Physics 201, Lecture 5physics.wisc.edu/undergrads/courses/fall2017/201/phy201_lect5.pdfPhysics 201,...
Transcript of Physics 201, Lecture 5physics.wisc.edu/undergrads/courses/fall2017/201/phy201_lect5.pdfPhysics 201,...
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Physics 201, Lecture 5 Today’s Topics
n Motion in 2D (Chap 4.1-4.3):
n 2D Kinematical Quantities (sec. 4.1) n 2D Kinematics with Constant Acceleration (sec. 4.2) n 2D Projectile (Sec 4.3)
n Expected from Preview:
n Displacement, velocity, acceleration in vector form n Kinematics equations in 2D form n Projectile motion in 2D.
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Review: Kinematical Quantities in 1D q Displacement: change of position from (x1,t1 ) (x2,t2)
q Velocity: rate of position change.
q Acceleration: rate of velocity change.
12 xxx −=Δ
dtd
txv xvavg =
Δ
Δ= ,
dtd
tva vaavg =
Δ
Δ= ,
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Kinematical Quantities in Vector Form q Displacement:
q Velocity (average and instantaneous):
q Acceleration (average and instantaneous):
if rrr −=Δ
dtvd
tva
tva
t
=Δ
Δ=
Δ
Δ=
=Δ 0lim, avg
dtrd
trv
trv
t
=Δ
Δ=
Δ
Δ=
=Δ 0lim, avg
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Kinematics from 1-D to 2/3-D q Constant Position:
q Constant Velocity:
q Constant Acceleration:
000 === avxx , , 000 === avrr , ,
0000 =+== atvxxvv , , 0000 =+== atvrrvv , ,
2000
000
21 tatvxx
tavvaa
++=
+== ,
2000
000
21 tatvrr
tavvaa
++=
+== ,
No change in physics, only new mathematical treatment!
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Kinematical Quantities in (x,y,z) q Like all vectors, kinematical quantities x, v, a, can be
decomposed into their x,y,z components:
k j ia
k j iv
k j ir
ˆˆˆ
ˆˆˆ
ˆˆˆ
zyx
zyx
zyx
aaa
vvv
rrr
++=
++=
++=
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Kinematics in x, y (,z) Coordination q Kinetics with constant acceleration:
§ Original Vector Form
§ In x,y (,z) system:
(After class exercise: write down equation for z component) q Important Note: For constant a, kinematics for x,y,z are
independent of each other. The only common parameter is time t.
2000000 2
1 tatvxxtavvaa xxxxxxx ++=+== , ,
2000000 2
1 tatvrrtavvaa ++=+== , ,
2000000 2
1 tatvyytavvaa yyyyyyy ++=+== , ,
x
y
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Practical Technique: Decompose Kinematic Parameters
q Decomposition
q Inversely:
i (x)
j (y)
θ
= x i +y j r
r
x
y
x = r cosθ y = r sinθ
v = vx i +vy j
i (x)
j (y)
θv
v vy
vx
vx = v cosθv vy = v sinθv
a = ax i +ay j
i (x)
j (y)
θa
a ay
ax
ax = a cosθa ay = a sinθa
xyyxr
=
+=
θtan
22
x
yv
yx
vv
vvv
=
+=
θtan
22
x
ya
yx
aa
aaa
=
+=
θtan
22
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Two Dimensional Free Fall q Free fall: Motion under sole influence of gravity
q Two dimensional free fall (projectile motion): vi has a horizontal component
q Quiz:During the flight of a projectile, its acceleration: § always ax=0, ay=-g § ax=g/2, ay=-g/2 § ax=g/sqrt(2), ay=-g/sqrt(2) § depends on the projection angle
(neglect air friction)
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Projectile horizontal: constant v ax=0, vx=vxi, x=x0+vxit vertical: constant a ay=-g, vy=vyi - gt y=y0+vyit - ½ gt2
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Demo: Ballistic car
The ball comes back down into where it was launched !
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Demo: Horizontal and Vertical Components of a Projectile Trajectory
Also demo: Jumping up in a car
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Exercise: Projectile Motion q A projectile is shot at an initial speed vi at an angle θ. After which, it
is in motion only under gravitational force. Find position at any time, air time, the range, maximum height.
Step 1: decompose vi vi =(vix,viy) = (vicosθi, visinθi), ax=0, ay= - g § Position at any t (Treat x, and y separately ) : x(t) = 0 + vixt = vicosθi t y(t) = 0 + viyt + ½ ay t2 = visinθi t - ½ gt2 § air time (think vertically Δy=0) : at tB , Δy=0 tB= 2 visinθi /g = Tair
§ Range (Think horizontally) R= vicosθi Tair = 2 vi
2sinθicosθi /g § Maximum height ( think vertically vy_A=0): at A: vy=0 tA= visinθi /g (= ½ tB !), h= yA = viy tA – ½ gtA
2 =½ vi2sin2θi /g
(or use 02 = viy2 – 2hg)
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Quiz: Which Ship Gets Hit First q A battle ship simultaneously fires at two target ships at
different distance with identical canons. Which ship gets hit first? (ignore ship height). A, B, same
t=2visinθi /g
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Quiz’: Which Ship Gets Hit First q A battle ship simultaneously fires at two target ships at
different distance (note: the two canons may not be identical) Which ship gets hit first? (ignore ship height). A, B, same, not enough information
A B
hint: Think vertically!
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Exercise: Jumping Over a Stream q A bridge that was 5.9 m long has been washed out by the rain several
days ago. How fast must a car be going to successfully jump the stream? Although the road is level on both sides of the bridge, the road on the far side is 3 m lower than the road on this side.
q Solution:
Ø Set up axes as shown
Ø Think vertically: Δy =-3m, ay =-g, viy=0 , use Δy= viy t + ½ ayt2 t=sqrt(2*Δy/ay) = 0.782s Ø Now think horizontally: vx = Δx/t = 5.9/0.782 = 7.54m/s
x
y
3m
5.9m
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Same Exercise, Different Axes q A bridge that was 5.9 m long has been washed out by the rain several
days ago. How fast must a car be going to successfully jump the stream? Although the road is level on both sides of the bridge, the road on the far side is 3 m lower than the road on this side.
q Solution’:
Ø Choose another orientation
Ø Think vertically: Δy =+3m, ay =+g, viy=0 , use Δy= viy t + ½ ayt2 t=sqrt(2*Δy/ay) = 0.782s Ø Now think horizontally: vx = Δx/t = 5.9/0.782 = 7.54m/s
x
y
3m
5.9m
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Demo: Projectile’s Maximum Range q At what shooting angle the canon can achieve maximum range?
R= 2 vi2sinθicosθi /g = vi
2sin2θi/g R=Rmax when θi= 45o (Rmax = vi
2/g)
See demo
75o and 15o have same range 60o and 30o have same range In general: θ and 90o – θ have same range
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Non-Symmetrical Projectile Motion
q A stone is thrown from the top of a building as shown. What is the horizontal distance x from the building to the landing position?
q Procedures: Step 1: vxi = v0 cos30o , vyi = v0 sin30o Step 2: think vertically. Δy = viy t – ½ gt2 to get flight time t. (need to solve quadratic eq.) Step 3: think horizontally. x = vix t using t from step 2.
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End-of-Lecture Quizzes q Consider two free fall processes A and B as shown.
q If thrown at the same time, which hit the ground first? (hint: sin30o=0.5) A, B, same time q Which one hits the ground at higher speed? A, B, same
Vi=20 m/s θi=30o