physics 101 lecture 4 motion in 2d and 3d · 2018. 12. 18. · motion of a turtle notice, you can...
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Physics 101Lecture 4
Motion in 2D and 3D
Assist. Prof. Dr. Ali ÖVGÜNEMU Physics Department
www.aovgun.com
February 5-8, 2013
Vector and its componentsq The components are the
legs of the right triangle whose hypotenuse is A
yx AAA!!!
+=
2 2 1tan yx y
x
AA A A and
Aθ − ⎛ ⎞
= + = ⎜ ⎟⎝ ⎠
)sin()cos(
⎩⎨⎧
==
θθ
AAAA
y
x
Or,( ) ( )
( )⎪⎪⎩
⎪⎪⎨
⎧
⎟⎟⎠
⎞⎜⎜⎝
⎛==
+=
−
x
y
x
y
yx
AA
AA
AAA
1
22
tanor tan θθ
!
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q Kinematic variables in one dimensionn Position: x(t) mn Velocity: v(t) m/sn Acceleration: a(t) m/s2
q Kinematic variables in three dimensionsn Position: mn Velocity: m/sn Acceleration: m/s2
q All are vectors: have direction and magnitudes
Motion in two dimensions
kvjvivtv zyxˆˆˆ)( ++=!
y
x
z
ij
k
x
kzjyixtr ˆˆˆ)( ++=!
kajaiata zyxˆˆˆ)( ++=!
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q In one dimension
q In two dimensionsn Position: the position of an object is
described by its position vector --always points to particle from origin.
n Displacement:
x1 (t1) = - 3.0 m, x2 (t2) = + 1.0 mΔx = +1.0 m + 3.0 m = +4.0 m
Position and Displacement
)(tr!
12 rrr !!! −=Δ
jyix
jyyixx
jyixjyixr
ˆˆ
ˆ)(ˆ)(
)ˆˆ()ˆˆ(
1212
1122
Δ+Δ=
−+−=
+−+=Δ!
)()( 1122 txtxx −=Δ
12 rrr !!! −=Δ
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q Average velocity
q Instantaneous velocity
q v is tangent to the path in x-y graph;
Average & Instantaneous Velocity
dtrd
trvv
tavg
!!!! =
ΔΔ=≡
→→ 00tlimlim
jvivjtyi
txv yavgxavgavg
ˆˆˆˆ,, +=
ΔΔ+
ΔΔ=!
trvavg Δ
Δ≡!
!
jvivjdtdyi
dtdx
dtrdv yx
ˆˆˆˆ +=+==!
!
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Motion of a Turtle
A turtle starts at the origin and moves with the speed of v0=10 cm/s in the direction of 25° to the horizontal.
(a) Find the coordinates of a turtle 10 seconds later.
(b) How far did the turtle walk in 10 seconds?
February 5-8, 2013
Motion of a TurtleNotice, you can solve the equations independently for the horizontal (x) and vertical (y) components of motion and then combine them!
yx vvv !!! +=0
0 0 cos 25 9.06 cm/sxv v= =!q X components:
q Y components:
q Distance from the origin:
0 90.6 cmxx v tΔ = =
0 0 sin 25 4.23 cm/syv v= =! 0 42.3 cmyy v tΔ = =
cm 0.10022 =Δ+Δ= yxd
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q Average acceleration
q Instantaneous acceleration
q The magnitude of the velocity (the speed) can changeq The direction of the velocity can change, even though the
magnitude is constantq Both the magnitude and the direction can change
Average & Instantaneous Acceleration
dtvd
tvaa
tavg
!!!! =
ΔΔ=≡
→→ 00tlimlim
jaiajtv
itva yavgxavg
yxavg
ˆˆˆˆ,, +=
ΔΔ
+ΔΔ=!
tvaavg Δ
Δ≡!
!
jaiajdtdv
idtdv
dtvda yx
yx ˆˆˆˆ +=+==!
!
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q Position
q Average velocity
q Instantaneous velocity
q Acceleration
q are not necessarily same direction.
Summary in two dimensionjyixtr ˆˆ)( +=!
jaiajdtdv
idtdv
dtvd
tvta yx
yx
tˆˆˆˆlim)(
0+=+==
ΔΔ=
→
!!!
jvivjtyi
tx
trv yavgxavgavg
ˆˆˆˆ,, +=
ΔΔ+
ΔΔ=
ΔΔ=!
!
jvivjdtdyi
dtdx
dtrd
trtv yxt
ˆˆˆˆlim)(0
+=+==ΔΔ=
→
!!!
dtdxvx ≡ dt
dyvy ≡
2
2
dtxd
dtdva x
x =≡ 2
2
dtyd
dtdv
a yy =≡
)( and ),( , tatv(t)r !!!
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Motion in two dimensions
tavv !!! += 0
q Motions in each dimension are independent componentsq Constant acceleration equations
q Constant acceleration equations hold in each dimension
n t = 0 beginning of the process;n where ax and ay are constant;n Initial velocity initial displacement ;
221
0 tatvrr !!!! +=−
tavv yyy += 0
221
00 tatvyy yy +=−
)(2 02
02 yyavv yyy −+=
tavv xxx += 02
21
00 tatvxx xx +=−
)(2 02
02 xxavv xxx −+=
jaiaa yxˆˆ+=!
jvivv yxˆˆ
000 +=! jyixr ˆˆ000 +=!
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q 2-D problem and define a coordinate system: x- horizontal, y- vertical (up +)
q Try to pick x0 = 0, y0 = 0 at t = 0q Horizontal motion + Vertical motionq Horizontal: ax = 0 , constant velocity
motionq Vertical: ay = -g = -9.8 m/s2, v0y = 0 q Equations:
Projectile Motion
221 gttvyy iyif −+=tavv yyy += 0
221
00 tatvyy yy +=−
)(2 02
02 yyavv yyy −+=
tavv xxx += 0
221
00 tatvxx xx +=−
)(2 02
02 xxavv xxx −+=
Horizontal Vertical
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q X and Y motions happen independently, so we can treat them separately
q Try to pick x0 = 0, y0 = 0 at t = 0q Horizontal motion + Vertical motionq Horizontal: ax = 0 , constant velocity
motionq Vertical: ay = -g = -9.8 m/s2
q x and y are connected by time tq y(x) is a parabola
Projectile Motion
gtvv yy −= 0
221
00 gttvyy y −+=xx vv 0=
tvxx x00 +=Horizontal Vertical
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q 2-D problem and define a coordinate system.q Horizontal: ax = 0 and vertical: ay = -g.q Try to pick x0 = 0, y0 = 0 at t = 0.q Velocity initial conditions:
n v0 can have x, y components.n v0x is constant usually.n v0y changes continuously.
q Equations:
Projectile Motion
000 cosθvv x =
Horizontal Vertical
000 sinθvv x =
gtvv yy −= 0
221
00 gttvyy y −+=xx vv 0=
tvxx x00 +=
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q Initial conditions (t = 0): x0 = 0, y0 = 0 v0x = v0 cosθ0 and v0y = v0 sinθ0
q Horizontal motion:
q Vertical motion:
q Parabola;n θ0 = 0 and θ0 = 90 ?
Trajectory of Projectile Motion
221
00 gttvy y −+=
xx v
xttvx0
0 0 =⇒+=
2
000 2 ⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟⎟⎠
⎞⎜⎜⎝
⎛=
xxy v
xgvxvy
2
022
00 cos2
tan xvgxy
θθ −=
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q Initial conditions (t = 0): x0 = 0, y0 = 0 v0x = v0 cosθ0 and v0x = v0 sinθ0, then
What is R and h ?
Horizontal Vertical
221
000 gttv y −+=tvx x00+=
gv
gvvtvxxR x
0200000
002sinsincos2 θθθ ===−=
gv
gv
t y 000 sin22 θ==
2
02
21
00 222⎟⎠⎞⎜
⎝⎛−=−=−= tgtvgttvyyh yhhy
gvh2sin 0
220 θ=
yy
yyy vgv
gvgtvv 00
00
2−=−=−=
h
gtvv yy −= 0
221
00 gttvyy y −+=xx vv 0=
tvxx x00 +=
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Projectile Motion at Various Initial Angles
q Complementary values of the initial angle result in the same rangen The heights will be
different
q The maximum range occurs at a projection angle of 45o
gvR φ2sin20=
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q Position
q Average velocity
q Instantaneous velocity
q Acceleration
q are not necessarily in the same direction.
Summaryjyixtr ˆˆ)( +=!
jaiajdtdv
idtdv
dtvd
tvta yx
yx
tˆˆˆˆlim)(
0+=+==
ΔΔ=
→
!!!
jvivjtyi
tx
trv yavgxavgavg
ˆˆˆˆ,, +=
ΔΔ+
ΔΔ=
ΔΔ=!
!
jvivjdtdyi
dtdx
dtrd
trtv yxt
ˆˆˆˆlim)(0
+=+==ΔΔ=
→
!!!
dtdxvx ≡ dt
dyvy ≡
2
2
dtxd
dtdva x
x =≡ 2
2
dtyd
dtdv
a yy =≡
)( and ),( , tatv(t)r !!!
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q If a particle moves with constant acceleration a, motion equations are
q Projectile motion is one type of 2-D motion under constant acceleration, where ax = 0, ay = -g.
Summary
jtatvyitatvxjyixr yiyiixixiifffˆ)(ˆ)(ˆˆ 2
212
21 +++++=+=!
jtavitavjvivtv yiyxixfyfxfˆ)(ˆ)(ˆˆ)( +++=+=!
tavv i!!! +=
221 tatvrr iif!!!! ++=
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Example: 1
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Example: 2A movie stunt driver on a motorcycle speeds horizontally off a 50m high cliff. If the motorcycle will land 90m from the base of the cliff, (ignore any kind of friction or resistance) (a) Find the time of flight, (b) Find its initial speed in x-direction ,
(c) Find its acceleration vector just before
hitting the ground.
During volcanic eruptions, chunks of solid rock can be blasted out of the volcano; these projectiles are called volcanic bombs. The figure below shows a cross section of Mt. Fuji, in Japan. From the vent A to the foot of the volcano at B, the vertical distance is h = 3.30km and horizontal distance is d = 940m. Neglecting air resistance,
(a) calculate the time of flight, and (4 P)
(b) calculate the initial speed of the projectile. (2P)
February 5-8, 2013
A cliff diver is about to jump down a cliff of height 35.0m, at the bottom of the cliff there is a 5m wide rock bank next to the sea. Calculate the minimum horizontal initial velocity the cliff jumper has to push off. (No initial velocity component in y direction)
Example: 3
Problem:1
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A projectile is fired at an initial velocity of 35.0 m/s at an angle of 30.0 degrees above the horizontal from the roof of a building 30.0 m high, as shown. Find
a) The maximum height of the projectile
b) The time to rise to the top of the trajectory
c) The total time of the projectile in the air
d)The velocity of the projectile at the ground
e)The range of the projectile
Problem:2aA plane drops a package of supplies to a party of explorers. If the plane is traveling horizontally at 40 m/s and is 100 m above the ground. Where does the package strike the ground?
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Problem:3
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Problem:4
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Problem:5
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Problem:6At what initial speed must the basketball player in
Figure throw the ball, at angle u0 = 37° above the horizontal, to make the foul shot?
The horizontal distances are = 30 cm and , = 440cm and the heights are = 220 cm and . = 300 cm.
February 5-8, 2013
Uniform circular motion
Constant speed, or,constant magnitude of velocity
Motion along a circle:Changing direction of velocity
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Circular Motion: Observationsq Object moving along a
curved path with constant speedn Magnitude of velocity: samen Direction of velocity: changingn Velocity: changingn Acceleration is NOT zero!n Net force acting on the
object is NOT zeron “Centripetal force” amFnet
!!=
February 5-8, 2013
q Centripetal acceleration
q Direction: Centripetal
Uniform Circular Motion
rv
tva
rv
rvtr
tv
rrvv
rr
vv
r
2
2
so,
=ΔΔ=
=ΔΔ=
ΔΔ
Δ=ΔΔ=Δ
O x
y
riR
A Bvi
rf
vfΔr
vi
vf
Δv = vf - vi
February 5-8, 2013
Uniform Circular Motionq Velocity:
n Magnitude: constant vn The direction of the velocity is
tangent to the circleq Acceleration:
n Magnitude: n directed toward the center of
the circle of motionq Period:
n time interval required for one complete revolution of the particle
rvac2
=
rvac2
=
vrT π2=
vac!! ⊥
February 5-8, 2013
Problem:8The bobsled track contains turns
with radii of 33 m and 24 m.
Find the centripetal acceleration
at each turn for a speed of
34 m/s. Express answers as
multiples of .
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