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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
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Physics 111: Lecture 2, Pg 2
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:
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Physics 111: Lecture 2, Pg 3
Recall what you saw:Recall what you saw:
12
22
32
42
200 at
2
1tvxx ++=
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Physics 111: Lecture 2, Pg 4
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 =
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Physics 111: Lecture 2, Pg 5
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
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Physics 111: Lecture 2, Pg 6
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
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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
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8/37Physics 111: Lecture 2, Pg 8
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
/=
=
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9/37Physics 111: Lecture 2, Pg 9
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
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10/37Physics 111: Lecture 2, Pg 10
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
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11/37Physics 111: Lecture 2, Pg 11
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 !!
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12/37Physics 111: Lecture 2, Pg 12
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
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13/37Physics 111: Lecture 2, Pg 13
Vectors...Vectors...
q There are two common ways of indicating that something isa vector quantity:
Boldface notation:AA
Arrow notation:
AA = A
A
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14/37Physics 111: Lecture 2, Pg 14
Vectors...Vectors...
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 )
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15/37Physics 111: Lecture 2, Pg 15
Vectors...Vectors...
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.
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16/37Physics 111: Lecture 2, Pg 16
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
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17/37Physics 111: Lecture 2, Pg 17
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!!
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18/37Physics 111: Lecture 2, Pg 18
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
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19/37Physics 111: Lecture 2, Pg 19
Lecture 2,Lecture 2,Act 2Act 2
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}
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20/37Physics 111: Lecture 2, Pg 20
Lecture 2,Lecture 2,Act 2Act 2
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}
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21/37Physics 111: Lecture 2, Pg 21
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 = )
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Physics 111: Lecture 2, Pg 22
3-D Kinematics3-D Kinematics
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= ( )
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Physics 111: Lecture 2, Pg 23
3-D Kinematics3-D Kinematics
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)
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Physics 111: Lecture 2, Pg 24
2-D Kinematics2-D Kinematics
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
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Physics 111: Lecture 2, Pg 25
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
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Physics 111: Lecture 2, Pg 26
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
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Physics 111: Lecture 2, Pg 27
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
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Physics 111: Lecture 2, Pg 28
Problem...Problem...
q Use geometry to figure out v0x and v0y :
y
x
g
vv
v0x
v0y
Find v0x= |v| cos
.and v0y= |v| sin .
y0
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Physics 111: Lecture 2, Pg 29
Problem...Problem...
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!!
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Physics 111: Lecture 2, Pg 30
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
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Physics 111: Lecture 2, Pg 31
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
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Physics 111: Lecture 2, Pg 32
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
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Physics 111: Lecture 2, Pg 33
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 ! )
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Physics 111: Lecture 2, Pg 34
Shooting the Monkey...Shooting the Monkey...
q If there were no gravity, simply aimat the monkey
r= r0
r=v0t
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Physics 111: Lecture 2, Pg 35
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!
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Physics 111: Lecture 2, Pg 36
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
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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
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