Gerak 1 Dimensi&Vector

8/14/2019 Gerak 1 Dimensi&Vector

1/37

Physics 111: Lecture 2, Pg 1

Physics 111: Lecture 2Physics 111: Lecture 2

Todays AgendaTodays Agendaq Recap of 1-D motion with constant acceleration

q 1-D free fall

example

q Review of Vectors

q 3-D Kinematics

Shoot the monkey

Baseball

Independence ofxand ycomponents


2/37


Review:Review:

q For constant acceleration we found:

atvv0+=

200 at

2

1tvxx ++=

consta =

x

a

vt

t

t

v)(v2

1v

)x2a(xvv

0av

020

2

+=

=

q From which we derived:


3/37


Recall what you saw:Recall what you saw:

12

22

32

42

200 at

2

1tvxx ++=


4/37


1-D Free-Fall1-D Free-Fall

q

This is a nice example of constant acceleration (gravity):q In this case, acceleration is caused by the force of gravity:

Usually pick y-axis upward

Acceleration of gravity is down:

y

ay= g

gtvvy

0y =

2y00

tg2

1tvyy +=

y

a

vt

t

gay =


5/37


Gravity facts:Gravity facts:

q gdoes not depend on the nature of the material!

Galileo (1564-1642) figured this out without fancy clocks& rulers!

q demo - feather & penny in vacuum

q Nominally, g= 9.81 m/s2

At the equator g= 9.78 m/s2

At the North pole g= 9.83 m/s2

q More on gravity in a few lectures!

Penny& feather

Ball

w/ cup


6/37


Problem:Problem:

q The pilot of a hovering helicopterdrops a lead brick from a heightof1000 m. How long does it taketo reach the ground and how fastis it moving when it gets there?(neglect air resistance)

1000 m


7/37Physics 111: Lecture 2, Pg 7

Problem:Problem:

q First choose coordinate system. Origin and y-direction.

q Next write down position equation:

q Realize that v0y= 0.

20y0 gt

21tvyy +=

20 gt

2

1yy =

1000 m

y = 0

y



Problem:Problem:

q Solve for time twhen y = 0giventhat y0= 1000 m.

q Recall:

q Solve forvy:

y0= 1000 m

y

s314sm819

m10002

g

y2t

20 .

.=

==

20

gt2

1yy -=

y = 0

)( 02

y02y yya2vv -- =

sm140

gy2v 0y

/=

=



Lecture 2,Lecture 2,Act 1Act 1

1D free fall1D free fallq Alice and Bill are standing at the top of a cliff of heightAlice and Bill are standing at the top of a cliff of height

HH. Both throw a ball with initial speed. Both throw a ball with initial speedvv00, Alice straight, Alice straight

downdown and Bill straightand Bill straight upup. The speed of the balls when. The speed of the balls whenthey hit the ground arethey hit the ground arevvAA andandvvBB respectively.respectively. WhichWhich

of the following is true:of the following is true:

(a)(a) vvAA > vvBB

vv00

vv00

BillBillAliceAlice

HH

vvAA vvBB



Lecture 2,Lecture 2,Act 1Act 11D Free fall1D Free fall

q Since the motion up and back down is symmetric, intuitionshould tell you that v = v0

We can prove that your intuition is correct:

vv00

BillBill

HH

vv= v= v00

( ) 0HHg2vv 202

== )(Equation:Equation:

This looks just like Bill threwThis looks just like Bill threwthe ball down with speedthe ball down with speed vv00, so, so

the speed at the bottom shouldthe speed at the bottom shouldbe the same as Alices ball.be the same as Alices ball.

y = 0y = 0



Lecture 2, Act 1Lecture 2, Act 11D Free fall1D Free fall

q We can also just use the equation directly:We can also just use the equation directly:

( )H0g2vv 202

= )(Alice:Alice:

vv00

vv00

AliceAlice BillBill

y = 0y = 0

( )H0g2vv 202 = )(Bill:Bill:

same !!same !!



Vectors (review):Vectors (review):

q In 1 dimension, we could specify direction with a + or- sign.For example, in the previous problem ay= -getc.

q In 2 or3 dimensions, we need more than a sign to specify thedirection of something:

q To illustrate this, consider theposition vectorrr in 2 dimensions.

ExampleExample: Where is Chicago?

Choose origin at UrbanaUrbana

Choose coordinates ofdistance (miles), anddirection (N,S,E,W)

In this case rris a vector thatpoints 120 milesnorth.

Chicago

Urbana

rr



Vectors...Vectors...

q There are two common ways of indicating that something isa vector quantity:

Boldface notation:AA

Arrow notation:

AA = A

A




q The components ofrrare its (x,y,z) coordinates

rr= (rx,ry,rz) = (x,y,z)

q Consider this in 2-D (since its easier to draw):

rx= x = r cos

ry= y = r sin

y

x

(x,y)

where r = |rr|

rr =

arctan( y / x )




q The magnitude (length) ofrris found using the Pythagoreantheorem:

r = = +r x y2 2rr

y

x

q The length of a vector clearly does notnot depend on its direction.



Unit Vectors:Unit Vectors:

q A Unit VectorUnit Vectoris a vector having length 1and no units

q It is used to specify a direction

q Unit vectoruu points in the direction ofUU

Often denoted with a hat: uu =

q Useful examples are the Cartesianunit vectors [i, j, ki, j, k]

point in the direction of thex, yand zaxes

UU

x

y

z

i

jj

k



Vector addition:Vector addition:

q Consider the vectorsAA and BB. FindAA+BB.

AA

BB

AA BB AA BB

CC=AA+BB

q We can arrange the vectors as we want, as long as wemaintain their length and direction!!



Vector addition using components:Vector addition using components:

q ConsiderCC=AA + BB.

(a) CC = (Axii+ Ayjj) + (Bxii+ Byjj) = (Ax+ Bx)ii+ (Ay+ By)jj

(b) CC = (Cxii+ Cyjj)

q Comparing components of(a) and (b):

Cx= Ax + Bx

Cy= Ay+ By

CC

BxAA

ByBB

Ax

Ay




VectorsVectorsq VectorA = {0,2,1}

q VectorB = {3,0,2}

q VectorC = {1,-4,2}

What is the resultant vector, D, fromadding A+B+C?

(a)(a) {3,5,-1}{3,5,-1} (b)(b) {4,-2,5}{4,-2,5} (c)(c){5,-2,4}{5,-2,4}




SolutionSolution

D = (AXi+ AYj+ AZk) + (BXi+ BYj+ BZk) + (CXi+ CYj+ CZk)

= (AX + BX + CX)i+ (AY + BY+ CY)j+ (AZ + BZ + CZ)k

= (0 + 3 + 1)i+ (2 + 0 - 4)j+ (1 + 2 + 2)k

= {4,-2,5}



3-D Kinematics3-D Kinematics

q The position, velocity, and acceleration of a particle in 3dimensions can be expressed as:

rr= xii+ yjj+ zkk

vv= vxii+ vyjj+ vzkk (ii,jj,kkunit vectors ) aa = axii+ ayjj+ azkk

q We have already seen the 1-D kinematics equations:

adv

dt

d x

dt= =

2

2v

dx

dt= x x(t = )


22/37



q For 3-D, we simply apply the 1-D equations to each of thecomponent equations.

q Which can be combined into the vector equations:

rr= rr(t) vv= drr/ dt aa = d2rr/ dt2

ad x

dtx =

2

2

v

dx

dtx =

x x(t = )

ad y

dty =

2

2

v

dy

dty =

y y t= ( )

ad z

dtz =

2

2

v

dz

dtz =

z z t= ( )


23/37



q So for constant acceleration we can integrate to get:

aa = const

vv= vv0+ aa t

rr= rr0+ vv0 t +1/2aa t

2

(where aa, vv, vv0, rr, rr0, are all vectors)


24/37



q Most 3-D problems can be reduced to 2-D problems whenacceleration is constant:

Choose yaxis to be along direction of acceleration

Choosexaxis to be along the other direction of motion

q ExampleExample: Throwing a baseball (neglecting air resistance)

Acceleration is constant (gravity)

Choose yaxis up: ay = -g

Choosexaxis along the ground in the direction of thethrow

lost marbles


25/37


x and y components of motion are independent.

q A man on a train tosses a ball straight up in the air.

View this from two reference frames:

Reference frameon the moving train.

Reference frame

on the ground.

Cart


26/37


Problem:Problem:

q Mark McGwire clobbers a fastball toward center-field. Theball is hit 1 m (yo ) above the plate, and its initial velocity is

36.5 m/s(v) at an angle of30o () above horizontal. Thecenter-field wall is 113 m (D) from the plate and is 3 m (h)high.

What time does the ball reach the fence?What time does the ball reach the fence?

Does Mark get a home run?Does Mark get a home run?

vvh

D

y0


27/37


Problem...Problem...

q Choose yaxis up.

q Choosexaxis along the ground in the direction of the hit.

q Choose the origin (0,0) to be at the plate.

q Say that the ball is hit at t = 0, x = x0= 0

Equations of motion are:

vx= v0x vy= v0y - gt

x = vxt y = y0 + v0yt -1/2 gt

2


28/37



q Use geometry to figure out v0x and v0y :

y

x

g

vv

v0x

v0y

Find v0x= |v| cos

.and v0y= |v| sin .

y0


29/37



q The time to reach the wall is: t = D / vx (easy!)

q We have an equation that tell us y(t) = y0 + v0yt + a t2/ 2

q So, were done....now we just plug in the numbers:

q Find:

vx= 36.5 cos(30) m/s = 31.6 m/s

vy= 36.5 sin(30) m/s = 18.25 m/s

t= (113 m) / (31.6 m/s) = 3.58 s

y(t) = (1.0 m) + (18.25 m/s)(3.58 s) -(0.5)(9.8

m/s2)(3.58 s)2

= (1.0 + 65.3 - 62.8) m = 3.5 mm

Since the wall is 3 m high, Mark gets the homer!!


30/37


Lecture 2,Lecture 2,Act 3Act 3Motion in 2DMotion in 2D

q Two footballs are thrown from the same point on a flat field.Both are thrown at an angle of30o above the horizontal.

Ball 2has twice the initial speed ofball 1. Ifball 1 is caughta distance D1 from the thrower, how far away from the

throwerD2will the receiver ofball 2be when he catches it?

(a) D2= 2D1 (b) D2= 4D1 (c) D2= 8D1


31/37


Lecture 2,Lecture 2,Act 3Act 3SolutionSolution

q The distance a ball will go is simplyx= (horizontal speed) x (time in air) = v0x t

2y00

tg

2

1tvyy +=

q To figure out time in air, consider theequation for the height of the ball:

0tg2

1tv 2

y0 =q When the ball is caught, y = y0

0tg2

1vt

y0=

g

v2t y0=

t = 0 (time of throw)

(time of catch)

twosolutions


32/37


Lecture 2,Lecture 2,Act 3Act 3SolutionSolution

g

v2t

y0=q So the time spent in the air is proportional to v0y:

q Since the angles are the same, both v0y and v0x forball 2

are twice those ofball 1.

ball 1

ball 2

v0y ,1

v0x ,1

v0y ,2

v0x ,2

v0,1

v0,2

q Ball 2 is in the airtwice as long as ball 1, but it also has twicethe horizontal speed, so it will go 44 times as far!!

x = v0x t


33/37


Shooting the MonkeyShooting the Monkey(tranquilizer gun)(tranquilizer gun)

q Where does the zookeeperaim if he wants to hit the monkey?

( He knows the monkey willlet go as soon as he shoots ! )


34/37


Shooting the Monkey...Shooting the Monkey...

q If there were no gravity, simply aimat the monkey

r= r0

r=v0t


35/37


Shooting the Monkey...Shooting the Monkey...

rr= vv0t -1/2ggt

2

q With gravity, still aim at the monkey! rr= r0 -

1

/2ggt2

Dart hits themonkey!


36/37


Recap:Recap:Shooting the monkey...Shooting the monkey...

xx==xx00

yy= -1/2ggt2

q This may be easier to think about.

Its exactly the same idea!!

xx== vv00tt

yy= -1/2ggt2


37/37

Recap of Lecture 2Recap of Lecture 2

q Recap of 1-D motion with constant acceleration. (Text: 2-3)

q 1-D Free-Fall (Text: 2-3)

example

q Review of Vectors (Text: 3-1 & 3-2)

q 3-D Kinematics (Text: 3-3 & 3-4)

Shoot the monkey (Ex. 3-11)

Baseball problem

Independence ofxand ycomponents

q Look at textbook problems Chapter 3: # 35, 39, 69, 97Chapter 3: # 35, 39, 69, 97

Gerak 1 Dimensi&Vector

Documents

Transcript of Gerak 1 Dimensi&Vector