L12 Equilibrium PHYS101 - UNIVERSE OF ALI OVGUN€¦ · EMU Physics Department . 2/13/17 Static Equilibrium q Equilibrium and static equilibrium ... Conditions for Equilibrium qThe

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

2/13/17

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.

2/13/17

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

February 13, 2017

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

= =

⎛ ⎞= =⎜ ⎟⎝ ⎠

∑ ∑

∑

February 13, 2017

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

February 13, 2017

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

February 13, 2017

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=

February 13, 2017

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

February 13, 2017

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

−=== ∫∫

February 13, 2017

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= = × =

February 13, 2017

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 = + − +

⎛ ⎞ ⎛ ⎞= + − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

February 13, 2017

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

February 13, 2017

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=

February 13, 2017

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 =

February 13, 2017

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=

February 13, 2017

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.

February 13, 2017

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

February 13, 2017

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

February 13, 2017

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 ===== ⊥

February 13, 2017

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.

February 13, 2017

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

February 13, 2017

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: