Circular Motion

1
Circular Motion

• Examples Planets, atoms, cars on curves,
  CD-ROMs, propellers, etc.

2
Rotational Kinematics

• How do we describe an object moving in a circle?

3

• Measure angles relative to reference line

T 600
??
T 00
4
Express Angles in Radians

• 1 revolution 3600
• 3600 2p radians
• 1 radian 57.30

Position in circular motion is expressed as an
angle, preferably in radians.
x in linear motion, ? in circular motion.
5
Example

• A fan blade, with a diameter of 1 m, is rotating
  at 1 revolution every second. Through what angle
  does the tip of the blade turn in 0.3 seconds?
• 1 rev in 1 second
• 0.5 rev in 0.5 sec
• 0.3 rev in 0.3 sec, so
• 0.3 rev 1080 1.88 radians

6
Speed in Circular Motion

• Since were not using x to measure position, we
  cant define speed as ?x/?t.
• But, since ? replaces x, why not define
  angular speed by
• ??/?t (?f ?o)/(tf to)
• This is called angular speed and is given the
  symbol (omega), ?.
• ? ??/?t
• In units of radians/second, i.e. rad/s.

7
Angular Velocity

• ? ??/?t

8
Connection between Linear and Circular Motion

• q x/r
• v/r
• a a/r

9
Example

• What is the angular speed of a 33 1/3 rpm record?
• ? 33.33 rev/min 33.33 rev/60 sec
• ? 33.33 ( 2 p rad)/60 s 3.49 rad/s

10
Acceleration in Circular Motion

• Consider your rotating car tires as you
  accelerate from 25 mph to 55 mph. What is
  happening to the rotational speed of the tires? R
  33 cm.

?f
?o
V24.5 m/s
V 11 m/s
11
Linear and Angular Connection

• ? v/r
• So, ?o (11 m/s)/0.33 m 33 rad/s
• And ?f (24.5 m/s)/0.33 m 73.5 rad/s.
• Therefore the change in angular speed,
• ?f ?o 40.5 rad/s ? ?

12
Angular Acceleration

• When you have changing angular speeds, this means
  the object has an angular acceleration, a
  (alpha), which is calculated by
• a ? ?/?t
• In units of radians/second2 rad/s2

13
Characterizing Circular Motion

• Radius, r
• Angular position, ?
• Angular displacement, ??
• Angular speed, ???/?t
• Angular acceleration, a??/?t

14
Kinematics of Circular Motion

• ???/?t
• a??/?t
• ?ave(?f ?o)/2
• T ?avet
• ?f ?o at
• T To ?ot ½ at2
• ?f2 ?o2 2a??

15
Kinematics Example

• A flywheel of a machine is rotating at 12 rev/s.
  Through what angle will the wheel be displaced
  from its original position after 5 seconds?
• Angular speed, ? 12 rev/s 75 rad/s
• T ?avet 75 rad/s 5 s 375 rad
• 2148750. 59.6875 revolutions, so .6875
  revolutions from start position 247o.

16

• A turntable revolves at 33 1/3 rpm. It is shut
  off and slow to a stop in 6.3 seconds. What is
  the angular acceleration?
• Through what angle did it turn as it slow to a
  stop? ?f0, ?o 33.33 rpm 3.49 rad/s,
• t 6.3 s
• ?f ?o at
• T To ?ot ½ at2

17
Dynamics of Rotation

• Examine circular motion taking Newtons Laws into
  consideration.
• 1st Law-
• 2nd Law-
• 3rd Law-

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Dynamics of Rotation

• 1st Law
• Is Moon at rest?
• Is Moon moving in a straight line?
• Conclusion

MOON
EARTH
19
Dynamics of Rotation
1st Law Objects executing circular motion have a
net force acting on themeven if you cant see
the agent of the force. What force acts on the
Moon?
MOON
EARTH
20
Dynamics of Rotation

• 2nd Law
• FNET ma
• a is a vector defined by
• a ?v/?t
• ?v vf vo
• For circular motion the speeds are the same, but
  the directions arent.

MOON
EARTH
21
Dynamics of Rotation
MOON
Lets visually examine the change in velocity
-vo
EARTH
vf
?v vf-vo
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Dynamics of Rotation
vo
So, a ?v exists because the direction is
changing, not the magnitude. How do we find the
acceleration? Dv/v Dr/r Dv/v vDt/r Dv/Dt
v2/r ac v2/r
MOON
?v vf-vo
vf
EARTH
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Dynamics of Rotation
The acceleration toward the center is called
centripetal acceleration, ac, given by ac
v2/R In magnitude and directed toward center.
MOON
v
R
EARTH
v
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Dynamics of Rotation
So, Newtons 2nd Law for rotation becomes, F mac
mv2/R In magnitude and directed toward center.
MOON
v
R
EARTH
v
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Dynamics of Rotation
A physical statement that relates cause and
effect. Cause F, effect mv2/R F mv2/R The
right side is what you see, the left side is
why.
MOON
v
R
EARTH
v
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Dynamics of Rotation
What is the cause for the Moons motion? F
GMm/R2 Newtons Universal Law of Gravity. G
6.67 x 10-11 Nm2/kg2
MOON,m
v
R
EARTH,M
v
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• ULG-two objects of given masses separated by
  known distance exert a gravitational force of
  attraction on each other whose size is determined
  from
• F GMm/R2
• You are sitting next to a person whose mass is 55
  kg. Your mass is 75 kg. What is the force of
  attraction between you if you are 0.8 m apart
  (center to center)?
• F (6.67 x 10-11Nm2/kg2)(75kg)(55kg)/(0.8m)2
• F0.00000043 N 0.43 microNewtons

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How do we know the mass of the Earth?

• Using the ULG,
• F GMm/R2
• And 2nd Law,
• F mv2/R,
• Combine,
• GMm/R2 mv2/R
• M v2R/G
• M (?R)2R/G
• M ?2R3/G

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• M ?2R3/G
• R 380,000,000 m
• ? 2p rad /27.3 days 2.664 x 10-6 rad/s
• So, M (2.664 x 10-6 rad/s)2(380,000,000
  m)3/(6.67 x 10-11 Nm2/kg2)
• M 5.84 x 1024 kg.
• True value 5.98 x 1024 kg.

30

• Earth and Moon orbit the center of mass of the
  system.
• Located 1070 miles below the Earths surface or
  2880 miles from center of Earth.

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Problem Solving Strategy for Circular Motion
Problems

• Is it Kinematics or Dynamics
• Kinematics-You are trying to characterize the
  motion by its position, speed or acceleration.
  Click here.
• Dynamics-You are trying to relate the motion to
  its causes. Click here.

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• A tire of diameter 26 inches is spinning with a
  constant angular velocity of 2 rad/s. What is
  the centripetal acceleration of a point on the
  rim of the tire?
• R 0.33 m
• ? 2 rad/s
• Centripetal Accel. ac v2/R
• v ?R
• ac v2/R ?2 R
• (2 rad/s)20.33 m
• ac 1.32 m/s2, directed toward axle.
• Another example, click here.

33

• A dentists drill spins at 1800 rpm. If it takes
  6 seconds to stop when turned off, what is the
  angular acceleration of the drill?
• Initial angular speed,
• ?o 1800 rpm188 rad/s
• Final angular speed ?f 0 rad/s
• Time, t 6 s.
• Angular Accel,
• a ??/?t(0-188rad/s)/(6s)
• -31 rad/s2

34

• A car moving at 25 m/s rounds a curve of radius
  100 m and is just on the verge of slipping. So
  if that is the fastest that it can round this
  curve, what is the maximum speed it can travel on
  a curve of radius 300 m?
• In both cases the force keeping that car from
  slipping will be the same, i.e. static friction.
  So, F mv2/R is the equation we will apply to
  each case.
• v12/R1 v2 2/R2
• v2 v1(R2/R1)1/2
• Next Example

35

• A 0.50-kg mass is attached to the end of a 1.0-m
  string. The system is whirled in a horizontal
  circular path. If the maximum tension that the
  string can withstand is 350 N. What is the
  maximum speed of the mass if the string is not to
  break.
• M 0.5 kg, R 1.0 m, F(max) 350 N
• F mv2/R
• V ( FR/m)1/2 (3501.0/0.5)1/2 26.5 m/s
• Next Example

36

• A car goes around a flat curve of radius 50 m at
  a speed of 14 m/s. What must be the minimum
  coefficient of friction between the tires and the
  road for the car to make the turn?
• V 14 m/s, R 50 m.
• F mv2/R and f µN µmg, so
• µmg mv2/R
• µ v2/gR
• (14 m/s)2/(9.8 m/s250m)
• µ 0.4
• Next Example

37

• The hydrogen atom consists of a proton of mass
  1.67X10-27 kg and an orbiting electron of mass
  9.11X10-31 kg. In one of its orbits, the
  electron is 5.3X10-11 m from the proton. What is
  the mutual gravitational attractive forces
  between the electron and proton?
• F GM1M2/R2
• (6.67e-111.67e-279.11e-31)/(5.3e-11)2
• F 3.6 x 10-47 N

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• 3rd Law
• F(earth on moon) -F(moon on Earth)
• FEM -FME

-FME
FEM
39
Energy of Orbiting Objects

• Consider Moon.
• It has velocity, so it has Kinetic Energy.
• E K U.
• What is the potential energy of a bound object?

40
E
U
K
K is pos, U is neg, E is neg.