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Instructor: Profs. Acosta, Hamlin, Muttalib
PHYSICS DEPARTMENT
PHY 2048
Final Exam
Name (PRINT, last, first):
April 22, 2017
Signature:
On my honor, I have neither given nor received unauthorized aid on this examination.
YOUR TEST NUMBER IS THE 5-DIGIT NUMBER AT THE TOP OF EACH PAGE.
DIRECTIONS
(1) Code your test number on your answer sheet (use 76–80 for the 5-digit number). Code your name on your
answer sheet. DARKEN CIRCLES COMPLETELY. Code your student number on your answer sheet.
(2) Print your name on this sheet and sign it also.
(3) Do all scratch work anywhere on this exam that you like. At the end of the test, this exam printout is to be turned in.
No credit will be given without both answer sheet and printout with scratch work most questions demand.
(4) Blacken the circle of your intended answer completely, using a #2 pencil or blue or black ink. Do not
make any stray marks or the answer sheet may not read properly.
(5) The answers are rounded off. Choose the closest to exact. There is no penalty for guessing.
>>>>>>>>WHEN YOU FINISH <<<<<<<<
Hand in the answer sheet separately.
Where needed use g = 9.8 m/s2
Axis
R
Axis
Axis
Annular cylinder
(or ring) about
central axis
Hoop about
central axis
R1
Solid cylinder
(or disk) about
central axis
R2
L
R
1
1
I = 2 M(R 12 + R 22 )
I = MR 2
Axis
Solid cylinder
(or disk) about
central diameter
Axis
L
L
I = 2 MR2
Thin rod about
axis through center
perpendicular to
length
Axis
Solid sphere
about any
diameter
2R
R
1
1
Axis
2
1
I = 4 MR2 + 12 ML2
I = 12 ML2
Thin spherical
shell about
any diameter
Axis
R
I = 5 MR2
Hoop about
any diameter
Axis
Slab about
perpendicular
axis through
center
2R
b
a
2
I = 3 MR2
1
I = 2 MR2
1
I = 12 M(a2 + b2)
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Formula-sheet: Exam 1
• For constant acceleration ~a:
~v = ~v0 + ~at
2
vx2 = vx0
+ 2ax (x − x0 )
1
~r = ~r0 + ~v0 t + ~at2
2
2
vy2 = vy0
+ 2ay (y − y0 )
Acceleration due to gravity: g = 9.8 m/s2 = 32 ft/s2 vertically down.
• For force acting on a body of mass m: F~ = m~a
• Frictional forces: fs,max = µs N ; fk = µk N. N : normal force.
• For uniform circular motion: centripetal acceleration is ac =
• Kinetic energy: K =
v2
r
1
mv 2
2
• Work done by a constant force: W = F~ · d~ = F d cos (angle between F~ and d~ )
• Work-kinetic energy theorem: W = Kf − Ki
~ = îAx + ĵAy ;
• Vectors (2d): A
A=
q
A2x + A2y ;
~ and î);
Ax = A cos (angle between A
~ and î)
Ay = A sin (angle between A
Formula-sheet: Exam 2
• Potential Energy: ∆U = −W = −
Gravitational (y=up): Fy = −mg
Z
~
r2
~
r1
F~ · d~r
Fx = −
dU
dx
U (y) = mgy
Elastic (x from spring equilibrium): Fx = −kx
U (x) =
1 2
kx
2
• Mechanical Energy: Emec = K + U
Emec = constant for an isolated system with conservative forces
• Work-Energy: W (external) = ∆K + ∆U + ∆E(thermal)
• Center of Mass: ~rcom =
N
1 X
mi~ri
Mtot i=1
• Linear Momentum: p~ = m~v
d~
p
F~ =
dt
P~tot
Mtot =
N
X
mi
i=1
Impulse: J~ = ∆~
p=
Z
tf
F~ (t)dt = F~av ∆t
ti
d~
p
If F~ =
= 0 then p~ = constant and p~f = p~i
dt
N
X
dP~tot
= Mtot ~acom
F~net =
pi
~
= Mtot ~vcom =
dt
i=1
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• Elastic Collisions of Two Bodies, 1D
v1f =
m1 − m2
2m2
v1i +
v2i
m1 + m2
m1 + m2
• Rockets: Thrust = M a = vrel
2m1
m2 − m1
v1i +
v2i
m1 + m2
m1 + m2
Mi
∆v = vrel ln
Mf
v2f =
dM
dt
• Rotational Variables
angular position: θ(t)
angular velocity: ω(t) =
dθ(t)
dt
d2 θ(t)
dω(t)
=
dt
dt2
• For constant angular acceleration α:
angular acceleration: α(t) =
ω 2 = ω02 + 2α(θ − θ0 )
1
θ = θ0 + (ω + ω0 )t
2
ω = ω0 + αt
1
θ = θ0 + ω0 t + αt2
2
• Angular to linear relationships for circular motion
arc length: s = rθ
velocity: v = rω
tangential acceleration: aT = rα
centripetal acceleration: ac = rω 2
• Rotational Inertia: I =
N
X
mi ri2 (discrete)
i=1
R
I = r2 dm (uniform)
Parallel Axis: I = Icom + Mtot d2 (d is displacement from com)
1 2
1
1
2
Iω Kroll = M vcom
+ Icom ω 2
2
2
2
= rω acom = rα
• Rotational, Rolling Kinetic Energy: Krot =
Rolling without slipping: xcom = rθ
vcom
• Torque: ~τ = ~r × F~
(where k̂ = î × ĵ gives directions for cross product)
τ = rF sin (angle between ~r and F~ ) = rF⊥
~
dL
dt
L = rp sin (angle between ~r and ~
p) = rp⊥
~
dL
~ = constant and L
~f = L
~i
= 0 then L
If ~τnet =
dt
~ = ~r × ~
• Angular Momentum: L
p
~τ =
• Work done by a constant torque: W = τ ∆θ = ∆Krot =
1 2 1 2
Iω − Iω
2 2 2 1
• Power done by a constant torque: P = τ ω
• For torque acting on a body with rotational inertia I: ~τ = I~
α
• Stress and Strain (Y = Young’s modulus, B = bulk modulus)
F
∆L
Linear:
=Y
A
L
F
∆V
Volume: P =
= −B
A
V
• For torque acting on a body with rotational inertia I: ~τ = I~
α
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PHY2048 Exam 3 Formula Sheet
Law of Gravitation
2
Magnitude of Force: Fgrav = G m1rm
2
Potential Energy: Ugrav = −G m1rm2
Law of Periods: T 2 =
4π 2
GM
r3
G = 6.67 × 10−11 Nm2 /kg2
GM m
Total Mechanical Energy for circular orbit: E = −
2r
r
r
2GM
GM
Orbital Speed: v =
Escape Speed: vescape =
R
r
Ideal Fluids
F
Units: 1 Pa = 1 N/m2 ; 105 Pa = 1 bar ≃ 1 atm
A
Equation of Continuity: RV = Av = constant (volume flow rate) Rm = ρAv = constant (mass flow rate)
Pressure: P =
Bernoulli’s Equation (y-axis up): P1 + 12 ρv12 + ρgy1 = P2 + 21 ρv22 + ρgy2 = constant
Fluids at rest (y-axis up): P2 = P1 + ρg(y1 − y2 )
Buoyancy Force: FBuoy = Mf luid g
Simple Harmonic Motion (SHM) (angular frequency ω = 2πf = 2π/T )
x(t) = xmax cos(ωt + φ)
2
2
v(t) = −ωxmax sin(ωt + φ)
vmax = ωxmax
2
a(t) = −ω xmax cos(ωt + φ) = −ω x(t)
amax = ω xmax
r
r
r
k
g
mgh
Simple Pendulum: ω =
Physical Pendulum: ω =
Linear Harmonic Oscillator: ω =
m
L
I
r
κ
Torsion Oscillator: ω =
Damped harmonic oscillator: x(t) = e−bt/2m xmax cos(ωt + φ)
I
Sinusoidal Traveling Waves (frequency f = 1/T = ω/2π, wave number k = 2π/λ)
y(x, t) = ymax sin(Φ) = ymax sin(kx ± ωt + φ)
Phase:
Φ = kx ± ωt
Wave Speed: vwave =
Interference (Max Constructive): ∆Φ = 2nπ
ω
λ
=
= λf
k
T
n = 0, ±1, ±2, · · ·
Interference (Max Destructive): ∆Φ = (2n + 1)π
Average power (string wave): Pavg =
(− = right moving, + = left moving)
r
τ
Wave Speed (tight string): vwave =
µ
1
2
µvω 2 ym
2
n = 0, ±1, ±2, · · ·
∆d = nλ n = 0, ±1, ±2, · · ·
∆d = (n + 21 )λ
n = 0, ±1, ±2, · · ·
Standing Waves on a String (L = length, n = harmonic number)
y ′ (x, t) = 2ymax sin(kx) cos(ωt)
Allowed Wavelengths & Frequencies: λn = 2L/n
fn =
vwave
nvwave
=
λn
2L
n = 1, 2, 3 · · ·
Sound Waves (P = Power)
Sound wave displacement: s(x, t) = sm cos(kx ± ωt)
Sound wave pressure: ∆p(x, t) = ∆pm cos(kx ± ωt)
r
P
Psource
B
2
Intensity (W/m ): I =
Speed of sound: vsound =
Isotropic Point Source: I(r) =
2
A
4πr
ρ
vsound − vD
(fs = frequency of source, vs , vD = speed of source, detector)
Doppler Shift: fobs = fs
vsound − vS
Change −vD to +vD if the detector is moving opposite the direction of the propagation of the sound wave.
Change −vs to +vs if the source is moving opposite the direction of the propagation of the sound wave.
nv
Beat frequency: fbeat = f1 − f2
Resonance for pipe open on both ends: f =
n = 1,2,3,...
2L
nv
I
Resonance for pipe closed at one end: f =
n = 1,3,5,...
Sound level: β = (10 dB) log
4L
I0