Seventh Edition
CHAPTER
15
VECTOR MECHANICS FOR ENGINEERS:
DYNAMICS
Ferdinand P. Beer
E. Russell Johnston, Jr.
Lecture Notes:
J. Walt Oler
Texas Tech University
Kinematics of
Rigid Bodies
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Seventh
Edition
Vector Mechanics for Engineers: Dynamics
Contents
Introduction
Translation
Rotation About a Fixed Axis: Velocity
Rotation About a Fixed Axis: Acceleration
Rotation About a Fixed Axis:
Representative Slab
Equations Defining the Rotation of a Rigid
Body About a Fixed Axis
Sample Problem 5.1
General Plane Motion
Absolute and Relative Velocity in Plane
Motion
Sample Problem 15.2
Sample Problem 15.3
Instantaneous Center of Rotation in Plane
Motion
Sample Problem 15.4
Sample Problem 15.5
Absolute and Relative Acceleration in Plane
Motion
Analysis of Plane Motion in Terms of a
Parameter
Sample Problem 15.6
Sample Problem 15.7
Sample Problem 15.8
Rate of Change With Respect to a Rotating
Frame
Coriolis Acceleration
Sample Problem 15.9
Sample Problem 15.10
Motion About a Fixed Point
General Motion
Sample Problem 15.11
Three Dimensional Motion. Coriolis
Acceleration
Frame of Reference in General Motion
Sample Problem 15.15
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15 - 2
1
Seventh
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Vector Mechanics for Engineers: Dynamics
Introduction
• Kinematics of rigid bodies: relations between
time and the positions, velocities, and
accelerations of the particles forming a rigid
body.
• Classification of rigid body motions:
- translation:
• rectilinear translation
• curvilinear translation
- rotation about a fixed axis
- general plane motion
- motion about a fixed point
- general motion
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Vector Mechanics for Engineers: Dynamics
Translation
• Consider rigid body in translation:
- direction of any straight line inside the
body is constant,
- all particles forming the body move in
parallel lines.
• For any two particles in the body,
r
r r
rB = rA + rB A
• Differentiating with respect to time,
r
r r
r
r&B = r&A + r&B A = r&A
r
r
vB = v A
All particles have the same velocity.
• Differentiating with respect to time again,
&rr&B = &rr&A + &rr&B A = r&r&A
r
r
aB = a A
All particles have the same acceleration.
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Vector Mechanics for Engineers: Dynamics
Rotation About a Fixed Axis. Velocity
• Consider rotation of rigid body about a
fixed axis AA’
r
r
• Velocity vector v = dr dt of the particle P is
tangent to the path with magnitude v = ds dt
∆s = ( BP )∆θ = (r sin φ )∆θ
v=
ds
∆θ
= rθ& sin φ
= lim (r sin φ )
dt ∆t →0
∆t
• The same result is obtained from
r
r dr r r
v=
=ω ×r
dt
r
r
r
ω = ω k = θ&k = angular velocity
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Vector Mechanics for Engineers: Dynamics
Rotation About a Fixed Axis. Acceleration
• Differentiating to determine the acceleration,
r
r dv d v r
= (ω × r )
a=
dt dt
r
r
dω r r dr
=
×r +ω ×
dt
dt
r
dω r r r
=
×r +ω ×v
dt
r
dω r
= α = angular acceleration
•
dt
r
r
r
= α k = ω& k = θ&&k
• Acceleration of P is combination of two
vectors,
r r r r r r
a = α × r + ω × (ω × r )
r r
α × r = tangential acceleration component
r r r
ω × (ω × r ) = radial acceleration component
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Vector Mechanics for Engineers: Dynamics
Rotation About a Fixed Axis. Representative Slab
• Consider the motion of a representative slab in
a plane perpendicular to the axis of rotation.
• Velocity of any point P of the slab,
r r
r r r
v = ω × r = ωk × r
v = rω
• Acceleration of any point P of the slab,
r r r r r r
a = α × r + ω × (ω × r )
r r
r
= α k × r − ω 2r
• Resolving the acceleration into tangential and
normal components,
r r
r
at = αk × r
a t = rα
r
2r
an = −ω r
a n = rω 2
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Vector Mechanics for Engineers: Dynamics
Equations Defining the Rotation of a Rigid Body
About a Fixed Axis
• Motion of a rigid body rotating around a fixed axis is
often specified by the type of angular acceleration.
• Recall ω =
α=
dθ
dt
or
dt =
dθ
ω
dω d θ
dω
= 2 =ω
dt
dθ
dt
2
• Uniform Rotation, α = 0:
θ = θ 0 + ωt
• Uniformly Accelerated Rotation, α = constant:
ω = ω0 + αt
θ = θ 0 + ω 0t + 12 α t 2
ω 2 = ω 02 + 2α (θ − θ 0 )
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Vector Mechanics for Engineers: Dynamics
Sample Problem 5.1
SOLUTION:
• Due to the action of the cable, the
tangential velocity and acceleration of
D are equal to the velocity and
acceleration of C. Calculate the initial
angular velocity and acceleration.
• Apply the relations for uniformly
accelerated rotation to determine the
velocity and angular position of the
pulley after 2 s.
Cable C has a constant acceleration of 9
in/s2 and an initial velocity of 12 in/s,
both directed to the right.
• Evaluate the initial tangential and
normal acceleration components of D.
Determine (a) the number of revolutions
of the pulley in 2 s, (b) the velocity and
change in position of the load B after 2 s,
and (c) the acceleration of the point D on
the rim of the inner pulley at t = 0.
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Vector Mechanics for Engineers: Dynamics
Sample Problem 5.1
SOLUTION:
• The tangential velocity and acceleration of D are equal to the
velocity and acceleration of C.
(arD )t = arC = 9 in. s →
(vr ) = (vr ) = 12 in. s →
D 0
C 0
(vD )0 = rω0
(vD )0
ω0 =
r
=
(aD )t = rα
(aD )t
α=
12
= 4 rad s
3
r
=
9
= 3 rad s 2
3
• Apply the relations for uniformly accelerated rotation to
determine velocity and angular position of pulley after 2 s.
(
)
ω = ω 0 + αt = 4 rad s + 3 rad s 2 (2 s ) = 10 rad s
θ
= ω 0 t + 12 αt 2
(
)
= (4 rad s )(2 s ) + 12 3 rad s 2 (2 s )2
= 14 rad
⎛ 1 rev ⎞
N = (14 rad )⎜
⎟ = number of revs
⎝ 2π rad ⎠
vB = rω = (5 in.)(10 rad s )
∆yB = rθ = (5 in.)(14 rad )
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N = 2.23 rev
r
vB = 50 in. s ↑
∆yB = 70 in.
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Vector Mechanics for Engineers: Dynamics
Sample Problem 5.1
• Evaluate the initial tangential and normal acceleration
components of D.
(arD )t = arC = 9 in. s →
(aD )n = rDω02 = (3 in.)(4 rad s )2 = 48 in
(arD )t = 9 in.
s2 →
(arD )n = 48 in.
s2
s2 ↓
Magnitude and direction of the total acceleration,
(aD )t2 + (aD )2n
aD =
= 92 + 482
tan φ =
=
aD = 48.8 in. s 2
(aD )n
(aD )t
48
9
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φ = 79.4°
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Vector Mechanics for Engineers: Dynamics
General Plane Motion
• General plane motion is neither a translation nor
a rotation.
• General plane motion can be considered as the
sum of a translation and rotation.
• Displacement of particles A and B to A2 and B2
can be divided into two parts:
- translation to A2 and B1′
- rotation of B1′ about A2 to B2
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Vector Mechanics for Engineers: Dynamics
Absolute and Relative Velocity in Plane Motion
• Any plane motion can be replaced by a translation of an
arbitrary reference point A and a simultaneous rotation
about A.
r
r
r
vB = v A + vB A
r r
r
v B A = ω k × rB A
v B A = rω
r r
r
r
v B = v A + ω k × rB A
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Vector Mechanics for Engineers: Dynamics
Absolute and Relative Velocity in Plane Motion
• Assuming that the velocity vA of end A is known, wish to determine the
velocity vB of end B and the angular velocity ω in terms of vA, l, and θ.
• The direction of vB and vB/A are known. Complete the velocity diagram.
vB
= tan θ
vA
v B = v A tan θ
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vA
v
= A = cosθ
v B A lω
ω=
vA
l cosθ
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Vector Mechanics for Engineers: Dynamics
Absolute and Relative Velocity in Plane Motion
• Selecting point B as the reference point and solving for the velocity vA of end A
and the angular velocity ω leads to an equivalent velocity triangle.
• vA/B has the same magnitude but opposite sense of vB/A. The sense of the
relative velocity is dependent on the choice of reference point.
• Angular velocity ω of the rod in its rotation about B is the same as its rotation
about A. Angular velocity is not dependent on the choice of reference point.
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Vector Mechanics for Engineers: Dynamics
Sample Problem 15.2
SOLUTION:
• The displacement of the gear center in
one revolution is equal to the outer
circumference. Relate the translational
and angular displacements. Differentiate
to relate the translational and angular
velocities.
The double gear rolls on the
stationary lower rack: the velocity of
its center is 1.2 m/s.
• The velocity for any point P on the gear
may be written as
Determine (a) the angular velocity of
the gear, and (b) the velocities of the
upper rack R and point D of the gear.
© 2003 The McGraw-Hill Companies, Inc. All rights reserved.
r
r
r
vP = v A + vP
A
r r
r
= v A + ωk × rP
A
Evaluate the velocities of points B and D.
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Vector Mechanics for Engineers: Dynamics
Sample Problem 15.2
SOLUTION:
• The displacement of the gear center in one revolution is
equal to the outer circumference.
For xA > 0 (moves to right), ω < 0 (rotates clockwise).
xA
θ
=−
2π r
2π
y
x A = − r1θ
Differentiate to relate the translational and angular
velocities.
x
v A = − r1ω
ω =−
vA
1.2 m s
=−
r1
0.150 m
r
r
r
ω = ωk = −(8 rad s )k
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Vector Mechanics for Engineers: Dynamics
Sample Problem 15.2
r
r
r
• For any point P on the gear, vP = v A + vP
A
Velocity of the upper rack is equal to
velocity of point B:
r r
r
r
r
vR = vB = v A + ωk × rB A
r
r
r
= (1.2 m s )i + (8 rad s )k × (0.10 m ) j
r
r
= (1.2 m s )i + (0.8 m s )i
r
r
vR = (2 m s )i
r r
r
= v A + ωk × rP
A
Velocity of the point D:
r r
r
r
vD = v A + ωk × rD A
r
r
r
= (1.2 m s )i + (8 rad s )k × (− 0.150 m )i
r
r
r
vD = (1.2 m s )i + (1.2 m s ) j
vD = 1.697 m s
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Vector Mechanics for Engineers: Dynamics
Absolute and Relative Acceleration in Plane Motion
• Absolute acceleration of a particle of the slab,
r
r
r
aB = a A + aB A
r
• Relative acceleration a B A associated with rotation about A includes
tangential and normal components,
r
(arB A ) = α k × rrB A (a B A ) = rα
(
r
aB
t
t
)
A n
=
r
−ω 2 rB A
(a B A )n = rω 2
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Vector Mechanics for Engineers: Dynamics
Absolute and Relative Acceleration in Plane Motion
r
r
• Given a A and v A ,
r
r
determine a B and α .
r
r
r
aB = a A + aB A
r
r
r
= a A + (a B A ) + (a B
n
)
A t
r
• Vector result depends on sense of a A and the
relative magnitudes of a A and (a B A )
n
• Must also know angular velocity ω.
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Vector Mechanics for Engineers: Dynamics
Absolute and Relative Acceleration in Plane Motion
r
r
r
• Write a B = a A + a B
+
→ x components:
+ ↑
A
in terms of the two component equations,
0 = a A + lω 2 sin θ − lα cos θ
y components: − a B = −lω 2 cos θ − lα sin θ
• Solve for aB and α.
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Vector Mechanics for Engineers: Dynamics
Analysis of Plane Motion in Terms of a Parameter
• In some cases, it is advantageous to determine the
absolute velocity and acceleration of a mechanism
directly.
x A = l sin θ
y B = l cos θ
v A = x& A
= lθ& cosθ
v B = y& B
= lω cosθ
a A = &x&A
= −lθ& sin θ
= −lω sin θ
a B = &y&B
= −lθ& 2 sin θ + lθ&& cosθ
= −lθ& 2 cosθ − lθ&&sin θ
= −lω 2 sin θ + lα cosθ
= −lω 2 cosθ − lα sin θ
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Vector Mechanics for Engineers: Dynamics
Instantaneous Center of Rotation in Plane Motion
• Plane motion of all particles in a slab can always be
replaced by the translation of an arbitrary point A and a
rotation about A with an angular velocity that is
independent of the choice of A.
• The same translational and rotational velocities at A are
obtained by allowing the slab to rotate with the same
angular velocity about the point C on a perpendicular to
the velocity at A.
• The velocity of all other particles in the slab are the same
as originally defined since the angular velocity and
translational velocity at A are equivalent.
• As far as the velocities are concerned, the slab seems to
rotate about the instantaneous center of rotation C.
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Vector Mechanics for Engineers: Dynamics
Instantaneous Center of Rotation in Plane Motion
• If the velocity at two points A and B are known, the
instantaneous center of rotation lies at the intersection
of the perpendiculars to the velocity vectors through A
and B .
• If the velocity vectors are parallel, the instantaneous
center of rotation is at infinity and the angular velocity
is zero.
• If the velocity vectors at A and B are perpendicular to
the line AB, the instantaneous center of rotation lies at
the intersection of the line AB with the line joining the
extremities of the velocity vectors at A and B.
• If the velocity magnitudes are equal, the instantaneous
center of rotation is at infinity and the angular velocity
is zero.
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Vector Mechanics for Engineers: Dynamics
Instantaneous Center of Rotation in Plane Motion
• The instantaneous center of rotation lies at the intersection of
the perpendiculars to the velocity vectors through A and B .
v
v
v
ω= A = A
v B = ( BC )ω = (l sin θ ) A
l cosθ
AC l cosθ
= v A tan θ
• The velocities of all particles on the rod are as if they were
rotated about C.
• The particle at the center of rotation has zero velocity.
• The particle coinciding with the center of rotation changes
with time and the acceleration of the particle at the
instantaneous center of rotation is not zero.
• The acceleration of the particles in the slab cannot be
determined as if the slab were simply rotating about C.
• The trace of the locus of the center of rotation on the body
is the body centrode and in space is the space centrode.
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Seventh
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Vector Mechanics for Engineers: Dynamics
Sample Problem 15.4
SOLUTION:
• The point C is in contact with the stationary
lower rack and, instantaneously, has zero
velocity. It must be the location of the
instantaneous center of rotation.
• Determine the angular velocity about C
based on the given velocity at A.
The double gear rolls on the
stationary lower rack: the velocity
of its center is 1.2 m/s.
• Evaluate the velocities at B and D based on
their rotation about C.
Determine (a) the angular velocity
of the gear, and (b) the velocities of
the upper rack R and point D of the
gear.
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15 - 26
13
Seventh
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 15.4
SOLUTION:
• The point C is in contact with the stationary lower rack
and, instantaneously, has zero velocity. It must be the
location of the instantaneous center of rotation.
• Determine the angular velocity about C based on the
given velocity at A.
v
1.2 m s
ω= A=
v A = rAω
= 8 rad s
rA 0.15 m
• Evaluate the velocities at B and D based on their rotation
about C.
vR = vB = rBω = (0.25 m )(8 rad s )
r
r
vR = (2 m s )i
rD = (0.15 m ) 2 = 0.2121 m
vD = rDω = (0.2121 m )(8 rad s )
vD = 1.697 m s
r
r
r
vD = (1.2i + 1.2 j )(m s )
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15 - 27
Seventh
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 15.5
SOLUTION:
• Determine the velocity at B from the
given crank rotation data.
The crank AB has a constant clockwise
angular velocity of 2000 rpm.
For the crank position indicated,
determine (a) the angular velocity of
the connecting rod BD, and (b) the
velocity of the piston P.
• The direction of the velocity vectors at B
and D are known. The instantaneous
center of rotation is at the intersection of
the perpendiculars to the velocities
through B and D.
• Determine the angular velocity about the
center of rotation based on the velocity
at B.
• Calculate the velocity at D based on its
rotation about the instantaneous center
of rotation.
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Seventh
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Vector Mechanics for Engineers: Dynamics
Sample Problem 15.5
SOLUTION:
• From Sample Problem 15.3,
r
r
r
vB = (403.9i − 481.3 j )(in. s )
vB = 628.3 in. s
β = 13.95°
• The instantaneous center of rotation is at the intersection
of the perpendiculars to the velocities through B and D.
γ B = 40° + β = 53.95°
γ D = 90° − β = 76.05°
CD
8 in.
BC
=
=
sin 76.05° sin 53.95° sin50°
BC = 10.14 in. CD = 8.44 in.
• Determine the angular velocity about the center of
rotation based on the velocity at B.
vB = (BC )ω BD
ω BD =
vB 628.3 in. s
=
BC 10.14 in.
ω BD = 62.0 rad s
• Calculate the velocity at D based on its rotation about
the instantaneous center of rotation.
vD = (CD )ω BD = (8.44 in.)(62.0 rad s )
vP = vD = 523 in. s = 43.6 ft s
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15 - 29
Seventh
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Vector Mechanics for Engineers: Dynamics
Sample Problem 15.6
SOLUTION:
• The expression of the gear position as a
function of θ is differentiated twice to
define the relationship between the
translational and angular accelerations.
The center of the double gear has a
velocity and acceleration to the right of
1.2 m/s and 3 m/s2, respectively. The
lower rack is stationary.
• The acceleration of each point on the
gear is obtained by adding the
acceleration of the gear center and the
relative accelerations with respect to the
center. The latter includes normal and
tangential acceleration components.
Determine (a) the angular acceleration
of the gear, and (b) the acceleration of
points B, C, and D.
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15 - 30
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Seventh
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 15.6
SOLUTION:
• The expression of the gear position as a function of θ
is differentiated twice to define the relationship
between the translational and angular accelerations.
x A = − r1θ
v A = − r1θ& = − r1ω
ω =−
vA
1.2 m s
=−
= −8 rad s
r1
0.150 m
a A = − r1θ&& = − r1α
α =−
aA
3 m s2
=−
r1
0.150 m
r
r
(
)r
α = α k = − 20 rad s 2 k
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15 - 31
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Vector Mechanics for Engineers: Dynamics
Sample Problem 15.6
• The acceleration of each point
is obtained by adding the
acceleration of the gear center
and the relative accelerations
with respect to the center.
The latter includes normal and
tangential acceleration
components.
(
) (
)
r
r
r
r
r
r
a B = a A + aB A = a A + a B A + aB A
t
n
r r
r
r
= a A + α k × rB A − ω 2 rB A
r
r
r
r
= 3 m s 2 i − 20 rad s 2 k × (0.100 m ) j − (8 rad s )2 (− 0.100 m ) j
r
r
r
= 3 m s 2 i + 2 m s 2 i − 6.40 m s 2 j
(
(
) (
) (
)
) (
(
)
) (
)
r
r
r
aB = 5 m s 2 i − 6.40 m s 2 j
© 2003 The McGraw-Hill Companies, Inc. All rights reserved.
aB = 8.12 m s 2
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Vector Mechanics for Engineers: Dynamics
Sample Problem 15.6
r r
r
r
= a A + α k × rC A − ω 2 rC A
r
r
r
r
= 3 m s 2 i − 20 rad s 2 k × (− 0.150 m ) j − (8 rad s )2 (− 0.150 m ) j
r
r
r
= 3 m s 2 i − 3 m s 2 i + 9.60 m s 2 j
r
r
ac = 9.60 m s 2 j
r r
r
r
r
r
r
aD = a A + aD A = a A + α k × rD A − ω 2 rD A
r
r
r
r
= 3 m s 2 i − 20 rad s 2 k × (− 0.150 m ) i − (8 rad s )2 (− 0.150m ) i
r
r
r
= 3 m s 2 i + 3 m s 2 j + 9.60 m s 2 i
r
r
r
aD = 12.6 m s 2 i + 3 m s 2 j
aD = 12.95 m s 2
r
r
r
aC = a A + aC
A
(
(
) (
) (
(
(
) (
) (
)
(
) (
)
)
(
) (
) (
© 2003 The McGraw-Hill Companies, Inc. All rights reserved.
)
)
)
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Vector Mechanics for Engineers: Dynamics
Sample Problem 15.3
SOLUTION:
• Will determine the absolute velocity of
point D with
r
r
r
vD = vB + vD B
r
• The velocity v B is obtained from the
given crank rotation data.
velocity vrD
The crank AB has a constant clockwise • The directions of the absolute
r
and the relative velocity v D B are
angular velocity of 2000 rpm.
determined from the problem geometry.
For the crank position indicated,
• The unknowns in the vector expression
determine (a) the angular velocity of
are the velocity magnitudes v D and v D B
the connecting rod BD, and (b) the
which may be determined from the
velocity of the piston P.
corresponding vector triangle.
• The angular velocity of the connecting
rod is calculated from v D B .
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Vector Mechanics for Engineers: Dynamics
Sample Problem 15.3
SOLUTION:
• Will determine the absolute velocity of point D with
r
r
r
vD = vB + vD B
r
• The velocity vB is obtained from the crank rotation data.
ω AB = ⎛⎜ 2000
⎝
rev ⎞⎛ min ⎞⎛ 2π rad ⎞
⎟⎜
⎟⎜
⎟ = 209.4 rad s
min ⎠⎜⎝ 60 s ⎟⎠⎝ rev ⎠
vB = ( AB )ω AB = (3 in.)(209.4 rad s )
The velocity direction is as shown.
r
• The direction of the absolute velocity vD is horizontal.
r
The direction of the relative velocity vD B is
perpendicular to BD. Compute the angle between the
horizontal and the connecting rod from the law of sines.
sin 40° sin β
=
8 in.
3 in.
β = 13.95°
© 2003 The McGraw-Hill Companies, Inc. All rights reserved.
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Seventh
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 15.3
• Determine the velocity magnitudes vD and vD
from the vector triangle.
B
vD B
628.3 in. s
vD
=
=
sin 53.95° sin 50° sin76.05°
vD = 523.4 in. s = 43.6 ft s
r
r
r
vD = vB + vD
B
vD
B
= 495.9 in. s
vD
B
= lω BD
vD
495.9 in. s
B
=
8 in.
l
= 62.0 rad s
ω BD =
vP = vD = 43.6 ft s
© 2003 The McGraw-Hill Companies, Inc. All rights reserved.
r
r
ω BD = (62.0 rad s )k
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Seventh
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Vector Mechanics for Engineers: Dynamics
Sample Problem 15.7
SOLUTION:
• The angular acceleration of the
connecting rod BD and the acceleration
of point D will be determined from
r
r
r
r
r
r
aD = aB + aD B = aB + aD B + aD B
(
)t (
)n
• The acceleration of B is determined from
the given rotation speed of AB.
• The directions of the accelerations
r
r
r
a D , a D B , and a D B are
t
n
determined from the geometry.
Crank AG of the engine system has a
constant clockwise angular velocity of
2000 rpm.
(
For the crank position shown,
determine the angular acceleration of
the connecting rod BD and the
acceleration of point D.
)
(
)
• Component equations for acceleration
of point D are solved simultaneously for
acceleration of D and angular
acceleration of the connecting rod.
© 2003 The McGraw-Hill Companies, Inc. All rights reserved.
15 - 37
Seventh
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 15.7
SOLUTION:
• The angular acceleration of the connecting rod BD and
the acceleration of point D will be determined from
r
r
r
r
r
r
a D = a B + aD B = aB + aD B + aD B
(
)t (
)n
• The acceleration of B is determined from the given rotation
speed of AB.
ω AB = 2000 rpm = 209.4 rad s = constant
α AB = 0
2
aB = rω AB
=
(
(123 ft )(209.4 rad s)2 = 10,962 ft s2
)
r
r
r
aB = 10,962 ft s 2 (− cos 40°i − sin 40° j )
© 2003 The McGraw-Hill Companies, Inc. All rights reserved.
15 - 38
19
Seventh
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 15.7
(
r
r
• The directions of the accelerations aD , aD
determined from the geometry.
r
r
aD = m aDi
)
B t , and
(arD B )n are
From Sample Problem 15.3, ωBD = 62.0 rad/s, β = 13.95o.
2
8 ft )(62.0 rad s )2 = 2563 ft s 2
(aD B )n = (BD )ω BD
= (12
(arD B )n = (2563 ft s2 )(− cos13.95°ir + sin13.95° rj )
(aD B )t = (BD )α BD = (128 ft )α BD = 0.667α BD
The direction of (aD/B)t is known but the sense is not known,
r
r
r
aD B = (0.667α BD )(± sin 76.05°i ± cos 76.05° j )
(
)t
© 2003 The McGraw-Hill Companies, Inc. All rights reserved.
15 - 39
Seventh
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 15.7
• Component equations for acceleration of point D are solved
simultaneously.
r
r
r
r
r
r
a D = a B + aD B = aB + aD B + aD B
(
)t (
)n
x components:
− aD = −10,962 cos 40° − 2563 cos13.95° + 0.667α BD sin 13.95°
y components:
0 = −10,962 sin 40° + 2563 sin 13.95° + 0.667α BD cos13.95°
(
)r
r
r
aD = −(9290 ft s 2 ) i
r
α BD = 9940 rad s 2 k
© 2003 The McGraw-Hill Companies, Inc. All rights reserved.
15 - 40
20
Seventh
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 15.8
SOLUTION:
• The angular velocities are determined by
simultaneously solving the component
equations for
r
r
r
vD = vB + vD B
In the position shown, crank AB has a
constant angular velocity ω1 = 20 rad/s
counterclockwise.
• The angular accelerations are determined
by simultaneously solving the component
equations for
r
r
r
aD = aB + aD B
Determine the angular velocities and
angular accelerations of the connecting
rod BD and crank DE.
© 2003 The McGraw-Hill Companies, Inc. All rights reserved.
15 - 41
Seventh
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 15.8
SOLUTION:
• The angular velocities are determined by simultaneously
solving the component equations for
r
r
r
vD = vB + vD B
r
r
r
r
r
r
vD = ω DE × rD = ω DE k × (− 17i + 17 j )
r
r
= −17ω DE i − 17ω DE j
r
r
r
r
r
r
vB = ω AB × rB = 20k × (8i + 14 j )
r
r
= −280i + 160 j
r
r
r
r
r
r
vD B = ω BD × rD B = ω BD k × (12i + 3 j )
r
r
= −3ω BD i + 12ω BD j
x components:
− 17ω DE = −280 − 3ω BD
y components:
− 17ω DE = +160 + 12ω BD
r
r
ω BD = −(29.33 rad s )k
© 2003 The McGraw-Hill Companies, Inc. All rights reserved.
r
r
ω DE = (11.29 rad s )k
15 - 42
21
Seventh
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 15.8
• The angular accelerations are determined by
simultaneously solving the component equations for
r
r
r
aD = aB + aD B
r
r
r
2 r
aD = α DE × rD − ω DE
rD
r
r
r
r
r
= α DE k × (− 17i + 17 j ) − (11.29 )2 (− 17i + 17 j )
r
r
r
r
= −17α DE i − 17α DE j + 2170i − 2170 j
r
r
r
r
r
2 r
aB = α AB × rB − ω AB
rB = 0 − (20 )2 (8i + 14 j )
r
r
= −3200i + 5600 j
r
r
r
2 r
aD B = α BD × rB D − ω BD
rB D
r
r
r
r
r
= α B D k × (12i + 3 j ) − (29.33)2 (12i + 3 j )
r
r
r
r
= −3α B D i + 12α B D j − 10,320i − 2580 j
x components: − 17α DE + 3α BD = −15,690
y components: − 17α DE − 12α BD = −6010
r
r
r
r
α BD = − 645 rad s 2 k
α DE = 809 rad s 2 k
(
© 2003 The McGraw-Hill Companies, Inc. All rights reserved.
)
(
)
15 - 43
Seventh
Edition
Vector Mechanics for Engineers: Dynamics
Motion About a Fixed Point
• The most general displacement of a rigid body with a
fixed point O is equivalent to a rotation of the body
about an axis through O.
• With the instantaneous axis of rotation and angular
r
velocity ω , the velocity of a particle P of the body is
r
r dr r r
=ω×r
v=
dt
and the acceleration of the particle P is
r
r r r r r r
r dω
α=
.
a = α × r + ω × (ω × r )
dt
r
• The angularr acceleration α represents the velocity of
the tip of ω .
r
• As the vector ω moves within the body and in space,
it generates a body cone and space cone which are
tangent along the instantaneous axis of rotation.
• Angular velocities have magnitude and direction and
obey parallelogram law of addition. They are vectors.
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Seventh
Edition
Vector Mechanics for Engineers: Dynamics
General Motion
• For particles A and B of a rigid body,
r
r
r
vB = v A + vB A
• Particle A is fixed within the body and motion of
the body relative to AX’Y’Z’ is the motion of a
body with a fixed point
r
r
r r
v B = v A + ω × rB A
• Similarly, the acceleration of the particle P is
r
r
r
aB = a A + aB A
r
r r
r r r
= a A + α × rB A + ω × (ω × rB A )
• Most general motion of a rigid body is equivalent to:
- a translation in which all particles have the same
velocity and acceleration of a reference particle A, and
- of a motion in which particle A is assumed fixed.
© 2003 The McGraw-Hill Companies, Inc. All rights reserved.
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