Section 1'5 Energy Some background information on Energy

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Section 1.5 EnergySome background information on
Energy
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• Energy can not be either created or destroyed,
  just transferred from one form to another.
• Thus, instead of considering kinetic problems
  from the point of view of the equations of motion
  or Newtons laws, we can solve the motion of the
  problem by considering the energy balance of the
  situation.
• This is often a much easier way to solve
  mechanics problems.
• In this section we will consider only kinetic and
  potential energy.

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1.5.1 Work done
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• We will define a quantity called Work.
• Work is done only when an object is moved through
  some displacement.
• It is only the force in the same direction as the
  displacement that causes work.
• The work done by a constant force on an object is
  defined as the product of the
• component of the force along the direction of the
  displacement
• and the magnitude of the displacement.

F
FxFcos ?
?x
W(Fcos ?) ?x
3
Work done
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• The units of work are Nm which is the same as
  the units of Joules J.
• If there is more than one force acting on an
  object, then the total work done is the algebraic
  sum of the amount of work done by each force.
• Work is a SCALAR. Thus, it has a size, but no
  direction. Note, however, that it does have a
  sign, thus can be positive as well as negative,
  depending upon the signs of ?x and F.
• Work is negative if the component of the force is
  in the opposite direction to the displacement.

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1.5.2 Kinetic energy and the work-energy theorem
0
v0
v
Fnet
?x

• Consider an object moved through a displacement
  ?x by a force Fnet, which increases the objects
  velocity from v0 to v.
• The work done is equal to Fnet times ?x.
• It can be shown that the work done on the object
  is

5
1.5.2 Kinetic energy and the work-energy theorem
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• Thus the kinetic energy of an object is given by
• and is in units of Joules J
• Provided there is no loss of energy due to
  friction, the work done on an object is given by
  the work-kinetic energy equation.

6
1.5.3 Potential Energy
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• When an object falls under the influence of
  gravity, the gravitational field exerts a force
  on the object, causing it to accelerate, and
  increasing its kinetic energy.
• Thus we can associate a potential energy with an
  object in a gravitational field (or in general
  any field).
• This potential energy can be converted into
  kinetic energy.
• It can be shown that the gravitational potential
  energy of an object at height y is given by

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Reference levels for gravitational potential
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• The choice of the zero level for potential energy
  is completely arbitrary.
• Thus, we chose the zero level to be the most
  convenient one to solve the problem.
• This usually means choosing the zero level to be
  the lowest that the object can fall to.

PEmgy1
PEmgy2
y1
PE0
y2
PE0
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1.5.4 Potential energy stored in a spring
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• A spring can also have a potential energy
  associated with it.
• The force exerted by a compressed, or stretched
  spring is
• Fs-kx
• (where k is the spring constant)
• It can be shown that the potential energy
  associated with a spring is

uncompressed
x
compressed
Fs-kx
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1.5.5 Conservative and Non-conservative forces
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• Forces can be split into two categories
• Conservative
• Non-conservative
• Gravity is an example of a conservative force.
  For these types of forces the total kinetic
  energy plus total potential energy of the system
  is conserved (i.e. there is no energy lost to the
  surroundings).
• A non-conservative force would be friction. In
  this case energy is lost (through heat) to the
  surroundings.)

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Conservative forces
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• The definition of a conservative force is that
• the work the force does on an object moving
  between two points is independent of the path the
  object takes between the points (i.e. it depends
  only upon the initial and final positions).
• Thus, if a force is conservative one can
  associate a potential energy function with it.
• The work done by a conservative force is equal to
  the initial value of the potential minus the
  final value.
• WcPEi-PEf

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Non-conservative forces
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• A force is non-conservative if the work it does
  on an object depends upon the path the object
  takes between its initial and final positions.
• e.g. Friction, air resistance
• For non-conservative forces, energy is
  transferred into some other form (e.g. heat) and
  lost from the system.
• In reality there are both conservative and
  non-conservative forces acting on an object
• In dealing with such problems, we separate the
  work done due to non-conservative forces, and the
  work done by conservative forces.
• Let us consider conservative forces first

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Conservation of mechanical energy
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• We define the mechanical energy of a system as
• mechanical energy KE PE
• If no non-conservative forces are involved we can
  say that
• The total mechanical energy of a system (equal to
  the sum of the kinetic and potential energy)
  remains constant.
• In any isolated system of objects that interact
  only through conservative forces, the total
  mechanical energy of the system remains constant.
• KEiPEi KEf PEf

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Method for solving problems using conservation of
mechanical energy
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• Sketch the system.
• Decide where the zero level of gravitational
  potential energy is going to be.
• Determine whether non-conservative forces are
  involved.
• If only conservative forces are involved, then
  the total mechanical energy of the system remains
  constant, thus KEiPEiKEfPEf.
• Solve for the unknown(s).

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1.5.6 Work done by a non-conservative force
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• In most situations, non-conservative forces, such
  as friction also act on a system. In this case,
  the total mechanical energy of the system is NOT
  conserved.
• We can split the work done on an object into the
  work done by conservative and non-conservative
  forces. It can be shown that
• Wnc Wc ?KE
• This is the same as saying that the work done by
  the conservative force (Wc) is equal to the
  energy lost from the system.

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Work done by a non-conservative force
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• It can be shown that the work done by all the
  non-conservative forces is equal to the change in
  mechanical energy of the system.
• Thus, to solve problems where non-conservative
  forces are involved, work out the change in
  mechanical energy of the system.

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1.5.7 Power
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• As well as knowing the energy transferred to a
  system, it would be useful to know the rate at
  which this energy is transferred.
• Thus, we define power as the time rate of energy
  transfer.
• If a FORCE does a work W on an object, during a
  time interval ?t, then the average power is
• Since WF ?x, we have

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1.5.8 Work done by a varying force
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• The work done by a constant force is equal to the
  area under a Fx vs x graph.
• Consider a varying force. One can form a series
  of small rectangles representing, the area of
  which represent the work done over that small x
  displacement.
• In the limit of the width of the small rectangles
  going to zero, we have the area under the curve.
• Thus
• The work done by a varying force acting on an
  object that undergoes a displacement is equal to
  the area under the graph of Fx vs x.

Fx
Constant force
Area Fx?x
x
Fx
Varying force
x
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Work done by a varying force - Example
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• The work done between
• 0-10s
• 10-20s
• 20-25s

Fx
4
2
0
x(s)
10
20
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Section 1.6 Momentum and Collisions
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• In this section we will cover the physics of
  colliding objects.
• In general, when objects collide there is a
  complicated variation in the forces between the
  objects. This makes solving the problem using
  Newtons laws very difficult.
• Thus, we will take an alternative approach,
  considering the momentum of the colliding
  objects. So
• What do we need to know to solve the problems
  involving colliding objects?
• Is there anything that is conserved in collisions?

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1.6.1 Momentum and Impulse
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• We define the linear momentum (p) of an object
  of mass m, moving with velocity v to be
• Note that both the velocity and the momentum are
  vectors. Momentum has dimensions kg m/s.
• We can split the velocity into x and y
  components, thus we can do the same to the
  momentum.

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Momentum and impulse
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• Thus, the bigger the mass, or the bigger the
  velocity, the larger the momentum of the object
• Newtons second law can be restated as
• The time rate of change of momentum of an object
  is equal to the Force acting on the object

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Momentum and impulse
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• Note that if there is no net force on an object,
  there is no change in the momentum.
• Thus, if Fnet0 on a system, the momentum is
  conserved.
• We define the impulse of a constant force to be
  F?t. Thus,
• This is the impulse-momentum theorem
• (remember ?pmvf-mvi).

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Momentum and impulse
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• Note that impulse is a vector quantity in the
  same direction as the force.
• What if the force is not constant?
• Consider the force time graph shown opposite.
• The impulse is equal to the area under the Force
  time graph.
• Or we can define an average force, Fav,such that
  has the same area.

F
t
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1.6.2 Conservation of momentum
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• When no external forces act on a system of two
  objects when they collide with each other, the
  total momentum before the collision is equal to
  the total momentum after the collision.
• Note that the velocities are vectors.

BEFORE
m1
m2
v1i
v2i
AFTER
m1
m2
v1f
v2f
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Conservation of momentum
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• Momentum is no longer conserved if
• an external force outside of the system acts on
  any of the objects which are colliding
• Thus, we will only consider collisions involving
  isolated systems in this section, i.e. we will
  only consider collisions where momentum is
  conserved. The system involves all of the
  objects that are involved in the collision.
• In most of the cases considered here the initial
  and final velocities are all along the same axis,
  so one only needs to consider the signs of the
  velocities, not the angles they have (i.e. the
  collision is in 1-D).
• Later we will consider the case of glancing
  collisions.

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1.6.3 Collisions
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• In all collisions we will consider momentum will
  be conserved.
• However, the kinetic energy may not be conserved.
  Some of the energy may be converted into sound
  energy, or into work used to deform the colliding
  objects.
• Thus, we will define two types of collisions
• INELASTIC COLLISION
• Momentum IS conserved
• Kinetic energy IS NOT conserved
• ELASTIC COLLISION
• Momentum IS conserved
• Kinetic energy IS conserved

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Perfectly inelastic collisions
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• If when two objects collide, they stick together
  and move off together with some common velocity
  after the collision, we say that the collision is
  PERFECTLY INELASTIC.
• For such collisions we can use the conservation
  of momentum, and the fact that the two final
  velocities are the same.

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Elastic collisions
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• For an elastic collision, both momentum and
  kinetic energy are conserved. Thus,

(1)
(2)
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Elastic collisions
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• For problems involving elastic collisions, there
  are usually two unknowns which we need to solve
  for.
• Thus, we will need to solve equations (1) and (2)
  from the previous slide SIMULTANEOUSLY, to get a
  solution.
• Instead of using equation (2) it is easier to use
  the following equation (which can be shown to be
  equivalent to equation (2)).

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Strategy for solving collision problems
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• Draw the collision in a before and after picture.
• Chose a set of coordinate axes, define the
  positive and negative directions for the
  velocities. Draw all the velocity vectors.
• Write an equation for the conservation of
  momentum.
• If the collision is INELASTIC, use this equation
  to solve for the unknown quantities.
• If the collision if ELASTIC, use the conservation
  of kinetic energy to construct a second equation,
  which can be used along with the above equation
  to solve for the unknowns.

31
1.6.4 Glancing collisions
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• Consider the collision shown opposite.
• The objects move off at an angle after the
  collision.
• In this case, we split the momentum into x and y
  components. The momentum is each direction is
  conserved independently

BEFORE
m1
v1i
m2
v2i0
v1fy
v1f
AFTER
m1
v1fx
?
?
v2fx
m2
v2fy
v2f
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Glancing collisions
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• Thus, for glancing collisions
• In the x-direction we have
• And in the y-direction we have

Note the negative sign because v2fy is downwards.
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Glancing collisions
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• Solving the x and y equations for momentum, gives
  the unknowns.
• Note that if the glancing collision is also an
  elastic collision, then the kinetic energy is
  also conserved.
• However, kinetic energy is a scalar so there is
  no need to split the kinetic energy equation up
  into x and y components. Thus, we have a third
  equation to use

34
1.7 Rotational Motion and the Law of Gravity
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• In this section we are going to investigate
  circular motion. This will allow us to describe
  how planets orbit stars, how airplanes bank, cars
  corner, and will open up a whole host of other
  scenarios to our understanding.
• It is important that we deal with the angles in
  our problems carefully. We will no longer be
  working in degrees, but in radians.

35
Another way of describing angles - Radians
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• A angle in radians is defined as
• Note that radians are dimensionless.
• The number of radians in 360 degrees 2?r/r2?.
• Thus, to convert from degrees to radians, use

s
?
r
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1.7.1 Angular speed and angular acceleration
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• Consider a wheel which is spinning. One of the
  spokes on the wheel will rotate through a certain
  angle in a given time. The number of radians
  which the spoke rotates through per second is
  called the average angular velocity.
• Thus, the angular velocity (omega or ?) is
  measured in radians per second (rad s-1).
• The instantaneous angular speed is limit of the
  average angular speed, as the time tends to zero.

tf
ti
?f
?
?i
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Angular acceleration
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• If an object can have an angular speed, then if
  this angular speed changes over time, it must
  have an angular acceleration.
• We define the average angular acceleration to be
  change in angular speed divided by the time over
  which the change occurred.
• Angular acceleration (alpha or ?) is measured
  in rad s-2.
• The instantaneous angular acceleration is limit
  of the average angular acceleration, as the time
  tends to zero.

38
1.7.3 Relations between linear and angular
expressions.
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• If an object is moving in a circle, clearly it
  has all of the preceding angular quantities
  associated with it.
• However, it also has a change in displacement, a
  linear speed, and possibly a linear acceleration.
• We can relate the linear to the angular
  quantities of an object.

vt
?
s
?i
39
Analogous expressions for angular and linear
motion
0

• Note that all the linear motion expression we
  derived earlier have analogous angular quantities.

40
Equations of motion for constant angular
acceleration
0

• By analogy to the 1-D equations of motion, we
  have the equations for an object undergoing
  rotational motion about a fixed axis, with a
  constant angular acceleration

41
1.7.4 Centripetal acceleration
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• Consider a car moving at a constant speed round a
  circular path. Because the direction of its
  velocity is changing, it has an acceleration.
• It can be shown that the acceleration required to
  make an object perform circular motion is
  directed towards the center of the circle, and is
  given by

42
Centripetal acceleration
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• Centripetal and tangential acceleration are often
  confused.
• If an object is moving in a circle with a
  constant speed, then it has a centripetal
  acceleration, directed towards the center of the
  circle, but no tangential acceleration.
• If the object also has an angular acceleration,
  it is accelerating round the circle (i.e. its
  angular speed is changing), then it also has a
  tangential acceleration.
• That is, the tangential acceleration describes a
  change in speed, while a centripetal acceleration
  describes the change in direction.
• To find the magnitude of the total acceleration,
  use the facts that the accelerations are at right
  angles to each other and pythagoras theorem to
  give

43
1.7.5 Forces causing centripetal acceleration
0

• The force causing a centripetal acceleration is
  given by the product of the mass of the object
  times the centripetal acceleration (Newtons 2nd
  law).
• Thus, for an object moving in a circle with
  constant speed, the force causing the centripetal
  acceleration is given by

44
Accelerations in circular motion
0
?i

• We have encountered three different accelerations
  which are possible in circular motion. It is
  important not to get them confused.
• The centripetal acceleration, ac
• This makes object go in the circle (changes
  direction of the velocity).
• If the object has a constant speed during
  circular motion, then this is the only
  acceleration that it has.

?f
??
ac
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Accelerations in circular motion
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• If the object is also increasing its angular
  velocity as time goes by, then it clearly has an
  angular acceleration, ?.
• In this case, its tangential speed is also
  increasing, thus it has a tangential
  acceleration, at.
• These are related via at?r.
• The total acceleration is given by vector
  addition of at and ac.

?
at
ac
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1.7.5 Forces causing centripetal acceleration
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• For the moment let us just consider objects going
  in circles with a constant speed. That is, they
  only have a centripetal acceleration.
• The force causing a centripetal acceleration is
  given by the product of the mass of the object
  times the centripetal acceleration (Newtons 2nd
  law).
• Thus, for an object moving in a circle with
  constant speed, the force causing the centripetal
  acceleration is given by

47
Forces causing circular motion
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• Thus, if we have an object performing circular
  motion, there has to be a centripetal force
  providing the acceleration.
• So the key to doing circular motion problems is
  to identify the force providing the centripetal
  acceleration.

48
1.7.6 Newtons Law of Universal Gravitation
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• Newton discovered that every particle with mass,
  exerts a force of attraction on another particle
  with mass. The strength of the force is given by
• G is the constant of universal gravitation, and
  is equal to 6.673 x 10-11 Nm2/kg2.
• This is the force that makes planets orbit stars,
  and moons orbit planets.

49
Gravitation
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• One important simplification in considering the
  forces of gravity is that
• The gravitational force exerted by a spherical
  object on a particle outside the sphere is the
  same as if the entire mass of the sphere were
  concentrated at is center.
• Thus the force due to the earth on an object on
  its surface is

50
1.7.7 Gravitational Potential Energy
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• Just like Fmg is only valid near the Earths
  surface, the equation we derived earlier for the
  gravitational potential energy PEmgy is only
  valid near the surface of the Earth.
• The general equation for the gravitational
  potential energy can be shown to be
• Note the negative sign, this is because we chose
  the zero of gravitational potential energy to be
  at infinity.

51
1.7.8 Escape speed
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• If an object is fired upwards with sufficient
  speed, then it can escape the gravitational pull
  of the earth, and continue on into space. The
  speed required to achieve this is called the
  escape speed.
• We can show that the escape speed is given by

52
1.7.9 Keplers Laws
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• In the 1600s a German astronomer Johannes Kepler
  found various mathematical relationships between
  the orbital properties of the planets.
• His third law states that the square of the
  orbital period of any planet is proportional to
  the cube of the average distance from the planet
  to the sun.
• This can be shown to be a natural consequence of
  Newtons universal law of gravitation.

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1.8.1 Torque and equilibrium
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• The tendency of a force to rotate an object about
  some axis is measured by a quantity called the
  torque.
• We will use the Greek letter tau (?) to describe
  the torque.
• Fd

F
d

• Note that the torque applied to an object depends
  upon the axis of rotations (i.e. where it is
  rotating about) and where the force is applied.

54
Torque
0

• The torque is only produced by the component of
  the force which is at right angles to the line
  from the axis of rotation to the point where the
  force is applied.
• Thus, for the case shown opposite
• We define a positive torque as causing a
  counterclockwise rotation.
• The units of torque are Nm.

F
FyFsin?
?
FxFcos?
d
55
1.8.2 Torque and equilibrium
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• There are two conditions which must be satisfied
  for an object to be in equilibrium. Firstly,
• which is Newtons first law.
• The second condition is that the net external
  torque acting on the object must be zero.
• This is the condition for rotational equilibrium.
• If this is satisfied, then an object remains in
  its rotational state.
• If it is not rotating, then it remains without
  rotation.
• If it is rotating with constant angular velocity,
  it remains rotating with the same angular
  velocity.

56
1.8.3 Center of gravity
0
y

• To compute the torque due to the force of gravity
  on an object, one can think of all of the weight
  being concentrated at a single point.
• One then just considers the torque due to a mass
  at this single point.

(x2,y2)
m
(x1,y1)
(xcg,ycg)
m
3m
x
m
(x3,y3)
57
1.8.4 Relationship between torque and angular
acceleration.
0

• Consider an object which has a net torque acting
  on it.
• It will undergo an angular acceleration.
• The size of this acceleration is proportional to
  the net torque, and also proportional to a
  quantity we will call the moment of inertia (I).

58
Moment of inertia
0

• Note that the correspondence between linear and
  rotational motion continues, with
• F ? ?
• m ? I
• The moment of inertia of a single object at
  distance r from the axis of rotation is
• The moment of inertia of a group of objects is
  given by

59
Moment of inertia of extended objects
0

• So we can calculate the moment of inertia of a
  set of separate objects if we know their masses,
  and distances from the axis of rotation.
• However, for extended objects, like spheres and
  rods, it is much more difficult to calculate what
  the moment of inertia is. Thus, you will usually
  be given what these are.

• Other geometries are shown in chapter 8 of the
  book.

60
1.8.5 Rotational Kinetic energy
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• If an object is rotating, it must have an energy
  associated with this rotational velocity. We
  call this the rotational kinetic energy, and it
  is given by
• Thus, we have a new equation for the conservation
  of mechanical energy (if there are no
  non-conservative forces, and KEt is the
  translational kinetic energy)

61
1.8.6 Angular momentum
0

• If an object is rotating, it must have an
  equivalent of the linear momentum (pmv) we
  encountered in chapter 7.
• We define a new quantity, L, to be the angular
  momentum, and it is equal to
• Thus, we have an alternative expression for the
  torque

62
Conservation of angular momentum
0

• The angular momentum of a system is conserved if
  there is no net torque on the system (??0).
• Thus,
• Note that if the angular momentum is conserved,
  and I changes, then ? must change to keep L
  constant.
• Thus is the reason that when an ice-skater is
  spinning and pulls their arms closer to their
  body, they spin faster (they have decreased their
  I, so ? must increase).

63
Rotational kinetic energy and angular momentum
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• The equation for rotational kinetic energy is

• The equation for angular momentum, and
  conservation of angular momentum is

64
Example supernova explosion
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