PHY 113 A General Physics I
11 AM-12:15 PM TR Olin 101
Plan for Lecture 4:
Chapter 4 – Motion in two dimensions
1. Position, velocity, and acceleration
in two dimensions
2. Two dimensional motion with
constant acceleration; parabolic
trajectories
3. Circular motion
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PHY 113 A Fall 2013 -- Lecture 4
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Updated schedule
4.1,4.12,4.35,4.60
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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
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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
b
q1
d1
f=b+q1
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From email questions on Webassign #3 -- continued
Problem 1 (3.11)
A
q
2B
graphically.
(d) Find A − 2B
magnitude
m
direction
° counterclockwise from the +x axis
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From email questions on Webassign #3 -- continued
Problem 1 (3.29)
F1
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F2
6
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.
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j
vertical direction (up)
r (t ) = x(t )ˆi + y (t )ˆj
i
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PHY 113 A Fall 2013 -- Lecture 4
horizontal
direction
8
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
dx
vx =
dt
Acceleration: a(t) = ax(t) i + ay(t) j
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PHY 113 A Fall 2013 -- Lecture 4
dv x
ax =
dt
dy
vy =
dt
ay =
dv y
dt
9
Visualization of the position vector r(t) of a particle
r(t1)
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r(t2)
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Visualization of the velocity vector v(t) of a particle
dr
r (t 2 )  r (t1 )
vt  =
= lim
dt t2 t1 0 t 2  t1
v(t)
r(t2)
r(t1)
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Visualization of the acceleration vector a(t) of a particle
dv
v (t 2 )  v (t1 )
at  =
= lim
dt t2 t1 0 t 2  t1
v(t2)
v(t1)
r(t2)
r(t1)
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a(t1)
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Figure from your text:
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Visualization of parabolic trajectory from textbook
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The trajectory graphs of the motion of an object y
versus x as a function of time, look somewhat like a
parabola:
y ( x) = Ax 2 + Bx + C
iclicker exercise
A. This is surprising
B. This is expected
C. No opinion
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Projectile motion (near earth’s surface)
j
vertical direction (up)
r (t ) = x(t )ˆi + y (t )ˆj
v (t ) = v (t )ˆi + v (t )ˆj
x
y
a(t ) =  gˆj
g = 9.8 m/s2
i
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PHY 113 A Fall 2013 -- Lecture 4
horizontal
direction
16
Projectile motion (near earth’s surface)
r (t ) = x(t )ˆi + y (t )ˆj
dr
v (t ) =
= v x (t )ˆi + v y (t )ˆj
dt
dv
a(t ) =
=  gˆj
dt
 vt  = v i  gtˆj note that vt = 0  = v i
1 2ˆ
 r t  = ri + v i t  gt j note that r t = 0  = ri
2
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Projectile motion (near earth’s surface)
r (t ) = x(t )ˆi + y (t )ˆj
dr
v (t ) =
= v x (t )ˆi + v y (t )ˆj
dt
dv
a(t ) =
=  gˆj
dt
dr
v (t ) =
= v i  gtˆj
dt
2ˆ
1
 r t  = ri + v i t  2 gt j
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note that r t = 0  = ri
PHY 113 A Fall 2013 -- Lecture 4
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Projectile motion (near earth’s surface)
r t  = ri + v i t  gt ˆj
1
2
dv
a(t ) =
=  gˆj
dt
2
dr
v (t ) =
= v i  gtˆj
dt
v i = v xi ˆi + v yi ˆj
v xi = v i cos q i
v yi = v i sin q i
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Projectile motion (near earth’s surface)
Trajectory equation in vector form:
r t  = ri + v i t  gt ˆj
1
2
2
v (t ) = v i  gtˆj
Trajectory equation in component form:
xt  = xi + v xi t
v x (t ) = v xi
2
1


y t = yi + v yi t  2 gt
v y (t ) = v yi  gt
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.
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Projectile motion (near earth’s surface)
Trajectory equation in component form:
xt  = xi + v xi t = xi + vi cos q i t
y t  = yi + v yi t  gt = yi + vi sin q i t  gt
v x (t ) = v xi = vi cos q i
1
2
2
1
2
2
v y (t ) = v yi  gt = vi sin q i  gt
Trajectory path y(x); eliminating t from the equations:
x  xi
t=
vi cos q i


x  xi
1  x  xi 
y  x  = yi + vi sin q i
 2 g

vi cos q i
v
cos
q
i 
 i
 x  xi 

y  x  = yi + tan q i  x  xi   g 
vi cos q i 

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1
2
21
2
Projectile motion (near earth’s surface)
Summary of results
xt  = xi + vi cos q i t
y t  = yi + vi sin q i t  12 gt 2
v x (t ) = vi cos q i
v y (t ) = vi sin q i  gt
 x  xi 

y  x  = yi + tan q i  x  xi   g 
 vi cos q i 
2
1
2
iclicker exercise:
These equations are so beautiful that
A. 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.
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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.
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h=7.1m
qi=53o
d=24m=x(2.2s)
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Problem solving steps
1. Visualize problem – labeling variables
2. Determine which basic physical principle(s) apply
3. 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=7.1m
qi=53o
d=24m=x(2.2s)
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h=7.1m
qi=53o
d=24m=x(2.2s)
xt  = xi + vi cos q i t
d = x(2.2) = vi cos 53o 2.2  = 24
24m
 vi =
= 18.12698m / s
o
cos 53 2.2 s 
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Review
Position x(t), Velocity y(t), Acceleration a(t)
in one dimension:
t
dx
v(t ) =
dt
dv
a(t ) =
dt

x(t ) =  v(t ' )dt '
t0
t

v(t ) =  a(t ' )dt '
t0
Special case of constant acceleration a(t)=a0:
dv
Suppose :
= a0
and v( 0 ) = v0 , x( 0 ) = x0
dt
Then :
v(t ) = v0 + a0t
x(t ) = x0 + v0t + 12 a0t 2
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Review – continued:
Special case of constant acceleration a(t)=a0:
dv
Suppose :
= a0
and v( 0 ) = v0 , x( 0 ) = x0
dt
Then :
v(t ) = v0 + a0t
initial
initial
velocity
position
x(t ) = x0 + v0t + 12 a0t 2
Relationship between position, velocity, and acceleration :
2a0  x(t )  x0  = v(t )   v0
2
2
v(t )  v0
Result derived using algebra : t =
a0
 v(t )  v0  1  v(t )  v0 
 + 2 a0 

x(t ) = x0 + v0 
a0 
a0 


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Summary of equations –
one-dimensional motion with constant acceleration
v(t ) = v0 + a0t
x(t ) = x0 + v0t + 12 a0t 2
2a0  x(t )  x0  = v(t )   v0
2
2
iclicker question:
Why did I show you part of the derivation of the last
equation?
A. Because professors like to torture physics students
B. Because you will need to be able to prove the
equation yourself
C. Because the “proof” helps you to understand the
meaning of the equation
D. All of the above
E. None of the above
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Review: Motion in two dimensions:
j
vertical direction (up)
r (t ) = x(t )ˆi + y (t )ˆj
i
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PHY 113 A Fall 2013 -- Lecture 4
horizontal
direction
30
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
dx
vx =
dt
Acceleration: a(t) = ax(t) i + ay(t) j
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PHY 113 A Fall 2013 -- Lecture 4
dv x
ax =
dt
dy
vy =
dt
ay =
dv y
dt
31
Projectile motion (constant acceleration)
(reasonable approximation near Earth’s surface)
j
vertical direction (up)
r (t ) = x(t )ˆi + y (t )ˆj
v (t ) = v (t )ˆi + v (t )ˆj
x
y
a(t ) =  gˆj
i
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PHY 113 A Fall 2013 -- Lecture 4
horizontal
direction
32
Projectile motion (near earth’s surface)
r t  = ri + v i t  gt ˆj
1
2
dv
a(t ) =
=  gˆj
dt
2
dr
v (t ) =
= v i  gtˆj
dt
v i = v xi ˆi + v yi ˆj
v xi = v i cos q i
v yi = v i sin q i
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Projectile motion (near earth’s surface)
Summary of component functions
xt  = xi + vi cos q i t
y t  = yi + vi sin q i t  12 gt 2
v x (t ) = vi cos q i
v y (t ) = vi sin q i  gt
 x  xi 

y  x  = yi + tan q i  x  xi   g 
 vi cos q i 
2
1
2
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Uniform circular motion – another example
of motion in two-dimensions
animation from
http://mathworld.wolfram.com/UniformCircularMotion.html
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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 acceleration
B. There is acceleration tangent to the circle
C. There is acceleration in the radial direction of
the circle
D. There is not enough information to conclude
that there is acceleration or not
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Uniform circular motion – continued
If vi = v f  v, then the acceleration
in the radial direction and the
centripetal acceleration is :
v2
a c =  rˆ
r
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Uniform circular motion – continued
r
v2
a c =  rˆ
r
 2  ˆ
a c = 
 rr
T 
2
a c = 2f  rrˆ
2
In terms of time period T for one cycle:
2r
v=
T
In terms of the frequency f of complete cycles:
1
f = ;
T
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v = 2πfr
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 2  ˆ
a c = 
 rr
T 
T = 24 hr  60 min/hr  60 s/min = 86400 s
2
 2 
3
2
a c = 
 637110 m/s rˆ
 86400 
=  0.03 m/s 2 rˆ
2
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
PHY 113 A Fall 2013 -- Lecture 4

39
iclicker exercise:
During circular motion, what happens when
the speed of the object changes?
A. There is only tangential acceleration
B. There is only radial acceleration
C. There are both radial and tangential
acceleration
D. I don’t need to know this because it never
happens.
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Circular motion
Assume : vt  = vq t θˆ
r̂
θ̂
r̂
ac
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2
vq
ac = 
rˆ
r
dvq ˆ
at =
θ
dt
θ̂
at
a = ac + at
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