dynamics lecture4 cylindrical components

7/28/2019 Dynamics Lecture4 Cylindrical Components

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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



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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



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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



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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



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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



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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



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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



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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.



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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



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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



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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


)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.


)(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
http://localhost/var/www/apps/conversion/tmp/scratch_4/ppt-qns/12.20.ppthttp://localhost/var/www/apps/conversion/tmp/scratch_4/ppt-qns/12.20.ppt


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Since v = 1.2 m/s

srad

rrv

/4

22

2

222

/18)2()(

sradrrrra

EXAMPLE 12.20

dynamics lecture4 cylindrical components

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