Transcript Chapter 7: Forces and Motion in 2D

Chapter 7: Forces and Motion
in 2D
Dr. Zalesinsky
www.YourJediMaster.com
Parts of the Chapter
7.1
Forces n
2D
Equilibrium and
Inclined Planes
7.2
Projectile
Motion
Independence of 2D Motion,
Projectiles Horizontally Launched and
Launched at an Angle
7.3
Circular
Motion
Centripetal Acceleration,
Uniform Circular Motion, and
Torque
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7.1
FORCES IN 2D
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4.11 Equilibrium Application of Newton’s Laws of Motion
Definition of Equilibrium
An object is in equilibrium when it has zero
acceleration.
F
x

0
Fy  0
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4.11 Equilibrium Application of Newton’s Laws of Motion
Reasoning Strategy
• Select an object(s) to which the equations of equilibrium are
to be applied.
• Draw a free-body diagram for each object chosen above.
Include only forces acting on the object, not forces the object
exerts on its environment.
• Choose a set of x, y axes for each object and
•resolve all forces
in the free-body diagram into components that point
along these
axes.
• Apply the equations and solve for the
•unknown quantities.
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4.11 Equilibrium Application of Newton’s Laws of Motion
 T1 sin 35  T2 sin 35  0


 T1 cos35  T2 cos35  F  0


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4.11 Equilibrium Application of Newton’s Laws of Motion
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4.11 Equilibrium Application of Newton’s Laws of Motion
Force

T1

T2

W
x component
y component
 T1 sin 10.0
 T1 cos10.0
 T2 sin 80.0
 T2 cos80.0
0


W
W  3150 N
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4.11 Equilibrium Application of Newton’s Laws of Motion

F
Fx   T1 sin 10.0  T2 sin 80.0  0
y
  T1 cos 10 .0  T2 cos 80 .0  W  0
The first equation gives


 sin 80 .0  
T
T1  
  2
 sin 10 .0 
Substitution into the second gives
 sin 80.0 




T cos10.0  T2 cos80.0  W  0
  2
 sin 10.0 
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4.11 Equilibrium Application of Newton’s Laws of Motion
T2 
W
 sin 80.0 




cos
10
.
0

cos
80
.
0
 
sin
10
.
0


T2  582 N
T1  3.3010 N
3
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4.12 Nonequilibrium Application of Newton’s Laws of Motion
When an object is accelerating, it is not in equilibrium.
F
x

 ma x
Fy  ma y
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4.12 Nonequilibrium Application of Newton’s Laws of Motion
The acceleration is along the x axis so a y
0
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4.12 Nonequilibrium Application of Newton’s Laws of Motion
Force

T1

T2

D

R
x component
y component

 T1 cos30.0
 T1 sin 30.0

 T2 cos30.0
 T2 sin 30.0
D
0
R
0


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4.12 Nonequilibrium Application of Newton’s Laws of Motion
F
y
  T1 sin 30.0  T2 sin 30.0  0

 T1  T2
F
x
  T1 cos30.0  T2 cos30.0  D  R

 m ax
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4.12 Nonequilibrium Application of Newton’s Laws of Motion
T1  T2  T
ma x  R  D
5
T

1
.
53

10
N

2 cos 30.0
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4.11.1. Consider the following: (i) the book is at rest, (i
i) the book is moving at a constant velocity, (iii) the
book is moving with a constant acceleration. Under
which of these conditions is the book in equilibrium?
a) (i) only
b) (ii) only
c) (iii) only
d) (i) and (ii) only
e) (ii) and (iii) only
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4.11.1. Consider the following: (i) the book is at rest, (i
i) the book is moving at a constant velocity, (iii) the
book is moving with a constant acceleration. Under
which of these conditions is the book in equilibrium?
a) (i) only
b) (ii) only
c) (iii) only
d) (i) and (ii) only
e) (ii) and (iii) only
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4.11.2. A block of mass M is hung by ropes as shown.
The system is in equilibrium. The point O represents
the knot, the junction of the three ropes. Which of t
he following statements is true concerning the magni
tudes of the three forces in equilibrium?
a) F1 + F2 = F3
b) F1 = F2 = 0.5×F3
c) F1 = F2 = F3
d) F1 > F3
e) F2 < F3
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4.11.2. A block of mass M is hung by ropes as shown.
The system is in equilibrium. The point O represents
the knot, the junction of the three ropes. Which of t
he following statements is true concerning the magni
tudes of the three forces in equilibrium?
a) F1 + F2 = F3
b) F1 = F2 = 0.5×F3
c) F1 = F2 = F3
d) F1 > F3
e) F2 < F3
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4.11.3. A team of dogs pulls a sled of mass 2mPwith a f
orce . A second sled of mass m is attached by a r
T sled. The tension in t
ope and pulled behind the first
he rope is . Assuming frictional forces are too sma
P
Tdetermine the ratio of the magnitudes
ll to consider,
of the forces
and , that is, P/T.
a) 3
b) 2
c) 1
d) 0.5
e) 0.33
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4.11.3. A team of dogs pulls a sled of mass 2mPwith a f
orce . A second sled of mass m is attached by a r
T sled. The tension in t
ope and pulled behind the first
he rope is . Assuming frictional forces are too sma
P
Tdetermine the ratio of the magnitudes
ll to consider,
of the forces
and , that is, P/T.
a) 3
b) 2
c) 1
d) 0.5
e) 0.33
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Motion along an Inclined Plane
See pp. 152 – 154 in text
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7.2
PROJECTILE MOTION
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3.3 Projectile Motion
Under the influence of gravity alone, an object near the
surface of the Earth will accelerate downwards at 9.80m/s2.
ay  9.80m s
2
ax  0
vx  vox  constant
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3.3 Projectile Motion
Example 3 A Falling Care Package
he airplane is moving horizontally with a constant velocity o
115 m/s at an altitude of 1050m. Determine the time requir
for the care package to hit the ground.
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3.3 Projectile Motion
y
ay
-1050 m -9.80 m/s2
vy
voy
t
0 m/s
?
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3.3 Projectile Motion
y
ay
vy
-1050 m -9.80 m/s2
y  voyt  ayt
1
2
t
2y

ay
2
voy
t
0 m/s
?
y  ayt
1
2
2
2 1050 m 

14
.
6
s
2
 9.80 m s
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3.3 Projectile Motion
Example 4 The Velocity of the Care Package
hat are the magnitude and direction of the final velocity of
the care package?
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3.3 Projectile Motion
y
ay
-1050 m -9.80 m/s2
vy
voy
t
?
0 m/s
14.6 s
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3.3 Projectile Motion
y
ay
vy
voy
t
?
0 m/s
14.6 s
-1050 m -9.80 m/s2


v y  voy  a y t  0   9.80 m s 14.6 s 
2
 143m s
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3.3 Projectile Motion
Conceptual Example 5
I Shot a Bullet into the Air...
Suppose you are driving a convertible with the top down.
The car is moving to the right at constant velocity. You point
rifle straight up into the air and fire it. In the absence of ai
resistance, where would the bullet land – behind you, ahead
of you, or in the barrel of the rifle?
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3.3 Projectile Motion
Example 6 The Height of a Kickoff
placekicker kicks a football at and angle of 40.0 degrees an
he initial speed of the ball is 22 m/s. Ignoring air resistance
determine the maximum height that the ball attains.
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3.3 Projectile Motion
vo

voy
vox
voy  vo sin   22m ssin 40  14m s

vox  vo sin   22m scos40  17m s

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3.3 Projectile Motion
y
ay
vy
voy
t
?
-9.80 m/s2
0
14 m/s
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3.3 Projectile Motion
y
ay
vy
voy
?
-9.80 m/s2
0
14 m/s
v  v  2ay y
2
y
2
oy
y
t
v v
2
y
2
oy
2a y
0  14 m s 
y


10
m
2
2  9.8 m s
2


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3.3 Projectile Motion
Example 7 The Time of Flight of a Kickoff
What is the time of flight between kickoff and landing?
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3.3 Projectile Motion
y
ay
0
-9.80 m/s2
vy
voy
t
14 m/s
?
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3.3 Projectile Motion
y
ay
vy
0
-9.80 m/s2
voy
t
14 m/s
?
y  voyt  ayt
1
2

2

0  14m st   9.80m s t
1
2

2
2

0  214m s   9.80m s t
2
t  2.9 s
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3.3 Projectile Motion
Example 8 The Range of a Kickoff
Calculate the range R of the projectile.
x  vox t  a x t  vox t
1
2
2
 17 m s 2.9 s   49 m
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3.3 Projectile Motion
Conceptual Example 10
Two Ways to Throw a Stone
From the top of a cliff, a person throws two stones. The ston
have identical initial speeds, but stone 1 is thrown downwa
at some angle above the horizontal and stone 2 is thrown a
he same angle below the horizontal. Neglecting air resistan
which stone, if either, strikes the water with greater velocity
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3.3.1. A football is kicked at an angle 25 with respect to t
he horizontal. Which one of the following statements b
est describes the acceleration of the football during thi
s event if air resistance is neglected?
a) The acceleration is zero m/s2 at all times.
b) The acceleration is zero m/s2 when the football has rea
ched the highest point in its trajectory.
c) The acceleration is positive as the football rises, and it
is negative as the football falls.
d) The acceleration starts at 9.8 m/s2 and drops to some
constant lower value as the ball approaches the ground
.
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3.3.1. A football is kicked at an angle 25 with respect to t
he horizontal. Which one of the following statements b
est describes the acceleration of the football during thi
s event if air resistance is neglected?
a) The acceleration is zero m/s2 at all times.
b) The acceleration is zero m/s2 when the football has rea
ched the highest point in its trajectory.
c) The acceleration is positive as the football rises, and it
is negative as the football falls.
d) The acceleration starts at 9.8 m/s2 and drops to some
constant lower value as the ball approaches the ground
.
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3.3.2. A baseball is hit upward and travels along a parabolic
arc before it strikes the ground. Which one of the followi
ng statements is necessarily true?
a) The velocity of the ball is a maximum when the ball is at
the highest point in the arc.
b) The x-component of the velocity of the ball is the same t
hroughout the ball's flight.
c) The acceleration of the ball decreases as the ball moves u
pward.
d) The velocity of the ball is zero m/s when the ball is at the
highest point in the arc.
e) The acceleration of the ball is zero m/s2 when the ball is
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3.3.2. A baseball is hit upward and travels along a parabolic
arc before it strikes the ground. Which one of the followi
ng statements is necessarily true?
a) The velocity of the ball is a maximum when the ball is at
the highest point in the arc.
b) The x-component of the velocity of the ball is the same t
hroughout the ball's flight.
c) The acceleration of the ball decreases as the ball moves u
pward.
d) The velocity of the ball is zero m/s when the ball is at the
highest point in the arc.
e) The acceleration of the ball is zero m/s2 when the ball is
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3.3.3. Two cannons are mounted on a high cliff. Cannon
A fires balls with twice the initial velocity of cannon B.
Both cannons are aimed horizontally and fired. How d
oes the horizontal range of cannon A compare to that o
f cannon B?
a) The range for both balls will be the same
b) The range of the cannon ball B is about 0.7 that of can
non ball A.
c) The range of the cannon ball B is about 1.4 times that
of cannon
ball A.
d) The range of the cannon ball B is about 2 times that of
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3.3.3. Two cannons are mounted on a high cliff. Cannon
A fires balls with twice the initial velocity of cannon B.
Both cannons are aimed horizontally and fired. How d
oes the horizontal range of cannon A compare to that o
f cannon B?
a) The range for both balls will be the same
b) The range of the cannon ball B is about 0.7 that of can
non ball A.
c) The range of the cannon ball B is about 1.4 times that
of cannon
ball A.
d) The range of the cannon ball B is about 2 times that of
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0.5 that of can
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3.3.4. Which one of the following statements concernin
g the range of a football is true if the football is kicke
d at an angle  with an initial speed v0?
a) The range is independent of initial speed v0.
b) The range is only dependent on the initial speed v0.
c) The range is independent of the angle.
d) The range is only dependent on the angle.
e) The range is dependent on both the initial speed v0
and the angle.
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3.3.4. Which one of the following statements concernin
g the range of a football is true if the football is kicke
d at an angle  with an initial speed v0?
a) The range is independent of initial speed v0.
b) The range is only dependent on the initial speed v0.
c) The range is independent of the angle.
d) The range is only dependent on the angle.
e) The range is dependent on both the initial speed v0
and the angle.
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3.3.5. A bullet is aimed at a target on the wall a distance L away fr
om the firing position. Because of gravity, the bullet strikes the
wall a distance Δy below the mark as suggested in the figure.
Note: The drawing is not to scale. If the distance L was half as
large, and the bullet had the same initial velocity, how would Δ
y be affected?
a) Δy will double.
b) Δy will be half as large.
c) Δy will be one fourth
as large.
d) Δy will be four times larger.
e) It is not possible to determine unless numerical values are give
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n for the distances.
3.3.1. A bicyclist is riding at a constant speed along a h
orizontal, straight-line path. The rider throws a ball
straight up to a height a few meters above her head.
Ignoring air resistance, where will the ball land?
a) in front of the rider
b) behind the rider
c) in the same hand that threw the ball
d) in the opposite hand to the one that threw it
e) This cannot be determined without knowing the spe
ed of the rider and the maximum height of the ball.
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3.3.1. A bicyclist is riding at a constant speed along a h
orizontal, straight-line path. The rider throws a ball
straight up to a height a few meters above her head.
Ignoring air resistance, where will the ball land?
a) in front of the rider
b) behind the rider
c) in the same hand that threw the ball
d) in the opposite hand to the one that threw it
e) This cannot be determined without knowing the spe
ed of the rider and the maximum height of the ball.
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3.3.2. Football A is kicked at a speed v at an angle of 
with respect to the horizontal direction. If football B
is kicked at the same angle, but with a speed 2v, wh
at is the ratio of the range of B to the range of A?
a) 1
b) 2
c) 3
d) 4
e) 9
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3.3.2. Football A is kicked at a speed v at an angle of 
with respect to the horizontal direction. If football B
is kicked at the same angle, but with a speed 2v, wh
at is the ratio of the range of B to the range of A?
a) 1
b) 2
c) 3
d) 4
e) 9
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3.3.3. Balls A, B, and C are identical. From the top of a tall buil
ding, ball A is launched with a velocity of 20 m/s at an angle
of 45 above the horizontal direction, ball B is launched with
a velocity of 20 m/s in the horizontal direction, and ball C is
launched with a velocity of 20 m/s at an angle of 45 below t
he horizontal direction. Which of the following choices corre
ctly relates the magnitudes of the velocities of the balls just
before they hit the ground below? Ignore any effects of air r
esistance.
a) vA = vC > vB
b) vA = vC = vB
c) vA > vC > vB
d) vA < vC < vB
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3.3.3. Balls A, B, and C are identical. From the top of a tall buil
ding, ball A is launched with a velocity of 20 m/s at an angle
of 45 above the horizontal direction, ball B is launched with
a velocity of 20 m/s in the horizontal direction, and ball C is
launched with a velocity of 20 m/s at an angle of 45 below t
he horizontal direction. Which of the following choices corre
ctly relates the magnitudes of the velocities of the balls just
before they hit the ground below? Ignore any effects of air r
esistance.
a) vA = vC > vB
b) vA = vC = vB
c) vA > vC > vB
d) vA < vC < vB
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3.3.4. A basketball is launched with an initial speed of 8
.5 m/s and follows the trajectory shown. The ball en
ters the basket 0.92 s after it is launched. What are
the distances x and y? Note: The drawing is not to
scale.
a) x = 6.0 m, y = 0.88 m
b) x = 5.4 m, y = 0.73 m
c) x = 5.7 m, y = 0.91 m
d) x = 7.6 m, y = 1.1 m
e) x = 6.3 m, y = 0.96 m
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3.3.4. A basketball is launched with an initial speed of 8
.5 m/s and follows the trajectory shown. The ball en
ters the basket 0.92 s after it is launched. What are
the distances x and y? Note: The drawing is not to
scale.
a) x = 6.0 m, y = 0.88 m
b) x = 5.4 m, y = 0.73 m
c) x = 5.7 m, y = 0.91 m
d) x = 7.6 m, y = 1.1 m
e) x = 6.3 m, y = 0.96 m
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3.3.5. A physics student standing on the edge of a cliff throws a
stone vertically downward with an initial speed of 10.0 m/s.
The instant before the stone hits the ground below, it is trav
eling at a speed of 30.0 m/s. If the physics student were to
throw the rock horizontally outward from the cliff instead, wi
th the same initial speed of 10.0 m/s, what is the magnitude
of the velocity of the stone just before it hits the ground? Ig
nore any effects of air resistance.
a) 10.0 m/s
b) 20.0 m/s
c) 30.0 m/s
d) 40.0 m/s
e) The height of the cliff must be specified to answer
this quest
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3.3.5. A physics student standing on the edge of a cliff throws a
stone vertically downward with an initial speed of 10.0 m/s.
The instant before the stone hits the ground below, it is trav
eling at a speed of 30.0 m/s. If the physics student were to
throw the rock horizontally outward from the cliff instead, wi
th the same initial speed of 10.0 m/s, what is the magnitude
of the velocity of the stone just before it hits the ground? Ig
nore any effects of air resistance.
a) 10.0 m/s
b) 20.0 m/s
c) 30.0 m/s
d) 40.0 m/s
e) The height of the cliff must be specified to answer
this quest
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3.3.5. At time t = 0 s, Ball A is thrown vertically upward with an
initial speed v0A. Ball B is thrown vertically upward shortly a
fter Ball A at time t. Ball B passes Ball A just as Ball A is rea
ching the top of its trajectory. What is the initial speed v0B o
f Ball B in terms of the given parameters? The acceleration
due to gravity is g.
a)
v0 B 
v0 B 
v02A
v0B = v0A  (1/2)gt2
b) v0B = v0A  (1/2)gt
 g t  v0 A gt
1
2
2 2
c)
v0 A  gt
v0 A  12 g 2t 2
v0 A  gt
d)
e)
v0B = 2v0A  gt
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3.3.5. At time t = 0 s, Ball A is thrown vertically upward with an
initial speed v0A. Ball B is thrown vertically upward shortly a
fter Ball A at time t. Ball B passes Ball A just as Ball A is rea
ching the top of its trajectory. What is the initial speed v0B o
f Ball B in terms of the given parameters? The acceleration
due to gravity is g.
a)
v0 B 
v0 B 
v02A
v0B = v0A  (1/2)gt2
b) v0B = v0A  (1/2)gt
 g t  v0 A gt
1
2
2 2
c)
v0 A  gt
v0 A  12 g 2t 2
v0 A  gt
d)
e)
v0B = 2v0A  gt
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3.3.6. A toy rocket is launched at an angle of 45 with a spee
d v0. If there is no air resistance, at what point during th
e time that it is in the air does the speed of the rocket eq
ual 0.5v0?
a) when the rocket is at one half of its maximum height as i
t is going upward
b) when the rocket is at one half of its maximum height as i
t is going downward
c) when the rocket is at its maximum height
d) when the rocket is at one fourth of its maximum height a
s it is going downward
e) at no time during the flight
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3.3.6. A toy rocket is launched at an angle of 45 with a spee
d v0. If there is no air resistance, at what point during th
e time that it is in the air does the speed of the rocket eq
ual 0.5v0?
a) when the rocket is at one half of its maximum height as i
t is going upward
b) when the rocket is at one half of its maximum height as i
t is going downward
c) when the rocket is at its maximum height
d) when the rocket is at one fourth of its maximum height a
s it is going downward
e) at no time during the flight
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3.3.7. During a high school track meet, an athlete perfo
rming the long jump runs and leaps at an angle of 2
5 and lands in a sand pit 8.5 m from his launch poin
t. If the launch point and landing points are at the s
ame height, y = 0 m, with what speed does the athl
ete land?
a) 6 m/s
b) 8 m/s
c) 10 m/s
d) 2 m/s
e) 4 m/s
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3.3.7. During a high school track meet, an athlete perfo
rming the long jump runs and leaps at an angle of 2
5 and lands in a sand pit 8.5 m from his launch poin
t. If the launch point and landing points are at the s
ame height, y = 0 m, with what speed does the athl
ete land?
a) 6 m/s
b) 8 m/s
c) 10 m/s
d) 2 m/s
e) 4 m/s
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3.3.8. An airplane is flying horizontally at a constant velocity
when a package is dropped from its cargo bay. Assuming
no air resistance, which one of the following statements is
correct?
a) The package follows a curved path that lags behind the ai
rplane.
b) The package follows a straight line path that lags behind
the airplane.
c) The package follows a straight line path, but it is always v
ertically below the airplane.
d) The package follows a curved path, but it is always vertic
ally below the airplane.
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3.3.8. An airplane is flying horizontally at a constant velocity
when a package is dropped from its cargo bay. Assuming
no air resistance, which one of the following statements is
correct?
a) The package follows a curved path that lags behind the ai
rplane.
b) The package follows a straight line path that lags behind
the airplane.
c) The package follows a straight line path, but it is always v
ertically below the airplane.
d) The package follows a curved path, but it is always vertic
ally below the airplane.
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3.3.9. In making a movie, a stuntman has to jump fro
m one roof onto another roof, located 2.0 m below.
The buildings are separated by a distance of 2.5 m.
What is the minimum horizontal speed that the stunt
man must have when jumping from the first roof to
have a successful jump?
a) 3.9 m/s
b) 2.5 m/s
c) 4.3 m/s
d) 4.5 m/s
e) 3.1 m/s
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3.3.9. In making a movie, a stuntman has to jump fro
m one roof onto another roof, located 2.0 m below.
The buildings are separated by a distance of 2.5 m.
What is the minimum horizontal speed that the stunt
man must have when jumping from the first roof to
have a successful jump?
a) 3.9 m/s
b) 2.5 m/s
c) 4.3 m/s
d) 4.5 m/s
e) 3.1 m/s
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3.3.10. When a projectile is launched at an angle  from a height h1 and t
he projectile lands at the same height, the maximum range, in the abs
ence of air resistance, occurs when  = 45. The same projectile is th
en launched at an angle  from a height h1, but it lands at a height h2
that is higher than h1, but less than the maximum height reached by t
he projectile when  = 45. In this case, in the absence of air resistan
ce, does the maximum range still occur for  = 45? All angles are me
asured with respect to the horizontal direction.
a) Yes,  = 45 will always have longest range regardless of the height h2.
b) No, depending on the height h2, the longest range may be reached for
angles less than 45.
c) No, depending on the height h2, the longest range may be reached for
angles greater than 45.
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3.3.10. When a projectile is launched at an angle  from a height h1 and t
he projectile lands at the same height, the maximum range, in the abs
ence of air resistance, occurs when  = 45. The same projectile is th
en launched at an angle  from a height h1, but it lands at a height h2
that is higher than h1, but less than the maximum height reached by t
he projectile when  = 45. In this case, in the absence of air resistan
ce, does the maximum range still occur for  = 45? All angles are me
asured with respect to the horizontal direction.
a) Yes,  = 45 will always have longest range regardless of the height h2.
b) No, depending on the height h2, the longest range may be reached for
angles less than 45.
c) No, depending on the height h2, the longest range may be reached for
angles greater than 45.
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3.3.11. Packages A and B are dropped from the same height simultaneous
ly. Package A is dropped from an airplane that is flying due east at co
nstant speed. Package B is dropped from rest from a helicopter hoveri
ng in a stationary position above the ground. Ignoring air friction effe
cts, which of the following statements is true?
a) A and B reach the ground at the same time, but B has a greater velocit
y in the vertical direction.
b) A and B reach the ground at the same time; and they have the same v
elocity in the vertical direction.
c) A and B reach the ground at different times because B has a greater ve
locity in both the horizontal and vertical directions.
d) A and B reach the ground at different times; and they have the same v
elocity in the vertical direction.
e) A reaches the ground first because it falls straight down, while B has to
travel much further than A.
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3.3.11. Packages A and B are dropped from the same height simultaneous
ly. Package A is dropped from an airplane that is flying due east at co
nstant speed. Package B is dropped from rest from a helicopter hoveri
ng in a stationary position above the ground. Ignoring air friction effe
cts, which of the following statements is true?
a) A and B reach the ground at the same time, but B has a greater velocit
y in the vertical direction.
b) A and B reach the ground at the same time; and they have the same v
elocity in the vertical direction.
c) A and B reach the ground at different times because B has a greater ve
locity in both the horizontal and vertical directions.
d) A and B reach the ground at different times; and they have the same v
elocity in the vertical direction.
e) A reaches the ground first because it falls straight down, while B has to
travel much further than A.
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3.3.5. A bullet is aimed at a target on the wall a distance L away from the
firing position. Because of gravity, the bullet strikes the wall a distanc
e Δy below the mark as suggested in the figure. Note: The drawing is
not to scale. If the distance L was half as large, and the bullet had the
same initial velocity, how would Δy be affected?
a) Δy will double.
b) Δy will be half as large.
c) Δy will be one fourth
as large.
d) Δy will be four times larger.
e) It is not possible to determine unless numerical values are given for th
e distances.
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7.3 : Motion Characteristics for
Circular Motion
Speed and Velocity

Any moving object can be described using the kine
matic concepts discussed in Unit 1. The motion of a
moving object can be explained using either Newton'
s Laws (Unit 2) and vector principles (Unit 3) or by m
eans of the Work-Energy Theorem (Ei + Wext = Ef ) .
The same concepts and principles used to describe
and explain the motion of an object can be used to d
escribe and explain the parabolic motion of a project
ile
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
In this unit, we will see that these same concepts and principl
es can also be used to describe and explain the motion of obj
ects which either move in circles or can be approximated to b
e moving in circles. Kinematic concepts and motion principles
will be applied to the motion of objects in circles and then ext
ended to analyze the motion of such objects as roller coaster
cars, a football player making a circular turn, and a planet orb
iting the sun. We will see that the beauty and power of physic
s lies in the fact that a few simple concepts and principles ca
n be used to explain the mechanics of the entire universe. Le
sson 1 of this study will begin with the development of kinem
atic and dynamic ideas can be used to describe and explain t
he motion of objects in circles.
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
Suppose that you were driving a car with the steerin
g wheel turned in such a manner that your car follow
ed the path of a perfect circle with a constant radius.
And suppose that as you drove, your speedometer
maintained a constant reading of 10 mi/hr. In such a
situation as this, the motion of your car would be des
cribed to be experiencing uniform circular motion. U
niform circular motion is the motion of an object in
a circle with a constant or uniform speed.
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
Uniform circular motion - circular motion at a consta
nt speed - is one of many forms of circular motion. A
n object moving in uniform circular motion would cov
er the same linear distance in each second of time.
When moving in a circle, an object traverses a dista
nce around the perimeter of the circle. So if your car
were to move in a circle with a constant speed of 5
m/s, then the car would travel 5 meters along the pe
rimeter of the circle in each second of time.
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
The distance of one complete cycle around the perimeter of a
circle is known as the circumference. At a uniform speed of
5 m/s, if the circle had a circumference of 5 meters, then it wo
uld take the car 1 second to make a complete cycle around th
e circle. At this uniform speed of 5 m/s, each cycle around th
e 5-m circumference circle would require 1 second. At 5 m/s,
a circle with a circumference of 20 meters could be made in 4
seconds; and at this uniform speed, every cycle around the 2
0-m circumference of the circle would take the same time peri
od of 4 seconds. This relationship between the circumference
of a circle, the time to complete one cycle around the circle, a
nd the speed of the object is merely an extension of the aver
age speed equation stated in Unit 1.
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Calculating Circular Speed
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
The circumference of any circle can be computed using from
the radius according to the equation
Circumference = 2*pi*Radius
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
Combining these two equations above will lead to a new
equation relating the speed of an object moving in unifor
m circular motion to the radius of the circle and the time t
o make one cycle around the circle (period).
R represents the radius of the circle and T represents the p
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
This equation, like all equations, can be used as a algebraic r
ecipe for problem solving. Yet it also can be used to guide ou
r thinking about the variables in the equation relate to each ot
her. For instance, the equation suggests that for objects movi
ng around circles of different radius in the same period, the o
bject traversing the circle of larger radius must be traveling wi
th the greatest speed. In fact, the average speed and the radi
us of the circle are directly proportional. A twofold increase in
radius corresponds to a twofold increase in speed; a threefol
d increase in radius corresponds to a three--fold increase in s
peed; and so on.
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
Objects moving in uniform circular motion will have a con
stant speed. But does this mean that they will have a con
stant velocity? Recall from Unit 1 that speed and velocity
refer to two distinctly different quantities. Speed is a scal
ar quantity and velocity is a vector quantity. Velocity, bei
ng a vector, has both a magnitude and a direction. The
magnitude of the velocity vector is merely the instantane
ous speed of the object; the direction of the velocity vect
or is directed in the same direction which the object mov
es. Since an object is moving in a circle, its direction is c
ontinuously changing. At one moment, the object is movi
ng northward such that the velocity vector is directed nor
thward. One quarter of a cycle later, the object would be
moving eastward such that the velocity vector is directed
eastward. As the object rounds the circle, the direction of
the velocity vector is different than it was the instant befo
re. So while the magnitude of the velocity vector may be
constant, the direction of the velocity vector is changing.
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
The best word that can be used to describe the direction of th
e velocity vector is the word tangential. The direction of the v
elocity vector at any instant is in the direction of a tangent line
drawn to the circle at the object's location. (A tangent line is a
line which touches the circle at one point but does not interse
ct it.) The diagram at the right shows the direction of the velo
city vector at four different point for an object moving in a cloc
kwise direction around a circle. While the actual direction of t
he object (and thus, of the velocity vector) is changing, it's dir
ection is always tangent to the circle.
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
To summarize, an object moving in uniform circular
motion is moving around the perimeter of the circle
with a constant speed. While the speed of the object
is constant, its velocity is changing. Velocity, being a
vector, has a constant magnitude but a changing dir
ection. The direction is always directed tangent to th
e circle and as the object turns the circle, the tangen
t line is always pointing in a new direction. As we pro
ceed through this unit, we will see that these same p
rinciples will have a similar extension to noncircular
motion.
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Example
Check your understanding
Example 1

A spiraled tube lies fixed in its horizontal position (i.e., it has b
een placed upon its side upon a table). When a marble is roll
ed through it curves around the tube, draw the path of the ma
rble after it exits the tube.
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Answer 1

The ball will move along a path which is tangent to the circle
at the point where it exits the tube. At that point, the ball will n
o longer curve or spiral, but rather travel in a straight line in th
e tangential direction.
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Lesson 1: Motion Characteristi
cs for Circular Motion
Acceleration

As mentioned earlier in Lesson 1, an object moving i
n uniform circular motion is moving in a circle with a
uniform or constant speed. The velocity vector is co
nstant in magnitude but changing in direction. Becau
se the speed is constant for such a motion, many stu
dents have the misconception that there is no accele
ration. "After all," they might say, "if I were driving a
car in a circle at a constant speed of 20 mi/hr, then t
he speed is not decreasing or increasing; therefore t
here must not be an acceleration."
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
At the heat of this common student misconceptio
n is the wrong belief that acceleration has to do
with speed and not with velocity. But the fact is t
hat an accelerating object is an object which is c
hanging its velocity. And since velocity is a vecto
r which has both magnitude and direction, a cha
nge in either the magnitude or the direction cons
titutes a change in the velocity. For this reason, i
t can be boldly declared that an object moving in
a circle at constant speed is indeed accelerating.
It is accelerating because its velocity is changing
its directions.
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
To understand this at a deeper level, we will have to com
bine the definition of acceleration with a review of some
basic vector principles. Recall from Unit 1 that accelerati
on as a quantity was defined as the rate at which the vel
ocity of an object changes. As such, it is calculated using
the following equation:
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
where vi represents the initial velocity and vf repr
esents the final velocity after some time of t. The
numerator of the equation is found by subtractin
g one vector (vi) from a second vector (vf). But t
he addition and subtraction of vectors from each
other is done in a manner much different than th
e addition and subtraction of scalar quantities. C
onsider the case of an object moving in a circle a
bout point C as shown in the diagram below. In a
time of t seconds, the object has moved from poi
nt A to point B. In this time, the velocity has chan
ged from vi to vf. The process of subtracting vi fr
om vf is shown in the vector diagram; this proces
s yields the change in velocity.
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
Note in the diagram below that there is a velocity change for
an object moving in a circle with a constant speed. Furthermo
re, note that this velocity change vector is directed towards th
e center. An object moving in a circle at a constant speed fro
m A to B experiences a velocity change and therefore an acc
eleration; this acceleration is directed towards point C - the c
enter of the circle.
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
The acceleration of an object is often measured using a devic
e known as an accelerometer. A simple home- made acceler
ometer involves a lit candle centered vertically in the middle o
f a open-air glass. If the glass is held level and at rest (such t
hat there is no acceleration), then the candle flame extends i
n an upward direction. However, if the glass is held at the en
d of an outstretched arm as you spin in a circle at a constant
rate (such that the flame experiences an acceleration), then t
he candle flame will no longer extend vertically upwards. Inst
ead the flame deflects from its upright position.
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
This signifies that there is an acceleration when the flam
e moves in a circular path at constant speed. The deflect
ion of the flame will be in the direction of the acceleration
. This is because the hot gases of the flame are less mas
sive (on a per mL basis) and thus have less inertia than t
he cooler gases which surround. Subsequently, the hotte
r and lighter gases of the flame experience the greater a
cceleration and will lurch towards the direction of the acc
eleration.
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
A careful examination of the flame reveals that the flame will
point towards the center of the circle, thus indicating that not
only is there an acceleration; but that there is an inward accel
eration. Objects moving in a circle at a constant speed experi
ence an acceleration which is directed towards the center of t
he circle.
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Cork in water demonstration

A further demonstrations of this principle was perfor
med in class using a cork accelerometer. A cork was
submerged in a sealed flask of water. The flask was
then held in an outstretched arm and moved in a circ
le at a constant rate of turning. Thus, the flask with b
oth the water and the cork were moving in uniform ci
rcular motion. Again, the least massive of the two ob
jects will lean in the direction of the acceleration.
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
In the case of the cork and the water, the cork is least massiv
e (on a per mL basis) and thus it experiences the greater acc
eleration. As the cork-water combination spun in the circle, th
e cork leaned towards the center of the circle. Once more, th
ere is proof that an object moving in circular motion at consta
nt speed experiences an acceleration which directed towards
the center of the circle.
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
So thus far, we have seen a geometric proof and
two real-world demonstrations of this inward acc
eleration. At this point it becomes the decision of
the student to believe or not to believe. Is it sensi
ble that an object moving in a circle experiences
an acceleration which is directed towards the ce
nter of the circle? Can you think of a logical reas
on to believe in say no acceleration or even an o
utward acceleration experienced by an object m
oving in uniform circular motion? In the next part
of Lesson 1, additional logical evidence will be p
resented to support the notion of an inward force
for an object moving in circular motion.
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Examples

1. The initial and final speed of a ball at two different points in
time is shown below. The direction of the ball is indicated by t
he arrow. For each case, indicate if there is an acceleration.
Explain why or why not. Indicate the direction of the accelerat
ion.
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.
Acceleration:
Yes or No? If there is an acceleration, then what direction is it?
Explain.
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Answer (a)

Since the velocity did not change, there is no acceleration.
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Acceleration:
Yes or No?
Explain.
If there is an acceleration, then what direction
is it?
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Answer b


Since the velocity changes (speeds up) there is acceleration.
If an object moving rightward speeds up then the acceleration
is also rightward.
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Acceleration:
Yes or No?
Explain.
If there is an acceleration, then what
direction is it?
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Answer c


Since there is a change in velocity there is an acceleration.
A rightward moving object that slows down has a leftward acc
eleration.
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Acceleration: Yes
or No? Explain.
If there is an acceleration, then what
direction is it?
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Answer d

Since there is a change in velocity (decrease in speed) then t
here is acceleration.
 A rightward moving object that slows down has a leftward acc
eleration.
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Acceleration:
Yes or No?
Explain.
If there is an acceleration, then what direction
is it?
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Answer e

Even though the initial and final speeds are the same, there h
as been a change in direction, so there is an acceleration.
 The object was moving in a rightward direction until it slowed
to 0 m/s then it changed to a leftward movement, so the accel
eration is leftward.
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
2. Explain the connection between your answers to the above
questions and the reasoning used to explain why an object m
oving in a circle at constant speed can be said to experience
an acceleration.
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Answer 2

An object that either experiences a change in magnitude or di
rection of the velocity vector can be said to be accelerating. T
his explains why an object moving in a circle with constant sp
eed can be said to accelerate—the velocity changes.
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Ex. 3

3. Dizzy Smith and Hector Vector are still discussing #1e. Diz
zy says that the ball is not accelerating because its velocity is
not changing. Hector says that since the ball has changed its
direction, there is an acceleration. Who do you agree with? A
rgue a position by explaining the discrepancy in the other stu
dent's argument.
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Answer 3

Agree with Hector. A change in direction constitutes a chang
e in velocity
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
4. Identify the three controls on an automobile which allow th
e car to be accelerated.
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Answer 4

Accelerator allows the car to speed up.
 Brake allows for the car to slow down.
 Steering wheel allows for change in direction.
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
For questions #5-#8: An object is moving in a clockwise
direction around a circle at constant speed. Use your un
derstanding of the concepts of velocity and acceleration t
o answer the next four questions. Use the diagram show
n at the right.
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
5. Which vector below represents the direction of the velocity
vector when the object is located at point B on the circle?
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Answer 5


Answer D
The velocity vector is directed tangent to the circle that would
be downward when at point B.
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
6. Which vector below represents the direction of the acceler
ation vector when the object is located at point C on the circle
?
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Answer 6

B The acceleration vector would be directed towards the cent
er that would be up and to the right when at point C
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
7. Which vector below represents the direction of the velocity
vector when the object is located at point C on the circle?
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Answer 7

A The velocity vector would be directed tangent to the circle a
nd that would be upwards at point C.
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
8. Which vector below represents the direction of the acceler
ation vector when the object is located at point A on the circle
?
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Answer 8

D the acceleration vector would be directed towards the cent
er and that would be straight down when the object is at point
A
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Lesson 1: Motion Characteristi
cs for Circular Motion
The Centripetal Force Requirement

As mentioned earlier in this lesson, an object movin
g in a circle is experiencing an acceleration. Even if
moving around the perimeter of the circle with a con
stant speed, there is still a change in velocity and su
bsequently an acceleration. This acceleration is dire
cted towards the center of the circle. And in accord
with Newton's second law of motion, an object which
experiences an acceleration must also be experienci
ng a net force; and the direction of the net force is in
the same direction as the acceleration.
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
So for an object moving in a circle, there must be an
inward force acting upon it in order to cause its inwa
rd acceleration. This is sometimes referred to as the
centripetal force requirement. The word "centripet
al" (not to be confused with the F-word "centrifugal")
means center-seeking. For object's moving in circula
r motion, there is a net force acting towards the cent
er which causes the object to seek the center.
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
To understand the ne
ed for a centripetal fo
rce, it is important to
have a sturdy underst
anding of the Newton'
s first law of motion the law of inertia. Th
e law of inertia states
that ...
"... objects in motion tend to
stay in motion with the
same speed and the same
direction unless acted upon
by an unbalanced force."
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
According to Newton's first law of motion, it is the na
tural tendency of all moving objects to continue in m
otion in the same direction that they are moving ... u
nless some form of unbalanced force acts upon the
object to deviate the its motion from its straight-line
path. Objects will tend to naturally travel in straight li
nes; an unbalanced force is required to cause it to tu
rn. The presence of the unbalanced force is required
for objects to move in circles.
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
The idea expressed by Newton's law of inertia shoul
d not be surprising to us. We experience this pheno
menon of inertia nearly everyday when we drive our
automobile. For example, imagine that your are a pa
ssenger in a car at a traffic light. The light turns gree
n and the driver "steps on the gas." The car begins t
o accelerate forward, yet relative to the seat which y
ou are on, your body begins to lean backwards. You
r body being at rest tends to stay at rest. This is one
aspect of the law of inertia - "objects at rest tend to s
tay at rest."
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
As the wheels of the car spin to generate a forw
ard force upon the car to cause a forward accele
ration, your body tends to stay in place. It certain
ly might seem to you as though your body were
experiencing a backwards force causing it to acc
elerate backwards; yet you would have a difficult
time identifying such a backwards force on your
body. Indeed there isn't one. The feeling of bein
g thrown backwards is merely the tendency of yo
ur body to resist the acceleration and to remain i
n its state of rest. The car is accelerating out fro
m under your body, leaving you with the false fe
eling of being thrown backwards.
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
Now imagine that you're driving along at constant speed and
then suddenly approach a stop sign. The driver steps on the
brakes. The wheels of the car lock and begin to skid across t
he pavement. This causes a backwards force upon the forwa
rd moving car and subsequently a backwards acceleration on
the car. However, your body being in motion tends to continu
e in motion while the car is slowing to a stop. It certainly migh
t seem to you as though your body were experiencing a forw
ards force causing it to accelerate forwards; yet you would on
ce more have a difficult time identifying such a forwards force
on your body
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
Indeed there is no physical object accelerating you f
orwards. The feeling of being thrown forwards is mer
ely the tendency of your body to resist the decelerati
on and to remain in its state of forward motion. This i
s the second aspect of Newton's law of inertia - "an
object in motion tends to stay in motion with the sam
e speed and in the same direction... ." The unbalanc
ed force acting upon the car causes it to slow down
while your body continues in its forward motion.
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These two driving scenarios are summ
arized by the following graphic.
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
In each case - the car starting from rest and the moving car b
raking to a stop - the direction which the passengers lean is o
pposite the direction of the acceleration. This is merely the re
sult of the passenger's inertia - the tendency to resist acceler
ation. The passengers lean is not an acceleration in itself but
rather the tendency to maintain whatever state of motion they
have while the car does the acceleration. The tendency of ou
r body to maintain its state of rest or motion while the surroun
dings (the car) accelerate is often misconstrued as an acceler
ation. This becomes particularly problematic when we consid
er the third possible inertia experience as a passenger in a m
oving automobile - the left hand turn.
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
Suppose you continue driving and suddenly round a sharp tur
n to the left at constant speed. During the turn, the car travels
in a circular-type path; that is, the car sweeps out one- quart
er of a circle. The unbalanced force acting upon the turned w
heels of the car cause an unbalanced force upon the car and
a subsequent acceleration. The unbalanced force and the ac
celeration are both directed towards the center of the circle a
bout which the car is turning. Your body however is in motion
and tends to stay in motion. It is the inertia of your body - the
tendency to resist acceleration - which causes it to continue i
n its forward motion.
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
While the car is accelerating inward, you continue in a straigh
t line. If you are sitting on the passenger side of the car, then
eventually the outside door of the car will hit you as the car tu
rns inward. This phenomenon might cause you to think that y
ou were being accelerated outwards away from the center of
the circle. In reality, you are continuing in your straight-line in
ertial path tangent to the circle while the car is accelerating o
ut from under you. The sensation of an outward force and an
outward acceleration is a false sensation. There is no physica
l object capable of pushing you outwards. You are merely ex
periencing the tendency of your body to continue in its path ta
ngent to the circular path along which the car is turning.
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
Any object moving in a circle (or along a circular path) experi
ences a centripetal force; that is there must be some physic
al force pushing or pulling the object towards the center of th
e circle. This is the centripetal force requirement. The word "c
entripetal" is merely an adjective used to describe the directio
n of the force. We are not introducing a new type of force but
rather describing the direction of the net force acting upon the
object which moves in the circle. Whatever the object, if it mo
ves in a circle, there is some force acting upon it to cause it t
o deviate from its straight-line path, accelerate inwards and m
ove along a circular path. Three such examples of centripetal
force are shown below.
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As a car makes a turn, the force of frict
ion acting upon the turned wheels of t
he car provide the centripetal force req
uired for circular motion.
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As a bucket of water is tied to a string
and spun in a circle, the force of tensio
n acting upon the bucket provides the
centripetal force required for circular
motion.
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As the moon orbits the Earth, the force
of gravity acting upon the moon provid
es the centripetal force required for cir
cular motion.
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Lesson 2
Newton’s 2nd Law Revisited

Newton's second law states that the acceleration of an object
is directly proportional to the net force acting upon the object
and inversely proportional to the mass of the object. The law i
s often expressed in the form of the following two equations.
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
In Unit 2, Newton's second law was used to analyze
a variety of physical situations. The idea was that if a
ny given physical situation is analyzed in terms of th
e individual forces which are acting upon an object, t
hen those individual forces must add up to the net fo
rce. Furthermore, the net force must be equal to the
mass times the acceleration. Subsequently, the acc
eleration of an object can be found if the mass of the
object and the magnitudes and directions of each in
dividual force are known.
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
And the magnitude of any individual force can be det
ermined if the mass of the object, the acceleration of
the object, and the magnitude of the other individual
forces are known. The process of analyzing such ph
ysical situations in order to determine unknown infor
mation is dependent upon the ability to represent the
physical situation by means of a free-body diagram.
A free-body diagram is a vector diagram which depic
ts the relative magnitude and direction of all the indiv
idual forces which are acting upon the object.
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Free Body Diagrams
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Determine the net force
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
The three major equations which will b
e useful are:
– the equation for net force (Fnet = m * a),

– the equation for gravitational force (Fgrav = m * g), and

– the equation for frictional force
(Ffrict = µ * Fnorm).

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Finding the individual forces
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
In this Lesson, we will use Unit 2 principles (freebody diagrams, Newton's second law equation,
etc.) and circular motion concepts in order to an
alyze a variety of physical situations involving th
e motion of objects in circles or along curved pat
hs. The mathematical equations discussed in Le
sson 1 and the concept of a centripetal force req
uirement will be applied in order to analyze roller
coasters and other amusement park rides, vario
us athletic movements, and other real-world phe
nomenon.
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
To illustrate how circular motion principles can be combi
ned with Newton's second law to analyze a physical situ
ation, consider a car moving in a horizontal circle on a le
vel surface. The diagram below depicts the car on the lef
t side of the circle.
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
Applying the concept of a centripetal force requirem
ent, we know that the net force acting upon the obje
ct is directed inwards. Since the car is positioned on
the left side of the circle, the net force is directed rig
htward. An analysis of the situation would reveal that
there are three forces acting upon the object - the fo
rce of gravity (acting downwards), the normal force o
f the pavement (acting upwards), and the force of fri
ction (acting inwards or rightwards). It is the friction f
orce which supplies the centripetal force requiremen
t for the car to move in a horizontal circle.
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
Without friction, the car would turn its wheels but
would not move in a circle (as is the case on an i
cy surface). This analysis leads to the free-body
diagram shown below. Observe that each force i
s represented by a vector arrow which points in t
he specific direction which the force acts; also n
otice that each force is labeled according to type
(Ffrict, Fnorm, and Fgrav). Such an analysis is th
e first step of any problem involving Newton's se
cond law and a circular motion.
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Now consider the following tw
o problems pertaining to this p
hysical scenario of the car mak
ing a turn on a horizontal surfa
ce
The two problems will be solve
d using the same general princ
iples. Yet because the given an
d requested information is diff
erent in each, the solution met
hod will be slightly different.
Sample Problem #1

A 900-kg car makes a 180-degree turn with a speed of 10.0
m/s. The radius of the circle through which the car is turning i
s 25.0 m. Determine the force of friction and the coefficient of
friction acting upon the car.
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
Sample problem #1 provides kinematic information (v and R)
and requests the value of an individual force. As such the sol
ution of the problem will demand that the acceleration and th
e net force first be determined; then the individual force value
can be found by use of the free-body diagram.
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
Known Information:
 m = 900 kg v = 10.0 m/s
 R = 25.0 m
 Requested Information:
 Ffrict = ??? mu = ????
 ("mu" - coefficient of friction)
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
The mass of the object can be used to determine th
e force of gravity acting in the downward direction. U
se the equation
 Fgrav = m * g
 where g can be approximated as 10 m/s/s. Knowing
that there is no vertical acceleration of the car, it can
be concluded that the vertical forces balance each o
ther. Thus, Fgrav = Fnorm= 9000 N. This allows us
to determine two of the three forces identified in the f
ree-body diagram. Only the friction force remains un
known.
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Free-body Diagram
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
Since the force of friction is the only horizontal force, it m
ust be equal to the net force acting upon the object. So if
the net force can be determined, then the friction force is
known. To determine the net force, the mass and the kin
ematic information (speed and radius) must be substitute
d into the following equation:
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
Substituting the given values yields a net force of 3600 Newt
ons. Thus, the force of friction is 3600 N.
 Finally the coefficient of friction ("mu") can be determined usi
ng the equation which relates the coefficient of friction to the f
orce of friction and the normal force.
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
Substituting 3600 N for Ffrict and 9000 N for Fnorm yields a
coefficient of friction of 0.400.
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Sample Problem 2

The coefficient of friction acting upon a 900-kg car is 0.850. T
he car is making a 180-degree turn around a curve with a rad
ius of 35.0 m. Determine the maximum speed with which the
car can make the turn.
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
Sample problem #2 provides information about the i
ndividual force values (or at least information which
allows for the determination of the individual force va
lues) and requests the value of the maximum speed
of the car. As such, its solution will demand that indi
vidual force values be used to determine the net forc
e and acceleration; then the acceleration can be use
d to determine the maximum speed of the car.
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




Known Information:
m = 900 kg "mu" = 0.85 (coefficient of friction)
R = 35.0 m
Requested Information:
v = ??? (the minimum speed would be the speed achiev
ed with the given friction coefficient)
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
The mass of the car can be used to determine th
e force of gravity acting in the downward directio
n. Use the equation
 Fgrav = m * g
 where g can be approximated as 10 m/s/s. Kno
wing that there is no vertical acceleration of the c
ar, it can be concluded that the vertical forces ba
lance each other. Thus, Fgrav = Fnorm= 9000
N. Since the coefficient of friction ("mu") is given,
the force of friction can be determined using the
following equation:
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Free Body Diagram
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
The net force acting upon any object is the vector su
m of all individual forces acting upon that object. So i
f all individual force values are known (as is the case
here), the net force can be calculated. The vertical f
orces add to 0 N. Since the force of friction is the onl
y horizontal force, it must be equal to the net force a
cting upon the object. Thus, Fnet = 7650 N.
 Once the net force is determined, the acceleration c
an be quickly calculated using the following equation
.
 Fnet = m*a
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
Substituting the given values yields an acceleration of 7.65 m
/s/s. Finally, the speed at which the car could travel around th
e turn can be calculated using the equation for centripetal acc
eleration
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
Substituting the known values for a and R into this equation a
nd solving algebraically yields a maximum speed of 16.4 m/s.

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Suggested Method of Solving
Circular Motion Problems
Step 1

From the verbal description of the physical situation, construc
t a free-body diagram. Represent each force by a vector arro
w and label the forces according to type.
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Step 2

Identify the given and the unknown information (express in te
rms of variables such as m = , a = , v = , etc.).
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Step 3

If any of the individual forces are directed at angles, then use
vector principles to resolve such forces into horizontal and ve
rtical components.
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Step 4

Determine the magnitude of any known forces and label on th
e free-body diagram.
(For example, if the mass is given, then the Fgrav can be det
ermined. And as another example, if there is no vertical accel
eration, then it is known that the vertical forces or force comp
onents balance, allowing for the possible determination of on
e or more of the individual forces in the vertical direction.)
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Step 5

Use circular motion equations to determine any unknown info
rmation.
(For example, if the speed and the radius are known, then th
e acceleration can be determined. And as another example, if
the period and radius are known, then the acceleration can b
e determined.)
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Step 6 (2 parts)

Use the remaining information to solve for the requested infor
mation.
a) If the problem requests the value of an individual force
, then use the kinematic information (R, T and v) to determine
the acceleration and the Fnet ; then use the free-body diagra
m to solve for the individual force value.
b) If the problem requests the value of the speed or radiu
s, then use the values of the individual forces to determine th
e net force and acceleration; then use the acceleration to det
ermine the value of the speed or radius.
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
The method prescribed above will serve you well as you
approach circular motion problems. However, one cautio
n is in order. Every physics problem differs from the previ
ous problem. As such, there is no magic formula for solvi
ng every one. Using an appropriate approach to solving
such problems (which involves constructing a FBD, ident
ifying known information, identifying the requested inform
ation, and using available equations) will never eliminate
the need to think, analyze and problem-solve. For this re
ason, make an effort to develop an appropriate approach
to every problem; yet always engage your critical analysi
s skills in the process of the solution. If physics problems
were a mere matter of following a foolproof, 5-step formu
la or using some memorized algorithm, then we wouldn't
call them "problems."
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Examples
Check your understanding
Example 1

1. A 1.5-kg bucket of water is tied by a rope and whirled i
n a circle with a radius of 1.0 m. At the top of the circular
loop, the speed of the bucket is 4.0 m/s. Determine the a
cceleration, the net force and the individual force values
when the bucket is at the top of the circular loop.
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
m = 1.5 kg
 a = ________ m/s/s
 Fnet = _________ N
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Answer 1

Fgrav = 15 N
 a = 16 m/s/s
 Fnet = 24 N
 Ftens = 24 N – 15 N = 9N
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Example 2

2. A 1.5-kg bucket of water is tied by a rope and whirled i
n a circle with a radius of 1.0 m. At the bottom of the circ
ular loop, the speed of the bucket is 6.0 m/s. Determine t
he acceleration, the net force and the individual force val
ues when the bucket is at the bottom of the circular loop.
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
m = 1.5 kg
 a = ________ m/s/s
 Fnet = _________ N
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Answer 2

F grav = 15 N
 a = 36 m/s/s
 F net = 54 N Up
 F tens = 69 N
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Now we are going to investigate
how to use this with
1.
Roller Coasters
2. Sports