.8= (5 20.3)/9.8= 2.58s,-1.56s

b) V just before impact
V= (-5m/s+2*9.8m/s²(21m)= 20.9 *only positive is correct=20.9m/s*

#74
a) rock is dropped into well: sound is heard 2.4s later. How deep is the well?
t₁+t₂=2.4
x-x_o=v_ot+1/2at²
d=0+1/2(9.8m/s²)t₁²
-d=-4.9m/s²t₁²
d=336m/s²t₂

#60
the acceleration of a marble in a fluid is proportional to v²
a=-3v² v_o=1.5m/s when is v=1.5m/s/2? a=dv/dt (to get v with respect to time: integrate this equasion)
-3v²=dv/dt
-3v²/v²=dv/(v²dt)
-3dt=1/v² * dv
-3t=-1/v+C 3t+c=1/v
v=1/(3t+c) v_o=1/c V=1/(3t+1/v_o)
v_o/2=1/(3t+1/v_o)= 3tv_o=1 t=1/(3*1.5)

75: train travels straight between 1 and 2
accellerate like this:
t₁+t₂+t₃=5mins
2t₁+t₂=5min

A to B
x-x_o=v_ot+1/2at²
d=0+1/2at₁²
d=1/2at₁²

B to C
d=v_bt₂
1/2at₁²=vbt₂ a=v_b/t₁ 1/2(vb/t₁)t₁²=vbt₂ t₁/2=t₂
2t₁+t₁/2=5 2.5t₁=5 t₁=2


Chapter 3. Vectors
Coordinates sytems:

r=x'
ø=0

r= (x²+y²)
hyp=r
sinø=(y/r) =(opp/hyp)
cosø=(adj/hyp)
tanø=(opp/adj)
y=hyp sinø
x=r cosø
Questions like: How much of a vector is in the x(or y) direction

vector has direction & magnitude and scalar just has magnitude
scalar: 50 people, 42
vector: 10 m/s up

Vector: A or A
Scalar: A

A+B |A|+|B|
D=A+B+C
Angles between vectors are measured when the vectors are put tail end to tail end and the angle LESS THAN OR EQUAL TO 180
|A_x|= |A|cosø=A_x
|A_y|= |A|sinø=A_y
vectors without arrows or bold mean the
magnitude of the vecor
î or x = unit vector in x direction
j or y = unit vector in y direction

A_x=A_xî Acosøî
A_y=A_yj Asinøj

A=A_xî+A_yj = Acosøî+Asinøj
We want final
Find R=A+B =A_xî+A_yj + B_xî+B_yj (A_x+B_x)î + (A_y+B_y)j =R_xî+R_yj |R|= (R_x² +R_y²)
tan^-1g=Ry/Rx


Ch 3 ex:
#2 vector between (2,y) and (r,30 )
r= (x² +y²)
tan 30 =y/x y=xtan 30 = 2tan30

#14
a force of f1 that has a magnitude of 6 acts in direction of 30
f₂ magnitude 5 in direction of y axis
what is magnitude and direction total force R?
6cos30 î+6sin30 j +5cos90 î + 5sin90 j
=5.2i+8j (5.2² +8²)=9.5=|R|
tan^-18/5.2 57


#16 R=A+B+C
=0i-3.5j+8.2cos30 i+8.2sin30 j-15i=-7.90i+.6j
|R|= (7.9² +.6² )=7.92 @ tan^-1(.6/7.9)=-4.3 (add 180 )=175.7 #32: A=3i+3j
B=i-4j
C=-2i+5j
D=A+B+C = 2i+4j |D|= (4+16)=4.47 tan^-1(4/2)=63.4
E=-A-B+C = -6i+6j |E|= (6² +6²)=8.5 180+tan^-1(6/-6)=

#50: F₁=120N @ 60
F₂=80N @ 75 from -x axis
F_total=F₁+F₂=(120cos60 -80cos75 )i + (120sin60 +80sin75 )j=39.29Ni+181.19Nj
|F|= (39.29² +181.19² )=185.4N @ tan^-1(39.9 /181.2 )= 12.2 from +Y axis

Ch 4 Motion in two dimensions
How to work with two dimensional motion. Work with 1D motion twice
3D sometimes Work with 1D motion 3 times

Horizontal motion is INDEPENDENT of vertical motion

Position vector (r)=xi+yj

Displacement vector: r=r_f-r_i r_i+r=r_f x_fi+y_fj-(xii+yij)=(x_f-x_i)i+(y_f-y_i)j=xi+yj
Origin doesn't matter as much as the Change in position

Average velocity vector: v=(r/t) (x/t)i+(y/t)j
= v_xi-v_yj

Instantaneous Velocity==lim_n->0(r/t)=dr/dt=d/dt(xi+yj)= dxi/dt+dyi/dt
=dx/dt i + dy/dt j v=v_xi+v_yj

Average acceleration==a=v/t=(v_f-v_i)/t = [ (vx_fi+vy_fj)-(vx_ii+vy_ij)]/t
=[(v_f-v_i)i+(v_f-v_i)j]/t = (v_xi+v_yi)/t = (v_x/t)i+(v_y/t)j
A=a_xi+a_yj
acceleration in x direction is independent of acceleration in y direction

Instantaneous Acceleration lim_t->0 v/t=dv/dt=(dv_x/dt)i+(dv_y/dt)j= a=a_xi+a_yj
Constant acceleration means components don't change

Solve if a is constant
a=dv/dt
a dt=dv/dt dt
(a_xi+a_yj)dt=d(v_xi+v_yj)
(a_xi)dt+(a_yj)dt=d(v_xi)+(v_yj)
(a_x)dt=d(v_x) = a_xt+C=v_x
1: a_xt+v_ox=v_x velocity in x direction is proportional to acceleration in x direction and time
(a_y)dt=(v_y) = a_yt+C=v_y = a_yt+v_oy=v_y = velocity in y direction is proportional to acceleration in y direction and time
at + v_o= v vectors have x and y components
v_ox=v_ocosø v_oy=v_osinø
v=dr/dt vdt=dr/dt * dt (v_o+at)dt=r v_ot+1/2at² +C=r
r=r_o+v_ot+1/2at²
2: x=x_o+v_oxt+1/2a_xt² y=y_o+v_oyt+1/2a_yt²
t=(v_x-v_ox)/a_x plug t in:
3: v^2x =v^2ox + 2a_x(x-x_o) v^2y =v^2oy + 2a_y(y-y_o)

4: x-x_o=[(v_x+v_ox)/2]t y-y_o=[(v_y+v_oy)/2]t

Time is a Scalar and can be used in either component (eg: solve for t in x direction and use it in the y direction)

Projectile motion: a=g down=9.8m/s² down

if a=g and coordinate system is +x-> and +y
v_x=v_ocosø
v_y=v_osinø -gt

x-x_o=v_ocosøt
y-y_o=v_osinø-1/2gt²

Uniform circular motion (contant speed)
a=v²/r inward

Tangential & radial acceleration
a=a_r(radial)+a_t(tangential) = (dv/dt)ø + (v²/r)(-r )

Relative Velocity & Relative acceleration


x=v_oxt+1/2at²
ut=vector between first and second observer
r=r'+ut second moves at constant speed

dr/dt=dr'/dt+u IN EACH DIRECTION (x,y)
how the position of changes with respect to first=how the position of changes with respect to second + speed of second
v=v'+u
dv/dt=dv'/dt+0
a=a' Both observers see the same acceleration

Exam question:
Non contant acceleration: must use integration. NOT CONSTANT ACCELERATION EQUATIONS

Laws of Motion
Law of Inertia
1. An object at rest will remain at rest and an object in motion will continue in motion with a constant velocity unless is experiences an exernal force.

Mass: measurement of inertia
Inertia: resistance to change


2. a=f/m acceleration vector = net force vector / mass = F_net=ma
Force: newtons=kg*m/s² =N
F->F_net->a->v->r

Acceleration is in the direction of the net force
contact force:
a push, pull or hit

field force:
gravity, charge, magnetic

3. If two bodies interact, the force exerted on body 1 by body 2 is equal to and opposite to the force exerted on body 2 by body 1.


3 Laws are only valid:
Inertial Frames: a non-accelerating frame
when V is << C

5 Steps for solving a problem involving the 3 Laws.

1. Draw a neat diagram of what's going on
2. Draw free-body diagram
Isolate a single object and draw the single object then all forces acting on the object without forces acting on any other object
3. establish a convienient coordinate system and find coordinates
4. solve equation using second law
5. check answer



Weight vs Mass vs Apparent Weight
F_net=W=mg Amount of inertia How much tension is on a string on a
units: N kg scale which is Weight if F_net is 0
Weightless: apparent weight is 0 NOT real weight=0
Friction force: Solid-to-Solid friction

Static friction:
f_s=F
f_s &#181;_sn

Kinetic friction: &#181;_kn
&#181;_k and &#181;_s are constants: cooefficient of kinetic and static friction table: Page 126

a=F_net/mass
a_x=F_{x net}/mass
a_y=F_{y net}/mass


Ch 6
Newton's Laws continued:
Uniform circular motion
a_r=is the Total acceleration
F_net=(mv²)/r
F=ma
F_y=T-mg=mv²/r
F_x=ma_x=0

All forces add up to mv² /r towards the center of the circle

If the force is non-uniform: the speed can change: there is a tangential force
F_x=m(d|v|/dt)


Solid to Fluid Friction
Resistive forces


For a dropped object
F_y=ma_y=mg-bv
a_y=g-(b/m)v


Terminal speed: 0=g-b/m v_t v_t=(mg)/b
Any V: a_y=g-(b/m)v
dv/dt=g-(b/m)v
dv/dt=g-(b/m)v=(-b/m)(v-mg/b)
(dv/dt)/(v-mg/b)=(-b/m)
(dv/dt)/(v-mg/b)=(-b/m)
(dv/dt)* (1/(v-mg/b)) dt=(-b/m)dt
ln(v-mg/b)=(-b/m)t + C
v=e^{((-b/m)t + C)}+mg/b
v=e^{((-b/m)t e C}+mg/b
v=C₂e^((-b/m)t +mg/b
V(t=0)=0=C₂e⁰+mg/b C²=-mg/b
v=(-mg/b)e^((-b/m)t + mg/b=(mg/b)(1-e^((-b/m)t )=v
If we released an object at above the terminal speed:

If v_o = 2(mg/b)
C₂e⁰+mg/b C₂=mg/b
mg/b(1+e^-(b/m)t )

take limit as t-> , v=mg/b

If speeds are great: Frictional force is quadratic
rather than linear relative to current V.


at terminal speed: a=0 a=g-(B/m)v² g=(B/m)v² (mg/b)=v^t
At High speeds AND released
v at any time: v(t)= (mg/b)[(1-e^{-2 (mg/b)t})/(1+e^{-2 (mg/b)t})]
This equation for the graph of the speed after some cut off speed and the other equation is valid for low speeds.

Chapter 7 Work and Energy
Work: The work done by a constant force is defined as the product of the component of the force in the direction of the displacement and the magnitude of the displacement.
Work=force*distance
w=fd

To calculate work: use force that we are asking about.


W=Fcosø*distance
W=F_{| |} *distance (F parallel * distance)
W=F*distance_{| |}

Units:
W=NM=Joule = J = kgm²/s²

Work done by friction: W=fcos180 * s
W_f=-fs

Math review: vector multiplication Vector Scalar Product (dot product, inner product)
2 vectors multiplied together to get a scalar
A B = |A|*|B|*cosø if we know the angle between the two vectors
A component of A is parallel to B and a component of A is perpendicular to B or vice versa

A_II B=A B_II if we know a parallel component
AxBx+AyBy if we know x,y components
A B=(Axi+Ayj) (Bxi+Byj)
Axi Bxi + Axi Byj + Ayj Bxi + Ayj Byj
AxBxi i + AxByi j + AyBxj i + AyByj j
AxBx + 0+0+ AyBy
the vectors have a magnitude of 1 and multiply by the angle between them
AxBx + AyBy

Mathmatical properties A A=AxAx+AyAy=Ax²Ay² =|A|²
A B = B A
A (B+C) = A B + A C

W_net =F_net S
=(F₁+F₂....) S
= F₁ S+F₂ S....
= W_f1 + W_f2...

If the force is varying, work will vary.

w=FXcosø
W_total=W
= lim_x->0Fxcosø
= Fcosø dx
=F¥ds if F is in the direction of S
s is the direction the box moved
if F is a constant and ø is a constant
W=F ds=F S
if just ø is constant
W=Fdscosø=cosøFds

Springs:
Hooke's Law: F x F is proportional to X F=-kx
Force varies with distance pulled or pushed
Linear relationship between force and distance pulled or pushed
-k is the spring constant Units: N/m
-k means how many Newtons you get for every meter pulled
- sign means that force is in opposite direction of movement

F_s=-kx W_s=Fds =Fdscosø
=kxdx cos 180 no minus sign on K because this is the -kxdx=-k(x²/2)|x_f^xi magnitude of the force
W_s=(1/2kx_i²) - 1/2kx_f² The spring has done negative
work: something has done work on the spring
W_h=Fdscosø=kxdxcos0 = kxdx
W_h=(1/2kx_f² )-(1/2kx_i²)
W_s=Fds = -kxi dxi = -kxdxii=-kxdx dot product taken in terms of unit vectors

Work Energy Theorem

W_net=F_netds = F_xds=ma dx = madx=mdv/dt dx =mvdv = (mv²/2)|_viv^f =
(1/2)mv_f² - (1/2)mv_f²
1/2mv² =kinetic energy
W_net = KE : Both in joules
Velocity must be less than 10% of C and all observers must be in inertial reference frames

KE=mC² (1/ (1-(v²/C²))-1) for any speed
Math: Taylor series: f(x)= f(a) + ( (x-a)f'(a))/1! + ((x-a)² f"(a))/2!+.....
KE=mC² (1/ (1-(v²/C²))-1)
f(v)=1/ (1-(v²/C²))
f'(v)=-1/2(((2v/c²)/(1-v²/C²)^3/2 ))

Power: The rate of energy transfer
P=dw/dt Units of watts 1watt=1joule/sec

lim_t->0w/t=Fx/t cosø = P=Fvcosø=F V

Chapter 8: potential energy & Conservation of energy

1: A force is conservative if the work done by that force acting on a particle moving between the two points is independent of the path the particle takes between the points.
(path independant)

W_PQ =Fds
It may not actually move along the path that we choose to integrate over.


W_PQ (along path 1)=W_PQ (along path 2)=-W_QP(along path 1 or 2)

2: The work done by a conservative force on a particle is 0 when the particle moves around any closed path and returns to it's initial point (d=0,w=0)

Gravity & Spring: conservative forces
Friction: Non-conservative force

Potential energy is defined as the negative of the work done by a conservative force
Potential energy=U=U_f-U_i = -_i^fFds
if U_i=0, U_f=U

Work - Energy theorem with conservative forces
W_net=K = W_c work done by a conservative force
-U=K
0 = K + U = (K+U) change in the sum of the two energies (use up potential to get kinetic and vice versa)
(K_f+U_f) - (K_i+U_i) = 0 = K + U
Law of Conservation of Mechanical Energy

GPE: gravitation potential energy (near earth's surface)
U= -_p^qFds =-_p^qmg(-j) (dxi+dyj)=-_p^q(0i + -mgdy)=_p^qmgdy = mgy|_p^q =mgh_q-mgh_p
Horizontal movement is irrelevant U_f
U_i
(K_f+U_f) - (K_i+U_i) = E_f-E_i
E_f=E_i starting energy = ending energy .5mv_f² +mgh_f=.5mv_i² +mgh_i

v_f²=2g(h_i-h_f) if object is dropped from rest.

Spring potential energy
Us=-Fds = -Ws = -(.5kx_i² -.5kx_f² )
change in PE of a spring = -Force distance = - work = -(.5spring constant * original position² - final position2 )

U_fs = .5 kx_f²
U_is = .5k_i²

Non conservative forces:
W_nc + W_c=K
W_nc-U=K
W_nc = K + U
W_nc = E_f-E_i
-f_s + E_i =E_f
-friction of a spring + initial Energy = final Energy
E_f + Heat = E_i
Heat = f_s solid to solid friction

Conservation of Energy
Energy can never be created or destroyed. Energy may be transformed from one form to another

Relationship of forces & potential energy
V_f-V_i = -Fdx - Fdy + V_i
dV_f/dx|_y=constant = -fx
Quantization of Energy
E=MC²

Chapter 9: Linear momentum & Collisions
Conservation of momentum
Define Momentum = p=mv Bold=vectors
P_x=mv_x
P_y=mv_y
units kgm/s=kgm/s² * s=Ns
Newton's second law: F_net=ma =m(dv/dt) = d(mv)/dt=dp/dt
F_net=dp/dt the net force on an object changes it's momentum
if F_net=0, then there is no change in momentum (P is constant)
= conservation of linear momentum (not spinning)
if F_net 0 then momentum is not constant. F_net=dp/dt
dp=F_netdt _i^fdp=_i^fF_netdt dp=p =_i^fF_netdt="Impulse"
Impulse = I =_i^fF_netdt = Impulse-momentum theorem
if F_net=Constant, then impulse is F_net*t
if F_net Constant, the impulse is average F * t
Average force = (1/t)F_netdt = sum of the forces divided by the contact time
F_net=dp/dt =d(mv)/dt
Force is directly proportional to mass and velocity
The Chage in momentum is not always equal to the momentum
A bouncing ball has a Larger change in momentum than a ball that sticks
Force is inversely proportional to the change in time

A two particle system (no external forces)

F₁₂ =dp/dt
F₂₁=dp/dt
F₂₁=-F₁₂
F₂₁+F₁₂=0
The net force for the system is 0. d(P₁+P₂)/dt=0
(P₁+P₂+...+P_n)= constant = total momentum of the system = P_total
Add all force generating objects to the system
if F_net = 0, p =0 which means momentum is constant

N-particle system with no external forces:
F₁₂ + F₁₃ + + F_1n + F₂₁ + F₂₃ +...+ F_2n +...+ F_n1 + F_n2 + ...+ F_nn-1 = 0
All pairs coorespond and cancel out to be 0 for the whole system
All forces on object 1, 2, n : dP₁/dt + dP₂/dt + dP_n/dt = (P₁+P₂+...+P_n)
(P₁+P₂+...+P_n)= constant = total momentum of the system = P_total
P_i=P_f if there are no external forces
P_ix=P_fx P_iy=P_fy

In a collision:
Is the kinetic energy conserved?
Elastic collision: Yes
Inelastic collision: No (some is converted (possibly to heat) )
Perfectly inelastic collision: two objects hit and stick: most KE is converted to heat
half of the kinetic energy goes to heat when two masses of the same size hit and stick and one was moving and one was not.
all KE is lost when initial total momentum is 0

Perfectly inelastic collision:
P_i=P P_ix=P_fx
v_f=(m₁v_1i+m₂v_2i)/(m₁+m₂)

Elastic collision
after collision: how fast are the two objects moving?
P_ix=P_fx elastic means: K_i=K_f



m₁v_1i+m₂v_2i=m₁v_f+m₂v_2f .5m₁v_1i² +.5m₂v_2i² =.5m₁v_1i² +.5m₂v_2i²
m₁(v_1f-v_1i)=m₂(v_2f -v_2i)
m₁(v_1i-v_1f )(v_1i +v_1f ) = m₂(v_2f-v_2i)(v_2f+v_2i)





(m₁v_1i+m₂v_2i-m₁v_1f)/m₂=v_2f
v_1i+v_1f = v_2i + v_2f
v_1f=((m₁ -m₂)v_1i)(m₁+m₂)+ (2m(v_2i))/(m1+m₂)

Symetry: look at previously solved equations and look for similar concepts and variables
Call everything with a "1": "2"
(m₁v_1i+m₂v_2i-m₂v_2f)/m₁=v_1f
v_2f=((m₂-m₁)v_2i)(m₂+m₁)+ (2m(v_1i))/(m2+m₁)

Inelastic collision:
pi=pf
m₁v_1i+m₂v_2i = m₁v_1f+m₂v_2f
perfectly inelastic collision:
m₁v₁i+m₂v_2i = (m₁+m₂)v_2f
K_i=Kf+heat heat is unknown in general
equation can only be solved if % kinetic energy lost is given or v_f is given

Two dimensional collision

choose coordinate system
P_ix=P_fx P_iy=P_fy
m₁v₁i+0=m₁v_1fcosø +m₂v_2fcos 0=m₁v_1fsinø - m₂v_2fsina
if elastic:

ki=k_f
.5m₁v₁i2 +0=.5m₁v_1f2 +.5m₂v_2f²

Real objects have size and shape
n-objects

m external forces


F_net=F=F₁₂ +F₁₃+...+F_1a+F_1b+....F_1m =m₁a₁
+F_net=F=F₂₁+F₂₃ +...+F_2a +F_2b +....F_2m =m₂a₂
+:
+:
+F_net=F=F_n1+F_n2+...+F_na+F_nb+....F_nm =m_na_n

=m₁a₁ +m₂a₂ +m_na_n =m_ia_i

=(Internal=0 because internal forces pair up and cancel)+F_ext
the center of mass of a system is a good approximation for the movement of a whole system
M_total=Ms_ystem
F_net=F_ext =M_totala_system
a_cm a^system =(m_ia_i)/m_total =weighted average: the more weight one mass has, the more it affects the system
a_cm =dV_cm /dt=(m_ia_i)/m_total )=(m_idv/dt)/m_total =d(m_iv_i/m_total)/dt
dV_cm /dt=d(m_iv_i/m_total)/dt: integrate both sides
V_cm=m_iv_i/m_total
position: m_total=m_total/dt=d(mr_i/m_total)/dt
rm_total=mr_i/m_total
F=F_net->a_cm ->v_cm->rm_total
X_cm =m_ix_i/m_total
Y_cm=m_iy_i/m_total

Real objects are continuous
r is the position of each little mass (or volume)
r_cm =lim_Mi->0mr_i/m_total =(1/m_total)rdm
density=mass/volume=

=dm/dv
dv=dm=(1/m_total)rdv
x_cm=(1/m_total)xrdv=(/m_total)xy(x)tdx
(t/m_total)x(h/l)xdx
(th/m_totall)₀^lx²dx
th/m_totall)(l³/3):substitute density=mass/volume
(th(m_total/.5lht)/m_totall)(l³/3)=(2/3)l=x_cm
Ycm=(1/m_total)ydm=(1/m_total)ydv=(/m_total)ydv=(/m_total)yt(l-x)dy=(/m_total)(yl-yx)dy
(/m_total)₀^h(yl-yx)dy=(/m_total)[₀^hyldy-y(l/h)ydy]
y=(h/l)x x=l/h
(t/m_total)[(lh²)/2-lh²/3)=(t/m_total)(lh²)/2-(t/m_total)(lh²/3)=(tlh²/6m_total)
(((m_total).5lh²)tlh² /6m_total)=(1/3)h
t=thickness
=dm/dv

x_cm=1/Mxdm=1/Mxdv=/Mxdv=/Mxty(x)dx=/Mxy(x)dx=/M_-R^Rx (R²-x²)dx=0
Y= (R²-x²)
when there is an odd function over an even interval: the answer is 0
when there is an even function over an even interval: the answer is 2₀^Af(x)
(t/M)(-1/2(2/(3(R²-x²)^3/2)))=(t/M)(-1/(3(R²-x²)^3/2))|_-R^R =0-0=0

y_cm=1/Mydm=1/Mydv=/Myt2xdv=t/My2xdx=2t/My (R²-y²)dx=
2t/M(-1/3(R²-y²)^3/2 |₀^R =2t/M[((-1/3)0 +(1/3)(R²)^3/2]
V=.5t r²
2(M/.5t r²)t/3M(R)³]=(4/3 )R =42%R

#56:



16.
h=60m
bucket=.75kg
.25l/s
what does the scale read after 3 seconds of filling the bucket
F_net=m_bg+mwg+dp/dt
.75kg*9.8m/s²+.75kg*9.8m/s²+d(mv)/dt
1.5kg*9.8m/s²+mdv/dt+vdm/dt=1.5kg*9.8m/s²+mdv/dt+vdm/dt
0 because velocity never changes
( 2gh)(.25l/s)



Chapter 10: spin the objects

Rotation of a rigid objects about a fixed axis

Average angular velocity==ø/t=
instantaneous angular velocity==lim_t->0 ø/t=dø/dt=w
Average angular acceleration==/t=
instantaneous angular acceleration==d/dt= =d²ø/dt²

direction of angular velocity: perpendicular to the plane.
curl fingers of right hand in direction of rotation and thumb will point in directino of angular velocity
angular Acceleration will be perpendicular to the plane either in the same direction as velocity if positive or opposite the direction of velocity if negative.

Rotational kinematics: (x=ø,v=,a=)
dt=(d/dt)dt
dt= If is a constant:
=_o+t no position ø-ø_o=_ot+.5t² no
_o² + 2(ø-ø_o)=² getting rid of time
ø-ø_o=[(+_o)/2]t no acceleration
Relationship between angular and linear quantities
v_t: tangential velocity=ds/dt= change of arc length with time=d(rø)/dt=r(dø/dt)=r
v_t=r when there is a larger radius (and hence a larger circumference), there is a larger linear speed (duh)

a_t=tangential acceleration=dv_t/dt=d(r)/dt=rd/dt=r=a_t
a_r=radial acceleration=v_t²/r=r²w²/r=rw² a_r=r²
a_total =Total acceleration= (a_t²+a_r²)= (r²² +r²(w²)² )=r (²+⁴)

Moment of Inertia of a uniform solid cylinder
I(ring)=rdm => MR²
I(hollow cylinder)=I(ring) => MR²
I(solid cylinder)=I(hollow cylinder)=lim_m->0MR=r² dm=r² 2 hdr
r² 2 hdr=densityradius²(circumference)height*small radius
2 h*R⁴/4=(M/( R²h))(R⁴/4)=1/2MR²
P286: Moment of inertia of various objects
I(sphere)=I(cylinder)
40: spherical shells
I(spherical shell)=I(rings)

Parallel axis Theorem
I=Icm+MD²
I at the CM + mass * distance from CM
I(cylinder rim)=.5MR² +MR² =(3/2)MR²

a=F_net/M
=/I : force to rotate a body about some axis
t:tangential
F_t=Ma_t
rF_t=rma_t=rmra =mr²
rF_t=I
=rF_t/I=torque/Inertia
torque: tangential piece of the force * radius at which the force is applied

== rFsin is the angle between force and the radius
F_t=Fsin
Total:
rF_t=I
_net=I_neta

Rotational Work & Energy
W=F s
dW=F ds
is the angle between force and motion
dW=Fdscos + =90
=Fdscos(90-)
=Fdscos90cos-+sin90sin- =F ds sin
=Frsindø
dW=dø
W=dø

Power:
P=dW/dt=dø/dt=(dø/dt)=
P= =Nm
P=Nm/s=Watt

Work Energy theorem
Work done on an object is equal to the change in kinetic energy
W=K
W=dø
K=.5I_f²-.5I_i²

=I
=Id/dt=Id(/dø)(dø/dt)
=Id(/dø)(dø/dt)
dø=I(d/dø)(dø/dt)dø
dø=I(d/dø)dø
=Id
=I²/2=.5i² -.5I²
=.5i² -.5I²
P294: Table of linear equations | angular equations


Chapter 11: Rolling motion, angular momentum, torque

K=.5I_p² =.5(I_cm+MD²)² =.5I_cm² +.5MD2² = .5I_cm² +.5MVc²
KE @ center

Vc=velocity of center Vc=velocity of edge after we have changed our viewpoint to say the object is rotating about the center. Always remember to put in the translational velocity term as well.

Rolling object down incline

mgh=.5I_cm² +.5mVc² + no heat
mgh=.5IVc² /R² +.5mVc²
[2gh/(1+ImR²)]=Vcm
Vcylinder= [2gh/(1+.5)]= [(4/3)gh]
Vhooop= [2gh/(1+1)]= gh
Vsphere= [2gh/(1+2/5)]= [(10/7)gh]
Vshell= [2gh/(1+2/3)]= [(6/5)gh]

Cross product:

||=rFsinø
=r x F
L=r x p
|L|=|r||p|sinø
|L|=mvr for motion in a circle
C is prependicular to both A and B (in or out of board)
|C|=|A||B|sinø = |B||A|sinø
C=AxB= -BxA
AxA=0 (sinø=0)
If A is perpendicular to B, then |AxB|=AB (maximize torque by making angles =90)
Ax(B+C)=AxB+AxC
d(AxB)/dt=AxdB/dt+dA/dt x B = AxB'+A'xB
ixj=k ixk=-j jxk=i

3x3 determinant
AxB= i j k
Ax Ay Az =(AyBz-AzBy)î+(AzBx-AxBz)j+(AxBy-AyBx)k
Bx By Bz

Angular Momentum of a single particle
L=rxp=rpsinø=rmvsin90=mvr in a circle
p=linear momentum
=rxf
particle moving in a straight line has a constant angular momentum which is only 0 if the movement is along an axis

=rxf =r x dp/dt
L=rxp dL/dt=d(rxp)/dt=rxdp/dt + dr/dtxp
=+0
dL/dt= F=dp/dt

for a continuous object, take the sum of the formula for a particle

t=dL/dt
_net=dL/dt=dL_tot/dt
_net =dL_tot/dt
for the ith object:
L_i=m_iv_ir_i
L_i=m_ir_ir_i is the same for all objects in the continuous object
(m_ir_i² )
L_total=I p=mv
dL_total/dt=d(Iw)/dt
net = Id/dt
net = I F_net=ma
conservation of angular momentum:
if =0, L_total is a constant
I₁₁+0=(I₁+I₂)_f
Units for angular momentum: kgm²/s (if you multiply top and bottom by s, units become kgm²s/s² =Js
torque of a gyroscope changes direction of angular motion rather than magnitude



Chapter 12: not 12.4

Static Equilibrium: We want an object to be stationary and we want it to remain stationary.
Vc=0 dVc/dt=0 F=0
=0 d/dt=0 t=0
F=F₁+F₂+F₃+F₄=0
=r1xF₁+r₂xF₂+r₃xF₃+r₄xF₄=0
In a plane:
F_x=0
F_y=0
_z=0


remember gravity acts on the cm
sketch a diagram
draw free body diagram
resolve components
choose axis system and calculate torques
solve equations

Extra credit:
write the moment of inertia in terms of (X)MR²
Examples

Chapter 13
Simple Harmonic motion
x(t)=Acos(t+)
A: amplitude
: angular frequency
t: time
: phase constant - determines where occilations start

Period: time for object to go through one cycle: T: units: seconds
x(t)=x(t+cT)
Acos(t+)=Acos((t+T)+)=Acos(t++T)
Acos(t+)=A[cos(wt+)cos(wT) -sin(wt+)sin(wT)]
Acos(t+)cos(T) -Asin(t+)sin(T)
cos(T)=1 => n2 where n is an integer
sin(T)=0 => n where n is an integer
wT=n2 where n is a integer
T=n(2 /) if n=1(for first period), T=2¹/
frequency (non angular): f: number of cycles in one time unit
units: # of cycles per second= 1/s = Hertz=Hz
f=1/T
f=/2¹ 2 f=

x(t)=Acos(2 ft+)
v(t)=-Asin(t+)
vmax=-A
a(t)=-²Acos(wt+)=-² x(t)
amax=²A
Force = -kx for a spring

Horizontal spring
F_x=-kx =ma=md²x/dt²
(-k/m)x=d²x/dt²
x=Acos(t+)
dx/dt=-Asin(t+)
d²x/dt² =-w²Acos(wt+)
-k/m=² = (k/m)
T=2 /=2 m/k
f=1/T=(1/2 ) k/m

Vertical spring
y_o is distance spring has been extended to equilibrium with a mass
y is distance extended from new equilibruim
F_y=-ky-mg=ma=-k(y_o+y)-mg-ky=ma=(-ky_o-mg)-ky=ma
0 -ky=ma
-ky=ma=md²y/dt²
y(t)=Acos(t+)
y(t)=Acos( (k/m)t+)
= (k/m)
T=2 /=2 m/k
f=1/T=(1/2 ) k/m
A is max angle (amplitude)

Energy
E=k+U=.5mv² +.5kx²
.5m(-Asin(t+))² +.5k(Acos(t+))²
.5m²A²sin²(t+)+.5kA²cos²(wt+)
.5m(k/m)A²sin²(wt+)+.5kA²cos²(wt+)
.5kA²(sin²(wt+)+cos²(wt+))=.5kA²(1)
.5kA² total energy does not change
.5kA² =.5mv² +.5kx²
(kA² -kx² )/m=v²
[k/m(A² -x²)/m]=v Simple pendulum:
mass is concentrated

Ftangentially=Ft=-mgsinø=ma_t=md²s/dt²
minus becuase the direction of the force is opposite the direction of ø
=md²lø/dt²
=mld²ø/dt
(-mg/l)sinø=(ml/l)d²ø/dt²
(-g/l)sinø=d²ø/dt²
motion is independant of mass
Taylor Series:
f(x)= _n=0 [(x-a)ⁿfⁿ(a)]/n!
take sinx as our approximation
sinx=(x-0)⁰+f⁰(ø)/0!+(x-0)¹+f¹(ø)/1!+(x-0)²+f²(ø)/2!+(x-0)³+f³(ø)/3!+.....
since the denominator is growing, the function is shrinking in significance
0+x(1)/1+x²(0)/2!-x³/6+....
x-x³/6+x⁵/5!-....
if x<30 or .5 rad, sinx=x
ø(t)=Acos(t-)=Acos( (g/l)t-)
s(t)=lø(t)=lAcos( (g/l)t-), Scos( (g/l)t-)
v(t)=-Smax g/lsin( g/lt+ø)
= g/l
t=2 l/g
f=1/T=(1/2 ) (g/l)

Physical pendulum: non point mass:

_z=mgsinødsin90 90 is the ø between r and F, d is distance from cm to axis
_z=-Frsinø = Id²ø/dt²
if ø 30 then -mgdø/I=d²ø/dt²
ø(t)=Acos(wt+)=Acos( (mgd/I)t+)
= (mgd/I)
t=2 (I/mgd)
f=1/t=(1/2 ) (mgd/I)

T(1meter)=2 (1/3mL² / mg1/2L)=2 [(2/3)(L/g)]

Torsional pendulum: spinning something hanging from a spring
Hooke's Law => F=-kx streatching a spring
=> =-kø rotating a spring
= -kø = I = -kø =Id²ø/dt²
=(k/I)ø=d²ø/dt²
= (I/k)
T=2 I/k
f=1/T=(1/2 ) (k/I)
friction: period is less than theoretical, but it maintains it's period

Damped oscillations:


F=ma=-k(y_o+y)-mg+f
ma=(-ky_o-mg) -ky+f
0
ma=-ky+f
md²y/dt² =-ky-bv=-ky-bdy/dt
y(t)=Ae(-b/2m)t cos[ (k/m - (b/2m)² )t+]

if there is a Lot of friction: critically damped
b= 4km
(b/2m)² =4km/4m² =k/m
= (k/m-k/m)=0

Every particle in the universe attracts every other particle with a force that is directly proportional to the product of the masses and inversely proportional to the square of the distance between them.

F₁₂(g)=Gm₁m₂/r²
r: distance between centers of gravity
G: universal gravitational constant: 6.67e^-11Nm²/kg²
F₁₂=Gm₁m₂/r² r₁₂ r₁₂ is a vector in the direction between 1 and 2.

On the surface of the earth:
F=Gm₁m₂/r² =(6.667e^-11Nm²/kg² )(5.98*10²⁴)m_p/(Re+1)² =(6.667e^-11Nm²/kg² )(5.98*10²⁴)/(6.37*10⁶)²
=9.83m/s²*m^p

Above the earth's surface:

F=[Gm₁/(Re+h)²]m₂
Many objects: F_1total=F₁₂+F₁₃+F₁₄+...
F1=Gm₁m₂/r₁₂² r₁₂+Gm₁m³/r₁₃ r₁₃+...


Gravitational potential energy:

U=Uf+U_i=_i^fFds F=mg(-j).....mg(y_i-yf) at surface of the earth
-_i^fGm₁m₂/r² (-r₁₂)ds
(-r₁₂)ds=dr
=Gm₁m_2i^f1/r²dr
=Gm₁m₂ (-1/r_f - 1/r_i)
=-Gm₁m₂ (1/r_f - 1/r_i)
U=U_f-U_i=-(Gm₁m₂)/r_f -(Gm₁m₂)/r_i
If U_i=0, r_i->
Energy=K+U
E=.5mv² -Gm₁m₂/r

ex.
If a bullet is fired at 2000m/s vertically, how high will it go?

Ei=Ef
.5m_bv_i² -Gm_bm_e/r_i=.5mv_f-Gm₁m₂/r_f
.5m_b(2000m/2)² -(6.67e^-11Nm²/kg₂ *5.98*10²⁴kg)/6.37*10⁶=
0-6.67e^-11Nm²/kg² *5.98*10²⁴kgmb/(6.37*10⁶+h)
h=210,000m

Escape Velocity
E=K=U=.5m₁v_es² -Gm₁m₂/r_i 0
v_es (2Gm₁/r)

Earth: v_es= (2G5.98*10²⁴/6.37*10⁶)

Energy of a satellite
E=.5mv_s² -Gm₁m₂/r
F_c=mv_s²/r=Gm₁m₂/r² m=m₂=mass of satellite
v_s²=Gm₁/r
v_s= (Gm₁/r)
.5mGm₁/r -Gm₁m₂/r
E_total=-Gm₁m₂/2r

Kepler's laws p 296
1st law: The planets' orbits are eliptical
2nd law: Two areas swept out by an orbiting body has equal area in equal times
3rd law: the square of the orbiting period of a planet is proportional to the cube of the radius

ch14

2. a)
F_net=Gm₁m₃/.2²(-î)+Gm₂m₃/.2² (î)
=Gm₃/.2² (-m₁+m₂)î


b) what position will the 50kg mass have 0 net force

Ch 15

4 states of matter:
solid: same neighbors locked in place
liquid: change neighbors, still touching other particles,fluid
gas: no neighbors, not touching other particles, fluid
plasma: no neighbors(some electrons leave), charged particles, not touching, fluid

Density is often more important than mass becuase we are often concered about only part of the mass
Pressure==force/area
==limit_area>0 force/area
P=dF/dA
=N/m² =pascal=Pa

Variation of pressure w/ depth


F_y=0
Fup=Fdown
P(y+dy)A=P(y)A+w
[P(y+dy)-P(y)]A=w
P(y+dy)-P(y)=w/A=mg/A=Vg/A=Agdy/A
P(y+dy)-P(y)=gdy
[P(y+dy)-P(y)]/dy=g
dP/dy=g
P=_i^fdP=_i^fgdy
P=g_i^fdy
P=gy for a liquid

P(h)-P(0)=gh
P(h)=P(0)+gh :pressure at a certain depth (h)
Pa: atmospheric pressure
101,300N/m² = 14.7psi
P(h) absolute pressure
P(h)-Pa=gh increase above atmospheric.= guage pressure


Buoyant force: F₂-F₁=P₂A-P₁A=(Pa+gh₂)A-(Pa+gh₁)A=g(h₂-h₁)A=gV=(m_f/V_f)gV
Archmedies principle: the upward force is equal to the weight of the displaced fluid.
Case I: Totally Submerged:(m_f)g =weight of displaced fluid
F_net=Buoyant force-weight of object
Case II: Floating (partially submerged): volumes are not equal=(m_f/V_f)gV_o
V_f/Vo=_o/_f

E=.5mv² +mgh+.5kx² +heat
divide everything by volume: E/v_ol=.5mv²/v_ol +mgh/v_ol+.5kx²/v_ol+heat/v_ol=
(no friction) E/v_ol=.5v² +gh+.5kx²/v_ol=constant
(no friction) E/v_ol=.5v² +gh+pressure=constant

energy is stored as pressure and can be released into kinetic energy

Bernouli's principle: P+.v² +gh=constant
if Vbefore=Vafter, Pbefore+.vbefore² +ghbefore=Pafter+.5vafter² +ghafter
,P=gh
P=.5(vb² -va²): if the velocity changes, the pressure changes

#6) total mass of earth's atmosphere: 5.27*10¹⁸kg
P_a=101,300Pa
=F/A F=P*A=101300*4 (6.37*10⁶)² =5.16*10¹⁹N=mg
m=5.16*10¹⁹N/9.8=5.27*10¹⁸kg

#14)
two meters of water in a tank that is 2m long. 1m high, there is a hinge on a door on the lower part.
Ftotal=
Pnet=Pa+gh-Pa=gh P=df/dt
F=df=PdA=P(2m)dy pressure*width of box(rectangle of water) times small y
12gy2mdy 1 to 2 meters deep g2mydy= g2m(y²/2|₁²)=
=Frsinø=Fr dt=PdA*r=gy(2m)dy(y-1) r=y-1
dF*r
t=dt=₀²gy(2m)(y-1)dy=g2₁²(y²-y)dy


#19)

P1=Pa+g(h1+h2+h3)
P2=Pa+gh1+gh3
Pa+g(h1+h2+h3)=Pa+gh1+gh3
g(h1+h2+h3)=g(h1+h3)
(H₂o)g(h1+h2+h3)=(Hg)g(h1+h3)

(H₂o)h1=[(Hg)-(H₂o)](h2)

Ch 16: Wave motion: the energy moves, not the material
Types of waves:
transverse: the vibrations are perpendicular to the direction of travel: water
longitudinal: the direction of vibration = direction of energy travel: spring

What if two (or more) waves are moving through the same medium?
(Superposition)
Ytotal=y1()+y2()

Reflection & Transmission of waves
Reflection occurs when the wave meets a boundary that the condition is not the same.
The reflection is inverted when the wave reaches more resistance
The transmission never inverts

=wavelength
y(x,t)=Asin(2 /)(x-vt)
2 / gives 2 when x =
2 / =K = wave number
=Asin(2 x/ -2 vt/ )=Asin(Kx -wt )
distance=vt frequency=f =vt=v/f f =v Kf=kv 2 f=w
ch 16:
#4:
y(x,t) = 5e^-(x+5t)2
5m/s left

#9: y₁(x,t)=5/((3x-4t)² +2)= 5/[(3(x-4/3t)²)+2] right at 4/3m/s
y₂(x,t)=-5/((3x+4t-6)² +2)= -5/[(3(x+4/3t-2)² )+2] left at 4/3m/s

ytotal=5/((3x-4t)² +2)-5/((3x+4t-6)² +2)=0
5/((3x-4t)² +2)=5/((3x+4t-6)² +2)

#20:
1.2= v=d/t=8*1.2/12sec=.8m/s
f=8/12=2/3Hz

#25:
y¹(x,t)=5sin(2x-10t)
y₂(x,t)=10cos(2x-10t)
ytotal=5sin(2x-10t)+10cos(2x-10t)
=5sin(2x-10t)+10sin(2x-10t+ /2)