Chapter 3 Kinematics of particles
14
Jump to first page 1 Chapter 3 Kinematics of particles Differentiation: the ratio between two increments when they are extremely small, e.g. velocity. Increment: distance traveled at the end of the 1 st second = 10 m distance traveled at the end of the 3 rd second = 14 m increment of distance = 4 m increment of time = 2 s Ratio between two increments = 4 m/2s = 2 m/s (average velocity) More detailed measurement: distance traveled at the end of 1.01 second, the distance traveled is 10.025 m, the ratio becomes 0.025 m / 0.01 s = 2.5 m/s (instantaneous velocity).
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Chapter 3 Kinematics of particles. Increment: distance traveled at the end of the 1 st second = 10 m distance traveled at the end of the 3 rd second = 14 m increment of distance = 4 m increment of time = 2 s - PowerPoint PPT Presentation
Transcript of Chapter 3 Kinematics of particles
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1
Chapter 3 Kinematics of particles
Differentiation: the ratio between two increments when they are extremely small, e.g. velocity.
Increment: distance traveled at the end of the 1st second = 10 m distance traveled at the end of the 3rd second = 14 m increment of distance = 4 m increment of time = 2 sRatio between two increments = 4 m/2s = 2 m/s (average velocity)More detailed measurement: distance traveled at the end of 1.01 second, the distance traveled is 10.025 m, the ratio becomes 0.025 m / 0.01 s = 2.5 m/s (instantaneous velocity).
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3
11
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constant. a is where,0
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yx
z
O
)(tr
)( ttr r
)()( where, velocity Average av trttrrt
rv
rdt
rd
t
rv
t
0
limInstantaneous velocity
rvdt
vda
Acceleration
Position )(tr
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Cartesian coordinate system
zzyyxxr va
zzyyxxrv
zzyyxxr
ˆˆˆ
ˆˆˆ
ˆˆˆ
x
z
y
r
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2D Polar coordination system
rrr ˆ
θ̂r̂
r
y
xO
: unit vector // to (radial) : unit vector to (in -direction)
r̂θ̂ r̂
r
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Velocity
)(ˆ ttr
)(ˆ trd)tt(θ̂
θ̂d
)(ˆd tr
)(ˆ tθ
d
,ˆˆ
ˆ
rrrrrv
rrr
θ̂ dr̂d
θ̂ θ̂td
d
td
)t(r̂d)t(r̂
Since
θrrrv ˆˆ θvrv θrrˆˆ
radialvelocity
tangential velocity
angular velocity
O)(tr
)(ˆ tr
)(ˆ ttθ
)( ttr
d
)Δ(ˆ ttr
)t(θ̂
θx
y
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θθrθrrθrra 2 ˆ)2(ˆ)(
centripetal acceleration.
θθ rθθ rθθ rrrrrva ˆˆˆˆˆ Acceleration
rθ trt
t
t
tθ rθ ˆˆor))(ˆ(
d
)(d
d
)(ˆdor)ˆ(dˆd
Since
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3s.atofand Find04.02.0 and 02.02.0
Example23
t B a vtrtt
θ
(m/s)ˆ0.414ˆ0.24rad/s 0.74 ,s/m 24.0 m, 0.56 3s,at
(2) 20.06 0.2
(1) 0.08
ˆˆSolution
θrv θrr t
t θ
trθθrrrv
)(m/s ˆ 0.557ˆ 0.227-
s/rad 36.0 s, 3at
0.12 ,2 /sm 0.08
ˆ2ˆ
2
2
θ r a
θ t
t θ r
θ)θrθ(rr)2θrr( a
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s/rad 10-3.63)/2sin(s/m 71.730cos2.9014.080000 cos
ˆsin ˆcos
ˆ ˆ
ˆ)2(ˆ)(
4-g
2o2g
2
gg
θr
2
rθr-θaθ θa-θrr
θθa r θa- θa ra
θθrθrrθrra
)s/m(ˆ014.01080ˆ102.1 ˆˆ 33 θrθθrrrv
.rad/s 014.0 and ,m/s10x2.1 ,km 80 ,30at and calculate
,s/m 2.9under g travellinisrocket a If Example
3
o
2g
θr rθθ r ,va
Solution v
ga
r
r
θ̂
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system. coordinatepolar applyingby 90 at and Find .0 and s rad5 90 At
ck.linear tra a andrack circular t a along moving is P :Exampleo1
,o θavθ
0.08m
0.04m
P
AO
r
r̂
θ̂
θθr θ rrv ˆˆˆ Solution For OAP, 0.082 = 0.042 +r2+2 x 0.04 r cosTime derivative:At = 90o, r = 0.06928 m
θθ)r(θrrr0 sin0.042cos0.0422
θrvθr
θrrr
ˆ346.0ˆ2.0)s/m(2.004.0
08.020
θ)θrθ(rr )θrr(a 2 ˆ2ˆ
θθrθθr θθrθθrθrrrr
sin08.0cos08.0sin08.0sin08.0cos08.0220
2
2
θra r ,θrrrr
ˆ2ˆ15.1)s/m(5774.0such that08.00 ,90At 2o
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zzθθrθrrθrr rvazzθθrrrrv
zzrrr
ˆˆ)2(ˆ)(ˆˆˆ
ˆˆ
2
3D Cylindrical coordination system
x
z
y
R
z
rr̂
r̂
θ̂
θ̂
z
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Normal and Tangential Coordinates (n-t) :
tvtt
t
tv ˆˆs
ˆ)s(lim
rlim
0s0t
tvtvtvt
a ˆˆ)̂(d
d
)(sr
)( ssr
s
t̂
n̂
radius of curvature
O’
β
O’
s)s(t̂
)ss(ˆ t
)s(t̂
)ss(ˆ tt̂
β
nvs
nv
s
tv
t
s
s
tt ˆ
1
d
ˆd
d
ˆd
d
d
d
ˆdˆ
ntva /v ˆˆ ρ)( 2
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Position of A relative to B :
Velocity of A relative to B :
Acceleration of A relative to B : BABA
BABABAA/B
BAA/B
aaa
vvv rrr
rrr
/
/or
A
B
o
Ar
Br
BAr /
x
y
