Rotation and Gravity

1
Rotation and Gravity
2
Torque
3
Consider the see saw

• A see saw is an example of a device that twists.
• A force that causes a twisting motion, multiplied
  by its distance from the point of rotation, is
  called a torque.
• The larger the torque, the easier it is for it to
  cause the twist.
• Torque is what makes a see saw fun.

4
Torque

• Consider a beam connected to a flat surface by a
  hinge. (This is a top view.)

Now consider a force F on the beam that is
applied a distance r from the hinge.
What happens? A rotation occurs due to the
combination of r and F. In this case, the
direction is clockwise.
5
Torque

• If we know the angle ? between F and r, we can
  calculate torque!
• ? r F sin ?
• ? is torque
• r is moment arm
• F is force
• ? is angle between F and r

?
6
Torque Units

• What are the SI units for torque?
• mN or Nm.
• Can you substitute Joule for Nm?
• No. Even though a Joule is a Nm, it is a scalar.
  Torque is a vector and cannot be ascribed energy
  units.

7
Sample Problem

• Consider the door to the classroom. We use torque
  to open it.
• Identify the following
• The point of rotation.
• The point of application of force.
• The moment arm (r).
• The angle between r and F (best guess).

8
Solution

• The hinge.
• Where your hand pushes the door.
• The distance from the hinge to the place your
  hand pushes the door.
• 90o

9
Sample Problem

• A crane lifts a load. If the mass of the load is
  500 kg, and the cranes 22-m long arm is at a 75o
  angle relative to the horizontal, calculate the
  torque exerted about the point of rotation at the
  base of the crane arm.

10
Solution

• t r F sin q
• r 22 m
• F mg (500 9.8)N
• 4900 N
• q 15o
• (22)(4900)(sin15o)
• 27900 Nm

11
Torque equation simplified

• If ? is 90o
• ? r F
• ? is torque
• r is moment arm
• F is force

?
F
12
Problem

• A standard door is 36 inches wide, with the
  doorknob located at 32 inches from the hinge.
  Calculate the torque a person applies when he
  pushes on the doorknob at right angles to the
  door with a force of 110 N. (Use 1 inch 2.54 cm
  to calculate the torque in SI units).

13
Solution

• Width of door is irrelevant.
• t r F sin q
• r 32 inches 81.28 cm 0.8128 m
• F 110 N
• 90o
• t (0.8128m)(110 N)(1) 89.4 Nm

14
Problem

• A double pulley has two weights hanging from it
  as shown.
• A) What is the net torque?
• B) In what direction will the pulley rotate?

3 cm
2 cm
2 kg
10 kg
15
Solution

• Counterclockwise torque
• ccw r F sin q (0.020m)(10 kg)(9.8m/s2)
• 1.96 Nm
• Clockwise torque
• cw r F sin q (0.030m)(2 kg)(9.8m/s2)
• 0.59 Nm
• Net torque
• tnet 1.96 0.59
• 1.37 Nm (ccw)

16
Now consider a balanced situation
40 kg
40 kg

• tccw tcw
• This is called rotational equilibrium!

17
Now consider a balanced situation
40 kg
40 kg

• If the weights are equal, and the moment arms are
  equal, then the clockwise and counterclockwise
  torques are equal and no net rotation will occur.
  The kids can balance!

18
Sample Problem

• A 5.0-meter long see saw has a 45-kg child sits
  all the way on one end and a 60-kg child all the
  way on the other end.
• What is the magnitude of the net torque?
• In what direction will the see saw rotate?

19
Solution

• A) tnet tB tL
• mBgr mLgr (mB mL)gr
• (60 45)(9.8)(2.5)
• 368 Nm
• B) The system will rotate downward where the
  big child is seated and upward where the little
  child is seated.

20
Sample Problem

• A 5.0-meter long see saw is balanced on a fulcrum
  at the middle. A 45-kg child sits all the way on
  one end. Where must a 60-kg child sit if the
  see-saw is to be balanced?

21
Solution

• You wish the net torque to be zero.
• 0 tB tL
• Therefore, the torques are equal in magnitude.
• tB tL
• mBgrB mLgrL
• (60)(9.8) rB (45)(9.8)(2.5)
• rB (45)(2.5) / 60 1.88 m from the fulcrum.

22
Sample Problem

• A 5.0-meter long see saw is balanced on a fulcrum
  at the middle. A 45-kg child sits all the way on
  one end. And a 60-kg child sits all the way on
  the other end. If the see saw has a mass of 100
  kg, where must the fulcrum be placed to attain a
  balanced situation?

23
Solution

• tccw tcw
• mBgrB mLgrL msgrs
• All the gs cancel
• rL 2.5 rs and rB 2.5 - rs
• (60)(2.5 - rs ) (45)(2.5 rs) (100)rs
• 150 - 60 rs 112.5 45 rs 100 rs
• 37.5 205 rs
• rs 0.18 m (just 18 cm!)

24
Sample Problem

• A 10-meter long wooden plank of mass 209 kg rests
  on a flat roof with 2.5 meters extended out
  beyond the roofs edge. How far out on the plank
  can an 80-kg man walk before he is in danger of
  falling?

25
Solution
m

• The moment arm of the beam is 2.5 m from the
  point of rotation at the edge of the roof
  (location of center of mass)
• tccw tcw
• mBgrB mmgrm
• rm mBrB/mm
• rm 6.5 m
• He can go all the way to the end

b
26
Torque Lab I

• Find the mass of your unknown by constructing a
  torque balance using the ruler.
• Rules
• The ruler must be balanced at least 15 cm from
  midpoint.
• You must use at least two mass hangers ATTACHED
  WITH CLIPS on the ruler. The amount of mass on
  the hanger is up to you.
• The unknown MUST BE HUNG FROM A CLIP.
• Hints
• The masses of rulers, clips, etc., are written on
  them.
• Dont forget the masses of your clips in your
  calculations.
• What to turn in
• Nothing. When you think you know the unknown
  mass, go see your teacher and she will mass it.
  Your grade is based on how close you get to the
  actual value.

27
Torque Lab II

• Predict the reading on the spring scale of the
  device to which you and your group are assigned.
• Rules
• You may not touch the device.
• Hints
• Pulleys simply change the line of action of the
  force.
• The masses of all components are written on the
  components.
• What to turn in
• Your calculations.
• You will be graded on clarity and correctness.
• Clearly indicate what you think the reading on
  the spring scale is.
• Clearly indicate the device you are analyzing.

28
SpaceThe Final Frontier
29
Johannes Kepler (1571-1630)

• Kepler developed some extremely important laws
  about planetary motion.
• Kepler based his laws on massive amounts of data
  collected by Tyco Brahe.
• Keplers laws were used by Newton in the
  development of his own laws.

30
Keplers Laws

• Planets orbit the sun in elliptical orbits, with
  the sun at a focus.
• Planets orbiting the sun carve out equal area
  triangles in equal times.
• The planets year is related to its distance from
  the sun in a predictable way.

31
Keplers Laws

• Lets look at a simulation of planetary motion at
  http//surendranath.tripod.com/Applets.html

32
Newtons Laws and Space

• Newtons Laws govern space travel if speeds are
  not near the speed of light.
• Some of the formulas we use involving
  gravitational force have to be re-examined when
  we travel in space.
• The formulas weve been using near the surface of
  the earth are approximations of the true
  formulas, so we need to learn the rest of the
  story.

33
The Universal Law of Gravity

• Newtons famous apple fell on Newtons famous
  head, and lead to this law.
• It tells us that the force of gravity objects
  exert on each other depends on their masses and
  the distance they are separated from each other.

34
The Force of Gravity

• Remember Fg mg?
• Weve use this to approximate the force of
  gravity on an object near the earths surface.
• This formula wont work for planets and space
  travel.
• It wont work for objects that are far from the
  earth.
• For space travel, we need a better formula.

35
The Force of Gravity

• Fg -Gm1m2/r2
• Fg Force due to gravity (N)
• G Universal gravitational constant
• 6.67 x 10-11 N m2/kg2
• m1 and m2 the two masses (kg)
• r the distance between the centers of the masses
  (m)
• The Universal Law of Gravity ALWAYS works,
  whereas F mg only works sometimes.

36
Sample Problem

• How much force does the earth exert on the moon?
• How much force does the moon exert on the earth?

37
Solution

• F GMm/r2
• 6.67E-115.97E247.35E22 / 3.84E82
• 1.98E20 1.98 x 1020 N
• Same force as earth exerts on moon, but opposite
  direction.

38
Sample Problem

• What would be your weight if you were orbiting
  the earth in a satellite at an altitude of
  3,000,000 km above the earths surface? (Note
  that even though you are apparently weightless,
  gravity is still exerting a force on your body,
  and this is your actual weight.)

39
Solution

• The your weight is the force of gravity on your
  body. Near the earths surface, that is mg.
  However, far from the earth, you must use
• F GMm/r2
• 6.67E-115.97E24(your mass) / 3.00E92
  
• 4.42 x 10-5 (your mass) (unit is Newtons)

40
Sample Problem

• Sally, an astrology buff, claims that the
  position of the planet Jupiter influences events
  in her life. She surmises this is due to its
  gravitational pull. Joe scoffs at Sally and says
  your Labrador Retriever exerts more
  gravitational pull on your body than the planet
  Jupiter does. Is Joe correct? (Assume a 100-lb
  Lab 1.0 meter away, and Jupiter at its farthest
  distance from Earth).

41
Solution

• For the laborador
• F Gm1m2/r2
• 6.67E-11(your mass)45.5 / 12
• 3.03E-9 (your mass)
• For Jupiter
• F Gm1m2/r2
• 6.67E-11(your mass)1.90E27 / (9.28E11)2
• 1.47E-7 (your mass)
• So the Jupiter exerts far more gravitational
  force on your body than the laborador 1 m away
  would.

42
Sample Problem (not in packet)

• Using Newtons Law of Universal Gravitation,
  derive a formula to show how the period of a
  planets orbit varies with the radius of that
  orbit. Assume a nearly circular orbit.

43
Solution

• GMm/r2 mv2/r (the mass of the satellite
  cancels)
• GM/r2 v2/r (but v equals circumference divided
  by period of revolution)
• GM/r2 (2pr/T)2/r
• GM/r2 4p2r2/T2r
• GM/r2 4p2r/T2 (cross multiply to solve for T2)
• T2 4p2r3/(GM)
• So the period squared is proportional to the
  radius of revolution cubed and inversely
  proportional to the mass of the central body
  causing the orbit.

44
Gravitational Potential Energy

• Remember Ug mgh?
• This is also an approximation we use when an
  object is near the earth.
• This formula wont work when we are very far from
  the surface of the earth.
• For space travel, we need another formula.

45
Gravitational Potential Energy

• Ug -Gm1m2/r
• Ug Gravitational potential energy (J)
• G Universal gravitational constant
• 6.67 x 10-11N m2/kg2
• m1 and m2 the two masses (kg)
• r the distance between the centers of the masses
  (m)
• Notice that the theoretical value of Ug is
  always negative.
• This formula always works!

46
Sample Problem

• What is the gravitational potential energy of a
  satellite that is in orbit about the Earth at an
  altitude equal to the earths radius? Assume the
  satellite has a mass of 10,000 kg.

47
Solution

• Ug -GMm/r
• Ug -6.67E-115.97E2410,000/(26.37E6)
• Ug -3.13E11 J

48
Escape Velocity

• Calculation of miniumum escape velocity from a
  planets surface can be done by using energy
  conservation.
• Assume the object gains potential energy and
  loses kinetic energy, and assume the final
  potential energy and final kinetic energy are
  both zero.
• U1 K1 U2 K2
• -GMm/r ½mv2 0
• v (2GM/r)1/2

49
Sample Problem

• What is the velocity necessary for a rocket to
  escape the gravitational field of the earth?
  Assume the rocket is near the earths surface.

50
Solution

• U1 K1 U2 K2
• -GmM/R ½ mv2 0
• ½ mv2 GmM/R
• v (2GM/R)½
• v ?(26.67E-115.97E24 / 6.37E6)
• v 11,200 m/s

51
Sample Problem

• Suppose a 2500-kg space probe accelerates on
  blast-off until it reaches a speed of 15,000 m/s.
  What is the rockets kinetic energy when it has
  effectively escaped the earths gravitational
  field?

52
Sample Problem

• Suppose a 2500-kg space probe accelerates on
  blast-off until it reaches a speed of 15,000 m/s.
  What is the rockets kinetic energy when it has
  effectively escaped the earths gravitational
  field?

53
Acceleration due to gravity

• Remember g 9.8 m/s2?
• This works find when we are near the surface of
  the earth.
• For space travel, we need a better formula!

54
Acceleration due to gravity

• g GM/r2
• This formula lets you calculate g anywhere if you
  know the distance a body is from the center of a
  planet.
• We can calculate the acceleration due to gravity
  anywhere!

55
Sample Problem

• What is the acceleration due to gravity at an
  altitude equal to the earths radius? What about
  an altitude equal to twice the earths radius?

56
Solution

• g GM/r2 (and we know its 9.8 m/s2 at surface)
• g GM/(1r)2 9.8/1 at surface
• g GM/(2r)2 9.8/4 at altitude of r
• g GM/(3r)2 9.8/9 at altitude of 2r

57
Acceleration and distance
58
Surface gravitational acceleration depends on
mass and radius.


59
Sample Problem

• What is the acceleration due to gravity at the
  surface of the moon?

60
Satellites
61
Orbital speed

• At the earths surface, if an object moves 8000
  meters horizontally, the surface of the earth
  will drop by 5 meters vertically.
• That is how far the object will fall vertically
  in one second (use the 1st kinematic equation to
  show this).
• Therefore, an object moving at 8000 m/s will
  never reach the earths surface.

• At any given altitude, there is only one speed
  for a stable circular orbit.
• From geometry, we can calculate what this orbital
  speed must be.

62
Some orbits are nearly circular.
63
Some orbits are highly elliptical.
64
Announcements 12/6/2005

• Lunch bunch HW due today.
• Regular class HW due tomorrow.
• Momentum exam corrections for makeup exam due
  tomorrow.
• Toy Day project due tomorrow.
• Get out your clickers.

65
Centripetal force and gravity

• The orbits we analyze mathematically will be
  nearly circular.
• Fg Fc
• (centripetal force is provided by gravity)
• GMm/r2 mv2/r
• The mass of the orbiting body cancels out in the
  expression above.
• One of the rs cancels as well
• GM/r v2

66
Earth satellites

• Satellites that orbit the earth can have circular
  or elliptical orbits, and they can be at
  different altitudes.
• A simulation is available at http//surendranath.t
  ripod.com/Applets.html

67
Sample Problem

• What velocity does a satellite in orbit about the
  earth at an altitude of 25,000 km have?
• What is the period of this satellite?

68
Sample Problem

• A geosynchronous satellite is one which remains
  above the same point on the earth. Such a
  satellite orbits the earth in 24 hours, thus
  matching the earth's rotation. How high must must
  a geosynchronous satellite be above the surface
  to maintain a geosynchronous orbit?
• Mearth 5.98x1024 kg
• Rearth 6.37 x 106 m