Physics 121,Lecture 5,Pg 1
Physics 121,Sections 9,10,11,and 12
Lecture 5
Today’s Topics:
? Homework 2,Due Friday Sept,16 @ 6:00PM
? Ch.3,# 2,11,18,20,25,32,36,46,50,and 56.
? Chapter 3,Forces and motion along a line
? Motion with constant acceleration
? Falling objects
? Apparent weight
? Chapter 4,Motion in 2-D
? Vectors
? Kinematics
? Projectile motion
Physics 121,Lecture 5,Pg 2
Recap,constant acceleration in 1-D
? For constant acceleration:
atvv 0 ??
200 at21tvxx ???
a c o n s t?
? From which we know:
v)(v21v
)x2a ( xvv
0av
0
2
0
2
??
???
Physics 121,Lecture 5,Pg 3
Lecture 5 ACT 1
1D Freefall
? Alice and Bill are standing at the top of a cliff of height
H,Both throw a ball with initial speed v0,Alice straight
down and Bill straight up,The speed of the balls when
they hit the ground are vA and vB respectively.
v0
v0
BillAlice
H
vA vB
Which of the following is true:
(a) vA < vB (b) vA = vB (c) vA > vB
Physics 121,Lecture 5,Pg 4
About Air resistance?
? When a body moves through a fluid
?A drag force due to friction take place
?Like friction,its direction is opposite to the motion
?This force increases dramatically with speed
?It is proportional to v2
? For an object falling through air
?As gravity accelerates it,v increases
?When the magnitude of Fd equals the weight mg
?Not net force ? v becomes constant
?The object has reached its terminal speed vt
Fd
mg
v
Physics 121,Lecture 5,Pg 5
Air resistance
? When a body reaches vt
?Both forces are equal in magnitude
?So the coefficient b is
?And
? The terminal speed vt varies mass,shape,size
?Feather,0.5 m/s
?Raindrop,7 m/s
?Skydiver,50-60 m/s (spread-eagle)
?Skydiver,100 m/s (diving)
Fd
mg
vt
Physics 121,Lecture 5,Pg 6
Apparent weight
? The weight is read by a scale
?It is given by the magnitude of the
normal force
?The block pushes on the scale with a
force equal but opposite to N (action-
reaction)
? If platform (and object+scale) is not accelerated
?Net force is zero
N mg
-N? The normal is found from the net force
a
Physics 121,Lecture 5,Pg 7
Apparent weight
? If the,elevator” is accelerated
?Net force ? 0
?Choosing axis as illustrated
a N
mg
? Apparent weight is
?W’ >mg if a>0 (upward):,heavier”
?W’ <mg if a>0 (downward):,lighter”
a
N mg
-N
Physics 121,Lecture 5,Pg 8
Chapter 4,Vectors
? In 1 dimension,we can specify direction with a + or - sign,
? In 2 or 3 dimensions,we need more than a sign to specify the
direction of something:
? To illustrate this,consider the position vector r in 2 dimensions.
Example,Where is Boston?
?Choose origin at New York
?Choose coordinate system
?Boston is 212 miles northeast
of New York
or
Boston is 150 miles north and
150 miles east of New York
Boston
New York
r
Physics 121,Lecture 5,Pg 9
Vectors...
? There are two common ways of indicating that something is
a vector quantity:
?Boldface notation,A
?“Arrow” notation:
A =
A
A
Physics 121,Lecture 5,Pg 10
Vectors,definition
? A vector is composed of a magnitude and a direction
?examples,displacement,velocity,acceleration
?magnitude of A is designated |A|
?usually carries units
? A vector has no particular position
? Two vectors are equal if their directions and magnitudes
match,
A
B C
A = C
A = B,B = C
Physics 121,Lecture 5,Pg 11
Vectors and scalars:
? A scalar is an ordinary number.
?a magnitude without a direction
?may have units (kg) or be just a number
?usually indicated by a regular letter,no bold face and no
arrow on top.
Note,the lack of specific designation of a scalar can lead
to confusion
? The product of a vector and a scalar is another vector in the
same direction but with modified magnitude.
A BA = -0.75 B
Physics 121,Lecture 5,Pg 12
Lecture 5,ACT 2
Vectors and Scalars
A) my velocity (3 m/s)
C) my destination
(the pub - 100,000 m)
B) my acceleration
downhill (30 m/s2)
D) my mass (150 kg)
Which of the following is not a vector?
(For bonus points,which answer has a reasonable
magnitude listed?)
While I conduct my daily run,several quantities
describe my condition
Physics 121,Lecture 5,Pg 13
Converting Coordinate Systems
? In circular coordinates the vector r = (r,q)
? In Cartesian the vector r = (rx,ry) = (x,y)
? We can convert between the two as
follows:
? r=|r| is the magnitude of the vector (does not depend on direction)
rx = x = r cos q??
ry = y = r sin q q???arctan( y / x )
22 yxr ??
y
x
(x,y)
q
rr
y
rx
r ? ? ?r x y2 2
r y
x
? The magnitude (length) of r
is found using Pythagoras’
theorem:
Physics 121,Lecture 5,Pg 14
Vector addition:
? The sum of two vectors is another vector.
A = B + C
B
C A
B
C
Physics 121,Lecture 5,Pg 15
Vector subtraction:
? Vector subtraction can be defined in terms of addition.
B - C
B
C
B
-C
B - C
= B + (-1)C
Physics 121,Lecture 5,Pg 16
Unit Vectors:
? A Unit Vector is a vector having length 1
and no units.
? It is used to specify a direction.
? Unit vector u points in the direction of U.
?Often denoted with a,hat”,u = ?
U
?
x
y
z
i
j
k
? Useful examples are the cartesian
unit vectors [ i,j,k ]
?point in the direction of the
x,y and z axes.
R = rxi + ryj + rzk
Physics 121,Lecture 5,Pg 17
Vector addition using components:
? Consider C = A + B.
(a) C = (Ax i + Ay j ) + (Bx i + By j ) = (Ax + Bx )i + (Ay + By )j
(b) C = (Cx i + Cy j )
? Comparing components of (a) and (b):
? Cx = Ax + Bx
? Cy = Ay + By C
BxA
ByB
Ax
Ay
Physics 121,Lecture 5,Pg 18
Lecture 5,ACT 3
Vector Addition
? Vector A = {0,2}
? Vector B = {3,0}
? Vector C = {1,-4}
What is the resultant vector,D,from
adding A+B+C?
(a) {3,-4} (b) {4,-2} (c) {5,-2}
Physics 121,Lecture 5,Pg 19
Review (1-D):
? For constant acceleration we found:
x
a
v t
t
t
v v at? ?0
200
2
1 attvxx ???
consta ?
? A few other useful formulas,
)x2 a ( xvv
v)(v
2
1v
0
2
0
2
0av
???
??
vav
Physics 121,Lecture 5,Pg 20
2-D Kinematics
? For 2-D,we simply apply the 1-D equations to each
of the component equations.
t
va x
x ?
??
t
xv
x ?
??
t
yv
y ?
??
t
va y
y ?
??
if xxx ??? if yyy ???
? Which can be combined into the vector equations:
??
?r ? r f ? r i
??
v ? ?r ?t
??
a ??v ?t
Physics 121,Lecture 5,Pg 21
2-D Kinematics
? So for constant acceleration we get:
?a = const
?v = v0 + a t
?r = r0 + v0 t + 1/2 a t2
(where a,v,v0,r,r0,are all vectors)
Physics 121,Lecture 5,Pg 22
3-D Kinematics
? Most 3-D problems can be reduced to 2-D problems when
acceleration is constant;
?Choose y axis to be along direction of acceleration.
?Choose x axis to be along the,other” direction of
motion.
? Example,Throwing a baseball (neglecting air resistance).
?Acceleration is constant (gravity).
?Choose y axis up,ay = -g.
?Choose x axis along the ground in the direction of the
throw.
Physics 121,Lecture 5,Pg 23
“x” and,y” components of motion are
independent.
? A man on a train tosses a ball straight up in the air.
?View this from two reference frames:
Reference frame
on the ground.
Reference frame
on the moving train.
y motion,a = -g y
x motion,x = v0t
Physics 121,Lecture 5,Pg 24
Projectile Motion.
? If I set something moving near the earth,it reduces to a 2-D
problem we call projectile motion,
? Use a coordinate system with x along the ground,y vertical
with respect to the ground,(Notice no change in third
direction.)
? Equations of motion reduce to:
?X,?x = voxt ax = 0
?Y,y = yo + voyt – g t2 /2 y positive upwards
Physics 121,Lecture 5,Pg 25
Lecture 5,ACT 4
2-D Motion
? Alice and Bill are playing air hockey on a table with no
bumpers at the ends,Alice scores a goal and the puck goes
flying off the end of the table,Which diagram best describes
the path of the puck?
Alice Bill
A) B) C)
Physics 121,Lecture 5,Pg 26
Problem:
? Sammy Sosa clobbers a fastball toward center-field,You
are checking out your new fancy radar gun which can detect
ball velocity,i.e,speed and direction,You measure that the
ball comes off the bat with initial velocity is 36.5 m/s at an
angle of 30o above horizontal,Since Sammy was hitting a
high fastball,you estimate that he contacted the ball about
one meter off of the ground,You know the dimensions of
Wrigley field and the center-field wall is 371 feet (113m) from
the plate and is 10 feet (3m) high,You decide to
demonstrate your superfast math and physics skills by
predicting whether Sammy get a home run before the play is
decided,
Physics 121,Lecture 5,Pg 27
Problem:
1) We need to find how high the ball is at a distance of 113m
away from where it starts,
q
v
h
D
yo
Physics 121,Lecture 5,Pg 28
Problem:
2) This is a problem in projectile motion.
Choose y axis up.
Choose x axis along the ground in the direction of the hit.
Choose the origin (0,0) to be at the plate.
Say that the ball is hit at t = 0,x = xo = 0,y = yo = 1m
q
v
h
D
y
x
Physics 121,Lecture 5,Pg 29
Problem...
? Variables
?vo = 36.5 m/s
?yo = 1 m
?h = 3 m
?qo = 30o
?D = 113 m
?a = (0,ay) ? ay = -g
?t = unknown,
?Yf – height of ball when x=113m,unknown,
our target
Physics 121,Lecture 5,Pg 30
Problem...
3) For projectile motion,
? Equations of motion are:
vx = v0x vy = v0y - g t
x = vx t y = y0 + v0y t - 1/ 2 g t2
And,use geometry to find vox and voy
y
x
g
q
v
v0x
v0yy
0
Find v0x = |v| cos q.
and v0y = |v| sin q.
Physics 121,Lecture 5,Pg 31
Problem...4) Solve the problem,
? The time to reach the wall is,t = D / vx (easy!)
? Height at any time,y(t) = y0 + v0y t - g t2/ 2
? Combining the two gives,y(t) = y0 + v0y D/vox - g D2/ (2vox2)
? And substitute for vox and voy,y(t) = y0 + D tanq - g D2/ 2(vocosq)2
? All are known quantities,Solved.
? Numbers:
? y(t) = (1.0 m) + (113 m)(tan 30) -
(0.5)(9.8 m/s2)(113 m)2/(36.5 m/s cos 30)2
= (1.0 + 65.2 - 62.6) m = 3.6 m
Physics 121,Lecture 5,Pg 32
Problem...
5)Think about the answer,
? The units work out correctly for a height (m)
? It seems reasonable for the ball to be a little over 3m
high when it gets to the fence,
? But,we haven’t yet answered the question
? Since the wall is 3m high,and the ball is 3.26m high
when it gets there,Sammy gets a homer.
Physics 121,Lecture 5,Pg 33
Recap of today’s lecture
? Homework 2,Due Friday Sept,16 @ 6:00PM
?Ch.3,# 2,11,18,20,25,32,36,46,50,and 56.
? Chapter 3,Forces and motion along a line
?Motion with constant acceleration
?Falling objects
?Apparent weight
? Chapter 4,Motion in 2-D
?Vectors
?Kinematics
?Projectile motion
Physics 121,Sections 9,10,11,and 12
Lecture 5
Today’s Topics:
? Homework 2,Due Friday Sept,16 @ 6:00PM
? Ch.3,# 2,11,18,20,25,32,36,46,50,and 56.
? Chapter 3,Forces and motion along a line
? Motion with constant acceleration
? Falling objects
? Apparent weight
? Chapter 4,Motion in 2-D
? Vectors
? Kinematics
? Projectile motion
Physics 121,Lecture 5,Pg 2
Recap,constant acceleration in 1-D
? For constant acceleration:
atvv 0 ??
200 at21tvxx ???
a c o n s t?
? From which we know:
v)(v21v
)x2a ( xvv
0av
0
2
0
2
??
???
Physics 121,Lecture 5,Pg 3
Lecture 5 ACT 1
1D Freefall
? Alice and Bill are standing at the top of a cliff of height
H,Both throw a ball with initial speed v0,Alice straight
down and Bill straight up,The speed of the balls when
they hit the ground are vA and vB respectively.
v0
v0
BillAlice
H
vA vB
Which of the following is true:
(a) vA < vB (b) vA = vB (c) vA > vB
Physics 121,Lecture 5,Pg 4
About Air resistance?
? When a body moves through a fluid
?A drag force due to friction take place
?Like friction,its direction is opposite to the motion
?This force increases dramatically with speed
?It is proportional to v2
? For an object falling through air
?As gravity accelerates it,v increases
?When the magnitude of Fd equals the weight mg
?Not net force ? v becomes constant
?The object has reached its terminal speed vt
Fd
mg
v
Physics 121,Lecture 5,Pg 5
Air resistance
? When a body reaches vt
?Both forces are equal in magnitude
?So the coefficient b is
?And
? The terminal speed vt varies mass,shape,size
?Feather,0.5 m/s
?Raindrop,7 m/s
?Skydiver,50-60 m/s (spread-eagle)
?Skydiver,100 m/s (diving)
Fd
mg
vt
Physics 121,Lecture 5,Pg 6
Apparent weight
? The weight is read by a scale
?It is given by the magnitude of the
normal force
?The block pushes on the scale with a
force equal but opposite to N (action-
reaction)
? If platform (and object+scale) is not accelerated
?Net force is zero
N mg
-N? The normal is found from the net force
a
Physics 121,Lecture 5,Pg 7
Apparent weight
? If the,elevator” is accelerated
?Net force ? 0
?Choosing axis as illustrated
a N
mg
? Apparent weight is
?W’ >mg if a>0 (upward):,heavier”
?W’ <mg if a>0 (downward):,lighter”
a
N mg
-N
Physics 121,Lecture 5,Pg 8
Chapter 4,Vectors
? In 1 dimension,we can specify direction with a + or - sign,
? In 2 or 3 dimensions,we need more than a sign to specify the
direction of something:
? To illustrate this,consider the position vector r in 2 dimensions.
Example,Where is Boston?
?Choose origin at New York
?Choose coordinate system
?Boston is 212 miles northeast
of New York
or
Boston is 150 miles north and
150 miles east of New York
Boston
New York
r
Physics 121,Lecture 5,Pg 9
Vectors...
? There are two common ways of indicating that something is
a vector quantity:
?Boldface notation,A
?“Arrow” notation:
A =
A
A
Physics 121,Lecture 5,Pg 10
Vectors,definition
? A vector is composed of a magnitude and a direction
?examples,displacement,velocity,acceleration
?magnitude of A is designated |A|
?usually carries units
? A vector has no particular position
? Two vectors are equal if their directions and magnitudes
match,
A
B C
A = C
A = B,B = C
Physics 121,Lecture 5,Pg 11
Vectors and scalars:
? A scalar is an ordinary number.
?a magnitude without a direction
?may have units (kg) or be just a number
?usually indicated by a regular letter,no bold face and no
arrow on top.
Note,the lack of specific designation of a scalar can lead
to confusion
? The product of a vector and a scalar is another vector in the
same direction but with modified magnitude.
A BA = -0.75 B
Physics 121,Lecture 5,Pg 12
Lecture 5,ACT 2
Vectors and Scalars
A) my velocity (3 m/s)
C) my destination
(the pub - 100,000 m)
B) my acceleration
downhill (30 m/s2)
D) my mass (150 kg)
Which of the following is not a vector?
(For bonus points,which answer has a reasonable
magnitude listed?)
While I conduct my daily run,several quantities
describe my condition
Physics 121,Lecture 5,Pg 13
Converting Coordinate Systems
? In circular coordinates the vector r = (r,q)
? In Cartesian the vector r = (rx,ry) = (x,y)
? We can convert between the two as
follows:
? r=|r| is the magnitude of the vector (does not depend on direction)
rx = x = r cos q??
ry = y = r sin q q???arctan( y / x )
22 yxr ??
y
x
(x,y)
q
rr
y
rx
r ? ? ?r x y2 2
r y
x
? The magnitude (length) of r
is found using Pythagoras’
theorem:
Physics 121,Lecture 5,Pg 14
Vector addition:
? The sum of two vectors is another vector.
A = B + C
B
C A
B
C
Physics 121,Lecture 5,Pg 15
Vector subtraction:
? Vector subtraction can be defined in terms of addition.
B - C
B
C
B
-C
B - C
= B + (-1)C
Physics 121,Lecture 5,Pg 16
Unit Vectors:
? A Unit Vector is a vector having length 1
and no units.
? It is used to specify a direction.
? Unit vector u points in the direction of U.
?Often denoted with a,hat”,u = ?
U
?
x
y
z
i
j
k
? Useful examples are the cartesian
unit vectors [ i,j,k ]
?point in the direction of the
x,y and z axes.
R = rxi + ryj + rzk
Physics 121,Lecture 5,Pg 17
Vector addition using components:
? Consider C = A + B.
(a) C = (Ax i + Ay j ) + (Bx i + By j ) = (Ax + Bx )i + (Ay + By )j
(b) C = (Cx i + Cy j )
? Comparing components of (a) and (b):
? Cx = Ax + Bx
? Cy = Ay + By C
BxA
ByB
Ax
Ay
Physics 121,Lecture 5,Pg 18
Lecture 5,ACT 3
Vector Addition
? Vector A = {0,2}
? Vector B = {3,0}
? Vector C = {1,-4}
What is the resultant vector,D,from
adding A+B+C?
(a) {3,-4} (b) {4,-2} (c) {5,-2}
Physics 121,Lecture 5,Pg 19
Review (1-D):
? For constant acceleration we found:
x
a
v t
t
t
v v at? ?0
200
2
1 attvxx ???
consta ?
? A few other useful formulas,
)x2 a ( xvv
v)(v
2
1v
0
2
0
2
0av
???
??
vav
Physics 121,Lecture 5,Pg 20
2-D Kinematics
? For 2-D,we simply apply the 1-D equations to each
of the component equations.
t
va x
x ?
??
t
xv
x ?
??
t
yv
y ?
??
t
va y
y ?
??
if xxx ??? if yyy ???
? Which can be combined into the vector equations:
??
?r ? r f ? r i
??
v ? ?r ?t
??
a ??v ?t
Physics 121,Lecture 5,Pg 21
2-D Kinematics
? So for constant acceleration we get:
?a = const
?v = v0 + a t
?r = r0 + v0 t + 1/2 a t2
(where a,v,v0,r,r0,are all vectors)
Physics 121,Lecture 5,Pg 22
3-D Kinematics
? Most 3-D problems can be reduced to 2-D problems when
acceleration is constant;
?Choose y axis to be along direction of acceleration.
?Choose x axis to be along the,other” direction of
motion.
? Example,Throwing a baseball (neglecting air resistance).
?Acceleration is constant (gravity).
?Choose y axis up,ay = -g.
?Choose x axis along the ground in the direction of the
throw.
Physics 121,Lecture 5,Pg 23
“x” and,y” components of motion are
independent.
? A man on a train tosses a ball straight up in the air.
?View this from two reference frames:
Reference frame
on the ground.
Reference frame
on the moving train.
y motion,a = -g y
x motion,x = v0t
Physics 121,Lecture 5,Pg 24
Projectile Motion.
? If I set something moving near the earth,it reduces to a 2-D
problem we call projectile motion,
? Use a coordinate system with x along the ground,y vertical
with respect to the ground,(Notice no change in third
direction.)
? Equations of motion reduce to:
?X,?x = voxt ax = 0
?Y,y = yo + voyt – g t2 /2 y positive upwards
Physics 121,Lecture 5,Pg 25
Lecture 5,ACT 4
2-D Motion
? Alice and Bill are playing air hockey on a table with no
bumpers at the ends,Alice scores a goal and the puck goes
flying off the end of the table,Which diagram best describes
the path of the puck?
Alice Bill
A) B) C)
Physics 121,Lecture 5,Pg 26
Problem:
? Sammy Sosa clobbers a fastball toward center-field,You
are checking out your new fancy radar gun which can detect
ball velocity,i.e,speed and direction,You measure that the
ball comes off the bat with initial velocity is 36.5 m/s at an
angle of 30o above horizontal,Since Sammy was hitting a
high fastball,you estimate that he contacted the ball about
one meter off of the ground,You know the dimensions of
Wrigley field and the center-field wall is 371 feet (113m) from
the plate and is 10 feet (3m) high,You decide to
demonstrate your superfast math and physics skills by
predicting whether Sammy get a home run before the play is
decided,
Physics 121,Lecture 5,Pg 27
Problem:
1) We need to find how high the ball is at a distance of 113m
away from where it starts,
q
v
h
D
yo
Physics 121,Lecture 5,Pg 28
Problem:
2) This is a problem in projectile motion.
Choose y axis up.
Choose x axis along the ground in the direction of the hit.
Choose the origin (0,0) to be at the plate.
Say that the ball is hit at t = 0,x = xo = 0,y = yo = 1m
q
v
h
D
y
x
Physics 121,Lecture 5,Pg 29
Problem...
? Variables
?vo = 36.5 m/s
?yo = 1 m
?h = 3 m
?qo = 30o
?D = 113 m
?a = (0,ay) ? ay = -g
?t = unknown,
?Yf – height of ball when x=113m,unknown,
our target
Physics 121,Lecture 5,Pg 30
Problem...
3) For projectile motion,
? Equations of motion are:
vx = v0x vy = v0y - g t
x = vx t y = y0 + v0y t - 1/ 2 g t2
And,use geometry to find vox and voy
y
x
g
q
v
v0x
v0yy
0
Find v0x = |v| cos q.
and v0y = |v| sin q.
Physics 121,Lecture 5,Pg 31
Problem...4) Solve the problem,
? The time to reach the wall is,t = D / vx (easy!)
? Height at any time,y(t) = y0 + v0y t - g t2/ 2
? Combining the two gives,y(t) = y0 + v0y D/vox - g D2/ (2vox2)
? And substitute for vox and voy,y(t) = y0 + D tanq - g D2/ 2(vocosq)2
? All are known quantities,Solved.
? Numbers:
? y(t) = (1.0 m) + (113 m)(tan 30) -
(0.5)(9.8 m/s2)(113 m)2/(36.5 m/s cos 30)2
= (1.0 + 65.2 - 62.6) m = 3.6 m
Physics 121,Lecture 5,Pg 32
Problem...
5)Think about the answer,
? The units work out correctly for a height (m)
? It seems reasonable for the ball to be a little over 3m
high when it gets to the fence,
? But,we haven’t yet answered the question
? Since the wall is 3m high,and the ball is 3.26m high
when it gets there,Sammy gets a homer.
Physics 121,Lecture 5,Pg 33
Recap of today’s lecture
? Homework 2,Due Friday Sept,16 @ 6:00PM
?Ch.3,# 2,11,18,20,25,32,36,46,50,and 56.
? Chapter 3,Forces and motion along a line
?Motion with constant acceleration
?Falling objects
?Apparent weight
? Chapter 4,Motion in 2-D
?Vectors
?Kinematics
?Projectile motion
