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Dr Mohamed AICHOUNI
CHAPTER 12
KINEMATICS OF A PARTICLE
Sections 12‐8 and 12‐9
This presentation is an adapted version from : R. C. Hibbeler, « Engineering Mechanics –
Dynamics », 12th Ed.
12.8 Curvilinear Motion:
Cylindrical Components
• uθ extends from P in the direction that occurs when r is held fixed and θ is
increased
• Note these directions are perpendicular to each other
Position
• At any instant, position of the particle defined by the position vector
r
r
r = ru r
Velocity
• Instantaneous velocity v is obtained by the time
r
derivative of r
r
r r
r
v = r& = r&ur + ru&r
&r , note that ur changes only its direction with respect to time
• To evaluate u
since magnitude of this vector = 1
• During time ∆t, a change ∆r will not cause a change in the direction of u
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Curvilinear Motion: Cylindrical
Components
• However, a change ∆θ will cause ur to become ur’ where : ur’ = ur + ∆ur
• Time change is ∆ur
• For small angles ∆θ, vector has a magnitude of 1 and acts in the uθ direction
r
r
Δθ ⎞ r
Δur ⎛
u&r = lim
= ⎜ lim
⎟uθ
Δt →0 Δt
⎝ Δt →0 Δt ⎠
r
r
u&r = θ&uθ
r
r
r
v = vr ur + vθ uθ
vr = r&
where
vθ = rθ&
• Radical component vr is a measure of the rate of increase or decrease in the
length of the radial coordinate
• Transverse component vθ is the rate of motion along the circumference of a
circle having a radius r
Curvilinear Motion: Cylindrical Components
Acceleration
• Taking the time derivatives, for the instant acceleration,
a = v& = &r&ur + r&u&r + r&θ&uθ + rθ&&uθ + rθ&u&θ
• During the time ∆t, a change ∆r will not change the direction uθ although a
change in ∆θ will cause uθ to become uθ’
• For small angles, this vector has a magnitude = 1 and acts in the –ur
direction :
∆uθ= - ∆θur
where
a = ar ur + aθ uθ
ar = &r& − rθ& 2
aθ = rθ&& + 2r&θ&
• The term
is called the angular acceleration since it
θ&& = d 2θ / dt 2
measures the change made in the angular velocity during an instant of time
• Use unit rad/s2
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Curvilinear Motion: Cylindrical
Components
Cylindrical Coordinates
• If the particle P moves along a space, then its
location may be specified by the three cylindrical
coordinates
di t r, θ,
θ z
• z coordinate is similar to that used for
rectangular coordinates
• Since the unit vector defining its direction, uz, is constant, the time derivatives
of this vector are zero
• Position, velocity, acceleration of the particle can be written in cylindrical
coordinates as shown
r
r
r
rp = ru r + zu z
r
r
r
r
v = r&ur + rθ&uθ + z&u z
r
r
r
r
a = (&r& − rθ& 2 )ur + ( rθ&& + 2r&θ&)uθ + &z&u z
EXAMPLE 12.17
The amusement park consists of a chair that is rotating in a horizontal circular
path of radius r such that the arm OB has an angular velocity and angular
acceleration. Determine the radial and transverse components of velocity and
acceleration of the passenger.
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Solution - EXAMPLE 12.17
Coordinate System.
Since the angular motion of the arm is reported,
polar coordinates are chosen for the solution. θ is not related to r, since radius is
constant for all θ.
Velocity and Acceleration. Since r is constant,
r=r
r& = 0 &r& = 0
vr = r& = 0
vθ = rθ&
ar = &r& − rθ& 2 = −rθ& 2
aθ = rθ&& + 2r&θ& = rθ&&
This special case of circular motion happen to be collinear with r and θ axes
12.9 Absolute Dependent Motion
Analysis of Two Particles
• Motion of one particle will depend on the corresponding motion of another particle
• Dependency occur when particles are interconnected by the inextensible cords
which are wrapped around pulleys.
• For example (Example 1), the movement of block A downward along the inclined
plane will cause a corresponding movement of block B up the other incline
• Specify the locations of the blocks using position coordinate sA and sB
• Note each of the coordinate axes is (1) referenced from a fixed point (O) or fixed
datum line, (2) measured along each inclined plane in the direction of motion of
block A and block B and (3) has a positive sense from C to A and D to B
• If total cord length is lT, the position coordinate are elated by the equation
s A + lCD + sB = lT
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Absolute Dependent Motion Analysis
of Two Particles
• Here lCD is the length passing over arc CD
• Taking time derivative of this expression
expression, realizing that lCD and lT remain
constant, while sA and sB measure the lengths of the changing segments of the
cord
ds A dsB
+
=0
dt
dt
or
vB = −v A
• The negative sign indicates that when block A has a velocity downward in the
direction of p
position sA, it causes a corresponding
p
g upward
p
velocity
y of block B;; B
moving in the negative sB direction
• Time differentiation of the velocities yields the relation between accelerations
aB = - aA
Absolute Dependent Motion Analysis
of Two Particles
• Example 2 involving dependent motion of two
blocks
• Position
P i i off bl
block
k A is
i specified
ifi d b
by sA, and
d the
h
position of the end of the cord which block B is
suspended is defined by sB
• Chose coordinate axes which are (1) referenced
from fixed points and datums, (2) measured in the
direction of motion of each block, (3) positive to the
right (sA) and positive downward (sB)
• During the motion, the red colored segments of the
cord remain constant
• If l represents the total length of the cord minus
these segments, then the position coordinates can be
related by
2sB + h + s A = l
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Absolute Dependent Motion Analysis
of Two Particles
Since l and h are constant during the motion, the two time derivatives yields
2v B = − v A
2a B = − a A
• When B moves downward (+sB), A moves to left (sA) with two times the motion
• This example can also be worked by defining the
position of block B from the center of the bottom
pulley ( a fixed point)
2( h − s B ) + h + s A = l
• Time differentiation yields
2v B = v A
2aB = a A
EXAMPLE 12.22
Determine the speed of block A if block B has an upward speed of 2m/s.
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EXAMPLE 12.22
Position Coordinate Equation.
Positions of A and B are defined using coordinates sA and sB.
Since the system has two cords which change length, it is necessary to use a
third coordinate sC in order to relate sA to sB. Length of the cords can be
expressed in terms of sA and sC, and the length of the other cord can be
expressed in terms of sB and sC. The red colored segments are not considered
in this analysis.
EXAMPLE 12.22
For the remaining cord length,
s A + 2 sC = l1
sB + ( sB − sC ) = l2
Eliminating sC yields,
s A + 4 sB = 2l2 + l1
Time Derivative.
The time derivative gives
v A + 4v B = 0
so that vB = -2m/s (upward)
v A = −4VB =
4 * 2 = +8m / s = 8m / s ↓
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Additional Problems
EXAMPLE 12.18
Coordinate System.
Since time-parametric equations of the particle is given, it is not necessary to
relate r to θ.
Velocity and Acceleration.
r = 100t 2
r& = 200t
&r& = 200
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t =1s
t =1 s
t =1 s
= 100mm θ = t 3
t =1 s
= 200mm / s θ& = 3t 2
= 200mm / s 2 θ&& = 6t
= 1rad = 5.73°
t =1 s
t =1 s
= 2rad / s
= 6rad / s 2
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EXAMPLE 12.18
r
r
r
v = r&ur + rθ&uθ
r
r
= {200ur + 300uθ }mm / s
The magnitude of v is
v = 2002 + 3002 = 361mm / s
δ = tan −1⎛⎜
300 ⎞
o
⎟ = 56.3
⎝ 200 ⎠
δ + 57.3o = 114o
EXAMPLE 12.18
r
r
r
a = (&r& − rθ& 2 )ur + ( rθ&& + 2r&θ&)uθ
r
r
= {−700ur + 1800uθ }mm / s 2
The magnitude of a is
a = 7002 + 18002 = 1930mm / s 2
φ = tan −1⎛⎜
1800 ⎞
o
⎟ = 68.7
⎝ 700 ⎠
(180 − φ ) + 57.3o = 169o
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EXAMPLE 12.19
The searchlight casts a spot of light along the face of a wall that is located 100m
from the searchlight
searchlight. Determine the magnitudes of the velocity and acceleration
at which the spot appears to travel across the wall at the instant θ = 45°. The
searchlight is rotating at a constant rate of 4 rad/s
EXAMPLE 12.19
Coordinate System.
Polar coordinates will be used since the angular rate of the searchlight
is given. To find the time derivatives, it is necessary to relate r to θ.
r = 100/cos θ = 100sec θ
Velocity and Acceleration.
r& = 100(secθ tan θ )θ&
2
2
&r& = 100 secθ tan 2 θ (θ& ) + 100 sec3 θ (θ& ) + 100(secθ tan θ )θ&&
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EXAMPLE 12.19
Since
θ&= 4 rad/s = constant, θ&&= 0, when θ = 45°,
r = 141.4
r& = 565.7
&r& = 6788.2
r
r
r
v = r&ur + rθ&uθ
r
r
= {565.7ur + 565.7uθ }m / s
v = 800m / s
EXAMPLE 12.19
r
r
r
a = (&r& − rθ& 2 )ur + (rθ&& + 2r&θ&)uθ
r
r
= {4525.5ur + 4525.5uθ }mm / s 2
a = 6400mm / s 2
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EXAMPLE 12.20
Due to the rotation of the forked rod, ball A
travels across the slotted path
path, a portion of
which is in the shape of a cardioids, r = 0.15(1 –
cos θ)m where θ is in radians. If the ball’s
velocity is v = 1.2m/s and its acceleration is
9m/s2 at instant θ = 180°, determine the angular
velocity and angular acceleration of the fork.
Solution : EXAMPLE 12.20
Coordinate System.
For this unusual path, use polar coordinates.
Velocity and Acceleration.
r = 0.15(1 − cosθ )
r& = 0.15(sin θ )θ&
&r& = 0.15(cosθ )θ&(θ&) + 0.15(sin θ )θ&&
Evaluating these results at θ = 180°
r = 0.3m r& = 0 &r& = −0.15θ& 2
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Solution : EXAMPLE 12.20
Since v = 1.2 m/s
v = (r& )2 + (rθ& )
θ& = 4rad / s
2
a = (&r& − rθ& 2 ) 2 + ( rθ&& + 2r&θ&) 2
θ&& = 18rad / s 2
EXAMPLE 12.23
Determine the speed with which block B rises if the end of the cord at A is pulled
down with a speed of 2m/s.
2m/s
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EXAMPLE 12.23
Position-Coordinate Equation.
The position of A is defined by sA, and the position of block B is specified by sB
since point E on the pulley will have the same motion as the block. Both
coordinates are measured from a horizontal datum passing through the fixed pin
at pulley D. Since the system consists of two cords, the coordinates sA and sB
cannot be related directly. By establishing a third position coordinate, sC, and the
length of the other cord in terms of sA, sB and sC.
EXAMPLE 12.23
Excluding the red colored segments of the cords, the remaining constant cord
lengths l1 and l2 (along the hook and link dimensions) can be expressed as
sC + sB = l1
(s A − sC ) + (sB − sC ) + sB = l2
Eliminating sC yields
sC + 4 sB = l2 + 2l1
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EXAMPLE 12.23
Time Derivative.
The time derivative give
v A + 4v B = 0
when vA = 2m/s (downward)
vB = −0.5m / s = 0.5m / s ↑
EXAMPLE 12.24
A man at A s hoisting a safe S by walking to
the right with a constant velocity vA =
0.5m/s. Determine the velocity and
acceleration of the safe when it reaches the
elevation at E. The rope is 30m long and
passes over a small pulley at D.
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Solution : EXAMPLE 12.24
Position Coordinate System.
Rope segment DA changes both direction and
magnitude. However, the ends of the rope, which
define the position of S and A, are specified by
means of the x and y coordinates measured from
a fixed point and directed along the paths of
motion of the ends of the rope. The x and y
coordinates may be related since the rope has a
fixed length l = 30m, which at all times is equal to
the length of the segment DA plus CD.
U i P
Using
Pythagorean
th
Th
Theorem,
I DA =
(15)2 + x 2
ICD = (15 − y )
l = lDA + lCD
30 =
(15)2 + x 2 + (15 − y )
y = 225 + x 2 − 15
(1)
Solution : EXAMPLE 12.24
Time Derivative.
Taking time derivative, using the chain rule where, vS = dy/dt and vA = dx/dt
⎤ dx
dy ⎡ 1
2x
=⎢
dt ⎣ 2 225 + x 2 ⎥⎦ dt
x
=
vA
225 + x 2
vS =
At y = 10 m, x = 20 m, vA = 0.5 m/s, vS = 400mm/s ↑
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(2)
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EXAMPLE 12.24
The acceleration is determined by taking the time derivative of eqn (2),
aS =
⎡ − x( dx / dt ) ⎤
⎡
⎤⎛ dx ⎞
1
=
xv
+
A
⎢
⎥
⎢
⎥ ⎜ ⎟v A
dt 2 ⎣ (225 + x 2 )3 / 2 ⎦
⎣ 225 + x 2 ⎦⎝ dt ⎠
d2y
⎡
⎤ dv A
1
225v 2A
+⎢
x
=
3/ 2
2 ⎥ dt
⎣ 225 + x ⎦
225 + x 2
(
)
At x = 20 m, with vA = 0.5 m/s,
aS = 3.6mm / s 2
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