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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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