Circular motion and the law of Gravity

1
Circular motion and the law of Gravity
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Angular displacement

• Difference between final and initial angles
• SI unit radian (rad)
• Counterclockwise rotation is positive by
  definition

3
Units

• Angles are sometimes expressed in degrees
• 180º p rad 3.1415rad
• 1 rad 180/p 57.295º
• 1 turn is 2p 360º
• The point can rotate by more than one turn, for
  example if it does 2.5 turns, the angular
  displacement is

4
Angular velocity

• Average angular velocity
• SI unit rad/s
• The angular velocity tells you how fast an object
  is rotating. It is sometimes expressed in rpm
  (rotation per minute)
• 1 rpm 2p/60 0.1047rad/s

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example

• An engine is rotating at 10.0 rpm. What is the
  angular displacement after 2.00 minutes of
  running?
• A) 5.00 rad
• B) 10.0 rad
• C) 126 rad
• D) 1200 rad

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example

• An engine is rotating at 10.0 rpm. What is the
  angular displacement after 2.00 minutes of
  running?
• A) 5.00 rad
• B) 10.0 rad
• C) 126 rad
• D) 1200 rad

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Angular acceleration

• Average angular acceleration
• SI unit rad/s2
• The angular acceleration tells you how much the
  angular velocity is changing through time. For
  example when you accelerate in your car, the
  wheels rotate more and more rapidly.

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example

• An engine rotates initially at 5000 rpm, it is
  shut off and its rotation speed decreases
  continuously until it comes to rest after 5.000
  minutes. What was the average angular
  acceleration?
• A) -1.745 rad/s2
• B) -16.67 rad/s2
• C) 0.000 rad/s2
• D) 16.67 rad/s2
• E) 1000 rad/s2

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example

• An engine rotates initially at 5000 rpm, it is
  shut off and its rotation speed decreases
  continuously until it comes to rest after 5.000
  minutes. What was the average angular
  acceleration?
• A) -1.745 rad/s2
• B) -16.67 rad/s2
• C) 0.000 rad/s2
• D) 16.67 rad/s2
• E) 1000 rad/s2

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Displacement and angular displacement

• For a point on a circle of radius r, the
  displacement Ds along the circular path is
• SI units
• Ds meter (m)
• r meter (m)
• Dq radian (rad)

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Tangential velocity and angular velocity

• tangential velocity and angular velocity are not
  the same thing
• An object moves on a circle of radius r2.0 m at
  an angular velocity w10 rad/s. Its velocity is

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Rotation of an object

• For a solid object rotating
• All the points of the object have equal angular
  velocity
• The points farther away from the axis of rotation
  have greater tangential velocity

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Geostationary satellite

• There is an orbit around earth for which a
  satellite rotates at the same angular velocity
  than the earth (both do one rotation in 24h).
• The earth has a radius RE6400 km and the
  geostationary orbit is at a radius Rg42 000 km.
  How much faster is the satellite speed compared
  to the speed of a point on earths surface?

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Acceleration and angular acceleration

• Tangential acceleration and angular acceleration
  are not the same thing

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Linear and rotational motions

• Linear

Rotational
position
angular position
velocity
angular velocity
acceleration
angular acceleration
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Units comparison

• Linear

Rotational
m
rad
m/s
rad/s
m/s2
rad/s2
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Multiply by r to go from angular quantities to
tangential quantities
m
rad
m/s
rad/s
m/s2
rad/s2
Ds is the displacement in meters of the point
moving on a circle (see next slides to understand
the meaning of tangential)
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Similarities with kinematics relations

• There is a direct parallel between linear motion
  and rotational motion, for example

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Lawn mower

• You turn on your lawn mower in your backyard. The
  blades accelerates continuously for Dt3.00s
  until they reach a rotational speed of 3000 rpm.
  What has been the total angular displacement
  during that time?
• A) 471 rad
• B) 942 rad
• C) 1.88103 rad

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Lawn mower

• You turn on your lawn mower in your backyard. The
  blades accelerates continuously for Dt3.00s
  until they reach a rotational speed of 3000 rpm.
  What has been the total angular displacement
  during that time?
• A) 471 rad
• B) 942 rad
• C) 1.88103 rad

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Rolling coin

• A coin has initial angular velocity wi10 rad/s
  and slows down with a-2.0 rad/s2 until it is at
  rest. What has been the angular displacement ? If
  the coin has a radius of 1.0 cm, how far did it
  roll?
• A) 20 rad 0.20 m
• B) 25 rad 0.25 m
• C) 40 rad 0.40 m

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Rolling coin

• A coin has initial angular velocity wi10 rad/s
  and slows down with a-2.0 rad/s2 until it is at
  rest. What has been the angular displacement ? If
  the coin has a radius of 1.0 cm, how far did it
  roll?

B) 25 rad 0.25 m
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Periodicity of repetitive circular motion

• A repetitive circular motion has a period of
  revolution, which menas the time (in seconds) it
  takes the object to do one full turn, it is
  usually noted with capital T letter
• Lets imagine an object does 5 rotations per
  second. Its angular velocity and period are

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Frequency of repetitive circular motion

• A repetitive circular motion has a period of
  revolution T, and a frequency f linked by
• T in seconds, f is in Hz (Hertz) which is
  nothing else than 1/s
• the frequency f tells us how many turns are being
  done every second
• In the example from the previous slide, the
  object was doing 5 turns per seconds so
• f 5Hz

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Circular motion

• We know that the acceleration is the change of
  the velocity vector through time
• If an object rotates on a circle at constant
  speed,
• the magnitude of v is constant
• but the direction of v changes and a?0

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Centripetal acceleration

• For circular motion at constant speed, what is
  the direction and magnitude of the acceleration ?
• Consider a small duration Dt
• Similar triangles

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Centripetal acceleration

• We found two things for circular motion at
  constant speed
• acceleration points toward the center of the
  circle (thats why its called centripetal)
• Magnitude of acceleration is v2/r

v
ac
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Example - ac

• A bicycle wheel is in the shop to service its
  brakes. The wheel is spinning freely at a
  constant speed.
• You estimate the wheel does half a rotation in
  one second
• What is the centripetal acceleration of a point
  on the tire if the wheel is 50 cm in radius?

A) 0.0 m/s2 B) 4.9 m/s2 C) 9.8 m/s2
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Example - ac
A) 0.0 m/s2 B) 4.9 m/s2 C) 9.8 m/s2
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Tangential centripetal acceleration

• When an object accelerates on a circular
  trajectory, its tangential velocity changes and
  we had found
• Where we used the subscript t for tangential to
  distinguish it from centripetal acceleration

• Object on a circular motion with centripetal and
  tangential acceleration (object can accelerate
  or decelerate along the circular trajectory )

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Example ac and at

• A friend is on merry-go-round initially rotating
  with wi2.0 rad/s and at r2.0 m
• You decide to speed-up the rotation and provide
  an angular acceleration of a1.0 rad/s2 for 1.0
  second
• What is the amplitude of the acceleration
  experienced by your friend after you are done?

• A) 4.0 m/s2
• B) 9.0 m/s2
• 18 m/s2
• 36 m/s2

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• A) 4.0 m/s2
• B) 9.0 m/s2
• 18 m/s2
• 36 m/s2

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Force causing ac

• For an object undergoing a circular motion at
  constant speed, Newtons 2nd law requires that a
  force provides the centripetal acceleration
  Fnetmacmv2/r . Example
• Gravitation force
• Tension in a rotating rope
• Friction between tires and road
• Normal force
• .

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Example - F

• A mass m12.0 kg is rotating at constant speed on
  a horizontal plane with radius R1.0 m. The mass
  is attached through a rope to another mass m21.0
  kg hanging vertically
• What is the tension in the rope and the speed of
  m1?

• A) 9.8 N 2.2 m/s
• B) 9.8 N 4.9 m/s
• 20 N 2.2 m/s
• 20 N 4.9 m/s

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• A) 9.8 N 2.2 m/s
• B) 9.8 N 4.9 m/s
• 20 N 2.2 m/s
• 20 N 4.9 m/s

T
Tm2g
Fg,2-m2g
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Conical pendulum

• The pendulum is rotating at constant speed around
  the vertical axis
• What is its speed if the length of the string is
  50 cm and the angle w.r.t. the vertical axis is
  45º?

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L

• The forces acting on the mass are gravity and the
  tension in the string
• The mass is not moving vertically
• The mass is rotating horizontally
• Speed is constant

Tsinq
Tcosq
T
-mg
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Car through a bend

• A car is on a level road and goes around a
  circular bend of radius R50 m at a speed of 30.0
  mi/h
• The centripetal force is provided by static
  friction between the tires and the road
• What is the minimum ms needed so that the car
  doesnt slide?

• 0.37
• 0.74
• C) 1.0

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• 0.37
• 0.74
• C) 1.0

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Car in a banked curve

• A car is turning in a frictionless banked curve
  with a bend radius R100m at an angle of 45º with
  the horizontal
• What must be the speed of the car to make it
  through on a circular path of radius R?

• Not possible
• 31 m/s
• C) 41 m/s
• D) 51 m/s

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• Not possible
• 31 m/s
• C) 41 m/s
• D) 51 m/s

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Car in a banked curve with friction

• A car is turning banked curve with a bend radius
  R100m at an angle of 45º with the horizontal
• The coefficient of static friction between the
  tires and the road is ms0.50
• What is the maximum speed of the car to avoid
  sliding in the curve?

• Not possible
• 34 m/s
• C) 44 m/s
• D) 54 m/s

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Fscosq
Fssinq
D) 54 m/s
Fs
Fs points down because the car goes faster than
the nominal 31 m/s
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Car in a banked curve with friction

• A car is turning banked curve with a bend radius
  R100m at an angle of 45º with the horizontal
• The coefficient of static friction between the
  tires and the road is ms0.50
• What is the minimum speed of the car to avoid
  sliding in the curve?

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Fs
Fssinq
Fs points up because the car goes slower than the
nominal 31 m/s and without friction the car would
slide down the bank
Fscosq
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Gravitation (take 2)

• At the surface of the Earth
• ME 6.0 1024 kg
• RE 6.4 106 m

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Approximation near earths surface

• Lets see how good was our approximation of a
  constant gravitational acceleration near the
  Earths surface
• Acceleration of a mass m at a height h above
  Earths surface is
• On Earth surface
• 100 m above surface
• 1000 m away from Earths surface
• 1000 km away from Earth surface
• Apprimation is good when h/RE very small (ltlt1)

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Synchronous orbit

• We had said that an object located at r42 000 km
  from the Earth center would always stay at the
  same position in the sky (so called synchronous
  or geostationary orbit), lets prove it!

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Gravitational Potential Energy (GPE)

• Near Earths surface we had define
• This formula is only a good approximation when h
  is small compared to Earths radius
• In general, GPE of an object at distance r from
  Earths center is
• Remember that important quantity is not GPE but
  DGPE, the change in GPE

50
Escape speed

• Now that we have a more precise formula for the
  gravitational potential energy, we can calculate
  the velocity that an object would need to have
  when launched from a planet to never fall back on
  it
• Initially, the object is on the planets surface,
  at the end it is infinitely far from it and at
  rest

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Escape speed

• We apply conservation of energy
• After some algebra we find
• Calculating for Earth we would find vesc11.2
  km/s

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Scaling tip

• In some problem you are asked how a quantity
  scales when you change some parameters by some
  coefficient
• For example one could ask, how does the escape
  velocity vary for a planetX which is 4 times as
  heavy as Earth and with a radius twice as big?
• The trick is to express these parameters as
• MX4ME
• RX2RE
• Then take general formula and replace, et voila

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Keplers laws

• Three Keplers laws describe the motion of
  planets around the Sun
• 1) All planets move on elliptical orbits with the
  sun as one focal point
• 2) A line drawn from the Sun to any planet sweeps
  out equal areas in equal time intervals
• 3) The square of the orbital period of any planet
  is proportional to the cube of the average
  distance form the planet to the sun
• Average distance from a planet to the sun on an
  elliptical orbit is equal to the semi-major axis
  of the ellipse

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Ellipses

• Circle distance to a given
• point is constant
• Ellipse Sum of distances from two points ( the
  two focal points) is constant

average distance from planet to focal
point (important for Keplers 3rd law)
pq constant
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Keplers laws

• First Law
• Consequence of force proportional to 1/r2
• Second Law
• AreaSABAreaSCD if
• DtABDtCD

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Keplers laws

• Third law
• Can be easily proven for circular orbit
• Assume planet of mass MP on a circular orbit of
  radius r around the Sun of mass MS

• T is the time the planet takes to do one
  revolution around the sun

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Homework (help)

• Using Keplers third law for this planet of mass
  MP and since orbit is at radius RP
• Rearranging to get
• The planet is a uniform sphere of density
• Use
  these eq. to find r