AP Physics C Review Mechanics - CHSN Review Project
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AP Physics C Review
Mechanics
CHSN Review Project
This is a review guide designed as preparatory information for the AP1 Physics C
Mechanics Exam on May 11, 2009. It may still, however, be useful for other purposes
as well. Use at your own risk. I hope you find this resource helpful. Enjoy!
This review guide was written by Dara Adib based on inspiration from Shelun Tsai’s
review packet.
This is a development version of the text that should be considered a work-in-
progress.
This review guide and other review material are developed by the CHSN Review
Project.
Copyright © 2009 Dara Adib. This is a freely licensed work, as explained in the Defi-
nition of Free Cultural Works (freedomdefined.org). Except as noted under “Graphic
Credits” on the following page, the work is licensed under the Creative Commons
Attribution-Share Alike 3.0 United States License. To view a copy of this license, visit
http://creativecommons.org/licenses/by-sa/3.0/us/ or send a letter to Creative
Commons, 171 Second Street, Suite 300, San Francisco, California, 94105, USA.
This review guide is provided “as is” without warranty of any kind, either expressed
or implied. You should not assume that this review guide is error-free or that it will
be suitable for the particular purpose which you have in mind when using it. In no
event shall the CHSN Review Project be liable for any special, incidental, indirect or
consequential damages of any kind, or any damages whatsoever, including, without
limitation, those resulting from loss of use, data or profits, whether or not advised of
the possibility of damage, and on any theory of liability, arising out of or in connec-
tion with the use or performance of this review guide or other documents which are
referenced by or linked to in this review guide.
1 AP is a registered trademark of the College Board, which was not involved in the production of, and does not endorse,
this product.
1“Why do we love ideal worlds? . . . I’ve been doing this for 38 years and school is an
ideal world.” — Steven Henning
Contents
Kinematic Equations 3
Free Body Diagrams 3
Projectile Motion 4
Circular Motion 4
Friction 5
Momentum-Impulse 5
Center of Mass 5
Energy 5
Rotational Motion 7
Simple Harmonic Motion 8
Gravity 9
Graphic Credits
• Figure 1 on page 3 is based off a public domain graphic by Concordia College and vectorized
by Stannered: http://en.wikipedia.org/wiki/File:Incline.svg.
• Figure 2 on page 3 is based off a public domain graphic by Mpfiz: http://en.wikipedia.
org/wiki/File:AtwoodMachine.svg.
• Figure 5 on page 7 is a public domain graphic by Rsfontenot: http://en.wikipedia.org/
wiki/File:Reference_line.PNG.
• Figure 6 on page 7 was drawn by Enoch Lau and vectorized by Stannered: http://en.
wikipedia.org/wiki/File:Angularvelocity.svg. It is licensed under the Creative Com-
mons Attribution-Share Alike 2.5 license: http://creativecommons.org/licenses/by-sa/
2.5/.
• Figure 7 on page 8 is based off a public domain graphic by Mazemaster: http://en.wikipedia.
org/wiki/File:Simple_Harmonic_Motion_Orbit.gif.
• Figure 8 on page 9 is a public domain graphic by Chetvorno: http://en.wikipedia.org/
wiki/File:Simple_gravity_pendulum.svg.
2Figure 1: Normal Force
Kinematic Equations
1
∆x = at2 + v0 t
2 Figure 2: Atwood’s Machine
∆v = at
(v)2 − (v0 )2 = 2a(∆x)
Figure 3: Draw a banked curve diagram
v0 + v
∆x = ×t
2
Pulled Weights
Free Body Diagrams F−f
a=
Σm
N Normal Force
f Frictional Force
T = ma
T Tension
mg Weight Elevator
F = ma Normal force acts upward, weight acts down-
ward.
In a particular direction: • Accelerating upward: N = |ma| + |mg|
• Constant velocity: N = |mg|
ΣF = (Σm)a
• Accelerating downward: N = |mg| − |ma|
Atwood’s Machine2
Banked Curve
|(m2 − m1 )|g Friction can act up the ramp (minimum velocity
a=
m1 + m2 when friction is maximum) or down the ramp
(maximum velocity when friction is maximum).
p
2 Pulley and string are assumed to be massless. videal = rg tan θ
3Range
s
rg(tan θ − µ) θ represents the smaller angle from the x-axis to
vmin =
µ tan θ + 1 the direction of the projectile’s initial motion.
Starting from a height of x = 0:
s
rg(tan θ + µ) (v0 )2 sin 2θ
vmax = xmax =
1 − µ tan θ g
Projectile Motion Circular Motion
Position Centripetal (radial)
Centripetal acceleration and force is directed to-
∆x = vx t
wards the center. It refers to a change in direc-
tion.
1
∆y = − gt2 + (vy )0 t v2
2 ac =
r
Velocity
mv2
θ represents the smaller angle from the x-axis to Fc = mac =
r
the direction of the projectile’s initial motion.
Tangential
(vx )0 = v0 cos θ
Tangential acceleration is tangent to the object’s
motion. It refers to a change in speed.
(vy )0 = v0 sin θ
d|v|
at =
dt
∆vx = 0
Combined
∆vy = −gt
q
atotal = (ac )2 + (at )2
Height
θ represents the smaller angle from the x-axis to Vertical loop
the direction of the projectile’s initial motion.
In a vertical loop, the centripetal acceleration is
Starting from a height of x = 0: caused by a normal force and gravity (weight).
(v0 sin θ)2
ymax =
2g
4Top Elastic
Kinetic energy is conserved.
F = ma
v2
N + mg = m × m1 v1 + m2 v2 = m1 v10 + m2 v20
r
mv2
N = − mg
r
−(v20 − v10 ) = v2 − v1
Bottom
Inelastic
F = ma Kinetic energy is not conserved.
v2
N − mg = m ×
r
mv 2 m1 v1 + m2 v2 = (m1 + m2 )v0
N = + mg
r
Center of Mass
Friction
Z Z
Σmr 1 1
Friction converts mechanical energy into heat. rcm = = rdm = xλdx
Static friction (at rest) is generally greater than Σm Σm Σm
kinematic friction (in motion).
dm M
λ= =
fmax = µN dx L
Z Z
Momentum-Impulse Σm = dm = λdx
p = mv
(Σm)vCM = Σmv = Σp
dp
F=
dt Fnet = (Σm)aCM
Z
I= Fdt = F∆t = ∆p = m∆v Energy
Work
Collisions
Z
Total momentum is always conserved when there W= Fdx = ∆K
are no external forces (F = dp
dt = 0).
5Power
W Fx
Pavg = =
t t
dW
Pinstant = = Fv
dt
Kinetic Energy
1 Linear Angular
K = mv2
2 dx ∆x dθ ∆θ
v= dt = ∆t ω= dt = ∆t
Potential Energy dv ∆v dω ∆ω
a= dt = ∆t α= dt = ∆t
∆x = 12 at2 + v0 t ∆θ = 12 αt2 + ω0 t
dU
F=− ∆v = at ∆ω = αt
dx
(v)2 − (v0 )2 = 2a(∆x) (ω)2 − (ω0 )2 = 2α(∆ω)
Z xf
∆U = − FC dx = −WC v0 + v ω0 + ω
xi
∆x = ×t ∆θ = ×t
2 2
F = ma τ = Iα
Z Z
UHooke
1
= − FHooke dx = − −kxdx = kx2 Rx Rθ
2 W = x0 Fdx Wrot = θ0 τdθ
W = 12 mv2 − 12 m(v0 )2 Wrot = 21 mω2 − 12 m(ω0 )2
Ug = mgh P = Fv Prot = τω
equilibrium point F = − du
dx = 0 (extrema) p = mv L = Iω
stable equilibrium U is a minimum dp dL
F= dt τ= dt
unstable equilibrium U is a maximum
Figure 4: Rotational Motion
Total
E = K+U
Ei + WNC = Ef
WNC represents non-conservative work that con-
verts mechanical energy into other forms of en-
ergy. For example, friction converts mechanical
energy into heat.
6Torque
τ = r × F = rF sin θ
τ = Iα
Moment of Inertia
Z
2
I = Σmr = r2 dm
Figure 5: Arc Length
I = Icm + Mh2
(h represents the distance from the center)
Values
1 2
rod (center) 12 ml
Figure 6: Angular Velocity 1 2
rod (end) 3 ml
Rotational Motion hollow hoop/cylinder mr2
1 2
solid disk/cylinder
2 mr
The same equations for linear motion can be mod- hollow sphere 2 mr2
3
ified for use with rotational motion (Figure 4 on 2
the previous page). solid sphere 5 mr2
Angular Motion Atwood’s Machine
s |(m2 − m1 )|g
θ= a=
r m1 + m2 + 12 M
v
ω= Angular Momentum
r
at
α= L = Iω
r
L = r × p = rp sin θ = rmv sin θ
p
at = r α2 + ω4
dL
ac = ω2 r τ=
dt
1 1 Total angular momentum is always conserved
Krolling = Iω2 + mv2
2 2 when there are no external torques (τ = dL
dt = 0).
7ω = 2πf
r v 2
0
A= (x0 )2 +
ω
−v0
φ = arctan
ωx0
1
E = kA2
2
Figure 7: Simple Harmonic Motion Spring
Simple Harmonic Motion Fs = −kx
r
m
Simple harmonic motion is the projection of uni- Ts = 2π
form circular notion on to a diameter. Likewise, k
uniform circular motion is the combination of r
simple harmonic motions along the x-axis and k
ωs =
y-axis that differ by a phase of 90◦ . m
amplitude (A) maximum magnitude of displace-
ment from equilibrium Pendulum
cycle one complete vibration Simple
period (T ) time for one cycle
s
frequency (f) cycles per time L
T = 2π
angular frequency (ω) radians per time g
r
x = Acos(ωt + φ) g
ω=
L
v = −ωA sin(ωt + φ) Compound
A cable with a moment of inertia swings back
and forth. d represents the distance from the
a = −ω2 A cos(ωt + φ) = −ω2 x
pendulum’s pivot to its center of mass.
s
2π 1 I
T= = T = 2π
ω f mgd
r
mgd
1 ω ω=
f= = I
T 2π
8frictionless pivot Energy
amplitude θ −Gm1 m2
U=
massless rod R
−GMm
E=
2r
bob's
trajectory massive bob 2πR
equilibrium v=
position T
Figure 8: Simple Pendulum r
2GM
vescape =
re
Torsional
For orbits around the earth, re represents the ra-
A horizontal mass with a moment of inertia is dius of the earth.
suspended from a cable and swings back and
forth.
r
I
T = 2π
k
k
ω=
I
Gravity
−Gm1 m2
F=
R2
Nm2
G ≈ 6.67 × 10−11
kg2
Kepler’s Laws
1. All orbits are elliptical.
2. Law of Equal Areas.
4π 2
3. T 2 = GM R3 = Ks R3 , where Ks is a uniform
constant for all satellites/planets orbiting
a specific body
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