Physics 101Lecture 12
Equilibrium and Angular Momentum
Ali ÖVGÜNEMU Physics Department
www.aovgun.com
2/13/17
Static Equilibriumq Equilibrium and static
equilibriumq Static equilibrium
conditionsn Net external force must
equal zeron Net external torque must
equal zeroq Center of gravityq Solving static equilibrium
problems
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Static and Dynamic Equilibrium
q Equilibrium implies the object is at rest (static) or its center of mass moves with a constant velocity (dynamic)
q We will consider only with the case in which linear and angular velocities are equal to zero, called “static equilibrium” : vCM = 0 and ω = 0
q Examplesn Book on tablen Hanging signn Ceiling fan – offn Ceiling fan – onn Ladder leaning against wall
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Conditions for Equilibriumq The first condition of
equilibrium is a statement of translational equilibrium
q The net external force on the object must equal zero
q It states that the translational acceleration of the object’s center of mass must be zero
0==∑= amFF extnet!!!
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Conditions for Equilibriumq If the object is modeled as a
particle, then this is the only condition that must be satisfied
q For an extended object to be in equilibrium, a second condition must be satisfied
q This second condition involves the rotational motion of the extended object
0=∑= extnet FF!!
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Conditions for Equilibriumq The second condition of
equilibrium is a statement of rotational equilibrium
q The net external torque on the object must equal zero
q It states the angular acceleration of the object to be zero
q This must be true for any axis of rotation
0==∑= αττ !!! Iextnet
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Conditions for EquilibriumqThe net force equals zero
n If the object is modeled as a particle, then this is the only condition that must be satisfied
qThe net torque equals zeron This is needed if the object cannot be
modeled as a particleqThese conditions describe the rigid objects
in the equilibrium analysis model
0=∑Fr
0τ =∑ r
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Static Equilibriumq Consider a light rod subject to the two
forces of equal magnitude as shown in figure. Choose the correct statement with regard to this situation:
(A) The object is in force equilibrium but not torque equilibrium.
(B) The object is in torque equilibrium but not force equilibrium
(C) The object is in both force equilibrium and torque equilibrium
(D) The object is in neither force equilibrium nor torque equilibrium
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Equilibrium Equationsq For simplicity, We will restrict the applications
to situations in which all the forces lie in the xy plane.
q Equation 1:q Equation 2: q There are three resulting equations
0 0 0 :0 ,,, ====∑= znetynetxnetextnet FFFFF!!
0 0 0 :0 ,,, ====∑= znetynetxnetextnet τττττ !!
0
0 0
,,
,,
,,
=∑=
=∑==∑=
zextznet
yextynet
xextxnet
FFFF
ττ
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q EX: A seesaw consisting of a uniform board of mass mpland length L supports at rest a father and daughter with masses M and m, respectively. The support is under the center of gravity of the board, the father is a distance d from the center, and the daughter is a distance 2.00 m from the center.
q A) Find the magnitude of the upward force n exerted by the support on the board.
q B) Find where the father should sit to balance the system at rest.
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gmMgmgngmMgmgnF
pl
plynet
++=
=−−−= 0,
m 00.22
000,
<=⎟⎠⎞⎜
⎝⎛=
==++−=
+++=
Mmd
Mmx
MgxmgdMgxmgd
nplfdznet τττττ
0
0 0
,,
,,
,,
=∑=
=∑==∑=
zextznet
yextynet
xextxnet
FFFF
ττ
A) Find the magnitude of the upward force n exerted by the support on the board.B) Find where the father should sit to balance the system at rest.
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Axis of Rotationq The net torque is about an axis through any
point in the xy planeq Does it matter which axis you choose for
calculating torques?q NO. The choice of an axis is arbitraryq If an object is in translational equilibrium and
the net torque is zero about one axis, then the net torque must be zero about any other axis
q We should be smart to choose a rotation axis to simplify problems
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Mmd
Mmx
MgxmgdMgxmgd
nplfdznet
2
000,
=⎟⎠⎞⎜
⎝⎛=
==++−=
+++= τττττ
0
0 0
,,
,,
,,
=∑=
=∑==∑=
zextznet
yextynet
xextxnet
FFFF
ττ
B) Find where the father should sit to balance the system at rest.
Mmd
Mmx
MgxmgddgmmgMggdmMgxMgd
ndgdmxdMg
plpl
pl
nplfdznet
2
0)(
0)(0,
=⎟⎠⎞⎜
⎝⎛=
=
=+++−−−
=+−+−=
+++= τττττ
OP
Rotation axis O Rotation axis P
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Center of Gravityq The torque due to the gravitational force on an
object of mass M is the force Mg acting at the center of gravity of the object
q If g is uniform over the object, then the center of gravity of the object coincides with its center of mass
q If the object is homogeneous and symmetrical, the center of gravity coincides with its geometric center
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Where is the Center of Mass ?
q Assume m1 = 1 kg, m2 = 3 kg, and x1 = 1 m, x2 = 5 m, where is the center of mass of these two objects?A) xCM = 1 mB) xCM = 2 mC) xCM = 3 mD) xCM = 4 mE) xCM = 5 m
21
2211
mmxmxmxCM +
+=
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Center of Mass (CM)q An object can be divided into
many small particlesn Each particle will have a
specific mass and specific coordinates
q The x coordinate of the center of mass will be
q Similar expressions can be found for the y coordinates
i ii
CMi
i
m xx
m=∑∑
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Center of Gravity (CG)q All the various gravitational forces acting on all the
various mass elements are equivalent to a single gravitational force acting through a single point called the center of gravity (CG)
q If
q then
!
!
+++=+++=
333222111
321 )(xgmxgmxgm
xgmmmxMg CGCGCGCG
!=== 321 ggg
i
iiCG m
xmmmmxmxmxmx
∑∑=
++++++=!!
321
332211
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CG of a Ladderq A uniform ladder of
length l rests against a smooth, vertical wall. When you calculate the torque due to the gravitational force, you have to find center of gravity of the ladder. The center of gravity should be located at
C
A
B
D
E
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Ladder Exampleq A uniform ladder of length
l rests against a smooth, vertical wall. The mass of the ladder is m, and the coefficient of static friction between the ladder and the ground is µs = 0.40. Find the minimum angle θat which the ladder does not slip.
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Problem-Solving Strategy 1q Draw sketch, decide what is in or out the systemq Draw a free body diagram (FBD)q Show and label all external forces acting on the objectq Indicate the locations of all the forcesq Establish a convenient coordinate systemq Find the components of the forces along the two axesq Apply the first condition for equilibrium q Be careful of signs
0 0
,,
,,
=∑==∑=
yextynet
xextxnet
FFFF
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Which free-body diagram is correct?
q A uniform ladder of length l rests against a smooth, vertical wall. The mass of the ladder is m, and the coefficient of static friction between the ladder and the ground is µs = 0.40. gravity: blue, friction: orange, normal: green
A B C D
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q A uniform ladder of length l rests against a smooth, vertical wall. The mass of the ladder is m, and the coefficient of static friction between the ladder and the ground is µs = 0.40. Find the minimum angle θ at which the ladder does not slip.
mgnfPmgnfP
mgnFPfF
ssx
x
y
xx
µµ =====
=−=∑=−=∑
max,
00
mg
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Problem-Solving Strategy 2q Choose a convenient axis for calculating the net torque
on the objectn Remember the choice of the axis is arbitrary
q Choose an origin that simplifies the calculations as much as possiblen A force that acts along a line passing through the origin
produces a zero torqueq Be careful of sign with respect to rotational axis
n positive if force tends to rotate object in CCWn negative if force tends to rotate object in CWn zero if force is on the rotational axis
q Apply the second condition for equilibrium 0,, =∑= zextznet ττ
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Choose an origin O that simplifies the calculations as much as possible ?
q A uniform ladder of length l rests against a smooth, vertical wall. The mass of the ladder is m, and the coefficient of static friction between the ladder and the ground is µs = 0.40. Find the minimum angle.
mg mg mg mg
O
O
O
O
A) B) C) D)
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q A uniform ladder of length l rests against a smooth, vertical wall. The mass of the ladder is m, and the coefficient of static friction between the ladder and the ground is µs = 0.40. Find the minimum angle θ at which the ladder does not slip.
!51])4.0(2
1[tan)21(tan
21
22tan
cossin
0cos2
sin00
11min
minmin
min
minmin
===
====
=−++=
+++=∑
−−
s
ss
PgfnO
mgmg
Pmg
lmgPl
µθ
µµθ
θθ
θθ
τττττ
mg
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Problem-Solving Strategy 3q The two conditions of equilibrium will give a system of
equationsq Solve the equations simultaneouslyq Make sure your results are consistent with your free
body diagramq If the solution gives a negative for a force, it is in the
opposite direction to what you drew in the free body diagram
q Check your results to confirm
0
0 0
,,
,,
,,
=∑=
=∑==∑=
zextznet
yextynet
xextxnet
FFFF
ττ
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Horizontal Beam Exampleq A uniform horizontal beam with a
length of l = 8.00 m and a weight of Wb = 200 N is attached to a wall by a pin connection. Its far end is supported by a cable that makes an angle of φ = 53° with the beam. A person of weight Wp = 600 N stands a distance d = 2.00 m from the wall. Find the tension in the cable as well as the magnitude and direction of the force exerted by the wall on the beam.
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Horizontal Beam Exampleq The beam is uniform
n So the center of gravity is at the geometric center of the beam
q The person is standing on the beam
q What are the tension in the cable and the force exerted by the wall on the beam?
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Horizontal Beam Example, 2q Analyze
n Draw a free body diagram
n Use the pivot in the problem (at the wall) as the pivotn This will generally be
easiestn Note there are three
unknowns (T, R, θ)
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q The forces can be resolved into components in the free body diagram
q Apply the two conditions of equilibrium to obtain three equations
q Solve for the unknowns
Horizontal Beam Example, 3
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Horizontal Beam Example, 3
0sinsin0coscos
=−−+=∑=−=∑
bpy
x
WWTRFTRF
φθφθ
Nm
mNmNl
lWdWT
lWdWlT
bp
bpz
31353sin)8(
)4)(200()2)(600(sin
)2(
0)2())(sin(
=+=+
=
=−−=∑
!φ
φτ
NNTR
TTWWT
TWWRR
bp
bp
5817.71cos53cos)313(
coscos
7.71sin
sintan
sinsin
tancossin
1
===
=⎟⎟⎠
⎞⎜⎜⎝
⎛ −+=
−+==
−
!
!
!
θφ
φφ
θ
φφ
θθθ
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Rotational Kinetic Energyq An object rotating about z axis with an angular
speed, ω, has rotational kinetic energy q The total rotational kinetic energy of the rigid
object is the sum of the energies of all its particles
q Where I is called the moment of inertiaq Unit of rotational kinetic energy is Joule (J)
2 2
2 2 2
12
1 12 2
R i i ii i
R i ii
K K m r
K m r I
ω
ω ω
= =
⎛ ⎞= =⎜ ⎟⎝ ⎠
∑ ∑
∑
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Work-Energy Theorem for pure Translational motion
q The work-energy theorem tells us
q Kinetic energy is for point mass only, ignoring rotation.
q Work
q Power
∫∫→→
⋅== sdFdWWnet
22
21
21
ififnet mvmvKEKEKEW −=−=Δ=
→→→
→⋅=⋅== vF
dtsdF
dtdWP
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Mechanical Energy Conservation
q Energy conservation
q When Wnc = 0,
q The total mechanical energy is conserved and remains the same at all times
q Remember, this is for conservative forces, no dissipative forces such as friction can be present
f f i iK U U K+ = +
ffii mgymvmgymv +=+ 22
21
21
ncW K U= Δ +Δ
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Total Energy of a Systemq A ball is rolling down a rampq Described by three types of energy
n Gravitational potential energy
n Translational kinetic energy
n Rotational kinetic energy
q Total energy of a system 2 21 12 2CME Mv Mgh Iω= + +
212rK Iω=
U Mgh=21
2t CMK Mv=
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Work done by a pure rotation
q Apply force F to mass at point r, causing rotation-only about axis
q Find the work done by F applied to the object at P as it rotates through an infinitesimal distance ds
q Only transverse component of F does work – the same component that contributes to torque θτddW =
cos(90 )sin sin
dW F d s F dsF ds Fr d
ϕϕ ϕ θ
→ →= ⋅ = −
= =
o
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Work-Kinetic Theorem pure rotation
q As object rotates from θi to θf , work done by the torque
q I is constant for rigid object
q Powerτωθτ ===
dtd
dtdWP
∫∫∫∫∫ =====f
i
f
i
f
i
f
i
f
i
dIddtdIdIddWW
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
ωωθωθαθτ
22
21
21
if IIdIdIWf
i
f
i
ωωωωωωθ
θ
θ
θ
−=== ∫∫
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q Ex: An motor attached to a grindstone exerts a constant torque of 10 N-m. The moment of inertia of the grindstone is I = 2 kg-m2. The system starts from rest.n Find the kinetic energy after 8 s
n Find the work done by the motor during this time
n Find the average power delivered by the motor
n Find the instantaneous power at t = 8 s
1600 200 watts8avg
dWPdt
= = =
2 21 1600 40 rad/s 5 rad/s2f f f iK I J t
Iτω ω ω α α= = ⇐ = + = ⇐ = =
JdW if
f
i
160016010)( =×=−== ∫ θθτθτθ
θ21( ) 160 rad
2f i it tθ θ ω α− = + = 2 21 1 1600 J2 2f i f iW K K I Iω ω= − = − =
10 40 400 wattsP τω= = × =
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Work-Energy Theoremq For pure translation
q For pure rotation
q Rolling: pure rotation + pure translation
2 2, ,
1 12 2net cm cm f cm i f iW K K K mv mv= Δ = − = −
2 2, ,
1 12 2net rot rot f rot i f iW K K K I Iω ω= Δ = − = −
, , , ,
2 2 2 2
( ) ( )
1 1 1 12 2 2 2
net total rot f cm f rot i cm i
f f i i
W K K K K K
I mv I mvω ω
= Δ = + − +
⎛ ⎞ ⎛ ⎞= + − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
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Energy Conservationq Energy conservation
q When Wnc = 0,
q The total mechanical energy is conserved and remains the same at all times
q Remember, this is for conservative forces, no dissipative forces such as friction can be present
, , , ,rot f cm f f rot i cm i iK K U K K U+ + = + +
fffiii mgymvImgymvI ++=++ 2222
21
21
21
21 ωω
nc totalW K U= Δ +Δ
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Total Energy of a Rolling System
q A ball is rolling down a rampq Described by three types of energy
n Gravitational potential energy
n Translational kinetic energy
n Rotational kinetic energy
q Total energy of a system 22
21
21 ωIMghMvE ++=
212rK Iω=
U Mgh=
212tK Mv=
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A Ball Rolling Down an Incline
q A ball of mass M and radius R starts from rest at a height of h and rolls down a 30° slope, what is the linear speed of the ball when it leaves the incline? Assume that the ball rolls without slipping.
22
210
2100 ff IMvMgh ω++=++
2222
21
21
21
21
fffiii ImgymvImgymv ωω ++=++
222
222
51
21
52
21
21
fff
f MvMvRv
MRMvMgh +=+=
2
52MRI =
Rv f
f =ω
2/1)710( ghv f =
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Rotational Work and Energyq A ball rolls without slipping down incline A,
starting from rest. At the same time, a box starts from rest and slides down incline B, which is identical to incline A except that it is frictionless. Which arrives at the bottom first?
q Ball rolling:
q Box sliding
2222
21
21
21
21
fffiii ImgymvImgymv ωω ++=++
2 21 12 2f fmgh mv Iω= +
ffii mgymvmgymv +=+ 22
21
21
21sliding: 2 fmgh mv=
2 2 2 21 1 2 7( / )2 2 5 10f f fmv mR v R mv⎛ ⎞= + =⎜ ⎟⎝ ⎠
27rolling: 10 fmgh mv=
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Blocks and Pulleyq Two blocks having different masses m1 and
m2 are connected by a string passing over a pulley. The pulley has a radius R and moment of inertia I about its axis of rotation. The string does not slip on the pulley, and the system is released from rest.
q Find the translational speeds of the blocks after block 2 descends through a distance h.
q Find the angular speed of the pulley at that time.
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q Find the translational speeds of the blocks after block 2 descends through a distance h.
q Find the angular speed of the pulley at that time.
ghmghmvRImm f 12
2221 )(
21 −=++
000)()21
21
21( 21
222
21 ++=−+++ ghmghmIvmvm fff ω
, , , ,rot f cm f f rot i cm i iK K U K K U+ + = + +
2/1
221
12
/)(2
⎥⎦
⎤⎢⎣
⎡++
−=RImmghmmvf
2/1
221
12
/)(21
⎥⎦
⎤⎢⎣
⎡++
−==RImmghmm
RRvf
fω
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Angular Momentumq Same basic techniques that were used in linear
motion can be applied to rotational motion.n F becomes τn m becomes In a becomes αn v becomes ω n x becomes θ
q Linear momentum defined asq What if mass of center of object is not moving,
but it is rotating?q Angular momentum
m=p v
I=L ω
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Angular Momentum Iq Angular momentum of a rotating rigid object
n L has the same direction as ω *
n L is positive when object rotates in CCWn L is negative when object rotates in CW
q Angular momentum SI unit: kg-m2/sCalculate L of a 10 kg disk when ω = 320 rad/s, R = 9 cm = 0.09 mL = Iω and I = MR2/2 for diskL = 1/2MR2ω = ½(10)(0.09)2(320) = 12.96 kgm2/s
*When rotation is about a principal axis
I=L ω ω!
L!
February 13, 2017
Angular Momentum IIq Angular momentum of a particle
q Angular momentum of a particle
n r is the particle’s instantaneous position vectorn p is its instantaneous linear momentumn Only tangential momentum component contributen Mentally place r and p tail to tail form a plane, L is
perpendicular to this plane
( )m= =L r×p r×v
φφωω sinsin2 rpmvrrmvmrIL ===== ⊥
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Angular Momentum of a Particle in Uniform Circular Motion
q The angular momentum vector points out of the diagram
q The magnitude is L = rp sinθ = mvr sin(90o) = mvr
q A particle in uniform circular motion has a constant angular momentum about an axis through the center of its path
O
Example: A particle moves in the xy plane in a circular path of radius r. Find the magnitude and direction of its angular momentum relative to an axis through O when its velocity is v.
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Angular Momentum and Torque
q Rotational motion: apply torque to a rigid bodyq The torque causes the angular momentum to changeq The net torque acting on a body is the time rate of
change of its angular momentum
q and are to be measured about the same originq The origin must not be accelerating (must be an
inertial frame)
netddt
= Σ = pF F netddt
= Σ = Lτ τ
Στ L
February 13, 2017
a!
a!
Ex4: A Non-isolated SystemA sphere mass m1 and a block of mass m2 are connected by a light cord that passes over a pulley. The radius of the pulley is R, and the mass of the thin rim is M. The spokes of the pulley have negligible mass. The block slides on a frictionless, horizontal surface. Find an expression for the linear acceleration of the two objects. gRmext 1=∑τ
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Masses are connected by a light cord. Find the linear acceleration a. • Use angular momentum approach• No friction between m2 and table• Treat block, pulley and sphere as a non-isolated system rotating about pulley axis. As sphere falls, pulley rotates, block slides• Constraints:
for pulley
Equal 's and 's for block and sphere/
v av ωR α d dt a αR dv / dt
ω= == =
• Ignore internal forces, consider external forces only• Net external torque on system:
• Angular momentum of system:(not constant)
ωMRvRmvRmIωvRmvRmLsys2
2121 ++=++=
gRmτMR)aRmR(mαMRaRmaR mdt
dLnet
sys121
221 ==++=++=
1 about center of wheelnet m gRτ =
21
1 mmM
gma++
=∴ same result followed from earlier method using 3 FBD’s & 2nd law
I
a!
a!
P1:
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P2:
P3:
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P4:
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P5:
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P7:
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P8:
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P9: