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T
at #102 of
f Nicholson
N h l
Hall.
H ll
Application of kinematic equation:
a = const.
v = v0 + at
at 2
x = x0 + v0t +
2
v = v0 + 2a ( x − x0 )
2
2
Example: A red car and a green car, identical except for the color, move toward each
other in adjacent lanes and parallel to the x axis. At time t=0, the red car is at xr=0 and
the green car is at xg= 220 m
m. If the red car has a constant velocity of 20 km/h,
km/h the cars
pass each other at x =44.5 m, and if it has a constant velocity of 40 km/h, they pass each
other at x =76.6m. What are (a) the initial velocity and (b) the acceleration of the green
car??
Red car: a=0 so
x f (1) = v1t1
x f (2) = v1t2
20km 50
20k
v1 =
=
m/s
h
9
40km 100
v2 =
=
m/s
9
h
t1 = 8.0s
t 2 = 6.9s
Now we must simultaneously solve two equations for the green car.
at12
44.5 − 220 = v0t1 +
= −175.5
2
at 22
76.6 − 220 = v0t 2 +
= −143.4
2
This Gives!
a = −2.0m / s 2
v0 = −13.9m / s
Special Case: free-falling body motion
Close to the surface of the Earth all objects move toward the center of the Earth
with an acceleration whose magnitude is constant and equal to 9.8 m/s2. We use
the symbol g to indicate the acceleration of an object in free fall.
B
a
a = -g
y
A
v = v0 − gt
(eq. 1)
gt 2
y = yo + v0t −
(eq. 2)
2
v 2 − v02 = −2 g ( y − yo )
(eq. 3)
Question
A person standing at the edge of a cliff throws one
ball straight up and another ball straight down at
the same initial speed.
speed Neglecting air resistance
resistance,
which ball with the greater speed hits the ground
below the cliff?
1.
2.
3.
upward.
downward.
neither—they both hit at the
same speed.
Kinematics: Taking Advantage of Symmetry
v = v0 − gt
y = v0t − gt
1
2
2
v = v − 2 gy
2
y=
2
0
1
2
( v + v0 ) t
Graphical Integration in Motion Analysis (nonconstant acceleration)
When the acceleration of a moving object is not constant we must use
integration to determine the velocity v(t ) and the position x(t) of the object.
The integration can be done either using the analytic or the graphical approach:
a (t ) =
t1
t1
t1
t1
t1
dv
→ dv = a (t )dt → ∫ dv = ∫ a (t )dt → v1 − v0 = ∫ a(t )dt → v1 = v0 + ∫ a(t ) dt
dt
t0
t0
t0
t0
∫ a(t )dt
= [ Area under the a versus t curve between t0 and t1 ]
t0
t
t
1
1
dx
v(t ) =
→ dx = v(t )dt → ∫ dx = ∫ v(t )dt →
dt
t0
t0
t1
t1
t0
t0
x1 − x0 = ∫ vdt → x1 = x0 + ∫ vdt
t1
dt
∫ vdt
to
= [ Area
A under
d th
the v versus t curve bbetween
t
t0 and
d t1 ]
Example: Acceleration: (a) If the position of a particle is given by x= 20t – 5t3,
where x is in meters and t is in seconds, when, if ever, is the particle’s velocity zero?
(b) When is its acceleration a zero?
(c) For what time range (positive or negative) is a negative?
(d) For what time range (positive or negative) is a positive?
(e) Graph x(t),
x(t) v(t) and a(t).
a(t)
x(t) = 20t − 5t 3
dx
2
v(t) =
= 20 − 15t
dt
2
dv d x
a(t)
( )=
= 2 = −30t
30
dt dt
In physics we have parameters that can be completely described by a
number and are known as scalars. Temperature and mass are such
parameters.
Other physical parameters require additional information about
direction and are known as vectors. Examples of vectors are
displacement velocity
displacement,
velocity, and acceleration
acceleration.
This chapter covers the basic mathematical language to describe
vectors In particular we need to know the following:
vectors.
•
•
•
•
•
•
•
•
Geometric vector addition and subtraction
G m t i vector
Geometric
t addition
dditi n and
nd subtraction
bt
ti n
Resolving a vector into its components
The notion of a unit vector
Addition and subtraction vectors by components
Multiplication of a vector by a scalar
The scalar (dot) product of two vectors
The vector (cross) product of two vectors
Vector expressed by components
G
Vector a can be written with its components and unit vectors
G
a = a ˆi + a ˆj (two-dimensional case)
x
y
ax = a cos θ
and
a y = a sin θ .
The quantities ax ˆi and a y ˆj are called the vector components
a = a +a
2
x
2
y
and tan θ =
ay
ax
.
G
Vector a in three-dimensional case
G
a = axiˆ + a y ˆj + az kˆ
θz
ax = a cos θ x ; a y = a sin θ y ; az = a sin θ z
a = a +a +a
2
x
2
y
2
z
ay
ax
az
cos θ x = ; cosθ y = ; cosθ x =
a
a
a
G
a
k̂
θx
iˆ
θy
ĵ
Adding vectors by components
G
G
ˆ
ˆ
a = ax i + a y j ; b = bx ˆi + by ˆj.
j
G G G
r = a + b = rx ˆi + ry ˆj.
y
G
r
G
b
G
a
x
O
The components rx and ry are given by the equations
rx = ax + bx and ry = a y + by .
Subtracting vectors by components
y
G
d
G
b
G
a
O
G
G
a = ax ˆi + a y ˆj ; b = bx ˆi + by ˆj.
j
G G G
d = a − b = d x ˆi + d y ˆj.
The components d x and d y are given by the equations
x
d x = ax − bx and d y = a y − by .
Multiplying a Vector by a Scalar
G G
G
Multiplication
p
of a vector a by
y a scalar s results in a new vector b = sa.
The magnitude b of the new vector is given by b = | s | a.
G
G
If s > 0,, vector b has the same direction as vector a.
G
G
If s < 0, vector b has a direction opposite to that of vector a.
The Scalar Product of Two Vectors
G
G G
G
The scalar product a ⋅ b of two vectors a and b is given by
G G
a ⋅ b =ab cos φ. The scalar product of two vectors is also
known as the "dot" product. The scalar product in terms
of vector components is given by the equation
G G
a ⋅ b =axbx + a y by + az bz .
Application in physics:
G G
Work: W = F ⋅ d = Fd cos φ
Example
G
G
ˆ
ˆ
What is the angle φ between a = 3.0i − 4.0 j and b = −2.0iˆ + 3.0kˆ ?
G G
G
a ⋅b
G
Use a ⋅ b = ab cos φ such that cos φ =
ab
Since
a = 3.02 + (−4.0)2 = 5.0 and b = (−2.0)2 + 3.02 = 3.61
and
G G
a ⋅ b = axbx + a y by + az bz = axbx = −6.0
G G
a ⋅b
−6.0
cos φ =
=
and
d φ = 109°
ab 5.0 × 3.61
k̂ G
b
φ
G
a
ĵ
iˆ
The Vector Product of Two Vectors
G G G
G
The vector product c = a × b is a vector c.
G
The magnitude of c is given by the equation
c = abb sin
i φ.
G
G
G
c is perpendicular to the plane P defined by a and b .
G
The sense of the vector c is given by the right-hand rule:
G
G
a. Place the vectors a and b tail to tail.
G
b. Rotate a in the plane P along the shortest angle
G
so that it coincides with b .
c. Rotate the fingers of the right hand in the same direction.
G
d The
d.
h thumb
h b off the
h right
i h hand
h d gives
i
the
h sense off c .
The vector product of two vectors is also known as
th "cross
the
"
" prodduct.
t
G G G
The vector product c = a × b in terms of vector components
G
G
G
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
a = a x i + ay j + az k, b = b x i + by j + bz k, c = c x ˆi + cy ˆj + cz kˆ
G
The vector components of vector c are given by the equations
cx = a y bz − az by , cy = az bx − axbz , cz = axby − a y bx .
ay
or cx =
by
az
az
, cy =
bz
bz
ax
, cz =
bx
bx
ax
ay
by
Note: Those familiar with the use of determinants can use the expression
i
j k
G G
a × b = ax a y az
bx by bz
Note: The order of the two vectors in the cross product is important:
G G
G G
b × a = − a ×b .
Rotation axis
(
)
Application in physics:
Torque:
G G
τ = r ×F
G
G
r
G
F
Example
G G G G G G G G G
Show a ⋅ (b × c ) = b ⋅ (c × a ) = c ⋅ (a × b )
i j
G G G
a ⋅ (b × c ) = (axiˆ + a y ˆj + az kˆ) ⋅ bx by
cx
cy
k
bz
cz
ax
= ax (by cz − bz cy ) + a y (bz cx − bx c y ) + az (bx cy − by cx ) = bx
ay
by
az
bz
cx
cy
cz
bx
by
bz
= cx
ax
cy
ay
G G G
cz = b ⋅ ( c × a )
az
cx
cy
cz
= ax
bx
ay
by
G G G
az = c ⋅ ( a × b )
bz