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Kinematics of 3- or 2-dimensional motion. z. Position vector:. Average velocity:. Instantaneous velocity:. y. x. Average acceleration:. Instantaneous acceleration:. a || → magnitude of velocity a ┴ → direction of velocity. Equations of 3-D Kinematics for Constant Acceleration. - PowerPoint PPT Presentation
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Kinematics of 3- or 2-dimensional motionz
x y
v
1v
2v
Position vector: zyx iziyixr
Average velocity:t
r
tt
rrv
12
12
Instantaneous velocity:
zyx idt
dzi
dt
dyi
dt
dx
dt
rd
t
rv
lim
Average acceleration:t
v
tt
vva
12
12
Instantaneous acceleration: 2
2
limdt
rd
dt
vd
t
va
0t
0t
zyx idt
zdi
dt
ydi
dt
xda
2
2
2
2
2
2
aaa
||
a|| → magnitude of velocitya┴ → direction of velocity
Equations of 3-D Kinematics for Constant Acceleration
)1( tavv
0
)2( tvv
rr
2
00
)3( 200 2
1tatvrr
)4( )(2 020
2 rravv
Result: 3-D motion with constant acceleration is a superposition of three independent motions along x, y, and z axes.
a
Projectile Motionax=0 → vx=v0x=constay= -g → vy= voy- gtx = x0 + vox ty = yo + voy t – gt2/2v0x= v0 cos α0 v0y= v0 sin α0 tan α = vy / vx Exam Example 6: Baseball Projectile Data: v0=22m/s, α0=40o
x0 y0 v0x v0y ax ay x y vx vy t
0 0 ? ? 0 -9.8m/s2 ? ? ? ? ?Find: (a) Maximum height h;(b) Time of flight T;(c) Horizontal range R; (d) Velocity when ball hits the ground
Solution: v0x=22m/s·cos40o=+17m/s; v0y=22m/s·sin40o=+14m/s
(a)vy=0 → h = (vy2-v0y
2) / (2ay)= - (14m/s)2 / (- 2 · 9.8m/s2) = +10 m(b)y = (v0y+vy)t / 2 → t = 2y / v0y= 2 · 10m / 14m/s = 1.45 s; T = 2t =2.9 s (c)R = x = v0x T = 17 m/s · 2.9 s = + 49 m (d)vx = v0x , vy = - v0y
(examples 3.7-3.8, problems 3.12)
Motion in a Circle(a)Uniform circular motion: v = const
(b) Non-uniform circular motion: v ≠ const
Centripetal acceleration:
r
v
t
v
r
tv
r
r
v
v
r
v
r
r
t
v
dt
vdaa radc
2
2
lim
Magnitude: ac = v2 / rDirection to center: rr /
dt
vda
r
vaa radc
tan
2
;
0t
Exam Example 7: Ferris Wheel (problems 3.29)
Data: R=14 m, v0 =3 m/s, a|| =0.5 m/s2
Find:(a) Centripetal acceleration(b) Total acceleration vector(c) Time of one revolution T
Solution:(a) Magnitude: ac =a┴ = v2 / rDirection to center: rr /
(b)
)/(tan/tan ||1
||
22||||
aaaa
aaaaaa
||a
a
a
θ
(c)
||
||200
02||
||0
4
022
)2/(2
a
RavvT
RTvTa
TavTvTR
Relative Velocity
c
Flying in acrosswind
Correcting for a crosswind
Exam Example 8: Relative motion of a projectile and a target (problem 3.56)
Data: h=8.75 m, α=60o, vp0 =15 m/s, vtx =-0.45 m/s
0
y
x
0pv
tv
Find: (a) distance D to the target at the moment of shot, (b) time of flight t, (c) relative velocity at contact.
Solution: relative velocity(c) Final relative velocity:
(b) Time of flight(a) Initial distance
tp vvv
)(60cos 0
00 txptxxpx vvvvv ghvvghvvEqKinematic ppyyppy 2sin2)4.( 22
02
02
gvvgvt ppypy /)sin(/ 0 tvD x
Principles of Special Theory of Relativity (Einstein
1905):1. Laws of Nature are invariant for all inertial frames of reference. (Mikelson-Morly’s experiment (1887): There is no “ether wind” ! )2. Velocity of light c is the same for all inertial frames and sources.
Relativistic laws for coordinates transformation and addition of velocities are not Galileo’s ones:
Contraction of length:
Slowing down of time: Twin paradox
Slowing and stopping light in gases (predicted at Texas A&M)
y y’
x
x’
V
v
xvyv
22
2
22 /1
/,
/1 cV
cVxtt
cV
tVxx
2
22
2 /1
/1,
/1 cVv
cVvv
cVv
Vvv
x
yy
x
xx
22 /1 cVxx
22 /1 cVtt
Proved by Fizeau experiment (1851)of light dragging by water )/( ncv
Lorentztransformation
