CHAPTER 16 Planar Kinematics of a Rigid Body

Chapter 16
Planar Kinematics of a
Rigid Body
APPLICATIONS
Passengers on this amusement ride are subjected to
curvilinear translation since the vehicle moves in a circular
path but always remains upright.
If the angular motion of the rotating arms is known, how
can we determine the velocity and acceleration experienced
by the passengers?
Does each passenger feel the same acceleration?
15-2
PLANAR RIGID BODY MOTION
平移
Translation: Translation occurs if every line segment on
the body remains parallel to its original direction during the
motion. When all points move along straight lines, the
motion is called rectilinear translation. When the paths of
motion are curved lines, the motion is called curvilinear
translation.
15-3
PLANAR RIGID BODY MOTION
(continued)
Rotation about a fixed axis: In this case, all
the particles of the body, except those on
the axis of rotation, move along circular
paths in planes perpendicular to the axis of
rotation.
繞固定軸旋轉
General plane motion: In this case, the
body undergoes both translation and
rotation. Translation occurs within a
plane and rotation occurs about an axis
perpendicular to this plane.
一般平面運動
15-4
15-5
16.2 Translation (平移)
rB = r A + rB/ A
(平移時，相關位置不變 ∴ r B / A = const.)
⇒一次導數： v B = v A
⇒二次導數： a B = a A
平移時，所有的點有相 同的速度和加速度
15-6
16.3 Rotation About a Fixed Axis
θ : angular position
d θ : angular displacement
方向以 " 右手定則 " 決定
ω : 角速度，θ 對時間的變化率，即 ω =
dθ
dt
α : 角加速度，ω 對時間的變化率
dω d θ
即α =
=
2
dt
dt
d ω dθ
又 α=
⋅
dt dθ
d θ dω
=
⋅
dt dθ
dω
=ω
dθ
⇒ α dθ = ω dω
2
(compare ads = vdv)
15-7
Motion of Point P --Velocity
P 以 O 為心，r 為半徑，沿圓周運動
由極座標：v r = r&
vθ = rθ&
Q r = const. ∴ r& = 0
⇒ v = vθ = rθ&
又ω = θ&
∴ v = rω
由上圖的右手定則，決定 v 的方向
⇒ v = ω × r P (= ω × r )
(r P 為旋轉軸上任一點到 P 的位置向量 )
v = ω (rP sin φ ) = ω (r )
15-8
Motion of Point P --Acceleration
dv
v
a =
, a =
dt
r
dω
由 v = ωr , α =
dt
d (ωr ) dω
=
r = αr
得a =
dt
dt
(ωr)
a =
=ω r
r
2
t
n
t
2
2
n
vector
d v dω
dr
=
×r +ω×
dt
dt
dt
= α × r + ω × (ω × r )
a=
P
P
P
P
Q a = αr sin φ = αr
t
P
⇒a=a +a
t
n
=α ×r −ω r
2
15-9
Constant Angular Acceleration
(1) α = α = 0
⇒ uniform rotation
C
⇒ ω = const.
即 θ =θ +ωt
O
( 2) α = α ≠ 0
⇒ uniformly acceleration rotation
C
⎧ ω = ω +α t
⎪⎪
1
⎨ θ =θ +ω t + α t
2
⎪
⎪⎩ 消去 t ⇒ ω = ω + 2α (θ - θ )
O
C
2
O
O
C
2
2
O
C
O
與直線運動的關係相似
v = v + 2a ( s − s )
2
2
O
C
O
15-10
p. 323, 16-4
The torsional pendulum (wheel) undergoes oscillations in the horizontal
plane, such that the angle of rotation, measured from the equilibrium
position, is given by θ = (0.5 sin 3t)rad, where t is in seconds. Determine
the maximum velocity of point A located at the periphery of the wheel
while the pendulum is oscillating. What is the acceleration of point A in
terms of t?
15-11
15-12
p. 325, 16-15
The 50-mm-radius pulley A of the clothes dryer rotates with an
angular acceleration of αA = (27θA1/2) rad/s2 , where θA is in radians.
Determine its angular acceleration when t = 1 s, starting from rest.
15-13
15-14
15-15
p. 327, 16-28
For a short time, gear A of the automobile starter rotates with an
angular acceleration of αA = (50ω1/2) rad/s2 , where ω is in rad/s.
Determine the angular velocity of gear B after gear A has rotated 50
rev, starting from rest. The radii of gears A and B are 10 mm and 25
mm, respectively.
15-16
15-17
15-18
p. 333, 16-36
Rod CD presses against AB, giving it an angular velocity. If the
angular velocity of AB is maintained at ω = 5 rad/s, determine the
required magnitude of the velocity v of CD as a function of the
angle θ of rod AB.
15-19
15-20
p. 336, 16-52
If the wedge moves to the left with a constant velocity v, determine
the angular velocity of the rod as a function of θ.
15-21
15-22
16.5 Relative-Motion Analysis:
Velocity
r =r +r
A
B
A/ B
位置 : r = r + r
位移 : d r = d r + d r
B
A
B
B/ A
A
平移
B/ A
旋轉
(以A為心)
15-24
速度：　dr
dr
dr
=
+
dt
dt
dt
B
A
B/ A
⇒v =v +v
B
A
magnitude : v
B/ A
B/ A
= ωr
B/ A
15-25
vB/ A = ω × r B/ A
相對速度表示旋轉的效應
∴vB = v A + ω × r B/ A
v
B/ A
15-26
15-27
p. 346, 16-56
The gear rests in a fixed horizontal rack. A cord is wrapped around the
inner core of the gear so that it remains horizontally tangent to the inner
core at A. If the cord is pulled to the right with a constant speed of 0.6
m/s, determine the velocity of the center of the gear, C.
15-28
15-29
p. 349, 16-74
At the instant shown,the truck travels to the right at 3 m/s,while the
pipe rolls counterclockwise at ω = 8 rad/s without slipping at B.
Determine the velocity of the pipe’s center G.
15-30
15-31
16.6 Instantaneous Center of
Zero Velocity
瞬時旋轉軸 & 瞬心(IC)
任一瞬間，RB(Rigid Body)上所有點的角速度(ω) 皆相
同，如同繞著一固定軸旋轉，此固定軸稱 瞬時旋轉軸
瞬心(IC)－瞬時旋轉軸和運動平面的交點
15-33
求IC：
(1) 已知v及ω
r
A / IC
=
v
A
ω
15-34
求IC：
(2) 已知兩點的速度 (不平行)
–求速度向量垂線(作用線)的交點
15-35
求IC：
(3) 已知大小及方向的兩平行速度
反向
同向不等
15-36
求IC：
※ RB平移時的IC，半徑為無限大
vA
vB
ω =
=
= 0
∞
∞
(同向相等IC在無限遠處)
15-37
p. 360, 16-94
The wheel is rigidly attached to gear A, which is in mesh with gear
racks D and E. If D has a velocity of vD = 1.8 m/s to the right and
wheel rolls on track C withour slipping, determine the velocity of
gear rack E.
15-38
15-39
p. 361, 16-100, 101
If rod AB is rotating with an angular velocity ωAB = 3 rad/s,
determine the angular velocity of rod BC and CD at the instant
shown.
15-40
15-41
15-42
16.7 Relative Motion Analysis:
Acceleration
RELATIVE MOTION ANALYSIS: ACCELERATION
(Section 16-7)
The equation relating the accelerations of two points on the
body is determined by differentiating the velocity equation
with respect to time.
dvB
dvA
dvB / A
=
+
dt
dt
dt
These are absolute accelerations
of points A and B. They are
measured from a set of fixed
x,y axes.
This term is the acceleration
of B with respect to A.
It will develop tangential
and normal components.
The result is aB = aA + (aB/A)t + (aB/A)n
15-44
RELATIVE MOTION ANALYSIS: ACCELERATION
(continued)
Graphically:
aB = aA + (aB/A)t + (aB/A)n
=
+
The relative tangential acceleration component (aB/A)t is (α x rB/A)
and perpendicular to rB/A.
The relative normal acceleration component (aB/A)n is (-ω2 rB/A)
and the direction is always from B towards A.
15-45
RELATIVE MOTION ANALYSIS: ACCELERATION
(continued)
Since the relative acceleration components can be expressed
as (aB/A)t = α × rB/A and (aB/A)n = - ω2 rB/A the relative
acceleration equation becomes
aB = aA + α × rB/A - ω2 rB/A
Note that the last term in the relative acceleration equation is
not a cross product. It is the product of a scalar (square of
the magnitude of angular velocity, ω2) and the relative
position vector, rB/A.
15-46
APPLICATION OF RELATIVE ACCELERATION
EQUATION
In applying the relative acceleration equation, the two points used in the
analysis (A and B) should generally be selected as points which have a
known motion, such as pin connections with other bodies.
In this mechanism, point B is known to travel along a circular path, so
aB can be expressed in terms of its normal and tangential components.
Note that point B on link BC will have the same acceleration as point B
on link AB.
Point C, connecting link BC and the piston, moves along a straight-line
path. Hence, aC is directed horizontally.
15-47
BODIES IN CONTACT
Consider two bodies in contact with one another without slipping,
where the points in contact move along different paths.
In this case, the tangential components of acceleration will be the
same, i. e.,
(aA)t = (aA’)t (which implies αBrB = αCrC ).
The normal components of acceleration will not be the same.
(aA)n ≠ (aA’)n so aA ≠ aA’
15-48
ROLLING MOTION
Another common type of problem encountered in dynamics
involves rolling motion without slip; e.g., a ball or disk rolling
along a flat surface without slipping. This problem can be
analyzed using relative velocity and acceleration equations.
As the cylinder rolls, point G (center) moves along a straight line,
while point A, on the rim of the cylinder, moves along a curved
path called a cycloid (擺線). If ω and α are known, the relative
velocity and acceleration equations can be applied to these points,
at the instant A is in contact with the ground.
15-49
ROLLING MOTION
(continued)
• Velocity:
Since no slip occurs, vA = 0 when A is in contact
with ground. From the kinematic diagram:
vG = vA + ω x rG/A
vG i = 0 + (-ω k) x (r j)
vG = ωr or vG = ωr i
• Acceleration:
Since G moves along a straight-line path, aG is
horizontal. Just before A touches ground, its
velocity is directed downward, and just after
contact, its velocity is directed upward. Thus,
point A accelerates upward as it leaves the ground.
aG = aA + α x rG/A – ω2rG/A => aG i = aA j + (-α k) x (r j) – ω2(r j)
Evaluating and equating i and j components:
aG = αr and aA = ω2r or aG = αr i and aA = ω2r j
15-50
p. 372, 16-111, 112
The hoop is cast on the rough surface such that it has an angular
velocity ω = 4 rad/s and an angular acceleration α = 5 rad/s2. Also,
its center has a velocity vo = 5 m/s and a deceleration aO = 2 m/s2.
Determine the acceleration of points A and B at this instant.
15-51
15-52
15-53
p. 375, 16-128
At a given instant, the gear has the angular motion shown.
Determine the accelerations of points A and B on the link and the
link’s angular acceleration at this instant.
15-54
15-55
15-56
16.8 Relative Motion Analysis:
Rotating Axes
Position
x, y 不平行於 X, Y
rB/ A = rB − rA
= xB i + y B j
j
(measured to x ,y frame)
i
r B = r A + (r B / A ) xyz
15-58
Velocity
將上式微分 ⇒ v B = v A +
drB/A
dt
drB/A
d
=
( xB i + yB j)
dt
dt
=
d j
dx B
d i dy B
i + xB
j + yB
+
dt
dt
dt
dt
=(
d j
dx B
dy
di
i + B j) + ( xB
+ yB
)
dt
dt
dt
dt
(v B / A ) xyz
15-59
j' = j + d j
旋轉座標軸的角速度
i' = i + d i
d i dθ
j = Ω j = Ω×i
=
dt
dt
dθ
=
( −i ) = − Ω i = Ω × j
dt
dt
dj
單位長
d j = 1 ⋅ dθ = dθ 方向為 − i
d i = 1 ⋅ dθ = dθ 方向為 j
d r B/ A
= (v B / A ) xyz + ( x B Ω × i + y B Ω × j )
∴
dt
= (v B / A ) xyz + Ω × ( x B i + y B j )
= (v B / A ) xyz + Ω × r B / A
15-60
v B = v A + Ω × (r B / A ) xyz + (v B / A ) xyz
Ω : xyz 座標軸對 XYZ 的角速度
xyz 座標軸的運動 (由XYZ 看) B 的運動 (由xyz看)
15-61
vB
⎧
⎨ absolute velocity of B
⎩
vA
⎧ absolute velocity of origin
⎨
⎩ of x, y, z frame
Ω × r B/A
(v B / A ) xyz
⎫ motion of B observed
⎬
⎭ from the X, Y, Z frame
⎫
⎪
⎪ motion of x, y, z frame
⎪⎪
⎬ observed from the X, Y, Z
⎪ frame
⎧ angular velocity effect caused ⎪
⎪
⎨
by
rotation
of
x,
y,
z
frame
⎪⎭
⎩
⎧ relative velocity of B
⎨
⎩ with respect to A
⎫ motion of B observed
⎬
⎭ from the x, y, z frame
15-62
Acceleration
d r B / A d (v B / A ) xyz
d vB d v A d Ω
=
+
× rB/ A + Ω×
+
dt
dt
dt
dt
dt
& × r B / A + Ω × d r B / A + d (v B / A ) xyz
⇒ aB = a A + Ω
dt
dt
Ω × (v B / A ) xyz + Ω × (Ω × r B / A )
d( v B/A )xyz
dt
d (v B / A ) y
⎡ d (vB / A ) x
=⎢
i+
dt
dt
⎣
d j⎤
⎤ ⎡
di
j ⎥ + ⎢(v B / A ) x + ( v B / A ) y
dt
dt ⎥⎦
⎦ ⎣
= (a B / A ) xyz + Ω × (v B / A ) xyz
& × r B / A + Ω × (Ω × r B / A ) + 2Ω × (v ) + (a )
⇒ aB = a A + Ω
B / A xyz
B / A xyz
15-63
科氏加速度 Coriolis acceleration :
2Ω × (v B / A ) xyz
(1)在運動的平面上
(2) 在法線方向上
15-64
aB
⎧
⎨ absolute acceleration of B
⎩
⎫ motion of B observed
⎬
⎭ from the X, Y, Z frame
∥
aA
& ×r
Ω
B/ A
Ω × (Ω × r B / A )
⎧ absolute acceleration of origin⎫
⎨
⎪
⎩ of x, y, z frame
⎪
⎪
⎪
＋
⎪
⎪ motion of x, y, z
⎧ angular acceleration effect
⎪⎪
⎪
caused
by
rotation
of
x,
y,
z
⎬ frame observed from
⎨
⎪ the X, Y, Z frame
⎪ frame
⎩
⎪
⎪
＋
⎪
⎪
⎧ angular velocity effect caused ⎪
⎪
⎨
by
rotation
of
x,
y,
z
frame
⎪⎭
⎩
＋
2Ω × (v B / A ) xyz
⎧ combined effect of B moving
⎪
⎨ relative to x, y, z coordinate s
⎪ and rotation of x, y, z frame
⎩
⎫
⎪
⎬ interacting motion
⎪
⎭
＋
(aB/A ) xyz
⎧ relative acceleration of B with ⎫ motion of B observed
⎬
⎨
⎭ from the x, y, z frame
⎩ respect to A
15-65
p. 385, ex. 16.21
15-66
15-67
p. 387, 16-140
At the instant , link DC has an angular velocity of ωDC = 4 rad/s and
an angular acceleration of αDC = 2 rad/s2. Determine the angular
velocity and angular acceleration of rod AB at this instant. The
collar at C is pin connected to DC and slides freely along AB.
15-68
15-69
15-70
p. 390, 16-156
A ride in an amusement park consists of a rotating arm AB having a
constant angular velocity ωAB = 2 rad/s about point A and a car
mounted at the end of the arm which has a constant angular
velocity ω’ = {-0.5k} rad/s, measured relative to the arm. At the
instant shown, determine the velocity and acceleration of the
passenger at C.
Earth
Moon
Sun
15-71
15-72
p. 390, 16-158
The “quick-return” mechanism consists of a crank AB,
slider block B, and slotted link CD. If the crank has the
angular motion shown, determine the angular motion of
the slotted link at this instant.
15-73
15-74