Download APphysics chapter 1

Chapter 1: Mechanics
Before we start chapter 1, I want to be sure you al recall a few simple rules….
1. you are expected to know how to work with powers and run your calculator
2. simplifying equations and substituting one equation into another is a common
practice in AP physics
3. all students are expected to have a working knowledge of advanced algebra
and trigonometry
4. all tests will consist of 3 questions (15 pts. constructed response) from old AP
physics tests and 3-5 multiple choice (5 pts. each) and they will have a 40min
time limit to complete
5. all homework problems will be worth 5 pts. each assignment and they’re due
at the beginning of the hour the next day
6. you will have approximately 20 take home quizzes (valued at 15-25 pts.)
throughout the year
7. we will do a chapter about every 6 days throughout the year (30 chapters-38
weeks)
8. sharing/copying answers will lead to zeros for all involved, however
cooperative work is encouraged
9. you are expected to have a prior knowledge of mechanics- heat, waves, light,
etc. from last year
10. we work only in the SI system- meters, kilograms, seconds, etc.
11. you will be given a formula sheet at the beginning of the year and it will be
the only sheet you may use on quizzes/ tests in class
12. I do not care about sig figs- round answers to 3 places
13. UNITS MATTER!! please use them
14. dimensional analysis is the approved method for solving problems converting
units
15. if you can do it in your head, cool!
16. if you don’t convert units it is wrong
17. if you don’t use units in your answer you will lose points
 = proportionality
< = less than
> = greater than
 = less than or equal to
 = greater than or equal to
<< = much less than
 = approximately equal to
 = not only equal to
 b = sum of values from a  b
 x = change in x = xf – xi
the origin is the reference point 
(R,  ) = distance R from the origin at angle  in polar coordinates
(R,  )
AP Problem Solving Strategies
1.
2.
3.
4.
5.
6.
7.
read the problem
draw a diagram
identify the data given and the data needed
choose and write out an equation
solve the equation(s) showing all work
evaluate your answer to see if it’s reasonable
check your answer and your math
8. turn it in for a big, beautiful
Assignment: pg 18-21 multiple-choice #3,6,7 and problems #3,6,14,16;21,39,45
Chapter 2: Motion in One Dimention
2.1. Displacement
The study of the motion of an object and the relationship of this motion to force and
mass is called dynamics.
Describing motion wihout regard to its causes is called kinematics.
In chapter 2 we will focus on kinematics in one dimentional motion (in a straight
line).

displacement:  x which is the difference between the final and starting
position of an object
 distance: how far something moved
 vector: direction and magnitude of movement/force
 scalar: magnitude of movement
*our book used a bolded small a for an acceleration vector and a bolded small v
for a velocity vector
2.2.
v = average velocity =  x/  t
2.3.
v  lim  t  0
 x/  t = instantaneous velocity
velocity is speed in a scalar quantity
*look at the pg 29 application—auto speedometers
Example 2.1. on pg 30
A toy train: find i. v total trip
ii. v seconds 0-4
iii. v seconds 4-8
iv. instantaneous v at T= 2 sec
v. instantaneous v at T= 5 sec
i. v =  x/  T = 10cm/12sec = .83cm/sec
ii. v =  x/  T = 4cm/4sec = 1cm/sec
iii. v
=  x/  T
slope = zero, thus v = zero
iv. slope at T= 2sec is 4cm/4sec = 1cm/sec
*instantaneous v= 1 cm/sec
v. slope = zero and instantaneous v = 0cm/sec
Why is v equal to the instantaneous v in iv. and v.?
a =  v/  T = (m/s)/s = m/s2
instantaneous acceleration is a  lim  T  0  v/  T or the slope of the velocity-time
curve at that point
go over example 2.2. on pg 33-37
Assignment: multiple choice #1, 2, problems #2, 7, 8, 12, 21, 28 on pg 44
Open books to pg 46 and take 3 minutes in groups of 2. How deep is the well in
meters?
2.5. Motion Diagrams
Recall from last year…..
constant velocity
positive acceleration
negative acceleration
2.6. One Dimenional Motion with Constant Acceleration
because a = ainst when a is constant we will use a= (vf-vi)/T  Vf = Vi +aT
also… dx = Vi T+ (1/2) aT2 and Vf2 = Vi2 + 2ad
*d = x
Recall that these are the big three kinematic equations from last year!!!
Add: dx = 1/2( Vf + Vi)T
Example 2.4.
A car with a velocity of 30m/s passes a trooper. The trooper takes off after the car
exactly 1sec. after it passes. He moves with a Vi = 0m/s and a = 3m/s2. How long
does it take for the trooper to catch the speeding car?
speeding car
xc = 30m + 30m/s(T)
trooper
xt = ½ aT2 = ½(3m/s2)T2
Because the cop catches the car… dxc = dxt
30m + 30m/s(T) = ½(3m/s2)T2
Move all to one side…. 0 = 1.5m/s2 T2 – 30m/sT – 30m then factor or use quadratic
equation
You get a negative value which is not useful (extraneous root) and T= 21.0 sec. Thus
the cop catches up in 21sec.
Dx = (30m/s)(21sec)= 630m
Example 2.7.
If 2 skydivers dive out of a plane 3sec. apart connected by a bungie cord would the
tension on the cord remain the same, increase, or decrease? *answer on pg 42
Example: A golf ball is released from rest from the top of a tall building. Neglecting
air resistance, calculate its position at a) 1sec. b) 2sec. c) 3sec.
choose the roof at x=0 and y=0
Vi = 0m/s
a = -9.8m/s2
T = 1, 2, 3sec.
d = ???
a) v = (-9.8m/s2)(1sec.)= -9.8m/s2
dy = ½ aT2 = ½(-9.8 m/s2)(1sec.)2 = -4.9m
b) v = (-9.8m/s2)(2sec.)= -19.6m/s
dy = ½ aT2 = ½(-9.8 m/s2)(2sec.)2 = -19.6m
c) v = (-9.8m/s2)(3sec.)= 29.6m/s
dy = ½ aT2 = ½(-9.8 m/s2)(3sec.)2 = -44.1m
Assignment: pg48-49 #26 ,29 ,30, 34
TEST on chapters 1 and 2 will be in 2 days—tomorrow we will review
Oh no!!! I
hate tests!!!
Ahhh!!!
Pasco lab free fall, with computers. (twenty minutes at start of class)
Chapter 3: Vectors and 2 Dimentional Motion
3.1. Vectors Revisited
*remember a scalar has magnitude but no direction…. like a lost freshman!!!
*a vector has direction as well as magnitude
*an example of a vector quantity is a force
*scalars are added using simple arithmetic
*vectors are added by parts
3.1. Adding Vectors
drawn tip to tail
R=A+B
A
B
If you multiply a vector by a scalar you get a vector, same for division
ex. 3(vector A) = 3A
-3/(vector A) = -3/A *sign shows direction
3.3. Component Vector Addition
A= 30N @ 72
B= 60N @ 100
30 cos72
60 cos100
x =
R=√(  x² +  y²)
30 sin72
60 sin100
y=
tan-1(  y/  x) = 
Now look at example 3.3 page #58 using vector components.
Solving Vectors
1. sketch vectors on an x,y coordinate plane
2. break into x,y components
3. add the squares of the sums of the x and y
4. square root the result from #3
5. draw sum of x and sum of y on the x,y grid to locate the correct quadrant
6. use the appropriate trig function to find 
7. add to or subtract from  to find angle from 0
Assignment: pg 72 #1, 8, 11, 13, 16, 19
3.4. Velocity in 2 Dimensions
recall that v =  x/  T now… v =  R/  T
*you must remember that whenever a velocity changes, a velocity occurs
3.5. Projectile Motion (woohoo, time to throw stuff )
g = 9.8m/s2, air resistance is ignored, and the Earth’s rotation does not affect our motion
Ex. Upward from the ground, a projectile is shot at 20 with an instantaneous velocity =
11m/s. Find the time in the air, distance through the air, and the maximum height.
Vf = 0m/s at top
Vi = 11m/s (sin20)
g = 9.8m/s2
Time: Vf = Vi + gT
0 = (11m/s)(sin20)+(-9.8m/s2)(T)
T = .384sec.
*use 2T to find total time in the air
T(2) = 2(.384) = .768sec.
Distance: dx = Vi T
dx = (11m/s)(cos20)(.786sec.)
dx = 7.94m
Maximum Height: dy = Vi T+ ½ aT2
ymax = (11m/s)(sin20)(.384sec.)- ½ (9.8m/s2)(.384sec.)2
ymax = .722m
Now an Ex. at 0 straight out off of a 100m cliff at 40m/s
Vi = 40m/s = Vix
* velocity in x direction remains constant
* velocity in y direction = 0 at T = 0
-100m
dx
distance in x direction is dx = Vix (T)
distance in y direction is dy = Viy (T)+1/2g (T2)
-100 = (0m/s)(T)+1/2(-9.8m/s2)( T2)
-100 = -4.9m/s2 ( T2)
20.4sec2 = T2
4.51sec = T
solve for T:
dx = Vix (T)
dx = (40m/s)(4.51sec)
dx = 180m
3.6. Relative Velocity
Ex. A boat heads due north across a river with an initial velocity of 10km/hr. The river
has a velocity of 5km/hr east. Find the relative velocity and angle.
5km/hr
10km/hr

52+102 = R2
11.2km/hr = R
 = tan-1(5/10) = 26.6
The boat travels at 11.2km/hr at an angle of 26.6.
Assignment: pg 73 #20, 24, 26, 29, 32, 41
Extra Credit pg 78 #60, due on test day, next Tuesday covering chapters 1-3 all
Chapter 4
4.1. Concept of Force
*When two objects touch and one “forces” another to move it is called contact force.
*Field forces are forces that interact without contacting each other. Gravity attractions,
strong and weak nuclear forces, and electromagnetism are the four fundamental forces
and are all examples of field forces.
4.2. Newton’s First Law (Law of Inertia)
*Objects at rest remain at rest unless acted upon by an outside force. The same applies
for objects in motion which must experience an external net force.
*When the net external force on an object equals 0 the object’s velocity (if moving) will
be constant.
*The tendency of an object to resist a change of motion is its inertia. The greater the
mass, the higher its inertia.
4.3. Newton’s Second Law (f =ma)
*Force is directly proportionate to the product of mass and acceleration.
*The unit of force is kg(m/s2) = 1 Newton
1g(cm/s2) = 1 dyne
105 dynes = 1N = .225lb
*Weight is a force….
fw = mg
the acceleration here is gravity where a = g = 9. 8m/s2
4.4. Newton’s Third Law
If two objects interact, the force exerted by one object is equal and opposite to the force
exerted by the object. Forces come in opposite or equal pairs and the action force is equal
and opposite of the reaction force. If there is no movement because of these interactions
they are in equilibrium.
4.5. Applications of Newton’s Laws
The normal force pushes back upward, opposite of gravity.
To do force problems always do the following steps:
1. Draw a free body diagram
2. Label all x and y forces
3. Solve as a system of equations
Ex.
T1
T2
y1 = T1sin37
y2 = T2sin53
37
x1 = T1cos37
53
x2 = T2cos53
Fw=100N (a neg Y)
Find the tension on each cable…
x direction:
x1 = x2
in equilibrium, the light is not moving left of right
 Fx = -T1(cos37)+ T2(cos53) = 0
T2 = T1(cos37/ cos53) = 1.33T1 = T2
y direction:
 Fy = T1sin37+ T2sin53-100N = 0
T1sin37+(1.33T1)(sin53) = 100N
T1 = 60.1N
Put into 1.33T1 = T2
(1.33)(60.1) = T2
79.9N = T2
T3 = 100N (look at picture)
When would T1= T2? (when  1 =  2)
Ex. 2
A 77N sled is pulled up a 30 incline (frictionless) by a child.
a) find the force exerted by the child on the rope
b) find the force exerted on the sled by the hill
sled 
30
y
x
Fn
Ft
Fn = normal force is perpendicular to the plane
Ft = tension
Fg = force of gravity (aka fw = weight)
30
Fg = 77N
 = 30
 fx = T-77N(sin30) = 0
T = 38.5N
 fy = N-77N(cos30) = 0
N = 66.7N
Since the hill pushes up at 66.7N the sled pushes down with 66.7N.
Why is the downward force of 66.7N < Fw at 77N?
(The sled is on a 30 incline off of Fn.)
Ex.3 Moving a Box
A person pulls with a constant force of 20N a box on a dolly with a weight of 300N.
a) find the acceleration of the system
b) how far will it move in 2sec? (assume Vi = 0 and friction = 0)
w = mg  m = w/g = 300N/9.8m/s2 = 30.6kg
f = ma  a = f/m = 20N/30.6kg = .654m/s2
Because force is constant the acceleration is also constant. Vi = 0
dx = Vi T+ ½ aT2
dx = ½ aT2
dx = ½ (.654m/s2)(2sec.2)
dx = 1.31m
Assignment: pg 105-106 #11, 12, 17, 18, 2, 6, 10
T = 2sec.
4.6. Force of Friction
*recall two types of friction: static and kinetic
*Fs > Fk for all surfaces because to get an object moving you must overcome surface
adhesions
*all objects have a coefficient of friction for both static and kinetic—see chart on pg 97
*turn to pg 98 and go over examples 4.5. and 4.6.
fx = µk .3 on a 4kg object
4kg
Find the acceleration of the two objects and the
tension on the string.
N
7kg
4kg
Fk
Ft
mg
Top Object:
 fx = T – Fk = (4kg)(a)
 fy = N – (4kg)(g) = 0
because Fk = μkN and (4kg)(g) = 39.2N
Fk = (.3)(39.2N) = 11.8N
therefore T = Fk + (4kg)(a) = 11.8 + (4kg)(a) = equation #1
now do the 7kg object…
 fy = (7kg)(g) – T = (7kg)(a)
or (7kg)(9.8m/s2) – (7kg)(a) = T = equation #2
68.6N – 7a = T (equation #2)
-11.8N + 4a = T (equation #1)
56.8N – 11a = 0
56.8N = 11a
a = 5.16m/s2
put into equation #1
2
11.8N + 4kg(5.16m/s ) = T = 32.4N
Assignment: pg 105-106 #7, 15, 20, 25, 34, 38, 68
Chapter 5 Work and Energy
“S” = distance
work – an object undergoes a displacement of distance “S” along a straight line while
acted on by a constant force “F” that makes an angle of  with “S”
W = Fcos  ”S”
Work is a scalar with units force times length equals a Newton meter (aka Joule (J))
If you lift something up you do positive work. Put it down and you do negative work.
Together they equal zero, so by picking up 100lbs, holding it up for 30min, and dropping
it back down, you have done zero work!!!
Ex. A force of 50N is exerted at an angle of 30 to move an object a distance of 30
meters. Calculate the work done by the force.
W = Fcos  S
= (50N)(cos30)(30m)
= 1300J
5.2.
W = FS = maS = m (Vf2-Vi2)/2 = ½ m Vf2-1/2 m Vi2
thus… Wnet = KEf - KEi =  KE (the work done on an object is equal to the change in
its kinetic energy)
Ex. A 1400kg car has a net force of 4500N applied on it. If the car starts from rest on a
flat track, find its kinetic energy and velocity after 100m.
Vi = 0
Vf = ?
F = 4500N
m = 1400kg
s = 100m
Wnet = FS = (4500N)(100m)
= 4.5x105 J
since this Wnet =  KE
1/2mv2 = 4.5x105J
v = (4.5x10^5J ) /(1/ 2)(1400kg)
= 25.4m/s
5.3. Potential Energy (Gravitational Potential Energy)
PE = mgy
Wgravity = PEi – PEf
*pick a reference level to assign zero, usually your starting point
ex. A 60kg skier drops from point A to point B (10m). Find PE at A and the difference at
B (using B as zero).
PEA = mgy
= (60kg)(9.8m/s)(10m)
= 5880J
 PE = PEA since B= 0
5.4. Conservation and Non-conservation of Forces
The work done on an object by a conservative force depends only on the initial and final
positions of the object. It is a conservative force if the work it does on an object moving
through any closed path is zero.
Gravity is conservative
Wg = mgyi – mgyf
note that wg = 0 only if yi = yf
Non-conservative - any force that leads to a dissipation of mechanical energy. (thermal
pollution)
Section 5.5. Conservation of Mechanical energy
Wconservative = PEi – PEf
remember… Wnet = PEi - PEf
and… W =  KE = KEf – KEi
so… KEi + PEi = KEf + PEf
thus… ½mvi2+ mgyi = ½mvf2 + mgyf
Turn to page 129 and we will go over example #5.5.
Potential energy stored in a spring
F = kx = Hooke’s Law
and the work done by an applied force on a spring is… W = Fx = ½kx2
This equates to the  PE = ½kx2 now stored in the spring.
Since PE is proportional to x2, PE is always positive.
This new form of energy is included as another term in our equation for conservation of
mechanical energy.
(KE + PEg + PEs)i = (KE + PEg + PEs)f
where PEg = gravitational potential energy
Ex. A .5kg block is pressed against a spring with K=80N/m for a distance of 2cm, then
released. Find Vf of the block at x = 0.
(KE + PEg + PEs)i = (KE + PEg + PEs)f
zero
PEgi = PEgf no height involved
PEsi = KEf
½kxi2 = ½mvf2
cancel ½ from both sides…
kxi2 =mvf2
(80N/m)(.02m)2 = (.5kg)Vf2
.25m/s = Vf
Assignment: pg 141 #1, 4, 7, 13, 15, 17, 19
5.6. Non-conservative forces and the Work—Kinetic Energy Theorem
Since all the work done by non-conservative forces equals the change in mechanical
energy of the system we use…
Wnc = (KEf + PEf) – (KEi + PEi)
Ex. A 20kg child slides down an irregularly curved water slide with a height 6m. If the
child starts at the top with an initial velocity of zero, what is the kid’s speed at the
bottom? (assume no friction)
yf = 0 yi = 6m
(20kg)(9.8m/s2)(6m) = ½(20kg)(Vf)2
10.8m/s = Vf
(you could also have used Vf2 = Vi2 + 2ad because friction = 0
5.7. Conservation of Energy in General
Recall that energy can never be created or destroyed, only transformed. The total energy
in the universe is constant.
5.8. Power
Power is the transfer of energy over a specific time period.
P = w/  T = F(  S)/  T = F V
units are in Watts = 1W = 1J/s = 1kgm2/s3
1hp = 550ftlb/s = 746W
T = tension = a force
T-F-Mg = 0
A 1000kg elevator carries a max load of 800kg. A constant friction force of 4000N
retards its upward motion. What minimum power, in kilowatts and horsepower, must the
motor deliver to lift a fully loaded elevator at 3m/s?
T-F-Mg = 0 = T = F +Mg
= 4000N+ (1800kg)(9.8m/s2)
= 21600N
using P = FV = TV
= (21600N)(3m/s) = 64800W
= 64.8kW = 86.9hp
Assignment: pg142 Day 1 21, 25, 26, 27, 28
Day 2 35, 41, 43, 44, 54
Chapter 6
6.1. Momentum and Impulse
P = mV
P = momentum
m = mass
V = velocity
Momentum is a vector quantity. The rate of change of the momentum of a particle is
equal to the resultant force on that particle.
The impulse of a force equals the change in momentum of the particle on which the force
acts.
Under the Impulse Approximation, it’s assumed that one force (or one of the forces)
acting on a particle is of a short time duration but of much greater magnitude than any of
the other forces.
*Newton’s 2nd law applies here…
F =  P/  T = change in momentum/time interval during momentum change
F =  P/  T = MVf - MVi/  T = M(Vf-Vi)/  T
but the impulse momentum theorem states F  T =  P = MVf-MVi
graphically, the impulse is the area under the curve on a force-time curve = WORK?
Why?
Force
Time (sec.)
Example #5 on pg 170…
A1500kg car moving at 15m/s collides with a utility pole and is brought to a rest in .3sec.
Find the average force exerted on the car during the collision.
F = (1500kg)(0m/s-15m/s)/.3sec = -75000N
The negative sign means the force is in the opposite direction of the car’s velocity.
Go over example 6.2. on page 153.
6.2. Conservation of Momentum
If two particles of masses M1 and M2 form an isolated system then the total mass of that
system will remain constant throughout the collision. The result of this is known as the
conservation of momentum.
M1V1i+ M2V2i = M1V1f+ M2V2f
This only holds true when the system is closed (no external forces)
Ex. #9 on page 170
Looking at the graph you will see that the 2kg particle has a force that varies with time.
Find the impulse of the force.
total impulse = total area under curve
= area (A) + area (B) + area (C)
= ½ (2sec)(4N) + (1sec)(4N) + ½ (2sec)(4N)
= 12N = 12kg m/s
6.3. Collisions
Inelastic collisions: momentum is conserved but energy is not (when two objects collide
and stick together the collision is perfectly inelastic). The deformation of objects is what
causes the loss of kinetic energy.
Elastic collisions: both momentum and kinetic energy are conserved, like pool balls and
air molecules at a normal temperature (no deformations).
Most of these are limiting in cases. Even billiard balls dent so some deformation,
however small, does occur.
For perfectly inelastic collisions (stick together), use m1v1i + m2v2i = (m1 + m2)vf
We can also use KE = ½ mv2
Assignment: pg 170 #4, 10, 15, 17
6.4. Glancing Collisions
Until now we have only look at straight-line collisions, and nothing else.
Glancing collisions (colliding masses) rebound at an angle relative to the line of motion
of incident mass.
example #35 on pg 172…
A 2000kg car moving east at 10m/s collides with a 3000kg car moving north. They stick
together and move as a unit after the collision at a 40º angle north of east at 5.22m/s. Find
Vi for the 3000kg car.
since momentum is conserved…
 Pf =  Pi
so the sum of the vector components of x and y for both cars are the same before and
after the collision. and since the 3000kg car (car #2) was moving in the y direction, we
only need to look at the y components.
finding Vi of car #2 we get …
P1 + P2 = P1f + P2f
MVi car#1 + MVi car#2 = MVf car#1 + MVf car#2
but car #1 had P = 0 in the y direction
sooooooo…
0 + (3000kg)(Vi) = 5000kg(5.22m/s)(sin40º)/3000kg
Vi = 5.59m/s
Assignment: pg 171 #21, 25, 29, 31
Test Tomorrow!!! Chapters 4, 5 & 6 – pick partners today
Chapter 7
7.1. Angular Speed and Acceleration
Most equations in chapters 7 and 8 require you to use radians, not degrees!!!
s
R
 = s/R
Since the angle through which the disk rotates is   =  2-  1, we can define angular
displacement as   . Angular speed as  (omega) is the rotation of a ridged object
through that angular displacement   in the time interval  T.
sooo…  =  2-  1/T2-T1 =   /  T
then we could say that instantaneous angular speed is the limit of the average speed
Ex. on pg 210 #1
Tires on a compact car with a diameter of 2ft are warranted for 60000 miles. a)
Determine the angle through which a tire will rotate under warranty. b) How many
revolutions of the tire is the answer is the answer from part a?
a)  = s/R = (60000mi/1ft)(5280ft/ 1mi) = 3.2x108 radians
b)  = 3.2x108rad = 1rev/2  rad = 5x107 revolutions
7.2. Rotational Motion Under Constant Acceleration (Notice the Comparison?)
v =  x/  T and  =   /  T
Additionally, notice the similarities between…
a =  v/  T and  =   /  T
(average acceleration for angular motion)
Ex. A bike wheel rotates with a constant angular acceleration of 3.5rad/sec2. If  i =
2rad./sec. at T = 0, through what angle does the wheel rotate in 2 sec?
 =  i T+1/2  T2 = (2rad/s)(2sec) + 1/2(3.5rad/s2)(2sec) 2 = 11rad = 630 degrees
 at t= 2 sec
 f =  i +  t = 2 rad/s + 3.5 rad/s2(2sec) = 9 rad/s
also you can use  = v/R where v = velocity tangent to radians
which solves into velocity tangent … Vt = R
and acceleration tangent is… at = R 
(  = angular acceleration)
*the tangential acceleration of a point on a rotating object equals the distance of that
point from the axis of rotation multiplied by the angular acceleration.
#Ex 11 on pg. 210
A rotating wheel requires 3sec to rotate 37 revolutions. Its angular velocity at the end of
3sec is 98rad/sec. What is the constant angular acceleration of the wheel?
 =  /t = (37rev)(2  rad/1rev)/3sec = 77.5rad/sec
7.4. Centripetal Acceleration
Driving around a circular track at a constant speed you will still have acceleration. Why?
Because you are changing direction (vector change).
An acceleration going around a circle is called centripetal acceleration. (means center
seeking)
ac = Vt2/R = R2 2/R = R 2
In circular motion the centripetal acceleration is directed inward towards the center of the
circle and has a magnitude of either V2/R or R 2.
Vc
ac
Assignment: page 210 #2, 3, 5, 7, 12, 16, 17
7.5. Forces Causing Centripetal Acceleration
ac = Vt2/R
F = mac = mVt2/R
Example #21 on page 211
A sample of blood is placed in centrifuge with a radius of 15cm. The mass of the blood is
3x10-16 kg (red corpuscle) and the magnitude of the force required to make it settle out of
the plasma is 4x10-11 N. At how many revolutions per second should the centrifuge be
operating???
m = mass
R = radius of centrifuge
Fc = 4x10-11 N = mVt2/R = mR  2
so…  = (Fc/mR)1/2 = (4x10-11 N/(3x10-16 kg)(.15m))1/2 = 942.8rad/sec. = 150rev/sec.
Skip 7.6
7.7 Newton’s Universal Law of Gravitation
F = G(m1m2/R2)
R = distance between objects
G = 6.67x10-11 (N m2/kg2)
for any object on the earth’s surface use…
F = G(mEm/RE2)
using two equations we get…
V = (Gm/R)1/2
This explains why dark matter exists. Distant stars are too far away from other stars
(masses) to move as fast as they do without the existence of an unseen mass (a.k.a. dark
matter). Yes it’s vector time!!!!!
Ex. Two .3kg balls are placed on a table. One is .4m north and the other is .3m east of a
third .3kg ball. Find the resultant force on the third ball
m2
m1
m3
The force on m1 by m2 will be upward.
The force on m1 by m3 will be to the right.
Find the force of each and add the vectors.
F2on1 = G(m1m2/R2)= (6.67x10-11(Nm2/R2))((.3kg)(.3kg)/(.4m2))= 3.75x10-11N
F3on1 = G(m1m2/R2)= (6.67x10-11(Nm2/R2))((.3kg)(.3kg)/(.3m2))= 6.67x10-11N
Fr =
(f2on1)2+( f3on1)2
=
(3.75x10-11)2+( 6.67x10-11)2
= 7.65x10-11N @29˚
tan-1  = 3.75/6.67 = 29˚
7.8. Gravitational Potential Energy (again)
Recall that PE = mgh only works for objects near the earth’s surface. For objects far from
the surface use PE = -G(mEm/R2) where R= the distance of the satellite from the center of
the earth.
do #36 in class
Assignment: pg #212 #15, 18, 31, 32, 33, 35, 37
satellite