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Title: Chapter 7 Rotational Motion and the Law of Gravity


1
Chapter 7 Rotational Motion and the Law of
Gravity
2
71 Measuring Rotational Quantities
3
Rotational Motion
  • Motion of a body that spins about an axis.

4
Axis of rotation
  • Is the line about which the rotation occurs.

5
Angles can be measured in radians (rads).
  • Radians an angle whose arc length is equal to
    its radius, which approximately equal to 57.3
    degrees.

6
1 radian 57.3o
  • s
  • ----
  • r

7
1 radian 57.3o
  • s
  • ----
  • r

s arc length r length of radius q angle of
rotation
8
  • The angle of 360o is one revolution. (1 rev
    360o)
  • One revolution is equal to the circumference of
    the circle of rotation.

9
Circumference is 2prso,1 rev 360o 2p rads
  • Radius is half the diameter (d).

10
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11
Angular Displacement (Dq)
  • The angle through which a point, line, or body is
    rotated in a specified direction and about a
    specified axis.

12
Describes how far an object has rotated.
  • Change
  • Angular in arc length
  • Displacement --------------
  • radius

13
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14
  • Counterclockwise (CCW) rotation is considered ()
  • Clockwise (CW) rotation is considered (-)

15
  • Ex 1 John Glenn, in 1962, circled the earth 3
    times in less than 5 hours. If his distance from
    the center of the earth was 6560 km, what arc
    length did he travel through?

16
  • G Dq 3 rev 6p rads, r 6560km
  • U Ds ?
  • E Ds Dq r
  • S Ds (6p)(6560)
  • S Ds 123653 km

17
Angular Speed (w)
  • The rate at which a body rotates about an axis,
    usually expressed in radians per second.
    (rads/sec)

18
  • Angular
  • Average displacement
  • Angular -------------------
  • Speed time interval

19
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20
  • Ex 2 In 1975, an ultra-fast centrifuge attained
    an average angular speed of 2.65 x 104 rads/sec.
    What was the angular displacement after 1.5 sec?

21
  • G wavg 2.65 x 104 rads/sec, Dt 1.5 sec
  • U Dq ?
  • E Dq wavg Dt
  • S Dq (2.65 x 104)(1.5)
  • S Dq 39750 radians

22
How fast is the earth rotating?
  • The Earth rotates at a moderate angular velocity
    of 7.2921159 10 -5 rads/s

23
  • Ex 3 A water supply tunnel in NYC is 1.70 x 105
    m long and a radius of 2 m. If an electronic eye
    sensor moves around the perimeter of the tunnel
    at an average angular speed of 5.9 rads/sec. How
    long will it take the beetle to travel the
    equivalent of the length of the tunnel?

24
  • G wavg 5.9 rads/sec, r 2 m, Ds 1.70 x 105
    m
  • U Dt ?
  • E Dt Dq / wavg
  • We need to find Dq.

25
  • Dq Ds / r
  • Dq 1.70 x 105 m / 2 m
  • Dq 85000 rads

26
  • S Dt 85000 rads
  • 5.9 rads/sec

27
  • S Dt 85000 rads
  • 5.9 rads/sec
  • S Dt 14406.8 sec

28
Angular Acceleration (a)
  • The time rate of change of angular speed,
    expressed in rads/sec2 .

29
wf - wia -------- Dt
30
  • wf final angular speed
  • wi initial angular speed
  • a angular acceleration
  • Dt elapsed time

31
Equation is rearranged just like the linear
acceleration equation
  • ALWAYS multiply by Dt first! Then isolate what
    you are looking for.

32
Ex 4 A brake is applied to a flywheel for 5
seconds, reducing the angular velocity of the
wheel uniformly from 2.2 rads/sec to 1.8
rads/sec. What is the angular acceleration?
33
G Dt 5 sec, w2 1.8 rads/sec, and w1 2.2
rads/sec
  • U a ?
  • E w2 - w1
  • a --------
  • Dt

34
  • 1.8 rads/sec 2.2 rads/sec
  • a ------------------------
  • 5 sec
  • a - 0.08 rads/sec2

35
In comparing angular and linear quantities, they
are similar.
36
Looking back to Ch 2 equations
  • Dx viDt ½ a Dt2
  • Dx ½ (vi vf)Dt
  • vf vi aDt
  • vf2 vi2 2aDx

37
We apply angular notation to these equations and
get
  • Dq wiDt ½ a Dt2
  • Dq ½ (wi wf)Dt
  • wf wi aDt
  • wf2 wi2 2aDq

38
  • Ex 5 A wheel on a bike moves through 11.0 rads
    in 2 seconds. What is the angular acceleration of
    the initial angular speed is 2.0 rads/sec?

39
  • G Dq 11 rads, Dt 2 sec, wi 2 rads/sec
  • U a ?
  • E Dq wiDt ½ aDt2
  • a 2(Dq - wiDt) / Dt2
  • S a 211 (2)(2) /22
  • S a 3.5 rads/sec2

40
7-2 Tangential and Centripetal Acceleration
41
Objects in motion have a tangential speed.
  • Tangent a line that lies in a plane of circle
    that intersects at one point.

42
Tangent line
43
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44
Tangential Speed (vt)
  • The instantaneous linear speed of an object
    directed along the tangent to the objects path.

45
  • r distance from axis
  • angular speed
  • vt tangential speed

46
  • Ex 6 The radius of a CD is 0.06 m. If a microbe
    riding on the discs rim has a tangential speed
    of 1.88 m/s, what is the microbes angular speed?

47
  • G r 0.06 m, vt 1.88 m/s
  • U w ?
  • E w vt / r
  • S w 1.88 m/s / 0.06 m
  • S w 31.3 rads/sec

48
Tangential Acceleration (at)
  • The instantaneous linear acceleration of an
    object directed along the tangent to the objects
    circular path

49
  • at tangential acceleration
  • r distance from axis
  • a angular acceleration

50
  • Ex 7 To crack a whip requires its tip to move
    at supersonic speed. This was achieved with a
    whip 56.24 m. If the tip moved in a circle and it
    accelerated from 6.0 rads/sec to 6.3 rads/sec in
    0.6 sec. What would the tangential acceleration
    be?

51
  • G r 56.24 m, Dt 0.6 sec
  • wi 6 rads/sec,
  • wf 6.3 rads/sec,
  • U at ?
  • E at ra

52
We need to find angular acceleration (a)
  • E wf - wi
  • a --------
  • t

53
  • S 6.3 - 6.0
  • a --------
  • 0.6
  • S a 0.5 rads/sec2

54
  • S at (56.24)(0.5)
  • S at 28.12 m/s2

55
Centripetal Acceleration (ac)
  • Acceleration directed toward the center of a
    circular path.

56
  • r distance from axis (m)
  • ac centripetal accel. (m/s2)
  • vt tangential speed (m/s)

57
  • r distance from axis (m)
  • ac centripetal accel. (m/s2)
  • w angular speed (rads/sec)

58
  • Ex 8Calculate the orbital radius of the Earth,
    if its tangential speed is 29.7 km/s and the
    centripetal acceleration acting on the Earth is 5
    x 10-3 m/s2?

59
  • G vt29.7 km/s 29700 m/s
  • ac 5 x 10-3 m/s2
  • U r ?
  • E vt2
  • r ----
  • ac

60
  • S 297002
  • r ------------
  • 5 x 10-3
  • S r 1.764 x 10 11m

61
Tangential and centripetal accelerations are
perpendicular.
ac
at
atotal
62
The total acceleration can be found by using the
Pythagorean Theorem.
63
The total acceleration can be found by using the
Pythagorean Theorem.
64
If you need to find the angle (q) use the inverse
tangent function (tan-1)
65
7-3 Causes of Circular Motion
66
If a mass undergoes an acceleration, there must
be a force causing it. (Newtons 2nd Law)
67
A force directed toward the center is necessary
for circular motion.
68
This force causing circular motion is called
Centripetal Force (Fc)
Fc
r
m
v
69
With a centripetal force, an object in motion
will be accelerated and change its direction.
70
Work Done by the Centripetal Force
  • Since the centripetal force on an object is
    always perpendicular to the objects velocity,
    the centripetal force never does work on the
    object - no energy is transformed.

71
Without a centripetal force, an object in motion
continues along a straight-line path.
72
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73
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74
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75
Centrifugal Force
  • centrifugal force is a fictitious force - it is
    not an interaction between 2 objects, and
    therefore not a real force.Nothing pulls an
    object away from the center of the circle.

76
Centrifugal Force
  • What is erroneously attributed to centrifugal
    force is actually the action of the objects
    inertia - whatever velocity it has (speed
    direction) it wants to keep.

77
From Newtons 2nd Law F ma
  • Fc mac

78
From Newtons 2nd Law F ma
  • Fc mac

79
From Newtons 2nd Law F maFc mac
80
  • Ex 9 Joe attaches a 0.5 kg mass to a 1m rope.
    Joe swings the rope in a horizontal circle above
    his head with a speed of 4 m/s. What force is
    exerted to keep the mass moving in a circle?

81
  • G m 0.5 kg
  • vt 4 m/s
  • r 1 m
  • U Fc ?
  • E Fc m(vt2 / r)

82
  • S
  • Fc .5 kg x(4 m/s)2/(1 m)
  • S Fc 8 N

83
Gravitational Force
  • The mutual force of attraction between particles
    of matter.

84
This is a field force that always exists between
any two objects, regardless of their sizes.
85
The gravitational force (Fg) depends upon the
distance between the two masses.
86
Newtons Law of Universal Gravitation
87
  • m1 mass 1 (kg)
  • m2 mass 2 (kg)
  • r distance between center of masses. (m)
  • G Constant of Universal Gravitation.

88
Units will cancel out with others leaving just
force units of Newtons (N).
89
The universal gravitational constant (G) can be
used to find the gravitational force between any
two particles.
90
This is what is known as the inverse-square law.
  • The force between the two masses decreases as the
    masses move farther apart.

91
The acceleration due to gravity, g, decreases
with distance from the earths center as 1/d2.
92
As an object moves away from the center of the
Earth, by a distance equivalent to the earths
radius.
93
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94
  • Ex 10 What is the force of attraction between
    Adam and Jason, gravitational of course. If they
    are 12 m apart and Jason mass is 82 kg and Adam
    is 74 kg.

95
  • G m1 82 kg, m2 74kg, r 12 m,
  • G 6.673x 10-11 n-m2/kg2
  • U Fg ?

96
  • G m1 82 kg, m2 74kg, r 12 m,
  • G 6.673x 10-11 n-m2/kg2
  • U Fg ?
  • E

97
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98
  • S Fg 2.81 x 10 -9 N

99
  • Ex 11 The sun has a mass 2 x 10 30 kg and a
    radius of 7 x 10 5 km. What mass must be located
    on the suns surface for gravitational force of
    470 N to exist between the mass and the sun?

100
  • G m1 2 x 10 30 kg,
  • r 7 x 10 5 km7 x 108 m
  • G 6.673x 10-11 n-m2/kg2
  • Fg 470 N
  • U m2 ?

101
E
102
E
103
S m2 1.7 kg
104
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105
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