LEC 03 (Curveliniear 051

8/11/2019 LEC 03 (Curveliniear 051

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King Fahd University ofPetroleum & Minerals

Mechanical Engineering Dynamics ME 201

BY

Dr. Meyassar N. Al-Haddad Lecture # 3


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12.3 Rectangular Kinematics:

Erratic Motion

Omitted


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Section 12.4 in your text

Path is described in three dimensions Position, velocity, and acceleration are vectors


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Position

* S is a path function* The position of the particle

measured from a fixed point O is given by the position vector r = r(t)

Example :r = {sin (2t) i + cos (2t) j 0.5 t k}


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Displacement

The displacement D rrepresents the change

in the particles position D r = r - r


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Velocity Average velocity

Instantaneous velocity

As D t = 0 then D r = D s Speed

Since r is tangent to thecurve at P, then the velocity is

tangent to the curve

t r

vavg DD

dt dr

t r

t DD

Dlim

0

dt dsv


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Acceleration

Average acceleration:

Hodograph curve velocityarrowhead points

Instantaneous acceleration:

2

2

0lim dt

r d dt dv

t va

t DD

D

t

va avg

D

D

Hodograph


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Acceleration (con.)

a acts tangent to thehodograph

a is not tangent to the path of motion

a directed toward theinside or concave side


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12.5 Curvilinear Motion: Rectangular

Components Rectangular : x, y, z

frame


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Position Position vector r

r = x i + y j + z k The magnitude of r is always

positive and defined as

Unit vector

The direction cosines are

222 z y xr

k r r

jr

r

ir r

u z y x

r

r r

r

r

r r z y x coscoscos


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Velocity

Velocity is the first timederivative of r

Where

Magnitude of velocity

Direction is alwaystangent to the path

k jidt

dr v

z y x

z y x z y x

222 z y xv


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Problem

The position of a particle is described by r A= {2t i +(t 2-1) j} ft. where tis in seconds. Determine the position of the point and the speed at 2second.

ji jir A 34)12()2(2

2

ji jt idt

dr v A A 4222

ft/s47.442 22 Av speed


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Acceleration

Acceleration is the first timederivative of v

Where

Magnitude of acceleration

Direction is not tangent to the path

k a jaiadt dv

a z y x

x x xa

aaa z y xa 222

y y ya

z z z a


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Example 12.9

The distance of the balloon from A at 2 sec

The magnitude and direction of velocity at 2 sec The magnitude and direction of acceleration at 2 sec

t x 8A

10

2 x y

Positionft16)2(8 x

ft6.2510/)16( 2

yft2.30)6.25()16( 22r

Velocity

ft/s8)8( t dt d

x x

ft/s6.2510/)8)(16(210/2)10/( 2 x x xdt

d y y

ft8.26)6.25()8( 22

o

x

y 6.728

6.25tantan 11

Accelerationft/s0)8(

dt d

xa x

2ft/s8.1210/)(210/)(2)10/2( x x x x x xdt d

ya y

222 ft/s8.12)8.12()0( a

o

x

ya a

a90

08.12

tantan 11

X=8t


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Example 12.10

k )75.0(2.0 j)5.1cos(5.0i)5.1sin(5.0)75.0(r

)180( orad t in second arguments in radians

At t = 0.75 s find location, velocity, and acceleration

k 2.0 j)2cos(5.0i)2sin(5.0 t t t r

k 15.0 j0354.0.499i0r

m522.0)15.0()0354.0()499.0( 222r

0.287k -0.0678ji955.0k 522.0 15.0 j522.00354.0i0.522.4990r u

o2.17)955.0(cos 1

o1.86)0678.0(cos 1 o

107)287.0(cos 1

Note: Put your calculator in Rad Mode


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LEC 03 (Curveliniear 051

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