Physics 201, Lecture 5physics.wisc.edu/undergrads/courses/fall2017/201/phy201_lect5.pdfPhysics 201,...

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.

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= ,

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

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!

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

++=

++=

++=

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

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

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)

Projectile horizontal: constant v ax=0, vx=vxi, x=x0+vxit vertical: constant a ay=-g, vy=vyi - gt y=y0+vyit - ½ gt2

Demo: Ballistic car

The ball comes back down into where it was launched !

Demo: Horizontal and Vertical Components of a Projectile Trajectory

Also demo: Jumping up in a car

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)

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

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!

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

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

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

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.

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