9/5/2013users.wfu.edu/natalie/f13phy113/lecturenote/Lecture4for...9/5/2013 2 9/5/2013 PHY 113 A Fall...
Transcript of 9/5/2013users.wfu.edu/natalie/f13phy113/lecturenote/Lecture4for...9/5/2013 2 9/5/2013 PHY 113 A Fall...
9/5/2013
1
9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 1
PHY 113 A General Physics I11 AM-12:15 PM TR Olin 101
Plan for Lecture 4:
Chapter 4 – Motion in two dimensions1. Position, velocity, and acceleration
in two dimensions2. Two dimensional motion with
constant acceleration; parabolic trajectories
3. Circular motion
9/5/2013PHY 113 A Fall 2013 -- Lecture 4 2
Updated schedule
4.1,4.12,4.35,4.60
9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 3
Tentative exam dates:
1. September 26, 2013 – covering Chap. 1-8
2. October 31, 2013 – covering Chap. 9-13, 15-17
3. November 26, 2013 – covering Chap. 14, 19-22
Final exam: December 12, 2013 at 9 AM
9/5/2013
2
9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 4
From email questions on Webassign #3
Problem 4 (3.24)
Note: In this case the angle f is actually measured as north of east.
f d1b q1
f=b+q1
9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 5
From email questions on Webassign #3 -- continued
Problem 1 (3.11)
magnitude m
direction ° counterclockwise from the +x axis
graphically.
(d) Find A − 2B
A
2Bq
9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 6
From email questions on Webassign #3 -- continued
Problem 1 (3.29)
F1 F2
9/5/2013
3
9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 7
In the previous lecture, we introduced the abstract notion of a vector. In this lecture, we will use that notion to describe position, velocity, and acceleration vectors in two dimensions.
iclicker exercise:
Why spend time studying two dimensions when the world as we know it is three dimensions?
A. Because it is difficult to draw 3 dimensions.B. Because in physics class, 2 dimensions are hard
enough to understand.C. Because if we understand 2 dimensions,
extension of the ideas to 3 dimensions is trivial.D. On Thursdays, it is good to stick to a plane.
9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 8
i
j vertical direction (up)
horizontaldirection
jir ˆ)(ˆ)()( tytxt +=
9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 9
Vectors relevant to motion in two dimenstions
Displacement: r(t) = x(t) i + y(t) j
Velocity: v(t) = vx(t) i + vy(t) j
Acceleration: a(t) = ax(t) i + ay(t) j
dtdx
x =vdtdy
y =v
dtdvx
x =a dtdvy
y =a
9/5/2013
4
9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 10
Visualization of the position vector r(t) of a particle
r(t1)r(t2)
9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 11
Visualization of the velocity vector v(t) of a particle
r(t1)r(t2)
12
12
0
)()(lim12 tt
ttdtdt
tt
==
rrrvv(t)
9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 12
Visualization of the acceleration vector a(t) of a particle
r(t1)r(t2)
12
12
0
)()(lim12 tt
ttdtdt
tt
==
vvvav(t1) v(t2)
a(t1)
9/5/2013
5
9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 13
Figure from your text:
9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 14
Visualization of parabolic trajectory from textbook
9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 15
The trajectory graphs of the motion of an object y versus x as a function of time, look somewhat like a parabola:
CBxAxxy ++= 2)(
iclicker exerciseA. This is surprisingB. This is expectedC. No opinion
9/5/2013
6
9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 16
Projectile motion (near earth’s surface)
i
j vertical direction (up)
horizontaldirection
ja
jiv
jir
ˆ)(
ˆ)(ˆ)()(
ˆ)(ˆ)()(
gt
tvtvt
tytxt
yx
=
+=
+=
g = 9.8 m/s2
9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 17
Projectile motion (near earth’s surface)
jirv
jir
ˆ)(ˆ)()(
ˆ)(ˆ)()(
tvtvdtdt
tytxt
yx +==
+=
iii
ii
tgttt
tgtt
gdtdt
rrjvrr
vvjvv
jva
==+=
===
==
0 that noteˆ21
0 that noteˆ
ˆ)(
2
9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 18
Projectile motion (near earth’s surface)
jva
jirv
jir
ˆ)(
ˆ)(ˆ)()(
ˆ)(ˆ)()(
gdtdt
tvtvdtdt
tytxt
yx
==
+==
+=
iii
i
tgttt
gtdtdt
rrjvrr
jvrv
==+=
==
0 that noteˆ
ˆ)(
221
9/5/2013
7
9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 19
Projectile motion (near earth’s surface)
ˆ)( jvrv gtdtdt i ==
jva ˆ)( gdtdt ==
jvrr ˆ221 gttt ii +=
iiyi
iixi
yixii
v
v
vv
q
q
sin
cos
ˆˆ
v
v
jiv
=
=
+=
9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 20
Projectile motion (near earth’s surface)Trajectory equation in vector form:
ˆ)( jvv gtt i = jvrr ˆ221 gttt ii +=
Aside: The equations for position and velocity written in this way are call “parametric” equations. They are related to each other through the time parameter.
Trajectory equation in component form:
2
21 gttvyty
tvxtx
yii
xii
+=
+=gtvtv
vtv
yiy
xix
==
)()(
9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 21
Projectile motion (near earth’s surface)
Trajectory path y(x); eliminating t from the equations:
Trajectory equation in component form:
2
212
21 sin
cos
gttvygttvytytvxtvxtx
iiiyii
iiixii
+=+=
+=+=
q
q
gtvgtvtvvvtv
iiyiy
iixix
====
qqsin)(
cos)(
2
21
2
21
costan
coscossin
cos
+=
+=
=
ii
iiii
ii
i
ii
iiii
ii
i
vxxgxxyxy
vxxg
vxxvyxy
vxxt
qqq
q
9/5/2013
8
9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 22
Projectile motion (near earth’s surface)Summary of results
2
21
221
costan
sin)(cos)(sincos
+=
==
+=+=
ii
iiii
iiyiix
iiiiii
vxxgxxyxy
gtvtvvtvgttvytytvxtx
qqqq
iclicker exercise:
These equations are so beautiful thatA. They should be framed and put on the wall.B. They should be used to perfect my
tennis/golf/basketball/soccer technique.C. They are not that beautiful.
9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 23
Diagram of various trajectories reaching the same height h=1 m:
y
x
iclicker exercise: Which trajectory takes the longest time?A. Brown B. Green C. Same time for all.
9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 24
h=7.1mqi=53o
d=24m=x(2.2s)
9/5/2013
9
9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 25
Problem solving steps1. Visualize problem – labeling variables2. Determine which basic physical principle(s) apply3. Write down the appropriate equations using the variables defined in
step 1.4. Check whether you have the correct amount of information to solve the
problem (same number of known relationships and unknowns).5. Solve the equations.6. Check whether your answer makes sense (units, order of magnitude,
etc.).
h=7.1mqi=53o
d=24m=x(2.2s)
9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 26
sms
mv
vxd
tvxtx
oi
oi
iii
/12698.182.253cos
24242.253cos)2.2(
cos
==
===
+= q
h=7.1mqi=53o
d=24m=x(2.2s)
9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 27
Review Position x(t), Velocity y(t), Acceleration a(t) in one dimension:
==
==
t
t
t
t
dttatvdtdvta
dttvtxdtdxtv
0
0
')'()()(
')'()()(
Special case of constant acceleration a(t)=a0:
202
100
00
000
)(
)(:Then
0,0and:Suppose
tatvxtxtavtv
x)x(v)v(adtdv
++=
+=
===
9/5/2013
10
9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 28
Review – continued:Special case of constant acceleration a(t)=a0:
202
100
00
000
)(
)(:Then
0,0and:Suppose
tatvxtxtavtv
x)x(v)v(adtdv
++=
+=
===
initialvelocity
initialposition
2
0
002
1
0
000
0
0
20
200
)()()(
)(:algebrausingderivedResult
)()(2
:onacceleratiand velocity,position,between ipRelationsh
+
+=
=
=
avtva
avtvvxtx
avtvt
vtvxtxa
9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 29
Summary of equations –one-dimensional motion with constant acceleration
20
200
202
100
00
)()(2
)(
)(
vtvxtxa
tatvxtxtavtv
=
++=
+=
iclicker question:Why did I show you part of the derivation of the last equation?
A. Because professors like to torture physics studentsB. Because you will need to be able to prove the
equation yourselfC. Because the “proof” helps you to understand the
meaning of the equationD. All of the above E. None of the above
9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 30
i
j vertical direction (up)
horizontaldirection
jir ˆ)(ˆ)()( tytxt +=
Review: Motion in two dimensions:
9/5/2013
11
9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 31
Vectors relevant to motion in two dimenstions
Displacement: r(t) = x(t) i + y(t) j
Velocity: v(t) = vx(t) i + vy(t) j
Acceleration: a(t) = ax(t) i + ay(t) j
dtdx
x =vdtdy
y =v
dtdvx
x =a dtdvy
y =a
9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 32
Projectile motion (constant acceleration)(reasonable approximation near Earth’s surface)
i
j vertical direction (up)
horizontaldirection
ja
jiv
jir
ˆ)(
ˆ)(ˆ)()(
ˆ)(ˆ)()(
gt
tvtvt
tytxt
yx
=
+=
+=
9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 33
Projectile motion (near earth’s surface)
ˆ)( jvrv gtdtdt i ==
jva ˆ)( gdtdt ==
jvrr ˆ221 gttt ii +=
iiyi
iixi
yixii
v
v
vv
q
q
sin
cos
ˆˆ
v
v
jiv
=
=
+=
9/5/2013
12
9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 34
Projectile motion (near earth’s surface)Summary of component functions
2
21
221
costan
sin)(cos)(sincos
+=
==
+=+=
ii
iiii
iiyiix
iiiiii
vxxgxxyxy
gtvtvvtvgttvytytvxtx
qqqq
9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 35
Uniform circular motion – another example of motion in two-dimensions
animation from http://mathworld.wolfram.com/UniformCircularMotion.html
9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 36
iclicker question:Assuming that the blue particle is moving at constant speed around the circle what can you say about its acceleration?
A. There is no accelerationB. There is acceleration tangent to the circleC. There is acceleration in the radial direction of
the circleD. There is not enough information to conclude
that there is acceleration or not
9/5/2013
13
9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 37
Uniform circular motion – continued
ra ˆ
:ison acceleratilcentripeta theanddirection radialin the
onaccelerati then the,If
2
rv
vvv
c
fi
9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 38
Uniform circular motion – continued
r
ra
ra
ra
ˆ2
ˆ2
ˆ
2
2
2
rf
rT
rv
c
c
c
Trv 2
In terms of time period T for one cycle:
In terms of the frequency f of complete cycles:πfrv
Tf 2;1
9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 39
r
ra
ra
ˆm/s03.0
ˆm/s10637186400
2
s86400s/min 60min/hr 60hr 24
ˆ2
2
232
2
c
c
T
rT
9/5/2013
14
9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 40
iclicker exercise:During circular motion, what happens when the speed of the object changes?A. There is only tangential accelerationB. There is only radial accelerationC. There are both radial and tangential
accelerationD. I don’t need to know this because it never
happens.
9/5/2013 PHY 113 A Fall 2013 -- Lecture 4 41
r̂ θ̂
Circular motion
θa
ra
θv
ˆ
ˆ
ˆ:Assume2
dtdvrv
tvt
t
c
r̂ θ̂ca ta
tc aaa