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Physics 115 2013
Final Review
Calculus is about “rates of change”.
A TIME RATE is anything divided by time.
CHANGE is expressed by using the Greek letter, Delta, D.
For example: Average SPEED is simply the “RATE at which DISTANCE changes”.
The MEANING?
For example, if t = 2 seconds, using x(t) = kt3=(1)(2)3= 8 meters.
The derivative, however, tell us how our DISPLACEMENT (x) changes as a function of TIME (t). The rate at which Displacement changes is also called VELOCITY. Thus if we use our derivative we can find out how fast the object is traveling at t = 2 second. Since dx/dt = 3kt2=3(1)(2)2= 12 m/s
23
3)( ktdtktd
Derivative of a power function
Unit Vector Notation
direction z in the 1 r unit vecto-ˆdirectiony in the 1r unit vecto-ˆdirectionx in the 1 r unit vecto-ˆ
k
j
i
jiJ ˆ4ˆ2
The proper terminology is to use the “hat” instead of the arrow. So we have i-hat, j-hat, and k-hat which are used to describe any type of motion in 3D space.
How would you write vectors J and K in unit vector notation?
jiK ˆ5ˆ2
ExampleA boat moves with a velocity of 15 m/s, N in a river which flows with a velocity of
8.0 m/s, west. Calculate the boat's resultant velocity with respect to due north.
1.28)5333.0(
5333.0158
/17158
1
22
Tan
Tan
smRv
15 m/s, N
8.0 m/s, W
Rv
The Final Answer :
smjsmi
sm
WofNsm
/ˆ15/ˆ8
1.118/17
1.28@/17
Dot Products in PhysicsConsider this situation: A force F is applied to a moving object as it transverses over a frictionless surface for a displacement, d.
As F is applied to the object it will increase the object's speed!
But which part of F really causes the object to increase in speed?
It is |F|Cos θ ! Because it is parallel to the displacement d In fact if you apply the dot product, you get (|F|Cos θ)d, which happens to be defined as "WORK" (check your equation sheet!)
cos
cos
xFxFW
BABA
Work is a type of energy and energy DOES NOT have a direction, that is why WORK is a scalar or in this case a SCALAR PRODUCT (AKA DOT PRODUCT).
ExampleSuppose a person moves in a straight line from the lockers( at a position x = 1.0 m) toward
the physics lab(at a position x = 9.0 m). “To the right” is taken as positive, as shown below
The answer is positive so the person must have been traveling horizontally in the positive direction.
ExampleSuppose the person turns around!
The answer is negative so the person must have been traveling horizontally in the negative direction
What is the DISPLACEMENT for the entire trip?
What is the total DISTANCE for the entire trip?
mxxx initialfinal 00.10.1 D
m1688
Instantaneous VelocityInstantaneous velocity is a
measure of an object’s displacement per unit time at a particular point in time.
dtdxv
txv t
DD
D 0lim
Example: A body’s position is defined as:?)(,ˆ4ˆ7)( 3 tvj
tittx
dt
jt
itd
dtdxv
)ˆ4ˆ7( 3 smj
tittv /]ˆ4ˆ21[)( 2
2
Instantaneous AccelerationInstantaneous velocity is a measure
of an object’s velocity per unit time at a particular point in time.
dtdva
tva t
DD
D 0lim
?)(,/]ˆ4ˆ21[)( 22 tasmj
tittv
If the velocity of an object is defined as:
dt
jt
itd
dtdva
)ˆ4ˆ21( 22
ssmjt
it //]ˆ8ˆ42[ 3
What do the “signs”( + or -) mean?Quantity Positive NegativeDisplacement Your position has
changed toward the positive.
Your position has changed toward the negative.
Velocity You are traveling in the +x or +y direction.
You are traveling in the –x or –y direction.
Acceleration If moving in the positive direction, you are speeding up. If moving in the negative direction, you are slowing down.
If moving in the positive direction, you are slowing down. . If moving in the negative direction, you are speeding up.
The 3 Kinematic equations
There are 3 major kinematic equations than can be used to describe the motion in DETAIL. All are used when the acceleration is CONSTANT. )(2
21
22
2
oo
oo
o
xxavv
attvxx
atvv
Kinematics for the VERTICAL DirectionAll 3 kinematics can be used to analyze one dimensional
motion in either the X direction OR the y direction.
)(2)(22
12
1
2222
22
ooyyoox
oyooxo
oyyo
yygvvxxavv
gttvyyattvxx
gtvvatvv
ExamplesA stone is dropped at rest from the top of a cliff. It is
observed to hit the ground 5.78 s later. How high is the cliff?
What do I know?
What do I want?
voy= 0 m/s y = ?
g = -9.8 m/s2
yo=0 m
t = 5.78 s
Which variable is NOT given andNOT asked for?
Final Velocity!
2
21 gttvyy oyo
yy 2)78.5(9.4)78.5)(0(
-163.7 m
H =163.7m
ExamplesA pitcher throws a fastball with a velocity of 43.5 m/s. It is determined that
during the windup and delivery the ball covers a displacement of 2.5 meters. This is from the point behind the body when the ball is at rest to the point of release. Calculate the acceleration during his throwing motion.
What do I know?
What do I want?
vo= 0 m/s a = ?
x = 2.5 m
v = 43.5 m/s
Which variable is NOT given andNOT asked for?
TIME
)(222oo xxavv
aa )05.2(205.43 22
378.5 m/s/s
ExamplesHow long does it take a car at rest to cross a 35.0 m intersection after
the light turns green, if the acceleration of the car is a constant 2.00 m/s/s?
What do I know?
What do I want?
vo= 0 m/s t = ?
x = 35 m
a = 2.00 m/s/s
Which variable is NOT given andNOT asked for?
Final Velocity
t
t 2)2(21)0(035
2
21 attvxx oxo
5.92 s
ExamplesA car accelerates from 12.5 m/s to 25 m/s in 6.0 seconds.
What was the acceleration?
What do I know?
What do I want?
vo= 12.5 m/s a = ?
v = 25 m/s
t = 6s
Which variable is NOT given andNOT asked for?
DISPLACEMENT
atvv o
aa )6(5.1225
2.08 m/s/s
SummaryThere are 3 types of MOTION graphs• Displacement(position) vs. Time• Velocity vs. Time• Acceleration vs. Time
There are 2 basic graph models• Slope• Area
Summary
t (s) t (s) t (s)
x (m)v (m/s) a (m/s/s)
slope = v
slope = a
area = xarea = v
Comparing and Sketching graphsOne of the more difficult applications of graphs in physics is when given a certain type of graph and asked to draw a different type of graph
t (s)
x (m)
slope = v
t (s)
v (m/s)
List 2 adjectives to describe the SLOPE or VELOCITY1. 2.
The slope is CONSTANTThe slope is POSITIVE
How could you translate what the SLOPE is doing on the graph ABOVE to the Y axis on the graph to the right?
Example
t (s)
x (m)
t (s)
v (m/s)
1st line• •
2nd line•
3rd line• •
The slope is constant
The slope is constantThe slope is “+”
The slope is “-”The slope is “0”
Example – Graph Matching
t (s)
v (m/s)
t (s)
a (m/s/s)
t (s)
a (m/s/s)
t (s)
a (m/s/s)
What is the SLOPE(a) doing?
The slope is increasing
Force Diagrams
A pictorial representation of forces complete with labels.
W1,Fg1 or m1g
•Weight(mg) – Always drawn from the center, straight down•Force Normal(FN) – A surface force always drawn perpendicular to a surface.•Tension(T or FT) – force in ropes and always drawn AWAY from object.•Friction(Ff)- Always drawn opposing the motion.
m2g
T
TFN
Ff
Force Diagrams
mg
FNFf
New’s 1st Law and Equilibrium
Since the Fnet = 0, a system moving at a constant speed or at rest MUST be at EQUILIBRIUM.
TIPS for solving problems• Draw a FBD• Resolve anything into COMPONENTS• Write equations of equilibrium• Solve for unknowns
Example10-kg box is being pulled across the table to the right at a constant speed
with a force of 50N at an angle of 30 degrees above the horizontal.a) Calculate the Force of Friction
b) Calculate the Normal Force
mg
FN Fa
Ff30
NFFNFF
axf
aax
3.433.4330cos50cos
Fax
Fay
NF
FmgF
mgFFmgF
N
ayN
ayN
N
73
30sin50)8.9)(10(
!
Springs – Hooke’s Law
One of the simplest type of simple harmonic motion is called Hooke's Law. This is primarily in reference to SPRINGS.
kxorkxFkk
xF
s
s
N/m):nitConstant(U SpringalityProportion ofConstant
The negative sign only tells us that “F” is what is called a RESTORING FORCE, in that it works in the OPPOSITE direction of the displacement.
Hooke’s Law from a Graphical Point of View
x(m) Force(N)0 0
0.1 120.2 240.3 360.4 480.5 600.6 72
graph x vs.F a of Slope
kxFk
kxF
s
sSuppose we had the following data:
Force vs. Displacement y = 120x + 1E-14R2 = 1
0
10
20
30
40
50
60
70
80
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7
Displacement(Meters)
Forc
e(Ne
wto
ns)
k =120 N/m
ExampleA load of 50 N attached to a spring hanging vertically stretches the spring
5.0 cm. The spring is now placed horizontally on a table and stretched 11.0 cm. What force is required to stretch the spring this amount?
kkkxFs
)05.0(501000 N/m
s
s
s
FF
kxF)11.0)(1000(
110 N
Newton’s Second Law
The acceleration of an object is directly proportional to the NET FORCE and inversely proportional to the mass.
maFmFa
maFa
NETNET
NET
1 FFNET
Tips:•Draw an FBD•Resolve vectors into components•Write equations of motion by adding and subtracting vectors to find the NET FORCE. Always write larger force – smaller force.•Solve for any unknowns
Newton’s 2nd Law
A 10-kg box is being pulled across the table to the right by a rope with an applied force of 50N. Calculate the acceleration of the box if a 12 N frictional force acts upon it.
mg
FN Fa
Ff
2/8.3
101250
sma
a
maFFmaF
fa
Net
In which
direction, is this object accelerating?
The X direction!
So N.S.L. is worked out using the forces in the “x” direction only
Example
m1g
m2g
T
TFN
A mass, m1 = 3.00kg, is resting on a frictionless horizontal table is connected to a cable that passes over a pulley and then is fastened to a hanging mass, m2 = 11.0 kg as shown below. Find the acceleration of each mass and the tension in the cable.
amTamTgm
maFNet
1
22
2
21
2
122
122
212
/7.714
)8.9)(11()(
smmmgma
mmagmamamgmamamgm
Example (cont.)
amTamTgm
maFNet
1
22
NT 1.23)7.7)(3(
Where does the calculus fit in?
dtxdm
dtdvmamF
2
220.03)( tttv
There could be situations where you are given a displacement function or velocity function. The derivative will need to be taken once or twice in order to get the acceleration. Here is an example.
You are standing on a bathroom scale in an elevator in a tall building. Your mass is 72-kg. The elevator starts from rest and travels upward with a speed that varies with time according to:
When t = 4.0s , what is the reading on the bathroom scale (a.k.a. Force Normal)?
)4(40.03)4(
40.03)20.03( 2
a
tdt
ttddtdva
4.6 m/s/s
)6.4)(72()8.9)(72(N
NN
net
FmgmaFmamgF
maF
1036.8 N
TWO types of Friction• Static – Friction that keeps an object at rest and prevents it
from moving (no sliding)• Kinetic – Friction that acts during motion (two surfaces sliding)
Force of Friction• The Force of Friction is
directly related to the Normal Force.
Nkkf
Nssf
Nf
FF
FF
FF
friction oft coefficien alityproportion ofconstant
The coefficient of friction is a unitless constant that is specific to the material type and usually less than one.
ExampleA 1500 N crate is being pushed
across a level floor at a constant speed by a force F of 600 N at an angle of 20° below the horizontal as shown in the figure.
a) What is the coefficient of kinetic friction between the crate and the floor?
mg
FNFa
20
Ff
Fay
Fax
331.021.170582.563
21.17051500)20(sin600
1500sin
82.563)20(cos600cos
k
k
N
aayN
aaxf
Nkf
NF
FmgFF
NFFF
FF
ExampleIf the 600 N force is instead pulling the block
at an angle of 20° above the horizontal as shown in the figure, what will be the acceleration of the crate. Assume that the coefficient of friction is the same as found in (a)
mg
FN
Ff
20
Fa
Fax
Fay
2/883.0
1.15357.4288.5631.153)20sin6001500(331.020cos600
)sin(coscos
sma
aa
maFmgFmaFF
maFFmaF
aa
Na
fax
Net
Inclines
cosmg
sinmg
mg
FNFf
Tips•Rotate Axis•Break weight into components•Write equations of motion or equilibrium•Solve
Example
gmamTamgmTmaFNET
1111
m2
m1
Masses m1 = 4.00 kg and m2 = 9.00 kg are connected by a light string that passes over a frictionless pulley. As shown in the diagram, m1 is held at rest on the floor and m2 rests on a fixed incline of angle 40 degrees. The masses are released from rest, and m2 slides 1.00 m down the incline in 4 seconds. Determine (a) The acceleration of each mass (b) The coefficient of kinetic friction and (c) the tension in the string.
m1g
m2g
FNT
T
Ff
40
40 amTFgm f 22 )(sin m2gcos40
m2gsin40
Example
cossin
cossincossin
sin
)(sin
sin
2
2112
22112
21122
2112
2112
22
gmamgmamgm
gmamgmamgmamgmamgmgm
amgmamFgm
amgmamFgm
amTFgm
k
k
k
Nk
f
f
2
2
2
/125.0
)4(2101
21
sma
a
attvx ox
gmamTamgmTmaFNET
1111
amTFgm f 22 )(sin
NT 7.39)8.9(4)125(.4
235.057.67
125.12.395.07.56
k
Horizontally Launched ProjectilesTo analyze a projectile in 2 dimensions we need 2
equations. One for the “x” direction and one for the “y” direction. And for this we use kinematic #2.
212oxx v t at
oxx v t
Remember, the velocity is CONSTANT horizontally, so that means the acceleration is ZERO!
212y gt
Remember that since the projectile is launched horizontally, the INITIAL VERTICAL VELOCITY is equal to ZERO.
Horizontally Launched ProjectilesExample: A plane traveling with a
horizontal velocity of 100 m/s is 500 m above the ground. At some point the pilot decides to drop some supplies to designated target below. (a) How long is the drop in the air? (b) How far away from point where it was launched will it land?
What do I know?
What I want to know?
vox=100 m/s t = ?y = 500 m x = ?voy= 0 m/s
g = -9.8 m/s/s
2 2
2
1 1500 ( 9.8)2 2102.04
y gt t
t t
10.1 seconds(100)(10.1)oxx v t 1010 m
Vertically Launched ProjectilesThere are several
things you must consider when doing these types of projectiles besides using components. If it begins and ends at ground level, the “y” displacement is ZERO: y = 0
Vertically Launched ProjectilesYou will still use kinematic #2, but YOU MUST use
COMPONENTS in the equation.
vo
vox
voy
oxx v t 212oyy v t gt
cossin
ox o
oy o
v vv v
ExampleA place kicker kicks a football with a velocity of 20.0 m/s and at an
angle of 53 degrees.(a) How long is the ball in the air?(b) How far away does it land?(c) How high does it travel?
v o=20.0 m
/s
53
cos20cos53 12.04 /
sin
20sin 53 15.97 /
ox o
ox
oy o
oy
v vv m sv v
v m s
ExampleA place kicker kicks a
football with a velocity of 20.0 m/s and at an angle of 53 degrees.
(a) How long is the ball in the air?
What I know What I want to know
vox=12.04 m/s t = ?voy=15.97 m/s x = ?y = 0 ymax=?g = - 9.8 m/s/s
2 2
2
1 0 (15.97) 4.9215.97 4.9 15.97 4.9
oyy v t gt t t
t t tt
3.26 s
Example
A place kicker kicks a football with a velocity of 20.0 m/s and at an angle of 53 degrees.
(b) How far away does it land?
What I know What I want to know
vox=12.04 m/s t = 3.26 svoy=15.97 m/s x = ?y = 0 ymax=?g = - 9.8 m/s/s
(12.04)(3.26)oxx v t 39.24 m
ExampleA place kicker kicks a football
with a velocity of 20.0 m/s and at an angle of 53 degrees.
(c) How high does it travel?
CUT YOUR TIME IN HALF!
What I know What I want to know
vox=12.04 m/s t = 3.26 svoy=15.97 m/s x = 39.24 my = 0 ymax=?g = - 9.8 m/s/s
2
2
12
(15.97)(1.63) 4.9(1.63)
oyy v t gt
yy
13.01 m
A special case…What if the projectile was launched from the ground at an angle and did
not land at the same level height from where it started? In other words, what if you have a situation where the “y-displacement” DOES NOT equal zero?
Let's look at the second kinematic closely!
Assuming it is shot from the ground. We see we have one squared term variable, one regular term variable, and a constant number with no variable. What is this?
A QUADRATIC EQUATION!
Circular Motion and New’s 2nd Law
Recall that according to Newton’s Second Law, the acceleration is directly proportional to the Force. If this is true: ForcelCentripetaF
rmvFF
rvamaF
c
cNET
cNET
2
2
Since the acceleration and the force are directly related, the force must ALSO point towards the center. This is called CENTRIPETAL FORCE.
NOTE: The centripetal force is a NET FORCE. It could be represented by one or more forces. So NEVER draw it in an F.B.D.
Examples
rgv
rmvmg
rmvF
FF
N
cf
2
2
2
What is the minimum coefficient of static friction necessary to allow a penny to rotate along a 33 1/3 rpm record (diameter= 0.300 m), whenthe penny is placed at the outer edge of the record?
mg
FN
Ff
Top view
Side view 187.0)8.9)(15.0(
)524.0(
/524.080.1
)15.0(22
sec80.1555.0
sec1sec555.0
sec60min1*
min3.33
22
rgv
smTrv
Trevrev
revrev
c
ExamplesThe maximum tension that a 0.50 m
string can tolerate is 14 N. A 0.25-kg ball attached to this string is being whirled in a vertical circle. What is the maximum speed the ball can have (a) the top of the circle, (b)at the bottom of the circle?
mgT
smvmmgTrv
mvmgTrrmvmgT
rmvmaFF ccNET
/74.525.0
))8.9)(25.0(14(5.0)(
)( 22
2
ExamplesAt the bottom?
smvmmgTrv
mvmgTrrmvmgT
rmvmaFF ccNET
/81.425.0
))8.9)(25.0(14(5.0)(
)( 22
2
mg
T
N.L.o.G – Putting it all together
221
2227
221
1067.6
Constant nalGravitatio UniversalGalityproportion ofconstant
rmmGF
kgNmxG
GrmmF
g
g
earth eLEAVING th areyou when thisUse
earth on the areyou when thisUse
221
rmmGF
mgF
g
g
Kepler’s 3rd Law – The Law of Periods"The square of the period of any planet is proportional to
the cube of the semi major axis of its orbit."
Gravitational forces are centripetal, thus we can set them equal to each other!
Since we are moving in a circle we can substitute the appropriate velocity formula!
Using algebra, you can see that everything in the parenthesis is CONSTANT. Thus theproportionality holds true!
The expression in the RED circle derived by setting the centripetal force equal to the gravitational force is called ORBITAL SPEED.
ExampleA 2-kg sliding puck whose initial velocity magnitude is v1 = 10 m/s strikes a wall at a 30 degree angle and bounces off. If it leaves the wall with a velocity magnitude of v2 = 10 m/s, and if the collision takes a total of 0.02 seconds to complete, what was the average force applied to the puck by thewall?
There is something you need to consider:Momentum is a VECTOR!!!
Let’s look at this problem using a X-Y axis for reference
Example cont’
If we did the same thing for the Y direction we would discover that the Force Net is equal to ZERO!
The temptation is to treat momentum as a SCALAR...DO NOT DO THIS! SIGNS COUNT!
Momentum is conserved!The Law of Conservation of Momentum: “In the absence
of an unbalanced external force, the total momentum before the collision is equal to the total momentum after the collision.”
smkgpsmkgpsmkgp
smkgp
smkgp
smkgmvp
total
car
truck
totalo
caro
otrucko
/*3300/*18005.4*400
/*15003*500
/*3300
/*800)2)(400(
/*2500)5)(500(
)(
)(
)(
Several Types of collisionsSometimes objects stick together or blow apart. In this
case, momentum is ALWAYS conserved.
2211)(
022011
2211022011
vmvmvmvmvmvmvmvmvmvm
pp
totalototal
totaltotal
afterbefore
When 2 objects collide and DON’T stick
When 2 objects collide and stick together
When 1 object breaks into 2 objects
Elastic Collision = Kinetic Energy is ConservedInelastic Collision = Kinetic Energy is NOT Conserved
WorkThe VERTICAL component of the force DOES NOT cause the block to move the right. The energy imparted to the box is evident by its motion to the right. Therefore ONLY the HORIZONTAL COMPONENT of the force actually does WORK.
When the FORCE and DISPLACEMENT are in the SAME DIRECTION you get a POSITIVE WORK VALUE. The ANGLE between the force and displacement is ZERO degrees. What happens when you put this in for the COSINE?
When the FORCE and DISPLACEMENT are in the OPPOSITE direction, yet still on the same axis, you get a NEGATIVE WORK VALUE. This negative doesn't mean the direction!!!! IT simply means that the force and displacement oppose each other. The ANGLE between the force and displacement in this case is 180 degrees. What happens when you put this in for the COSINE?When the FORCE and DISPLACEMENT are PERPENDICULAR, you get NO WORK!!! The ANGLE between the force and displacement in this case is 90 degrees. What happens when you put this in for the COSINE?
ExamplecosrFW
rFW
A box of mass m = 2.0 kg is moving over a frictional floor ( uk = 0.3) has a force whosemagnitude is F = 25 N applied to it at an angle of 30 degrees, as shown to the left. The box is observed to move 16 meters in the horizontal direction before falling off the table.
a) How much work does F do before taking the plunge?
JWNmW
W
rFWrFW
4.3464.346
30cos1625
cos
Example cont’What if we had done this in UNIT VECTOR notation?
JWNmW
W
rFrFWjiF
yyxx
4.3464.346
)05.12()1665.21(
)()(
ˆ5.12ˆ65.21
Example cont’
JW
FW
rFWrFW
N
0
90cos16
cos
Fn
How much work does the FORCE NORMAL do and Why?
There is NO WORK since “F” and “r” are perpendicular.
Ff
How much does the internal energy of the system increase?
34.08 J
Elastic Potential Energy
The graph of F vs.x for a spring that is IDEAL in nature will always produce a line with a positive linear slope. Thus the area under the line will always be represented as a triangle.
NOTE: Keep in mind that this can be applied to WORK or can be conserved with any other type of energy.
Elastic potential energy
20
2
00
21|
2|
)(
)()(
kxUWxkW
dxxkdxkxW
dxkxdxxFW
springxx
x
x
x
x
Elastic “potential” energy is a fitting term as springs STORE energy when there are elongated or compressed.
Energy is CONSERVED!
afterbefore
oo
oo
oo
EnergyEnergyUKUKUUKKUUKK
UKW
DD)(
ExampleA 2.0 m pendulum is released from rest when the support
string is at an angle of 25 degrees with the vertical. What is the speed of the bob at the bottom of the string?
L Lcos
h
h = L – Lcosh = 2-2cosh = 0.187 m
EB = EA
UO = Kmgho = 1/2mv2
gho = 1/2v2
1.83 = v2
1.35 m/s = v
Ballistic Pendulum
• Event includes a collision during which momentum conservation is the best model and post-collision motion during which energy conservation is the best model.
• Examples: last three problems on HW, Workshop 14,