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PHY 113 A General Physics I 9-9:50 AM MWF Olin 101 Plan for Lecture 5: Chapter 4 – Motion in two dimensions Position, velocity, and acceleration in two dimensions Two dimensional motion with constant acceleration. - PowerPoint PPT Presentation
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PHY 113 A Fall 2012 -- Lecture 5 19/7/2012
PHY 113 A General Physics I9-9:50 AM MWF Olin 101
Plan for Lecture 5:
Chapter 4 – Motion in two dimensions
1. Position, velocity, and acceleration in two dimensions
2. Two dimensional motion with constant acceleration
PHY 113 A Fall 2012 -- Lecture 5 29/7/2012
PHY 113 A Fall 2012 -- Lecture 5 39/7/2012
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 Fridays, it is good to stick to a plane.
PHY 113 A Fall 2012 -- Lecture 5 49/7/2012
i
j vertical direction (up)
horizontaldirection
jir ˆ)(ˆ)()( tytxt
PHY 113 A Fall 2012 -- Lecture 5 59/7/2012
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
dt
dvxx a
dt
dvyy a
PHY 113 A Fall 2012 -- Lecture 5 69/7/2012
Visualization of the position vector r(t) of a particle
r(t1)r(t2)
PHY 113 A Fall 2012 -- Lecture 5 79/7/2012
Visualization of the velocity vector v(t) of a particle
r(t1)
r(t2)
12
12
0
)()(lim
12 tt
tt
dt
dt
tt
rrrv
v(t)
PHY 113 A Fall 2012 -- Lecture 5 89/7/2012
Visualization of the acceleration vector a(t) of a particle
r(t1)
r(t2)
12
12
0
)()(lim
12 tt
tt
dt
dt
tt
vvva
v(t1) v(t2)
a(t1)
PHY 113 A Fall 2012 -- Lecture 5 99/7/2012
Animation of position vector components associated with trajectory motion
PHY 113 A Fall 2012 -- Lecture 5 109/7/2012
Animation of velocity vector and its components associated with trajectory motion
PHY 113 A Fall 2012 -- Lecture 5 119/7/2012
Animation of acceleration vector associated with motion along a path
PHY 113 A Fall 2012 -- Lecture 5 129/7/2012
Projectile motion (near earth’s surface)
i
j vertical direction (up)
horizontaldirection
ja
jiv
jir
ˆ)(
ˆ)(ˆ)()(
ˆ)(ˆ)()(
gt
tvtvt
tytxt
yx
PHY 113 A Fall 2012 -- Lecture 5 139/7/2012
Projectile motion (near earth’s surface)
jv
a
jir
v
jir
ˆ)(
ˆ)(ˆ)()(
ˆ)(ˆ)()(
gdt
dt
tvtvdt
dt
tytxt
yx
ii tgtt
gdt
dt
vvjvv
jv
a
0 that note ˆ
ˆ)(
PHY 113 A Fall 2012 -- Lecture 5 149/7/2012
Projectile motion (near earth’s surface)
jv
a
jir
v
jir
ˆ)(
ˆ)(ˆ)()(
ˆ)(ˆ)()(
gdt
dt
tvtvdt
dt
tytxt
yx
iii
i
tgttt
gtdt
dt
rrjvrr
jvr
v
0 that note ˆ
ˆ)(
221
PHY 113 A Fall 2012 -- Lecture 5 159/7/2012
Projectile motion (near earth’s surface)
ˆ)( jvr
v gtdt
dt i
jv
a ˆ)( gdt
dt
jvrr ˆ221 gttt ii
iiyi
iixi
yixii
v
v
vv
sin
cos
ˆˆ
v
v
jiv
PHY 113 A Fall 2012 -- Lecture 5 169/7/2012
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
)(
)(
PHY 113 A Fall 2012 -- Lecture 5 179/7/2012
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
gttvygttvyty
tvxtvxtx
iiiyii
iiixii
gtvgtvtv
vvtv
iiyiy
iixix
sin)(
cos)(
2
21
2
21
costan
coscossin
cos
ii
iiii
ii
i
ii
iiii
ii
i
v
xxgxxyxy
v
xxg
v
xxvyxy
v
xxt
PHY 113 A Fall 2012 -- Lecture 5 189/7/2012
Projectile motion (near earth’s surface) Summary of results
2
21
221
costan
sin)( cos)(
sin cos
ii
iiii
iiyiix
iiiiii
v
xxgxxyxy
gtvtvvtv
gttvytytvxtx
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.
PHY 113 A Fall 2012 -- Lecture 5 199/7/2012
h=7.1mqi=53o
d=24m=x(2.2s)
PHY 113 A Fall 2012 -- Lecture 5 209/7/2012
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)
PHY 113 A Fall 2012 -- Lecture 5 219/7/2012
sms
mv
vxd
tvxtx
oi
oi
iii
/12698.182.253cos
24
242.253cos)2.2(
cos
h=7.1mqi=53o
d=24m=x(2.2s)