Advanced Placement Physics
Kinematics in One Dimension
Kinematics
Study of Motion
Distance
Total distance traveled from start to finish.
Displacement Straight line distance between the start point and ending point of the problem.
Speed
A scalar quantity (no direction specified) that shows the rate that distance d is covered.
Instantaneous
The speed at an instant in time. Right now. Your speedometer reading when you glance it at.
Average
The total distance divided by the total time for the entire trip.
Constant
If the same speed is maintained over the entire trip
Velocity
A vector quantity consisting of magnitude and direction. Displacement x divided by time.
Acceleration Change in velocity (change in displacement and/or direction)
Kinematic Equations
You can only use the constant velocity equation when there is no acceleration. If acceleration is present (Question
contains terms such as: starts from rest, final velocity of, accelerates, comes to rest, etc.), then you must use the
three Kinematic equations in the highlighted boarder boxes below.
Speed
Constant Velocity
vs =
d
t
v=
x
t
or
x = vt
Average Velocity
v=
x − xo
t
Acceleration
a=
v − vo
t
v +v
v= o
2
Another Way of Looking at Average Velocity
One of the four Kinematic Equations. But it is mostly used in
conjunction with the above equations to derive the next three
equations. Occasionally it is useful in problems.
v= vo + at
Velocity
Rearranged the acceleration equation from above. Useful for
determining v, when a and t are given. However, if any three variables
are available and the fourth is needed rearrange this as necessary.
1
x = xo + vo t + at 2
2
v2 =
vo − 2a ( x − xo )
2
Position
Key equation to determine distance when a is involved. Used
extensively in falling body problems. Its derivative is the velocity
equation above.
When no time is given
When v, a, and/or x are known, but no information is given about t,
then this can be used to solve for the unknown variable.
x o initial position, x final position, v o initial velocity, v final velocity, a acceleration, t time
Problem Solving Strategy
1.
2.
3.
4.
Draw a picture (Mental or on Paper)
List known and unknown variables.
a) Caution; some may be extraneous, and are not necessary to solve the problem.
b) Often either the starting or ending point is at rest, meaning a value of zero.
Do necessary conversions.
Choose an equation that can be solved with the known variables.
a) This equation may or may not be the answer you are looking for.
b) It may provide a new variable for use in another equation.
c) This may lead to a succession of equations.
+ or – ????:
“ + ” & “– ” can be used to indicate direction, and/or acceleration (+) or deceleration (–).
2
–9.8 m/s Be careful here. Does this mean the object is decelerating (slowing) or does it mean that the object is
moving along a negative (perhaps the y) axis? It would depend on the problem. For an object moving on the x axis it
would mean decelerating. For an object falling along the y axis, due to gravity, it means the object is accelerating, but
2
in the downward direction (-+9.8). In forces it is easier to use 9.8 m/s as a positive number.
Study Guide and Review
Kinematics in One Dimension
9
Falling Bodies
Displacement:
Velocity, initial:
Velocity, final:
Acceleration:
yo = 0
y =0
y = +
y = −
vo = 0
vo = +
vo = −
=0
=+
=−
= vo
v
v
v
v
Initial position. We can choose the reference frame / coordinate axis.
If the object ends the problem at the same elevation it started at.
If the object ends the problem at a higher elevation than it started.
If the object ends the problem at a lower elevation than it started.
If it is dropped from rest.
If fired upward.
If fired downward.
At the moment it reaches maximum altitude, right before falling back to earth.
If it hits something on the way up and never reaches max altitude (Rare problem).
On the return trip.
If it lands at the same elevation that the problem began at.
g = −9.8 m s 2
Projectile Motion
Motion in two dimensions happens simultaneously.
• In the x direction the velocity is constant, with no acceleration occurring in this dimension.
• In the y direction the acceleration of gravity slows upward motion and enhances downward motion.
• Both happen simultaneously, however they can be analyzed separately using vector components.
∗
The following review of variables can be overwhelming to memorize. It is much easier if you think it
through or draw a pictorial representation.
Angles:
Displacement:
Velocity, initial:
Velocity, final:
Acceleration:
Study Guide and Review
All angles are measured from East. Above the horizon is positive, below negative.
xo = 0
yo = 0
x= +
y =0
y = +
y = −
If the object ends the problem at the same elevation it started at.
If the object ends the problem at a higher elevation than it started.
If the object ends the problem at a lower elevation than it started.
vo
Splits into components,
vo x = +
In every problem, we choose to fire it in the positive x direction.
vo y = 0
If fired horizontally.
vo y = +
If fired at a positive angle (above the horizon).
vo y = −
If fired at a negative angle (below the horizon).
vx = vo x
vo y = 0
Since there is constant velocity in the x direction, initial and final are the same.
vo y = +
If the object hits something on the way up. Not used in problems very often.
vo y = −
On the return trip.
v
Resultant from adding vectors
The x is always positive.
vo x = vo cos θ , vo y = vo sin θ
At the top of the trajectory
v x and v xo . Has an angle not a + or –.
g = −9.8 m s 2
Kinematics in One Dimension
10
Graphing Motion
Name __________________________
Kinematics WS ___
Date
_____________ Per _______
1. A car travels at a constant 20 m/s for 10 s. Fill in the table showing the cars displacement from the origin at the end of
each second. Then graph the motion.
t (s)
d (m)
1
2
3
4
5
6
d
7
8
9
10
t
2. What is the slope of the curve plotted? __________
3. How does this compare to the velocity, given, in number 1 above?
4. Complete the table showing the cars velocity at the end of each second. Then complete the velocity-time plot.
t (s)
v (m/s)
1
2
3
4
5
6
v
7
8
9
10
t
5. Find the area under the curve for the first 5 seconds. __________
6. How does this compare to the displacement, for the first 5 seconds, in number 1 above?
Study Guide and Review
Worksheet: Graphing Motion
11
7. The slope of the displacement-time graph is
____________________.
8. The slope of the velocity-time graph is
____________________.
9. The area under the acceleration-time graph is
____________________.
10. The area under the velocity-time graph is
____________________.
Answer the next series of questions using the following displacement-time graph.
20
15
d
(m)
10
5
5
10
t (s)
15
20
25
11. How far does the object travel during the first 5 seconds (1 to 5 s)?
__________
12. How far does the object travel during the second 5 seconds (5 to 10 s)?
__________
13. How far does the object travel during the third 5 seconds (10 to 15 s)?
__________
14. How far does the object travel during the fourth 5 seconds (15 to 20 s)?
__________
15. How far does the object travel during the last 10 seconds (20 to 30 s)?
__________
30
16. During which time interval(s) is the object standing still?
17. Does the car ever accelerate in this scenario?
18. Draw the velocity time graph for the above scenario.
Study Guide and Review
Worksheet: Graphing Motion
12
Answer the next series of questions using the following velocity-time graph.
20
C
15
v
D
(m)
10
B
E
5
A
5
10
t (s)
15
20
25
19. During which interval(s) is the object accelerating?
_______________
20. During which interval(s) is the acceleration the greatest?
_______________
21. During which interval(s) is the object standing still?
_______________
22. During which intervals does the object have the same speed?
_______________
23. What is the displacement during interval A?
_______________
24. What is the displacement during interval B?
_______________
25. What is the displacement during interval C?
_______________
26. What is the displacement during interval D?
_______________
27. What is the displacement during interval E?
_______________
30
28. Draw the acceleration-time graph for the above scenario.
Study Guide and Review
Worksheet: Graphing Motion
13
Complete the series of graphs: displacement-time, velocity-time, and acceleration-time.
29.
30.
31.
d
d
d
t
v
t
a
d
d
t
35.
37.
t
Study Guide and Review
d
t
v
t
t
a
a
t
t
t
v
a
t
40.
d
v
a
t
t
t
t
a
39.
d
v
v
t
t
t
t
a
38.
d
d
v
a
t
36.
t
t
a
t
d
v
t
a
t
t
v
t
a
t
34.
v
t
a
33.
t
v
t
t
d
t
v
t
32.
t
Worksheet: Graphing Motion
t
14
Advanced Placement Physics
Kinematics in Two Dimensions
X Component Equation
Equations Given on Test
1
x = xo + vo t + at 2
2
x = vo x t
1
x = xo + vo t + at 2
2
v=
vo x + at
x
vx = vo x
v= vo + at
v=
vo y + gt
y
vx =vo x + 2a ( x − xo )
v2 =
vo + 2a ( x − xo )
v y =vo y + 2 g ( y − yo )
2
Y Component Equations
y =yo + vo y t +
1 2
gt
2
=
y vo y t − 4.9t 2
2
2
2
vx = vo x
2
=
vy
X Variables
Common to both X & Y
vo y − 19.6 y
2
Y Variables
xo = 0
vo =
yo = 0
x=
θ=
y=
vo x = vo cos θ
t=
vo y = vo sin θ
=
vx v=
vo cos θ
ox
=
v
ax = 0
vx + v y
2
2
vy =
g = −9.8
Strategies that work most of the time.
∗
When no time is given. Finding time is the key to all falling body or projectile motion problems
vy =
vo y + 2 g ( y − yo )
solve for
nd
v=
vo y + gt
y
use
v y from above to get t
rd
x = vo x t
use
t from above to solve for range, x
st
1
2
3
2
(Alternative:
∗
y =yo + vo y t +
v y this can be +/- , but is usually minus
1 2
gt , and the quadratic, followed by x = vo x t )
2
When time or range x is given. This makes the problem easy since velocity is constant in the x direction.
st
1 x = vo x t
Given time solve for x . Given x solve for time.
nd
2
∗
2
y =yo + vo y t +
1 2
gt
2
Once you have time this is easy, and you don’t need the quadratic.
When an object is dropped or fired horizontally
v=
vo y + gt
y
y =yo + vo y t +
1 2
gt
2
vo y = 0 and yo = 0 .
becomes
v y = gt
becomes
y=
1 2
gt
2
These versions are time savers, particularly the last one, since it now no longer requires the quadratic formula.
Study Guide and Review
Kinematics in Two Dimensions
15
PROJECTILE MOTION
Time is ruled by gravity and height. Most problems require y variables and y equations to solve for time. From
time distance in the x direction and the final v can be determined.
vo x = vx
holds true for all projectile motion problems.
vo=
v=
vo
x
x
vo y = 0
voy only
Horizontal
vo x = vo cos θ
vo y = vo sin θ
Vox
θ=0
-voy since down angle means -θ
vo
vx
v
=
v
Downward
vx
-y
-vy
vo x = vo cos θ
-vy
v
vx + v y
2
-y
=
v
2
At top:
2
+/-vy
Going up
means 2 possible t’s at
altitudes above ground
Going down
+vy -vy
+θ
Max. y
v =
v + 2 g ( y − yo )
2
y
2
oy
Study Guide and Review
Lands above axis
+voy
vox
Solved by setting vy=0
2
Upward
vy=0, v=vx=vox
vo y = vo sin θ
vx + v y
Lands on axis
Lands on level ground
vy=-voy
Land below axis
vx
tup = tdown
for objects
returning to
ground level
+y
y=0
-y
-vy
=
v
vx + v y
Kinematics in Two Dimensions
2
2
16
Advanced Placement Physics
Force
st
Newton’s 1 Law
nd
Newton’s 2 Law
rd
Newton’s 3 Law
ΣF
Any push or pull
Law of inertia (Restatement of Galileo’s principle of inertia)
∑ F = ma
Equal and opposite forces. For every action force there is an equal & opposite reaction force.
Forces come in action - reaction pairs.
Key to all problems.
∑F =
ma
Force
Σ Fx
Σ F ||
Σ F in x direction on traditional coordinate axis.
Σ F parallel to a slope (direction of motion).
Σ Fy
Σ F⊥
Σ F in y direction.
Σ F perpendicular to slope.
Sum of force is Net Force. You may need to solve for a using the kinematic equations, then solve
for force, or given force you solve for a and then use it in the kinematic equations to find v, x, or t.
Strategy on force Problems:
1. Draw FBD.
2. Set direction of motion. What would the object do if it could? Considered this the positive direction.
3. Using the forces listed below write the ΣF equations relevant to the problem. In what direction is the problem
moving? What matters, the x or the y direction? The parallel or the perpendicular direction? Any force vectors in
the FBD pointing in the direction of motion are positive while any vectors the other way are negative.
4. Substitute known equation, (forces like F g become mg).
5. Substitute for ΣF. Ask yourself what the sum of force should be based on the chart below. Is the object standing still,
moving at constant velocity, or accelerating. Substitute zero or ma for ΣF.
1
v=0
a=0
∆v = 0
ΣF = 0
2
v = +/- a constant value
∆v = 0
ΣF = 0
a=0
3
v increasing or decreasing
a = +/- a constant value
∆v = +/- a constant value
ΣF = m a
6. Plug in and solve. (All values including 9.8 are entered as positives. The negative signs were decided when setting
up the sum of force equation. Plugging in – 9.8 will just turn a vector assigned as – F g into a positive. You decided its
sign based on the way it was pointing relative to the problems direction of motion. Don’t reverse it now!)
FP
Fg
FT
FN
F fr
F ar
Fc
FB
Push or Pull.
Force of gravity. Fg = mg
Tension is a rope, string, etc. This force has no equation. You either solve for it, or it cancels, or it’s given.
Force Normal. A contact force, always perpendicular to the surface. (On a tilted surface use Σ F | | & Σ F ⊥ )
Friction force. F fr = µ FN Always opposes motion. Static friction: not moving. Kinetic friction: object moving.
Force of air resistance. This force has no equation. You either solve for it, or it cancels, or it’s given.
Force Centripetal. It is the ΣF in circular motion problems. So F c can be any force that keeps an object in circular
motion. =
Fc =
Fg
Fc FN=
Fc F=
Fc F=
Fc FB
etc.
T
fr
It can also be two or more of these added together. The direction of motion is toward the center. So any force
directed toward the center is positive and any force directed outward is negative.
The key in using any of these equations is to ask yourself: 1. What is causing the circular motion? 2. Then set up
the equality. 3. Substitute known equations. 4. Solve.
Force due to a magnetic field. This force is perpendicular to the field and perpendicular to the velocity of the
particle. So any charged particle will move in a circle. Use the right hand rule for positive charges or positive
current, and use the left hand for negative charges or electron current.
F any subscript that make ssense to solve the problem
Normal force: Gravity pulls the object down the slope and into the slope. If we only consider the motion into the
FN = Fg cosθ
slope (perpendicular), the object has no perpendicular velocity. So the ΣF⊥= 0. Then the
surface must push upward, equal and opposite to the perpendicular gravity component.
Named the normal force, it is a contact force and operates perpendicular to any surface.
It must counter only the component of gravity perpendicular to the surface.
FN = Fg cosθ
θ
Fg
Where θ is the angle between F g and F g ⊥ . It is also the tilt angle
of the surface measured from the ground. Substituting mg for F g .
θ
Study Guide and Review
FN = mg cosθ
Flat surfaces θ = 0 ,
o
Force
FN = Fg
or FN = mg
17
Friction: opposes motion. Motion is always parallel to a surface, so friction always acts parallel.
Static Friction:
Friction that will prevent an object from moving. As long as the object is standing still the force of
friction must be equal to the push, pull, component of gravity or other force that attempts to move the object. (If
there is no force attempting to cause motion, then there can be no friction).
Static friction is the strongest type of friction since the surfaces have a stronger adherence when stationary.
Kinetic Friction: Friction for moving objects. Once an object begins to move breaking static frictions hold, then the
friction is termed kinetic. Kinetic friction is not as strong as static friction, but it still opposes motion.
Coefficient of friction:
µ
a value of the adherence or strength of friction.
µ k for kinetic friction and µ s for
static friction.
Ffr = µ FN
so
Ffr = µmg cosθ
Force Parallel: Motion on a slope is parallel to the slope. F g and F N are at an angle to each other leaving a gap of
magnitude Fg sin θ when these two vectors are added tip to tail. Fg sin θ is not a force by itself, it is the sum of
force when F g and F N are added together. It is not part of the FBD. It describes the motion of the object parallel to
the slope, if no other forces are acting on it. What if we sum the
forces in the direction of motion (which is parallel to the slope)?
FN = Fg cosθ
Fg sin θ is down the slope and positive, since objects generally want
Fg sin θ
to go down hill (direction of natural motion is positive). Any force
θ
opposing the natural downward motion is a retarding force and is
Fg
negative. So uphill is negative.
Fg sin θ
We need an overall sum of force in the F direction.
θ
=
ΣF Fg sin θ − Fretarding
What do you use for force retarding? It could be friction F fr , air resistance F ar , a rope holding up the slope F T ,
someone pushing up the slope F P , or a combination of forces. Substitute the appropriate F and solve.
Friction on the slope: Friction is the retarding force in the scenarios discussed above.
1. No friction.
(What will the object do? Accelerate ΣF = ma)
FN = Fg cosθ
=
ΣF Fg sin θ − Fretarding
FN
θ
=
ΣF Fg sin θ − 0
Fg
ma = mg sin θ
a = g sin θ
Fg
Fg sin θ
θ
Ffr
FN = Fg cosθ
2. v = 0 or v is constant.
(No acceleration ΣF = 0)
Ffr
=
ΣF Fg sin θ − Fretarding
FN
=
0 Fg sin θ − Ffr
θ
Fg
µ mg cos θ = mg sin θ
µ cos θ = sin θ
µ = tan θ
Fg
Fg sin θ
θ
Ffr
Ffr
FN = Fg cosθ
FN
θ
Fg
Fg sin θ
Study Guide and Review
Fg
θ
3. Accelerating with friction present.
(Accelerates so ΣF = ma)
=
ΣF Fg sin θ − Fretarding
=
ΣF Fg sin θ − Ffr
=
ma mg sin θ − µ mg cos θ
=
a g sin θ − µ g cos θ
=
a g ( sin θ − µ cos θ )
Force
18
Complex Force Problems
Set direction of motion as positive. If you are not sure what the direction of motion will be take a guess. If the
problem returns negative values for the final result, you were wrong, the problem went the opposite of your prediction.
Vertical & Horizontal
Direction
of Motion
A
Pulley
+FN
-Ffr
+FTA
mA
+
B
A
-Fg
+Fg
If it doesn’t say which is more massive, pick one. In this
case I picked B as the heavier object and used this to set
the direction of motion. Find what does not change, T, and
rearrange in terms of this. Set the equations as equal,
substitute and solve.
Tension is the same for both blocks. Rearrange to get
equations in terms of tension, then set them equal so
tension cancels. Then substitute and solve.
∑ FB = FgB − FT
FT =
∑ FA + FfrA
F=
FgB − ∑ FB
T
∑ FA = FT − FgA
∑ FB = FgB − FT
FT =
∑ FA + FgA
F=
FgB − ∑ FB
T
∑ FA + F=
FgB − ∑ FB
gA
∑ FA + FfrA
= FgB − ∑ FB
mA a + mA g = mB g − mB a
m g − mA g
a= B
mA + mB
mA a + µ mA g cos θ =
mB g − mB a
m g − µ mA g cos θ
a= B
mA + mB
Friction on horizontal surfaces
1. Friction is the only force in the horizontal direction.
ΣF =
Ffr
µ Fg
ΣF =
ma = µ mg
a = µg
+FgB
-FgA
mB
∑ FA = FT − FfrA
B
A
-FT
B
-FTB
+FT
Incline
+FT
Direction of
Motion
FN
mB
+
-Ffr
mA
mB
mA
-Fg
FN
-FT
θ
mA
Fg
Fg
2. When friction and the forward force are equal. Object
can be standing still or moving at constant velocity.
ΣF = FP − Ffr
=
0 FP − Ffr
I picked m A as moving down the slope, so m B moves up.
Tension prevents m A from sliding down the slope and is
therefore acting like friction. If there was friction it would be
another arrow opposing motion down the slope. Just
subtract it as well. F g and F N are at angles to each other
leaving a vector gap of F g sinθ (see previous page)
+FN
-Ffr
mA
FP = Ffr
FP = µ mg
+FP
-Fg
3. When friction is not strong enough to prevent the object
from accelerating anyway.
ΣF = FP − Ffr
ma
= FP − µ mg
F − µ mg
a= P
m
=
∑ F Fg sin θ − Fretarding
=
∑ FA FgA sin θ − FT
=
FT
+FN
-Ffr
gA
mA
B
A
gA
T
A
B
T
− FgB
B
+ FgB
gB
mA g sin θ − mA a = mB a + mB g
g ( mA sin θ − mB )
a=
( mA + mB )
+FP
-Fg
Study Guide and Review
∑ F= F
F sin θ − ∑ F
=
F ∑F
F sin θ − ∑ F = ∑ F + F
Force
19
Vertical Circular Motion
FT top
Fg
FT bottom
Horizontal Circular Motion
A ball at the end of a string is
swung in a vertical circle. Any
force pointing to the center is
positive centripetal force, while
force vectors pointing away from
the center are negative
centripetal force. Sum the
forces. Look for the force that is
the same, and set up an equality.
Fg
F=
Fg + FT top
c
Fc =
− Fg + FT bottom
FT top= Fc − Fg
FT bottom= Fc + Fg
v2
F=
m
− mg
T top
r
v2
FT bottom
= m + mg
r
Fc = Ffr
v2
m = µ mg
r
v = µ gr
Magnetic Field
Force on a charged particle
A charged particle moving in a magnetic field will
experience a force causing it to follow a curved path
and be deflected from its original course. If the force is
strong enough the particle can be made to follow a
circular path.
FB = qvB sin θ
or
Lawn Mower
o
Pushing with 90 N at 45
Constant speed. ∑ F =
FN
FP
0
Solve for the Retarding Force
Fret
∑ Fx = FP x − Fret .
Fg
=
0 FP x − Fret .
FP
FN
Solve for the Normal Force
∑ Fy =
− FP y + FN − Fg
θ is the angle between the velocity and magnetic field.
Fret.
FN =
∑ Fy + FP y + Fg
Fg
2


0 + 90 N sin 45 + (16kg ) 9.8 m s  =
220 N
FN =
Force on a current carrying wire
The magnetic field can also move a current carrying
wire. The wire can jump.
(
o
)
(
)
Solve for F p to accelerate from rest to 1.5 m/s in 2.5 s
FB = BI  sin θ
v x − v xo
1.5 − 0
2
=
= 0.6 m s
t
2.5
∑ Fx = ma x = (16kg ) 0.6 m s 2 = 9.6 N
v x = v xo + a x t
B is the magnetic field strength.
I is the current in the wire.
 is the charge on the particle
θ is the angle between the velocity and magnetic field.
ax =
(
)
You need this force to accelerate, but you still need to
overcome the retarding force.
The Right Hand Rule is used to determine the direction of
deflection of the charged particles in the top scenario and
the direction of movement of the wire in the bottom
scenario.
Study Guide and Review
v2
µ=
rg
o
F=
F=
90 N cos 45
=
63.6 N
ret .
Px
q is the charge on the particle. See constants table.
v is the velocity of the particle.
B is the magnetic field strength.
How do you choose the right equation?
q is for charged particles, and  length of wire.
A penny on a circular disk
rotating horizontally. What
keeps it from flying off?
Friction. Something must be
keeping it going in a circle.
Otherwise it would move in a straight line. Friction is the
only candidate. No force is pushing it out of the circle (If
friction let go the penny would move due to inertia in a
direction tangent to the disk. It would not move out from
the center of the circle, since no such force is present in
this problem.) Force centripetal is the sum of forces for
circular motion.
∑ Fx = FP x − Fret .
FP x =
∑ Fx + Fret . =
9.6 N + 63.6 N =
73.2 N
But you aren’t pushing in the x direction. You need the
o
push at 45 to generate 73.2 N in the x direction.
FP x
73.2 N
FP x = Fpush cos 45o
=
Fpush
=
= 104 N
o
cos 45
0.707
Force
20
Advanced Placement Physics
Circular Motion, Gravity, & Satellites
Frequency: How often a repeating event happens. Measured in revolutions per second. Time is in the denominator.
Period:
The time for one revolution. T = 1 Time is in the numerator. It is the inverse of frequency.
f
Speed:
Velocity:
Traveling in circles requires speed since direction is changing.
However, you can measure instantaneous velocity for a point on the curve. Instantaneous velocity in any
type of curved motion is tangent to the curve. Tangential Velocity.
Projectile Motion
Circular Motion
Satellite Motion
The equation for speed and tangential velocity is the same v =
2π r
T
Acceleration: Centripetal Acceleration. Due to inertia objects would follow the tangential velocity. But, they don’t.
The direction is being changed toward the center of the circle, or to the foci. In other words they are
being accelerated toward the center. ac =
Force:
v 2 Centripetal means center seeking.
r
Centripetal Force. If an object is changing direction (accelerating) it must be doing so because a
force is acting. Remember objects follow inertia (in this case the tangential velocity) unless acted upon
by an external force. If the object is changing direction to the center of the circle or to the foci it must be
forced that way.
Fc = mac
Fc = m
v2
r
1. As always, ask what the object is doing. Changing direction, accelerating, toward the center, force centripetal.
2. Set the direction of motion as positive. Toward the center is positive, since this is the desired outcome.
3. Identify the sum of force equation. In circular motion Fc is the sum of force. Fc can be any of the previous forces.
If gravity is causing circular motion then Fc = Fg . If friction is then Fc = Ffr . If a surface is then Fc = FN .
4. Substitute the relevant force equations and solve. For Fc substitute m
Gravity
Fg = G
m1m2
r2
and
v2
r
Fg = mg combined are mg = G
m1m2
r2
simplified is
g =G
m
r2
r is not a radius, but is the distance between attracting objects measured from center to center. Is the problem asking for
the height of a satellite above earth’s surface? After you get r from the equation subtract earth’s radius. Are you given
height above the surface? Add the earth’s radius to get r and then plug this in. Think center to center.
Inverse Square Law: If r doubles (x2), invert to get ½ and then square it to get ¼. Gravity is ¼ its original value so
F g is ¼ of what it was and g is ¼ of what it was. So multiply F g by ¼ to get the new weight, or multiply g by ¼ to get the
new acceleration of gravity. If r is cut to a (x 1/3), invert it to get 3 and square it to get 9. Multiply F g or g by 9.
Apparent Weight: This is a consequence of your inertia. When an elevator, jet airplane, rocket, etc. accelerates
upward the passenger wants to stay put due to inertia and is pulled down by gravity. The elevator pushes up and you feel
heavier. Add the acceleration of the elevator to the acceleration of gravity Fg apparent
= mg + ma . If the elevator is going
down subtract Fg apparent
=
g a=
so Fg apparent 0 . This same
= mg − ma . If the elevator is falling you will feel weightless
phenomenon works in circular motion. Your inertia wants to send you flying at the tangential velocity. You feel pressed
up against the side of the car on the outside of the turn. So you think there is a force directed outward. This false nonexistent force is really your inertia trying to send you out of the circle. The side of the car keeps you in moving in a circle
just as the floor of the elevator moves you up. The car is forced to the center of the turn. No force exists to the outside.
However, it feels like gravity, just like your inertia in the accelerating elevator makes you feel heavier. You are feeling g’s
similar to what fighter pilots feel when turning hard. It is not your real weight, but rather what you appear to weight,
apparent weight.
Study Guide and Review
Circular Motion, Gravity, & Satellites
21
Kepler’s Three Laws of Satellite Motion
1. Satellites move in elliptical orbits. The body they orbit about is located at one of the two foci.
2. An imaginary line from the central body to the orbiting body will sweep equal areas of space in equal times.
2
3.
 T1   r1 
  = 
 T2   r2 
3
Compares the orbit of one satellite to another (i.e. you can use the earth’s orbit to solve for any
other planet’s orbit. Remember, in this case r is not the radius of earth, but rather the earth sun distance.
Rotation
All parts of an object are rotating around the axis. All parts of the body have the same period
of rotation. This means that the parts farther from the central axis of rotation are moving
faster. So if we look at some of the tangential velocities diagramed at the right, we see that
they are in all directions and vary in magnitude. So we need a new measurement of velocity.
Collectively all the velocities are known as the angular velocity, which is a measure of the
radians turned by the object per second. Because the period is the same for the various
parts of the rotating object, they move through the same angle in the same time. In rotation
the parts of a rotating body on the outside move faster. They need to travel through the same number of degree or
radians in the same amount of time as the inner parts of the body, but the circumference near the edge of a spinning
object is longer than close to the center. So the outer edge must be moving faster to cover the longer distance in the
same time interval. (This differs from the circular motion of the planets, which are not attached, and therefore not a single
rotating body. The planets move in circular motion individually. Here the inner planets move faster. The planets closer to
the sun must move faster in order to escape the gravity of the sun. They also travel a shorter distance and therefore have
the shortest period of orbit).
All the equations for an object in circular motion hold true if we are looking at a single point and only a
specific point on a rotating object.
Rotating objects have rotational inertia and an accompanying angular momentum, meaning that a rotating object will
continue to rotate unless acted upon by an unbalanced torque, & a non-rotating object will not rotate unless acted upon
by an unbalanced torque.
Torque:
The force that causes rotation. In rotation problems we look at the sum of torque (not the sum
of force). But it is exactly the same methodology.
τ = rF sin θ
o
Strongest when the force is perpendicular to the lever arm (since sin 90 equals one).
Balanced Torque:
The sum of torque is zero. No rotation.
Unbalance Torque:
Adding all the clockwise and counterclockwise torque does not sum to zero. So there is
excess torque in either the clockwise or counterclockwise direction. This will cause the object to rotate.
1. As always, ask what the object is doing. Is it rotating or is it standing still?
2. Set the direction of motion as positive. It will either rotate clockwise or counterclockwise. If you pick the wrong
direction your final answer will be negative, telling you that you did thing in reverse. But, the answer will be correct
nonetheless. If it is not moving pick one direction to be positive, it really doesn’t matter. But the other must be
negative, so that the torque cancels.
3. Identify the sum of torque equation.
∑ τ = ∑ τ clockwsise − ∑ τ counterclockwsise
or
∑ τ = ∑ τ counterclockwsise − ∑ τ clockwsise
4. Substitute the relevant force equations and solve (examples assume clockwise was positive direction)
Rotating you get some + / Not Rotating
∑ τ = ∑ ( rF sin θ )clockwsise − ∑ ( rF sin θ )counterclockwsise
0 = ∑ ( rF sin θ )clockwsise − ∑ ( rF sin θ )counterclockwsise
Angular momentum: Depends on mass (like regular momentum) and it also depends on mass distribution. As an
ice skater brings their arms closer to the body they begin to spin faster, since the mass has a shorter distance to travel.
Angular momentum is conserved. The radius gets smaller, but angular velocity increases (vice versa as the
skater moves arms outward). A galaxy, solar system, star, or planet forms from a larger cloud of dust. As the cloud is
pulled together by gravity its radius shrinks. So the angular velocity must increase. These objects all begin to spin faster
and faster. That is why we have day and night.
Study Guide and Review
Circular Motion, Gravity, & Satellites
22
Advanced Placement Physics
Work, Energy, Momentum, and Oscillations
Understanding the relationships between All Forms of Energy, Conservation of Energy, and
Work Energy Theorem are extremely essential for success on the AP Exam. The review
here is very limited, since this critical information is given substantial emphasis in the
course overview. Often energy is either the only way to progress in an AP Free Response
problem, or it is the easiest (quickest) way to solve the problem. Students who have a
thorough understanding of energy will achieve success on the AP Exam and arrive at
college as a more accomplished physics student.
Energy is conserved: It cannot be created or destroyed, but it can change forms.
Can energy be lost? No! Lost energy goes to the environment. A car (system) looses energy due to air resistance, so air
molecules (environment) gain energy and move faster. Energy is conserved.
When we did kinematics, problems might have started at 205.65 m from where we were standing. But, to make it easier
we said the problem started at 0 m. For energy pretend the system has zero internal energy initially. Then only worry
about the other forms of energy. We can then solve for how much the internal energy changes in the problem.
Work An object or problem has a certain amount of energy starting the problem (potential energy due to position and/or
kinetic energy due to motion). Remember were pretending internal energy is zero. Think of work as the energy that
is added (+W) to the system or subtracted (-W) from the system. If you add a force to something that is standing still
it will begin to move a distance. This requires positive work, the product of the force used and distance moved.
W = Fs cosθ
Force applied over a distance. Force and distance must be parallel. Note: this does not mean the
x axis which cosθ usually goes along with. θ is the angle between direction of motion and applied force.
Work is the Area Under the Force Distance Curve: This is the integral of the force distance function in a
calculus based course. But, our functions will be simple enough to allow us to use geometry to find the area.
Kinetic Energy
K=
1 2
mv
2
Energy of moving matter. Note that doubling mass doubles kinetic
energy, but doubling velocity quadruples kinetic energy. So you car at
60 mph is 4 times more lethal than at 30 mph.
Potential Energy
Gravitational
U = mgh
Depends on height. Consider the lowest point in the problem to be zero
height. This isn’t correct, but who wants to add the radius of the earth to
every number in the problem. Radius factors out at the end anyway.
Energy of a particle experiencing an electric potential.
Electric
U E = qV
Spring
US =
1 2
kx
2
Energy of a compressed spring with spring constant k.
Capacitor
UC =
1
1
QV = CV 2
2
2
Energy of capacitor.
Energy of Photons E = hν = pc
Used in modern physics
Work Energy Theorem
Work put into a system = the change in energy of the system. If you do work on a system you add energy (+W). If
the system moves to a lower energy state (dropping a bowling ball on your toe), then the system does work on the
environment (-W). It can transfer energy to the environment. The bowling ball has –W while your toe gets +W (toe
gets energy)
W = ∆U
W = ∆U E
W = ∆U s
W=K
W = Qheat energy
etc.
But, what if the energy changes from zero to some amount or from some amount to zero.
W =U
W = UE
W = Us
W=K
W = Qheat energy
etc.
Work and work-energy theorem are great for changes in energy, when energy moves from one thing to another or is
added or subtracted. But what if a system doesn’t exchange energy with the environment or another system. What if
it has certain types of energy in the beginning of the problem, but it has a different amount of each energy at the end?
Study Guide and Review
Work, Energy, Momentum, & Oscillations
23
Conservation of Energy
Energy cannot be created or destroyed, but can change form and be transferred.
Internal Energyi + K i + U i + Any other energy
=
Energy f + K f + U f + Any other energy f
i
The big picture:
However, the problem may only talk about two forms of energy.
As an example: If the problem only involves Potential Energy and Kinetic Energy
1
1
then substitute known equations mghi + mv 2 i = mgh f + mv 2 f
U i + Ki = U f + K f
2
2
Here are some other possibilities: The first is accelerating charges, the second is for springs.
1
1
qVi + mv 2 i = qV f + mv 2 f
2
2
1 2 1 2 1 2
1
kx i + mv i = kx f + mv 2 f
2
2
2
2
etc.
The following formulas are specific short cuts usually applied when there are two extremes in the problem.
Gravity mgh = 1 mv 2
A mass m starts at the highest point and ends at the lowest point, or vice versa.
2
1 2
mv
2
1 2 1 2
Spring
kx = mv
2
2
1
Electrons
hν = mv 2
2
Electric
qV =
When a charge q is accelerated by charged plates with a potential difference V.
If a compressed spring extends to the equilibrium position, or vice versa.
When energy of a photon is transferred to an electron, or vice versa.
E1i + E2 i = E1 f + E2 f
Collisions
bK
1i
gd
Can be used by itself and with conservation of momentum below.
i
+ K2 i = K1 f + K2 f + Kdissipated In collisions total energy is conserved, but kinetic energy is not.
Unlike momentum, kinetic energy can decrease in collisions which are not perfectly elastic. But where does it go? The
deformation of colliding bodies turns into heat (internal energy). So if you take the Kinetic energy at the start, it will equal
the kinetic energy at the end plus the amount of kinetic energy dissipated. The energy dissipated is conserved: transfers
to internal energy.
Power: Rate at which work is done. Powerful machines do more work in the same time, or the same work in less time.
P=
W
t
P = Fv
Work or Energy delivered as a rate of time.
It involves work. Making this another of the very important concepts.
As an example you can go from energy to work to power then to voltage and current
P = IV
Energy and time: Think Power when you see energy and time, Joules and seconds.
p = mv inertia in motion. Measure of how difficult it is to stop an object.
Momentum
Ft = ∆p Trade off between time taken to stop and force needed to stop.
Impulse
Conservation of Momentum
Total momentum before a collision must match total momentum after. Not given
on the AP exam. One object might be standing still at the start or after.
m1v1i + m2 v2 i = m1v1 f + m2 v2 f
Completely Elastic Collision: Bounce off completely.
b
m1v1i + m2 v2 i = v f m1 + m2
g
Inelastic Collision: The objects stick together, mass adds, one velocity.
Oscillations
Period
T=
Frequency
Springs
Restoring force
Period of a spring
Pendulum
Study Guide and Review
1
f
Time for one revolution, measured in seconds
The number of revolution, turns, vibrations, oscillations, rotations per second.
F = − kx
displace a spring and it will return to equilibrium, center.
k is the spring constant, the minus sign is not mathematical
Ts = 2π
m
k
Depends on mass of object attached to spring and k.
Tp = 2π

g
Depends on length of the pendulum and g.
Work, Energy, Momentum, & Oscillations
24