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Curvilinear Motion: Cylindrical Components
If motion is restricted to the plane, polar coordinates rand are used
Polar Coordinates
Specify the location ofP using both the radial coordinater, which extends outward
from the fixed origin O to the particle and a traverse coordinate, which is thecounterclockwise angle between a fixed reference line and the raxis
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Angle usually measured in degrees or radians, where
1 rad = 180
Positive directions of the rand coordinates are
defined by the unit vectors urand u
urextends from P along increasing r, when is heldfixed
Curvilinear Motion: Cylindrical Components
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uextends from P in the direction that occurs when ris held fixed and is increased
Note these directions are perpendicular to each other
Position
At any instant, position of the particle defined by the position vector
rurr
Curvilinear Motion: Cylindrical Components
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Velocity
Instantaneous velocity v is obtained by the time derivative ofr
rr ururrv
To evaluate , note that urchanges only its direction with respect to time since
magnitude of this vector = 1
During time t, a change rwill not cause a change in the direction ofu
ru
Curvilinear Motion: Cylindrical Components
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However, a change will cause urto become ur where
ur = ur+ urTime change is urFor small angles , vector has a magnitude of 1 and acts in the udirection
uu
utt
uu
r
t
r
tr
00limlim
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rv
rv
uvuvv
r
rr
where
Radical componentvr is a measure of the rate of increase or decrease in the length of
the radial coordinate
Transverse componentv
is the rate of motion along the circumference of a circle
having a radius r
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Angular velocity indicates the rate of change of the angle
Sincevrandvare mutually perpendicular, the magnitude of the velocity or speed is
simply the positive value of
Direction ofv is tangent to the path at P
dtd /
22 rrv
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Acceleration
Taking the time derivatives, for the instant acceleration,
urururururva rr
During the time t, a change r will not change the direction ualthough a change
in will cause uto become u
For small angles, this vector has a magnitude = 1 and acts in theurdirection
u= - ur
Curvilinear Motion: Cylindrical Components
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rra
rra
uauaa
r
rr
2
2
where
The term is called the angular acceleration since it measures the
change made in the angular velocity during an instant of time
Use unit rad/s2
22 / dtd
Curvilinear Motion: Cylindrical Components
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Since arand aare always perpendicular, the magnitude of the acceleration is simply
the positive value of
222 2 rrrra Direction is determined from the vector addition of its components
Acceleration is not tangent to the path
Curvilinear Motion: Cylindrical Components
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Cylindrical Coordinates
If the particle P moves along a space, then its
location may be specified by the three cylindrical
coordinates r, , z
z coordinate is similar to that used for rectangularcoordinates
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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
zr
zr
zrp
uzurrurra
uzururv
uzurr
)2()( 2
Curvilinear Motion: Cylindrical Components
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Time Derivatives
Two types of problems usually occur
1) If the coordinates are specified as time parametric equations, r = r(t) and = (t),
then the time derivative can be formed directly.
2) If the time parametric equations are not given, it is necessary to specify the path r =
f() and find the relationship between the time derivatives using the chain rule of
calculus.
Curvilinear Motion: Cylindrical Components
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PROCEDURE FOR ANALYSIS
Coordinate System
Polar coordinate are used to solve problem involving angular motion of the radial
coordinate r, used to describe the particles motion
To use polar coordinates, the origin is established at a fixed point and the radial line r
is directed to the particle
The transverse coordinate is measured from a fixed reference line to radial line
Curvilinear Motion: Cylindrical Components
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Velocity and Acceleration
Once r and the four time derivatives have been evaluated at the instant
considered, their values can be used to obtain the radial and transverse components of
v and a
Use chain rule of calculus to find the time derivatives ofr = f()
Motion in 3D requires a simple extension of the above procedure to find
,,,rr
Curvilinear Motion: Cylindrical Components
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The amusement park consists of a chair that is rotating in a horizontal circular path of
radius rsuch that the arm OB has an angular velocity and angular acceleration.
Determine the radial and transverse components of velocity and acceleration of the
passenger.
EXAMPLE 12.17
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Coordinate System. Since the angular motion of the arm is reported, polar coordinates
are chosen for the solution. is not related to r, since radius is constant for all .
Velocity and Acceleration. Since ris constant,
00 rrrr
EXAMPLE 12.17
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rrra
rrra
rv
rv
r
r
2
0
22
This special case of circular motion happen to be collinear with rand axes
EXAMPLE 12.17
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The rod OA is rotating in the horizontal plane such that = (t3) rad. At the same time,
the collar B is sliding outwards along OA so that r= (100t2)mm. If in both cases, tis in
seconds, determine the velocity and acceleration of the collar when t= 1s.
EXAMPLE 12.18
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Coordinate System. Since time-parametric equations of the particle is given, it is not
necessary to relate rto .
Velocity and Acceleration.
22
2
3
1
2
/66/200200
/33/200200
3.571100100
11
11
1
sradtsmmr
sradtsmmtr
radtmmtr
stst
stst
stst
EXAMPLE 12.18
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smmuu
ururv
r
r
/}300200{
The magnitude ofv is
1143.57
3.56200
300
tan
/361300200
1
22
smmv
EXAMPLE 12.18
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2
2
/}1800700{
)2()(
smmuu
urrurra
r
r
The magnitude ofa is
1693.57)180(
7.68700
1800
tan
/19301800700
1
222
smma
EXAMPLE 12.18
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The searchlight casts a spot of light along the face of a wall that is located 100m from
the searchlight. Determine the magnitudes of the velocity and acceleration at which
the spot appears to travel across the wall at the instant = 45. The searchlight is
rotating at a constant rate of 4 rad/s
EXAMPLE 12.19
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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 rto .
r= 100/cos = 100sec
Velocity and Acceleration.
)tan(sec100sec100tansec100
)tan(sec100
2322
r
r
EXAMPLE 12.19
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Since = 4 rad/s = constant, = 0, when = 45,
2.6788
7.565
4.141
r
r
r
smv
smuu
ururv
r
r
/800
/}7.5657.565{
EXAMPLE 12.19
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2
2
2
/6400
/}5.45255.4525{
)2()(
smma
smmuu
urrurra
r
r
EXAMPLE 12.19
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Due to the rotation of the forked rod, ballA travels
across the slotted path, a portion of which is in the
shape of a cardioid, r= 0.15(1 cos )m where is
in radians. If the balls velocity is v = 1.2m/s and its
acceleration is 9m/s2 at instant = 180, determine
the angular velocity and angular acceleration of the
fork.
EXAMPLE 12.20
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Coordinate System. For this unusual path, use polar coordinates.
Velocity and Acceleration.
)(sin15.0)()(cos15.0
)(sin15.0
)cos1(15.0
r
r
r
Evaluating these results at = 180
2
15.003.0
rrmr
EXAMPLE 12.20
View Free Body Diagram
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Since v = 1.2 m/s
srad
rrv
/4
22
2
222
/18)2()(
sradrrrra
EXAMPLE 12.20