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Kinetics of Rigid Bodies inThree Dimensions



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Introduction

• The fundamental relations developed for

the plane motion of rigid bodies may also

be applied to the general motion of three

dimensional bodies.

I H G =• The relation which was used

to determine the angular momentum of a

rigid slab is not valid for general three

GG

H M

amF

=

=

∑

∑

dimensiona bodies and motion.

• The current chapter is concerned with

evaluation of the angular momentum and

its rate of change for three dimensional

motion and application to effectiveforces, the impulse-momentum and the

work-energy principles.



• Angular momentum of a body about its mass center,

( ) ( )[ ]∑∑==

′××′=×′=n

iiii

n

iiiiG mr r mvr H

11

∆

ω ∆

• The x component of the angular momentum,

( ) ( )[ ]∑=

′×−′×=

n

n

ii yii zii x mr zr y H

1

∆ω ω

Rigid Body Angular Momentum in Three Dimensions

( ) ∑∑∑===

=

−−+=

−−−=

n

iiii z

n

iiii y

n

iiii x

i ii xi zii yi xi

m x zm y xm z y

m z x z x y y

111

22

1

∆∆∆ ω ω ω

ω

dm zxdm xydm z y H z y x x ∫ ∫ ∫ −−+= ω ω ω 22

z xz y xy x x I I I −−+=

z z y zy x zx z

z yz y y x yx y

I I I H

I I I H

ω ω ω

ω

+−−=

−+−=



• Transformation of into is characterized

by the inertia tensor for the body,

G H

+−−

−+−

−−+

z zy zx

yz y yx

xz xy x

I I I

I I I

I I I

• With respect to the principal axes of inertia,

′


z z y zy x zx z

z yz y y x yx y

z xz y xy x x x

I I I H

I I I H

I I I H

ω ω ω

ω ω ω

ω

+−−=

−+−=

−−+=

′

′

z

y

I

I

00

00

z z z y y y x x x I H I H I H ′′′′′′′′′ ===

• The angular momentum of a rigid body

and its angular velocity have the same

direction if, and only if, is directed along a

principal axis of inertia.

G H



• The momenta of the particles of a rigid body can

be reduced to:

vm

L

=

= momentumlinear

G H G aboutmomentumangular=

−−=


z z y zy x zx z

z yz y y x yx y

z z y xy x x

I I I H

I I I H

ω ω ω

ω ω ω

+−−=

−+−=

• The angular momentum about any other given

point O is

GO H vmr H

+×=



• The angular momentum of a body constrained to

rotate about a fixed point may be calculated from

GO H vmr H

+×=

• Or, the angular momentum may be computed

directly from the moments and products of inertia

with respect to the Oxyz frame.


z z y zy x zx z

z yz y y x yx y

z xz y xy x x x

I I I H

I I I H

I I I H

ω ω ω

ω ω ω

ω ω ω

+−−=

−+−=

−−+=

( )

( )[ ]∑

∑

=

=

××=

×=

n

iiii

iiiO

mr r

mvr H

1

1

∆

∆

ω



Principle of Impulse and Momentum

• The principle of impulse and momentum can be applied directly to thethree-dimensional motion of a rigid body,

Syst Momenta1 + Syst Ext Imp1-2 = Syst Momenta2

• The free-body diagram equation is used to develop component and

moment equations.

• For bodies rotating about a fixed point, eliminate the impulse of the

reactions at O by writing equation for moments of momenta and

impulses about O.



• Kinetic energy of particles forming rigid body,

2(

∆

∆

222

212

21

1

2

212

21

1

2

212

21

y x xy z z y y x x

n

iii

n

iii

I I I I vm

mr vm

vmvmT

ω ω ω ω ω

ω

−+++=

′×+=

′+=

∑

∑

=

=

Kinetic Energy

x z zx z y yz −−

• If the axes correspond instantaneously with the

principle axes,

)(222

212

21

z z y y x x I I I vmT ′′′′′′ +++= ω ω ω

• With these results, the principles of work and

energy and conservation of energy may be applied

to the three-dimensional motion of a rigid body.



• Kinetic energy of a rigid body with a fixed point,

)22

2( 22221

x z zx z y yz

y x xy z z y y x x

I I

I I I I T

ω ω ω ω

ω ω ω ω ω

−−

−++=

Kinetic Energy

• e axes xyz correspon ns an aneous y w

the principle axes Ox’y’z’,

)(222

21

z z y y x x I I I T ′′′′′′ ++= ω ω ω



SOLUTION:

• Apply the principle of impulse and

momentum. Since the initial momenta

is zero, the system of impulses must be

equivalent to the final system of

momenta.

• Assume that the supporting cables

Problem: 1

Rectangular plate of mass m that is

suspended from two wires is hit at D in

a direction perpendicular to the plate.

Immediately after the impact,

determine a) the velocity of the mass

center G, and b) the angular velocity of

the plate.

and the rotation about an axis normal tothe plate is zero.

• Principle of impulse and momentum

yields to two equations for linear

momentum and two equations forangular momentum.

• Solve for the two horizontal components

of the linear and angular velocity

vectors.



SOLUTION:

Problem: 1

• Apply the principle of impulse and momentum. Since the initial momenta is zero,

the system of impulses must be equivalent to the final system of momenta.

• Assume that the supporting cables remain taut such that the vertical velocity and the

rotation about an axis normal to the plate is zero.

k vivv z x

+= ji y x

+=

Since the x, y, and z axes are principal axes of inertia,

jmaimb j I i I H y x y y x xG

ω ω ω ω 2

1212

121 +=+=



Problem: 1

• r nc p e o mpu se an momen um y e s wo equa ons or near momen um an

two equations for angular momentum.

• Solve for the two horizontal components of the linear and angular velocity vectors.

xmv=0

0=

xv

zvmt F =− ∆

mt F v z∆−=

( )k mt F v

∆−=

x

x

mb

H t bF

ω 2121

21

∆

=

=

mbt F x ∆6=

y

y

ma

H t aF

ω 2121

21

∆

=

=−

( )mat F y ∆6−=

( ) jbia

mab

t F +=

∆6ω



Problem: 1

( )k mt F v

∆−

( ) jbiamab

t F +=

∆6ω

jmaimb H y xG

ω ω 2

1212

121 +=



A homogeneous disk of mass m is

SOLUTION:

• The disk rotates about the vertical axis

through O as well as about OG.

Combine the rotation components forthe angular velocity of the disk.

• Compute the angular momentum of the

Problem: 2

moun e on an ax e o neg g e

mass. The disk rotates counter-

clockwise at the rate ω 1 about OG.

Determine: a) the angular velocity of

the disk, b) its angular momentum about

O, c) its kinetic energy, and d) the

vector and couple at G equivalent to the

momenta of the particles of the disk.

noting that O is a fixed point.

• The kinetic energy is computed from the

angular velocity and moments of inertia.

• The vector and couple at G are also

computed from the angular velocity and

moments of inertia.



SOLUTION:

• The disk rotates about the vertical axis through O as well

as about OG. Combine the rotation components for the

angular velocity of the disk.

ji

21 ω ω ω +=

Noting that the velocity at C is zero,

==

Problem: 2

( ) ( )( )

Lr

k r L

jr i L ji

12

12

210

ω ω

ω ω

ω ω

=

−=

−×+=

( ) j Lr i

11−=



• Compute the angular momentum of the disk using

principle axes of inertia and noting that O is a fixed point.

−

k I j I i I H z z y y x xO

ω ω ω ++=

( )( )

( ) 00212

12

412

1221

=+==

−+==

==

mr mL I H

Lr mr mL I H

mr I H

z z z

y y y

x x x

ω

ω ω

ω ω

Problem: 2

11

( ) j Lr r Lmimr H O

1241212

21 ω ω +−=

• The kinetic energy is computed from the angular velocity

and moments of inertia.

( )( )[ ]21

2

4122

12

21

22221

Lr r Lmmr

I I I T z z y y x x

ω ω

ω ω ω

−++=

++=

212

22

81 6 ω

+=

L

r mr T



( ) j Lr i

11 −=

• The vector and couple at G are also computed from the

angular velocity and moments of inertia.

k mr vm

1ω =

( ) j Lr mr imr

k I j I i I H z z y y x xG

ω ω

ω ω ω

−+=

++= ′′′

2

41

12

21

r

Problem: 2

−= j

L

imr H G2

12



The electric motor with an attached disk is running at a constant low speed of 120 rev/min

. .

assembly is next set is rotation about the vertical Z-axis at the constant rate N = 60rev/min with a fixed angle γ of 300. Determine (a) the angular velocity and angular

acceleration of the disk, (b) the velocity and acceleration of point A at the top of the disk

for the instant shwon.



A thin homogeneous square plate of mass ‘m’ and side ‘a’ is

welded to a vertical shaft AB with which it forms s an angle

of 45°. Knowing that the shaft rotates with a constant angularvelocity ω, determine the angular momentum HA of the plate

about point A.

Y’

Y

Y’Z’ 45°

Y’Z’ as principal axes

ω ’ = ω cos 45; ω ’ = ω sin 45

R. Ganesh Narayanan 25

A X’

Z

ωx’ = 0



To describe the motion of an aircraft, a set of axes Gxyz, fixed to the

aircraft is defined. In level flight Gx is forward, Gz is downward and Gxz is

the vertical plane of symmetry; G is the centre of gravity of the mass. The

moments of inertia of the aircraft about these axes are:

Kgm 400= Kgm 12000= Kgm 8000= Kgm2

xz

2

zz2

yy

2

I I I 4000= I xx

The aircraft flies at 400m/s toward north; it then follows a curve of 12Km

30° during this change of course, find the resulting inertia torques.

Answer: -1.92 Nm -0.333 Nm -0.192Nm







z (k )P

3

Angular Momentum (Moment of Momentum) of a rigid body

The relative angular momentum of a rigid body:

Assume a body to be rotating with angular velocity ω and consider a particle P of

mass m p positioned at r po relative to some point O so that:

po= xi + y jr + zk

and

Oy ( j )

x (i )

x

y

1ω 2ω

po

1 2 3= i + j + k ω ω ω ω

where i , j and k refer to arbitrarily chosenaxes Oxyz .

O fixed and Oxyz axes could be fixed and29



( ) ( ) (from (2)) p P po po poo po

= m = mh vr r r ω ∑ ∧ ∑ ∧ ∧

If we express h o in the form h = h 1 i + h 2 j + h 3 k , we find that:

( ) [ ] [ ]

[ ] ( ) [ ]

1 1 2 3

2 1 2 3

2 2P P P

2 2P P P

22

h = m + - m xy - m xz y z

h = - m yx + m + - m yz x z

ω ω ω

ω ω ω

∑ ∑ ∑

∑ ∑ ∑

3 1 2 3P P P- - x

are the Moments of inertia of the body about Oxyz and

( )2 2 xx P I m + y z= ∑ ( )2 2

yy P I m + x z= ∑ ( )22

zz P I m + y x=

∑

xy P I m xy= ∑ yz P

I m yz= ∑ zx P I m zx= ∑

are the products of inertia referred to Oxyz . 30



1 1 2 3

2 1 2 3

xx xy zx

yx yy yz

h = I - I - I

h = - I + I - I

h = - I - I + I

ω ω ω

ω ω ω

ω ω ω

(5)

zx yz yy

relative angular momentum of body about O relative to i , j , k

31



In general, if the reference axes Oxyz rotate w.r.t. the body (or vice versa) then the

products and moments of inertia will vary with time. By considering the axes to

rotate with the body the moments and products of inertia will be constant. Further,

if we choose Oxyz as principal axes of inertia then, by the definition of principal

axes, the products of inertia will be zero and the expression (5) for the relativeangular momentum simplifies to:

1 2 3 xx yy zzo= I i + I j + I k h ω ω ω (6)

Relative angular momentum

32

E i f M i



Equations of MotionConsider a system of n particles of which p is typical. r po defines the position of p

relative to O and v po the velocity of p relative to O . Oxyz rotates with angular

velocity and O is moving. The angular momentum of the system relative to O is:

z

z (k ) P

3Ω

povO does not necessarily coincide with thecentre of mass

Oy ( j )

x (i )

x

y

1Ω 2Ω

poOxyz rotating axes with angular

velocity

( ) ( by equation (2))n

p poo po

p=1

=h vr m ∧

∑ (7)

33



The forces acting on the particle may be separated into two groups: (i) due to

influences received from outside the system, called external forces F p

and (ii) due

to interaction between particles of the system, called internal forces F p .

From Newton's second law we may write, for a single particle,

( ) p p p p

d + = m vF F ′ (8)

where v p is the absolute velocity of p .

Equation (8) describes the linear motion of particle p . The corresponding equation

for the rotational motion of p about O follows from (8) by considering the moment

about O of the forces acting on the particle. Thus,

( ) p po p po p po p

d + = vr F r F r

mdt ′

∧ ∧ ∧34



Now, by Newton's third law is zero since it represents the

summation of the moment of equal and opposite pairs of reaction forces. Also

represents the sum of the moments about O of all the external forces. This is

the total externally applied moment, which we will denote by T o .

po p r F ′

∧∑ p po F r ∧∑

which, summing over all the particles becomes

( ) p p po po p po pd F + = vr r F r mdt

′ ∧ ∧ ∧ ∑ ∑ ∑

35

Thus

( ) po po p

d = vT r m

dt

∧

∑ (9)

Returning now to equation (7) we may write

( )o p po po

d h d = vr m

dt dt ∧ ∑



It can be shown that

( ) ( )o p p po po p o

d h d d = -v vr r m mdt dt dt

∧ ∧ ∑ ∑ (10)

From equation (9) it can be seen that the first term on the right hand side of

equation (9) is equal to T o . Noting also that dv o /dt = a o , the absolute acceleration

of O , equation (10) may be expressed as:

36

( )o

po po o

d h= - m aT r

dt ∧∑

where G is the centre of mass of the system of particles.

o Go o= - M aT r ∧



[ ])v m(r dt

d =

dt

hd po p po

o ∧∑

∧∑+

∧∑

dt

)v md( r )v m(

dt

r d =

dt

hd po p

po po p

poo

vdt

r d po

po =Thus

r d

po

dt

po p

and vvv o p po−=

∧∑

∧∑ )v m(

dt

d r -)vm(

dt

d r =

dt

hd o p po p p po

o

37



Thus,

( ) oo Go o

d h- M =aT r

dt ∧ (11)

In the special cases where either a o = 0 or O and G coincide (r Go = 0) equation(11) takes the form:

h+t

=dt

=T o

oo

o

∧Ω

∂

(12)

38



Scalar versions of the Equations of Motion

Let h o = h 1 i + h 2 j + h 3 k be defined in components along references which are rotating withangular velocity = 1 i + 2 j + 3 k .

It can be shown that (see handout) the equations of motion given by equation (12)(i.e. the special case when either a o = 0 or r Go = 0 ) may be written as:

1 1 2 3 3 2T h h h= − Ω + Ω

−

132 2 3 1 1 3

3 3 1 2 2 1T h h h= − Ω + Ω

where T 1, T 2, T 3 are the components of externally applied moment (T o ) along i , j andk respecively.

Equations (13) are referred to as the Momentum equations and have almostuniversal application in the study of gyroscopic phenomena and motion withrespect to moving axes.

39



In general, the components (h 1, h 2 , h 3 ) of angular momentum for a rigid body,

rotating with angular velocityω

, may be calculated from equations (5) and (6),noting that the body angular velocity must be specified in components along the

moving reference axes which, in general, are not fixed in the moving body. In

other words, and ω are not, in general, equal.

If however, in a special case the reference axes are fixed in the body alongprincipal axes, then = ω and if the principal moment of inertia along reference

axes Oxyz are A, B, C respectively, then equations (6) and (13) may be combined

to give:

( )

( )

( )

1 1 2 3

2 2 3 1

3 3 1 2

= A - B - C T

= B - C - AT

= C - A- BT

ω ω ω ω ω ω ω ω ω ω ω ω



(14)

Equations (14) are called the Euler dynamical Equations.

40



• Angular momentum and its rate of change are

taken with respect to centroidal axes GX’Y’Z’ of

fixed orientation.

• Convenient to use body fixed axes Gxyz where

moments and products of inertia are not time

• Transformation of into is independent of

the system of coordinate axes.

G H

Motion of Rigid Bodies in Three Dimensions

G H M

amF

=

=

∑

∑

dependent.

• Define rate of change of change of with

respect to the rotating frame,G H

k H j H i H H z y xGxyzG

++=

Then,

ω Ω Ω

=×+= GGxyzGG H H H



GGxyzGG H H M

×+=∑ Ω

• With and Gxyz chosen to correspond

to the principal axes of inertia,

ω Ω =

z y z y x x x I I I M

−−=

−−=∑

Euler’s Equations:

Euler’s Equation Motion and d’Alembert’s Principle

( ) y x y x z z z

x z z y y y

I I I M ω ω ω −−=∑

• System of external forces and effective forces

are equivalent for general three dimensional

motion.

• System of external forces are equivalent to

the vector and couple, .and G H am

M i Ab Fi d P i Fi d A i



• For a rigid body rotation around a fixed point,

( ) OOxyzO

OO

H H

H M

×+=

=∑

Ω

• For a rigid body rotation around a fixed axis,

z z yz y xz x I H I H I H −=−=−=

Motion About a Fixed Point or a Fixed Axis

( )( )

( ) ( ) 2ω α

ω ω

ω

i I j I k I j I i I

k I j I i I k

k I j I i I

yz xz z yz xz

z yz xz

z yz xz

OOxyzOO

+−++−−=

+−−×+

+−−=

α

ω α

ω α

z z

xz yz y

yz xz x

I M

I I M

I I M

=

+−=

+−=

∑

∑

∑2

2

M ti Ab t Fi d A i



α

ω α

ω α

z z

xz yz y

yz xz x

I M

I I M

I I M

=

+−=

+−=

∑

∑

∑2

2

• For a rigid body rotation around a fixed axis,

• If symmetrical with respect to the xy plane,

Motion About a Fixed Axis

α z z y x ===

• If not symmetrical, the sum of external moments

will not be zero, even if α = 0,

022

=== ∑∑∑ z xz y yz x M I M I M ω ω

• A rotating shaft requires both static and

dynamic balancing to avoid excessive

vibration and bearing reactions.

( )0=

( )0≠

P bl 3



• Expressing that the system of external

forces is equivalent to the system of

SOLUTION:

• Evaluate the system of effective forces

by reducing them to a vector

attached at G and couple

am

.G H

Problem: 3

Rod AB with weight W = 20 kg is

pinned at A to a vertical axle which

rotates with constant angular velocity

ω = 15 rad/s. The rod position is

maintained by a horizontal wire BC .

Determine the tension in the wire and

the reaction at A.

e ect ve orces, wr te vector express ons

for the sum of moments about A and thesummation of forces.

• Solve for the wire tension and the

reactions at A.

P bl 3



SOLUTION:• Evaluate the system of effective forces by reducing them

to a vector attached at G and coupleam

.G H

( )

( ) I

I L I r aa n

2

2

212

sm5.112

cos

−=

−=−== ω β ω

( ) ( ) I am

N22505.11220 −=−=

Problem: 3

k I j I i I H z z y y x xG ++=

0sincos

0 2

212

21

==−=

===

z y x

z y x mL I I mL I

ω β ω ω β ω ω

imL H G

β ω cos2

121−=

( ) ( )

( )k k mL

imL ji

H H H GGxyzGG

mN5.649cossin

cossincos0

22

121

2

121

⋅==

×+−+=

×+=

β β ω

β ω β ω β ω

ω

Problem: 3



• Expressing that the system of external forces is equivalentto the system of effective forces, write vector expressions

for the sum of moments about A and the summation of

forces.

( )eff A A M M ∑∑ =

K I J J I I T J

5.6492250886.02.1961732.1

−

+−×=−×+−×

Problem: 3

...

N1613=T

( )eff

F F ∑∑ =

I J I K A J A I A Z Y X

22502.1961613 −=−−++

( ) ( ) J I A

N96.21N976 +−=



• A gyroscope consists of a rotor with its mass center

fixed in space but which can spin freely about its

geometric axis and assume any orientation.

• From a reference position with gimbals and a

reference diameter of the rotor aligned, thegyroscope may be brought to any orientation

through a succession of three steps:

’a ro a on o ou er g m a roug a ou ,

b) rotation of inner gimbal through q about

c) rotation of the rotor through y about CC’.

• ϕ , θ , and ψ are called the Eulerian Angles and

spinof rate

nutationof rate

precessionof rate

=

=

=

Ψ

θ

φ



• The angular velocity of the gyroscope,

k jK

Ψ θ φ ω ++=

( )k ji

jiK

θ φ Ψ θ θ φ ω

θ θ

cossin

cossinwith

+++−=

+−=

• Equation of motion,

OOxyzOO =

( )

jK

k I j I i I H O

θ φ Ω

θ φ Ψ θ θ φ

+=

++′+′−= cossin

( ) ( )

( ) ( )

( )θ φ Ψ

θ φ Ψ θ φ θ θ φ θ

θ φ Ψ θ θ φ θ θ φ

cos

cossincossin

coscos2sin

2

+=

++−′=

+++′−=

∑

∑

∑

dt

d I M

I I M

I I M

z

y

x



Steady precession,

constantare,, ψ φ θ

k i

k I i I H

k i

zO

z

θ φ θ φ Ω

ω θ φ

ω θ φ ω

cossin

sin

sin

+−=

+′−=

+−=

( ) j I I

H M

z

OO

θ φ θ φ ω

Ω

sincos′−=

×=∑

Couple is applied about an axisperpendicular to the precession

and spin axes

When the precession and spin

axis are at a right angle,

j I M O

φ Ψ

θ

=

°=

∑

90

Gyroscope will precess about an

axis perpendicular to both the

spin axis and couple axis.



• Consider motion about its mass center of anaxisymmetrical body under no force but its own

weight, e.g., projectiles, satellites, and space craft.

constant0 == GG H H

• Define the Z axis to be aligned with and z in a

rotating axes system along the axis of symmetry.

The x axis is chosen to lie in the Zz plane.

G H

G x I H H θ ′=−= sin

I

G

x ′

−=ω s n

y y I H ′== 0 0= y

zG z I H H θ == cos I

H G z

θ ω

cos=

• θ = constant and body is in steady precession.

• Note: θ γ ω

tantan I

I

z

x

′==−



Two cases of motion of an axisymmetrical bodywhich under no force which involve no precession:

• Body set to spin about its axis of symmetry,

alignedareand0

G

x x

H H

ω ω ==

and body keeps spinning about its axis of

symmetry.

• Body is set to spin about its transverse axis,

alignedareand

0

G

z z

H

H

ω

==

and body keeps spinning about the given

transverse axis.



The motion of a body about a fixed point (or its masscenter) can be represented by the motion of a body

cone rolling on a space cone. In the case of steady

precession the two cones are circular.

• I < I’. Case of an elongated body. γ < θ and the

vector ω lies inside the angle ZGz. The space

cone and body cone are tangent externally; the

spin and precession are both counterclockwise

from the positive z axis. The precession is said tobe direct .

• I > I’. Case of a flattened body. γ > θ and the

vector ω lies outside the angle ZGz. The space

cone is inside the body cone; the spin and

precession have opposite senses. The precession

is said to be retrograde.

k h jhih=h 3210++



0

∧

321

3210

hhh

k ji =h ω ω ω ω

( ) ( ) ( )k hh jhhihh=h 2121313132320ω −+−−−∧

∧

321

3210

ω ω ω

ω ω ω ω

C B A

k ji

=h

3210

( ) ( ) ( )k B A j AC i BC =h 2121131332320ω −+−−−∧

54

Angular Momentum



g

( )Goo G Go= + M h h vr ∧

Angular Momentum (Rigid Body)

(4)

(5)

1 1 2 3

2 1 2 3

xx xy zx

yx yy yz

h = I - I - I h = - I + I - I

- -

ω ω ω ω ω ω

Angular Momentum (if Principle Axes used)

(6)

55

3 1 2 3 zx yz yy

1 2 3 xx yy zzo = I i + I j + I k h ω ω ω

Equations of Motion



Equations of Motion (if O and G coincide and/or

a 0=0)

h+t h =

dt hd =T

ooo

o ∧Ω∂

∂

E uations of Motion 12 ex ressed as Scalar Com onents

(11)

(12)

( )o

o Go o

d h- M =aT r dt ∧

See Equation (14) for the special case in which the reference axes are fixed

along the principle axes of the body.

(13)

56

1 1 2 3 3 2

2 2 3 1 1 3

3 3 1 2 2 1

T h h h

T h h h

T h h h

= − Ω + Ω

= − Ω + Ω

= − Ω + Ω





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