Using the

1
Using the Clicker

• If you have a clicker now, and did not do this
  last time, please enter your ID in your clicker.
• First, turn on your clicker by sliding the power
  switch, on the left, up. Next, store your student
  number in the clicker. You only have to do this
  once.
• Press the button to enter the setup menu.
• Press the up arrow button to get to ID
• Press the big green arrow key
• Press the T button, then the up arrow to get a U
• Enter the rest of your BU ID.
• Press the big green arrow key.

2
A figure skater

• A spinning figure skater is an excellent example
  of angular momentum conservation. The skater
  starts spinning with her arms outstretched, and
  has a rotational inertia of Ii and an initial
  angular velocity of ?i. When she moves her arms
  close to her body, she spins faster. Her moment
  of inertia decreases, so her angular velocity
  must increase to keep the angular momentum
  constant.
• Conserving angular momentum
• In this process, what happens to the skater's
  kinetic energy?

3
A figure skater

• When the figure skater moves her arms in closer
  to her body while she is spinning, what happens
  to the skaters rotational kinetic energy?
• It increases
• It decreases
• It must stay the same, because of conservation
  of energy

4
Kinetic energy

• The terms in brackets are the same, so the final
  kinetic energy is larger than the initial kinetic
  energy, because
• .
• Where does the extra kinetic energy come from?

5
Kinetic energy

• The terms in brackets are the same, so the final
  kinetic energy is larger than the initial kinetic
  energy, because
• .
• Where does the extra kinetic energy come from?
• The skater does work on her arms in bringing them
  closer to her body, and that work shows up as an
  increase in kinetic energy.

6
A bicycle wheel

• A person standing on a turntable while holding a
  bicycle wheel is an excellent place to observe
  angular momentum conservation in action.
  Initially, the bicycle wheel is rotating about a
  horizontal axis, and the person is at rest.
• The initial angular momentum about a vertical
  axis is zero.
• If the person re-positions the bicycle wheel so
  its rotation axis is vertical, the wheel exerts a
  torque on the person during the re-positioning
  that makes the person spin in the opposite
  direction. The angular momenta cancel, so L 0
  at all times about a vertical axis.
• Flipping the bike wheel over makes the person
  spin in the opposite direction.

7
Jumping on a merry-go-round

• Lets analyze a rotational collision.
• Sarah, with mass m and velocity v, runs toward a
  playground merry-go-round, which is initially at
  rest, and jumps on at its edge. Sarah and the
  merry-go-round (mass M, radius R, and I cMR2)
  then spin together with a constant angular
  velocity ?f. If Sarah's initial velocity is
  tangent to the circular merry-go-round, what is
  ?f?
• Simulation
• What concept should we use to attack this
  problem?

8
Jumping on a merry-go-round

• Lets analyze a rotational collision.
• Sarah, with mass m and velocity v, runs toward a
  playground merry-go-round, which is initially at
  rest, and jumps on at its edge. Sarah and the
  merry-go-round (mass M, radius R, and I cMR2)
  then spin together with a constant angular
  velocity ?f. If Sarah's initial velocity is
  tangent to the circular merry-go-round, what is
  ?f?
• Simulation
• What concept should we use to attack this
  problem?Conservation of angular momentum.

9
Jumping on a merry-go-round

• The system clearly has angular momentum after the
  completely inelastic collision, but where is the
  angular momentum beforehand?
• Its with Sarah. Much like a force gives rise to
  a torque, Sarahs linear momentum can be
  converted to an angular momentum relative to an
  axis through the center of the merry-go-round.
• In this case,
  .
• The angular momentum is directed counterclockwise.

10
Jumping on a merry-go-round

• Conserving angular momentum
• Lets define counterclockwise to be positive.
• We can treat Sarah as a point, a distance R from
  the center.
• Solving for the final angular speed

11
Newtons Law of Universal Gravitation

• Two objects of mass m and M, with their centers
  of gravity separated by a distance r, exert
  attractive forces on one another. The magnitude
  of this gravitational force is given by
• where G is the universal gravitational constant
• The direction of the force exerted on one object
  is toward the center of gravity of the second
  object the force is attractive.

12
Newtons Law of Universal Gravitation

• Newtons form of the equation for the force of
  gravity must be consistent with the mg we have
  been using up to this point in the course
• For an object of mass m at the surface of the
  Earth, this tells us that

13
At the center of the Earth

• If we bring an object from far away toward the
  Earth, the gravitational force increases. The
  closer it gets, the bigger the force. This is
  certainly true when the mass is outside the Earth
  - what happens if we bring it right to the
  surface and then keep going, tunneling into the
  Earth?
• What is the force of gravity on an object if is
  right at the center of the Earth?
• zero
• infinite

14
Inside the Earth

• The net gravitational force on an object at the
  center of the Earth is zero forces from
  opposite sides of the Earth cancel out.
• This is a consequence of Gauss' Law for Gravity.
  One implication of Gauss Law is that inside a
  uniform spherical shell, the force of gravity due
  to the shell is zero. Outside the shell, the
  force is exactly the same as that from a point
  object of the same mass as the shell, placed at
  the center of the shell.
• Newtons Law of Universal Gravitation applies as
  long as one object is not overlapping the other.

15
Superposition

• If an object experiences multiple forces, we can
  use
• The principle of superposition - the net force
  acting on an object is the vector sum of the
  individual forces acting on that object.

16
Rank these situations
A small blue ball has one or more large red balls
placed near it. The red balls are all the same
mass and the same distance from the blue one.
Rank the different cases based on the net
gravitational force experienced by the blue ball
due to the neighboring red ball(s). 1.
13gt2gt4 2. 3gt2gt1gt4 3. 4gt3gt2gt1 4. 2gt13gt4
5. 4gt123
17
Ranking the situations

• Does case 4 give the largest net force or the
  smallest?
• Is there another case with the same magnitude net
  force as case 1?
• Remember this kind of situation when we look at
  charged objects next semester. There are many
  similarities between gravitational interactions
  and interactions between charged objects.

18
Ranking the situations

• Does case 4 give the largest net force or the
  smallest?
• The smallest there is no net force in case 4.
• Is there another case with the same magnitude net
  force as case 1? Yes, case 3.
• Remember this kind of situation when we look at
  charged objects next semester. There are many
  similarities between gravitational interactions
  and interactions between charged objects.

19
Gravitational potential energy

• The energy of interaction (that is, the
  gravitational potential energy) of two objects of
  mass m and M separated by a distance r is
• The negative sign just tells us that the
  interaction is attractive.
• Note that with this equation the potential energy
  is defined to be zero when r infinity.
• What matters is the change in gravitational
  potential energy. For small changes in height at
  the Earths surface, the equation above gives the
  same change in potential energy as mgh.

20
Escape speed

• How fast would you have to throw an object so it
  never came back down? Ignore air resistance.
  Let's find the escape speed - the minimum speed
  required to escape from a planet's gravitational
  pull.
• How should we try to figure this out?
• Attack the problem from a force perspective?
• From an energy perspective?

21
Escape speed

• How fast would you have to throw an object so it
  never came back down? Ignore air resistance.
  Let's find the escape speed - the minimum speed
  required to escape from a planet's gravitational
  pull.
• How should we try to figure this out?
• Attack the problem from a force perspective?
• From an energy perspective?
• Forces are hard to work with here, because the
  size of the force changes as the object gets
  farther away. Energy is easier to work with in
  this case.

22
Escape speed

• Lets start with the conservation of energy
  equation.
• Which terms can we cross out immediately?

23
Escape speed

• Lets start with the conservation of energy
  equation.
• Which terms can we cross out immediately?
• Assume no resistive forces, so
• Assume the object barely makes it
• to infinity, so both Uf and Kf are zero.
• This leaves

24
Escape speed

• If the total mechanical energy is negative, the
  object comes back. If it is positive, it never
  comes back.
• The mass of the object, m, does not matter.
  Solving for the escape speed gives
• M is the mass of the planet R is the planets
  radius.
• For the Earth, we get vescape 11.2 km/s.

25
Orbits and Energy

• Simulation
• Consider an object in orbit around a much larger
  object.
• A circular orbit is a very special case,
  requiring a particular speed. A little faster, or
  a little slower for our object, and the orbit is
  elliptical.
• If the speed is simply times the speed for
  a circular orbit, the object goes off in a
  parabolic path and never comes back. If the speed
  is even larger, the path is hyperbolic, but the
  object still doesnt come back.

26
Whiteboard