Rotational Mechanics

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Part IA Natural Sciences Tripos 2013/14
Rotational Mechanics Special Relativity
Lecture 1
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Mechanics in Rotational Motion
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The centre of mass
MO 128 TM 149

• Suppose that we have a distribution of masses,
  all within one rigid body. Where should we put
  the fulcrum such that there is no rotation when
  suspended in a uniform gravitational field? This
  point is the centre of weight, but it also has
  special properties when there is no gravitational
  field and so is usually called the centre of
  mass. For the simple case of a uniform light
  rigid rod of length l connecting two point masses
  m1 and m2, it is obvious where the centre of mass
  must be

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y
l2
l1
m1
C
m2
Centre of mass
x
x1
x2
x0
We require that the total turning moment about C
is zero, so we define its position to be such
that m1l1 m2l2. Since l1 x0 x1, and l2 x2
x0, we can write this as

where M m1m2.
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• We can generalise this to a one-dimensional rigid
  body of N masses as follows
• For a three-dimensional rigid body, this
  expression must be satisfied in each dimension
  simultaneously, and we may write
• Now we can express this in vector form, writing
  the position vector of the centre of mass as R
  (x0, y0, z0), and of each contributing mass as ri
  (xi, yi, zi)

back
9

• We are now in a position to find the centre of
  mass of a continuous body using integration
  rather than summation. Write the summation for R
  using dmi mi

Hence, for continuous bodies, we can replace the
sum over i elements with an integral
In 1 dimension, dm is ?l dl, where ?l is the mass
per unit length and l is the length variable. In
2 dimensions, dm is ?a da, where ?a is the mass
per unit area and a is the area variable.
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In 3 dimensions, dm is ?v dv, where ?v is the
mass per unit volume and v is the volume
variable. We can also use the principle of
superposition to work out what happens if the
rigid body is lumpy. The centre of mass of a
composite system of several lumps of mass can be
calculated by first finding the centre of mass of
each lump separately, and then finding the centre
of mass of the whole considering each lump as a
point mass at its individual respective centre of
mass.
m2
m1



m3
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For discrete masses
For continuous bodies
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Example 1 find the centre of mass of a lamina
shaped like an isosceles triangle
Divide the lamina into thin strips, each having a
width of dx. Take moments about the origin. The
moment of the element shown is xdm, where dm is
the mass of the element. The area of the element
is 2ydx, and the area of the lamina is bh/2.
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Example 1 find the centre of mass of a lamina
shaped like an isosceles triangle
Divide the lamina into thin strips, each having a
width of dx. Take moments about the origin. The
moment of the element shown is xdm, where dm is
the mass of the element. The area of the element
is 2ydx, and the area of the lamina is bh/2.
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The lever balance

• We find experimentally, that the lever is
  balanced only when m1l1 m2l2

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• This is the case even though the sum of all the
  forces is zero wherever we put the fulcrum, i.e.
• We conclude that a second condition is necessary
  for equilibrium, that is that the sum of all the
  turning moments must be equal to zero.
• This is expressed in the law of the lever
• where li is the perpendicular distance and fi is
  the force.

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• The two conditions which must both be satisfied
  simultaneously for a body in static equilibrium
  are
• The vector sum of all the external forces acting
  on the body is exactly zero.
• The sum of all the turning moments about an axis
  through any point is exactly zero.

MO 149 TM 397
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Circular motion

• When a particle changes its position from P to P1
  relative to a fixed point, O, we can describe the
  change in position using position vectors. The
  incremental vector dr is added to the position
  vector, r, describing the position P, to find the
  new position P1
• This constitutes a complete description of the
  change in position. Note, however, that the

P1
r dr
dr
P
O
r
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• change has two parts a change in the radial
  distance from O, and a change in the angle about
  an axis OQ perpendicular to the plane of the
  triangle OPP1
• Here, we shall be concerned with the rotational
  aspects of the motion, i.e. things to do with

Q
P1
r dr
dr
?
P
O
r
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• changes in the angle, ?. Note that this always
  requires you to define an axis about which the
  rotation takes place, in this case the axis OQ.
  The rotation can be in two senses, clockwise and
  counter-clockwise (or anti-clockwise). We adopt
  the right-handed convention to describe the
  positive direction of the axis of rotation (as
  shown by the arrow from O to Q). When looking
  along the axis in the direction of the arrow, the
  positive rotation is clockwise. If you look along
  the axis in the other direction, such that the
  arrow is pointing directly at you, rotation is
  anticlockwise.

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• The angle is best measured in radians. If it is
  small, i.e. ltlt 1 radian, we can describe the
  rotation by a vector, ? , through O and along OQ,
  in the direction of OQ and of magnitude equal to
  ?.
• We can see that small rotations can be
  represented by vectors as follows

Q
?
?
O
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• For very small angles, the segment of the circle
  AB (on the surface of the sphere of radius a
  centred on O) becomes approximately straight, and
  can therefore be represented by vector s of
  length a?. Three small rotations can be performed
  such that the segments
• form a closed vector triangle. The rotation axes
  are perpendicular to each rotation, so straight
  lines proportional to the angles also form a
  closed triangle, i.e. they add up as vectors.

Q
a? 2
B
a? 3
?
s
a? 1
A
a
O
?
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• therefore describe the angular velocity by the
  vector
• Using a similar argument, the angular
  acceleration may also be represented by a vector
  which points along the axis about which the
  angular acceleration is happening

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The moment of a force as a vector

• We have already met the concept of the moment of
  a force, or the turning moment. We defined
  this as the force times the perpendicular
  distance of its point of action from the axis of
  rotation. We can represent the moment as a vector
  which, just as in the case of the angular
  velocity etc., points in a clockwise sense along
  the axis of rotation, and has a magnitude equal
  to the magnitude of the turning moment.
• We consider a force, F, acting at point P which
  is described by position vector r from origin O

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Force F acts at point P, which is at position
vector r from point O.

• The moment of the force about O is G Fr sin(? )
  about the axis through O perpendicular to both r
  and F, clockwise looking in the direction of the
  arrow.
• We define the moment vector G r ? F

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• Example 2 find a vector expression for the
  moment of a couple.
• A couple is a combination of two equal and
  opposite forces which are not in line with each
  other.

The forces act at position vectors r1 and r2 with
respect to an arbitrary origin, O.
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• Example 2 find a vector expression for the
  moment of a couple.
• A couple is a combination of two equal and
  opposite forces which are not in line with each
  other.

The forces act at position vectors r1 and r2 with
respect to an arbitrary origin, O.
The moment of the couple is independent of the
origin O, and depends only on the vector forces
and the position vector of the point of action of
one of them with respect to the other.
27
Part IA Natural Sciences Tripos 2013/14
Rotational Mechanics Special Relativity
Lecture 2
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Angular acceleration of a rigid body

• Newtons second law gives us the relationship
  between the linear force applied to a body of
  mass m and its acceleration. We have seen that
  the rotational equivalent of the force, F, is the
  moment of the force (or torque), G, and the
  rotational equivalent of the linear acceleration,
  , is the angular acceleration, . Can we find
  an equivalent to Newtons second law for
  rotational mechanics, and if so, what is the
  equivalent of the mass, m ?

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• To answer this question, consider a particle of
  mass m rotating about an axis OQ
• Its linear speed is v r?, but let us suppose
  that its speed is increasing, i.e. it has an
  acceleration in the direction of v. N2 tells us
  that there must therefore be a force acting on
  the particle in this

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• direction of magnitude F where
• The magnitude of the moment of this force about
  OQ is directed along OQ.
  For this particle we may therefore write
• Now we can consider a rigid body rotating about
  OQ as made up of the sum of N such elementary
  particles. Let the ith such particle be a
  distance ri from OQ and have a mass of mi. Then
  summing over the whole body, we have

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• where Gext is the total external vector moment
  acting on the body about OQ.
• If G is the rotational equivalent of F, and is
  the rotational equivalent of , then we must
  conclude that the rotational equivalent of the
  mass of the body is
• We call this quantity the moment of inertia, I,
  and then the rotational equivalent of N2 is

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Moment of inertia
MO 154 TM 293

• In linear mechanics, the mass is measure of a
  bodys reluctance to change its state of linear
  motion. The larger the mass, the slower the rate
  of change of velocity for a given applied force.
  In rotational mechanics, the moment of inertia
  takes the place of the mass, and it is a measure
  of a bodys reluctance to change its state of
  angular motion. The larger the moment of inertia,
  the slower the rate of change of angular velocity
  for a given applied moment of a force. Like the
  mass, the moment of inertia is a scalar quantity,
  usually

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• given the symbol I. (Do not confuse this with
  the vector impulse, .) Thus, for a rigid body
  which can be thought of as being composed from N
  particles
• In terms of I, we can write the rotational
  equivalent of N2 as

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• Example 3 an electric motor is attached to the
  axis of a massive flywheel of moment of inertia
  70 kg m2. When an electric current is switched
  on, the motor applies a constant torque of 150 N
  m. What is the rotation rate of the flywheel
  after 30 s?

The torque causes an angular acceleration which
increases the angular velocity of the flywheel.
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• Example 3 an electric motor is attached to the
  axis of a massive flywheel of moment of inertia
  70 kg m2. When an electric current is switched
  on, the motor applies a constant torque of 150 N
  m. What is the rotation rate of the flywheel
  after 30 s?

The torque causes an angular acceleration which
increases the angular velocity of the flywheel.
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Moments of inertia of continuous bodies

• The expression we have obtained for the moment of
  inertia, I, of a rigid body involves a summation
  over N elemental mass contributions
• For continuous bodies, it is more convenient to
  use an integral form. We imagine that the body is
  divided into a very large number of very small
  masses, dm, all joined together to make up the
  whole. A particular one is at distance r from the
  axis, so it contributes an amount dI to the total

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• where dI r2dm. Now go to the limit as dm tends
  to zero, and integrate over all the contributions
  to get the total moment of inertia, thus
• The trick in applying this expression is often to
  express dm in terms of the distance variable (r)
  using the density. You should also take advantage
  of the spatial symmetry of the problem to
  simplify the expression that you must integrate.
  Some examples follow.

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Example 4 find an expression for the moment of
inertia of a rod of length l about an axis
through one end perpendicular to the rod.
Let the rods density be ? per unit length.
Consider an element of the rod, length dx, at
distance x from one end. The mass of the element
is dm ? dx. The contribution to the total
moment of inertia from this element is dI, given
by
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Example 4 find an expression for the moment of
inertia of a rod of length l about an axis
through one end perpendicular to the rod.
Let the rods density be ? per unit length.
Consider an element of the rod, length dx, at
distance x from one end. The mass of the element
is dm ? dx. The contribution to the total
moment of inertia from this element is dI, given
by
rod
dx
x
l
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Example 5 find an expression for the moment of
inertia of a solid disc of mass M and radius a
about an axis through its centre perpendicular to
the plane of the disc.
Let the disc have a surface density of ? per unit
area. Consider a radial element of the disc at
radius r, thickness dr. This has mass dm which is
equal to 2?rdr?. It makes a contribution, dI, to
the total moment of inertia given by
r
a
dr
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Example 5 find an expression for the moment of
inertia of a solid disc of mass M and radius a
about an axis through its centre perpendicular to
the plane of the disc.
Let the disc have a surface density of ? per unit
area. Consider a radial element of the disc at
radius r, thickness dr. This has mass dm which is
equal to 2?rdr?. It makes a contribution, dI, to
the total moment of inertia given by
r
a
dr
Note that this formula also applies to a cylinder.
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Parallel axes theorem

• This theorem allows us to obtain the moment of
  inertia, I, of a rigid body about any axis, AB,
  parallel to an axis, OP, through the centre of
  mass about which the moment of inertia of the
  body is I0.

Consider the contribution from the element mi
B
rigid body
P
centre of mass
a
MO 155 TM 297
O
A
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• Let I be the moment of inertia about axis AB.
• Let I0 be the moment of inertia about an axis,
  OP, parallel to AB and through the centre of mass.

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• Let I be the moment of inertia about axis AB.
• Let I0 be the moment of inertia about an axis,
  OP, parallel to AB and through the centre of mass.

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Example 6 find an expression for the moment of
inertia of a rod of length l and mass M about an
axis through its centre perpendicular to the rod.
Let the moment of inertia about the axis through
the centre be I0. We have already calculated the
moment of inertia, I, about a parallel axis
through one end of a rod to be Ml2/3. Thus
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Example 6 find an expression for the moment of
inertia of a rod of length l and mass M about an
axis through its centre perpendicular to the rod.
Let the moment of inertia about the axis through
the centre be I0. We have already calculated the
moment of inertia, I, about a parallel axis
through one end of a rod to be Ml2/3. Thus
B
P
a l/2
A
O
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Perpendicular axes theorem

• This is another useful theorem which relates the
  moments of inertia about three perpendicular axes
  through any point in a lamina, one of which is
  perpendicular to the plane of the lamina.
• Consider the contribution by an elemental mass mi

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• Let the lamina be in the xy plane. Then the
  moment of inertia about the z axis is

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Example 7 find an expression for the moment of
inertia of a disc of radius a and mass M about an
axis through its centre in the plane of the disc.

• We have already found the moment of inertia of
  the disc about the z axis to be Ma2/2. Thus

z
a
y
x
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Example 7 find an expression for the moment of
inertia of a disc of radius a and mass M about an
axis through its centre in the plane of the disc.

• We have already found the moment of inertia of
  the disc about the z axis to be Ma2/2. Thus

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Sounds of Pulsars
032954 P0.71452s 0833-45
P0.089s 053121 P0.033s 0437-4715
P0.00575s 193721 P0.00167s

• Single pulses are usually very variable !

52
Part IA Natural Sciences Tripos 2013/14
Rotational Mechanics Special Relativity
Lecture 3
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53
Angular momentum
MO 103 TM 331

• We define the angular momentum, L, to be the
  moment of the momentum about a point.
• Suppose that a particle, A, of mass m and at
  position vector r relative to point B, has
  momentum P mv, where v is its velocity

The moment of the momentum (by direct analogy
with the moment of a force) is given by L r ?
P r ? mv mr ? v
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Conservation of angular momentum

• The angular momentum, like the linear momentum,
  is conserved in an isolated system. To show this,
  consider a system of N interacting particles. The
  angular momentum of the ith particle is Li miri
  ? vi. The total angular momentum of the system is
  therefore given by
• This is a vector addition of the angular momenta
  of all the particles about a given point. We can
  differentiate with respect to time to find the
  rate of change of L

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• Now the value of Gint is zero for the following
  reason

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• The internal interaction on the ith particle by
  the jth particle is in line and oppositely
  directed to the interaction on the jth particle
  by the ith particle by N3. The moment about O is
  ri ? Fij rj ? Fji
• (ri rj) ? Fij b ? Fij . This is zero since
  b is in line with Fij. All the internal
  interactions are in

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• similar pairs, each of which comes to zero.
  Hence Gint must be zero. Therefore
• and we conclude that the rate of change of the
  total angular momentum of a system is just the
  vector sum of the external moments applied to
  that system.
• When the system is isolated, Gext 0, so we
  conclude that the total angular momentum of an
  isolated system is constant.

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• Notes
• This statement is always true, no matter what
  dissipative forces there might be internally.
• This statement is always true about any axis, not
  just one through the centre of mass.
• The rotational equivalent of N2 is
  and we have just shown that .
  Therefore we can write

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Angular impulse

• We defined the impulse of a force as the integral
  of the force with respect to time, i.e.
• The action of the impulse was to change the
  linear momentum, thus
• Now the moment of the force is r ? F, and the
  angular momentum is r ? P. Taking the cross
  product with r in the above equation we get

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• Thus we see that when a moment of a force acts
  for a finite time, it causes the total angular
  momentum of the system to change, the change
  being equal to the integral of the angular moment
  with respect to time.

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Rotational kinetic energy

• The translational kinetic energy for a particle
  of mass m moving in a straight line at speed v is
  mv2/2. We can obtain the equivalent rotational
  quantity by considering the ith particle of a
  rigid object in rotation about an axis at at
  angular speed ?. Let the particle have mass mi
  and be at distance ri from the axis. Then

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• Summing over the entire body (N particles), we
  get
• Note that, for an individual point particle, we
  can consider its kinetic energy either as the
  translational quantity mv2/2, or as the
  rotational quantity mr2?2/2. However, in general
  the motion of a solid object must be analysed in
  terms of the linear motion of its centre of mass,
  plus the rotation around the centre of mass, so
  the total KE is the sum of the translational and
  rotational components.

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Rotational oscillations the physical pendulum

• Consider a rigid body of arbitrary shape that is
  suspended from, and free to rotate about, a
  horizontal frictionless axis (A). It is displaced
  slightly from equilibrium. What is the period of
  oscillation?

A
?
A
l
l? y
l
?
C
C
y
mg
?
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• Point C, the centre of mass, is raised relative
  to its equilibrium position by an amount y.
  Conserving energy we have
• Now
  Substituting and differentiating with respect
  to time gives us
• Hence the period of small oscillations is

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Example 8 the pendulum of a grandfather clock is
made from a brass disc of diameter 2a and mass M
with its centre attached to the end of a thin
metal rod having a mass of m so that the centre
of the disk is L below the point of suspension.
What is the period of the pendulum?

• The moment of inertia of the disc about an axis
  through its centre and perpendicular to its face
  is Ma2/2. Using the parallel axes theorem we
  have
• The moment of inertia of the rod is mL2/3, so the
  total moment of inertia is
• Now the centre of mass is at distance l below

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• the pivot such that

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Rotational Mechanics Special Relativity
Lecture 4
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68
The general motion of a rigid body

• We have seen already that the linear motion of a
  rigid body can be analysed by considering only
  the linear motion of its centre of mass which
  moves as if it carries all the mass and is acted
  upon by the sum of all the external forces. This
  is a required condition, but is not sufficient to
  describe the entire motion of the body because is
  does not take into account the rotational motion
  of the body about the centre of mass. We need a
  second statement, added to the first, to define
  the general motion of a body, linear plus
  rotational.

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• The additional statement is that there is
  rotation about an axis through the centre of mass
  of a body which is the result of the sum of the
  external moments of all the forces acting on the
  body about that axis, and the moment of inertia
  of the body about that axis. This second
  statement, added to the first, is sufficient to
  define the general motion of a body.
• We will clarify this with the example of a
  cylinder rolling down an inclined plane

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Example 9 what is the acceleration of a uniform
solid cylinder rolling, without slipping, down a
plane inclined at angle a to the horizontal?

• The cylinder is acted on by the body force, mg,
  by the normal reaction force, N, and by the
  frictional force, F, as shown in the diagrams.
  The net result is (a) the centre of mass
  accelerates down the plane, and (b) the cylinder
  rolls down the plane.

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• (a) The linear motion of the centre of mass is as
  if it is a point mass equal to the total mass of
  the cylinder, acted upon by the sum of the
  external forces. Resolving down the slope

(b) The rotation about the cylindrical axis is as
if acted upon by the sum of the moments of the
external forces where I is the moment of
inertia about the cylindrical axis. We also have
v ?a. Thus
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• Substituting for F gives us
• Now I ma2/2, so we deduce that

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• Substituting for F gives us
• Now I ma2/2, so we deduce that

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• In terms of energy
• (a) Rotational KE of the cylinder is

N
a
F
?
v
(b) The linear KE of the centre of mass is (c)
Thus when the centre of mass has dropped through
a distance h we have
mg
a
s
h
a
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• In terms of energy
• (a) Rotational KE of the cylinder is

N
a
F
?
v
(b) The linear KE of the centre of mass is (c)
Thus when the centre of mass has dropped through
a distance h we have
mg
a
s
h
a
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Rotating frames of reference

• Suppose that we have a particle, P, which is
  rotating at constant angular speed in a circle
  about O

y
?
y
P
r
x
P
O
? t
x
O
r
?
?
?
Oblique view
Plan view
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• We can represent the position of P with respect
  to O by the position vector r (x,y), where x is
  r cos(? t) and y is r sin(? t). We differentiate
  to find the velocity, and again to find the
  acceleration, thus

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• This tells us that the point P has a constant
  acceleration (since ? and r are constant) which
  is of magnitude ? 2 r and is directed along r in
  the negative direction, i.e. towards O. The
  actual speed of the particle is constant and
  equal to ? r in a tangential direction, but the
  acceleration arises from the fact that P is
  constantly changing direction towards the centre
  of the circle, so the vector velocity is
  constantly changing.
• Newtons second law of motion tells us that there
  must be a force associated with the acceleration.

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Centripetal force

• The centripetal force is the force on a particle
  which is directed towards the axis of rotation
  and which is required to maintain the rotational
  motion of the particle. From N2
• where is the unit vector along r. Note that
  the centripetal force does no work as the
  velocity and force are orthogonal to each other.

?
m
r
F
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Linear and rotational equivalents
We can identify quantities in linear mechanics
and rotational mechanics which behave in
equivalent fashions. If you are not sure what to
do in a rotational problem, think what you would
do in the equivalent linear problem, and then use
the table below.
Linear Rotational

mass m small displacement dr velocity v dr/dt acceleration a dv/dt d2r /dt2 linear momentum p mv force F dp/dt ma linear kinetic energy mv2/2 work done dW F?dr linear impulse dp Fdt moment of inertia I small angular displacement d? angular velocity ? d?/dt angular acceleration d? /dt d2? /dt2 angular momentum L I? moment G dL/dt Id2?/dt2 angular kinetic energy I?2/2 work done dW G?d? angular impulse dL Gdt
mass m small displacement dr velocity v dr/dt acceleration a dv/dt d2r /dt2 linear momentum p mv force F dp/dt ma linear kinetic energy mv2/2 work done dW F?dr linear impulse dp Fdt moment of inertia I small angular displacement d? angular velocity ? d?/dt angular acceleration d? /dt d2? /dt2 angular momentum L I? moment G dL/dt Id2?/dt2 angular kinetic energy I?2/2 work done dW G?d? angular impulse dL Gdt
mass m small displacement dr velocity v dr/dt acceleration a dv/dt d2r /dt2 linear momentum p mv force F dp/dt ma linear kinetic energy mv2/2 work done dW F?dr linear impulse dp Fdt moment of inertia I small angular displacement d? angular velocity ? d?/dt angular acceleration d? /dt d2? /dt2 angular momentum L I? moment G dL/dt Id2?/dt2 angular kinetic energy I?2/2 work done dW G?d? angular impulse dL Gdt
mass m small displacement dr velocity v dr/dt acceleration a dv/dt d2r /dt2 linear momentum p mv force F dp/dt ma linear kinetic energy mv2/2 work done dW F?dr linear impulse dp Fdt moment of inertia I small angular displacement d? angular velocity ? d?/dt angular acceleration d? /dt d2? /dt2 angular momentum L I? moment G dL/dt Id2?/dt2 angular kinetic energy I?2/2 work done dW G?d? angular impulse dL Gdt
mass m small displacement dr velocity v dr/dt acceleration a dv/dt d2r /dt2 linear momentum p mv force F dp/dt ma linear kinetic energy mv2/2 work done dW F?dr linear impulse dp Fdt moment of inertia I small angular displacement d? angular velocity ? d?/dt angular acceleration d? /dt d2? /dt2 angular momentum L I? moment G dL/dt Id2?/dt2 angular kinetic energy I?2/2 work done dW G?d? angular impulse dL Gdt
mass m small displacement dr velocity v dr/dt acceleration a dv/dt d2r /dt2 linear momentum p mv force F dp/dt ma linear kinetic energy mv2/2 work done dW F?dr linear impulse dp Fdt moment of inertia I small angular displacement d? angular velocity ? d?/dt angular acceleration d? /dt d2? /dt2 angular momentum L I? moment G dL/dt Id2?/dt2 angular kinetic energy I?2/2 work done dW G?d? angular impulse dL Gdt
mass m small displacement dr velocity v dr/dt acceleration a dv/dt d2r /dt2 linear momentum p mv force F dp/dt ma linear kinetic energy mv2/2 work done dW F?dr linear impulse dp Fdt moment of inertia I small angular displacement d? angular velocity ? d?/dt angular acceleration d? /dt d2? /dt2 angular momentum L I? moment G dL/dt Id2?/dt2 angular kinetic energy I?2/2 work done dW G?d? angular impulse dL Gdt
mass m small displacement dr velocity v dr/dt acceleration a dv/dt d2r /dt2 linear momentum p mv force F dp/dt ma linear kinetic energy mv2/2 work done dW F?dr linear impulse dp Fdt moment of inertia I small angular displacement d? angular velocity ? d?/dt angular acceleration d? /dt d2? /dt2 angular momentum L I? moment G dL/dt Id2?/dt2 angular kinetic energy I?2/2 work done dW G?d? angular impulse dL Gdt
81
Example 10 (Tripos 2000). A rod of mass M and
length L lies on a smooth horizontal table. A
small particle of mass m travels at speed v0 on
the table at 90 to the rod. It collides with the
end of the rod and sticks to it. Calculate the
speed of the centre of mass of the combined rod
and particle after the collision, and find the
new position of the centre of mass

• a) Linear momentum before linear momentum after

82
Example 10 (Tripos 2000). A rod of mass M and
length L lies on a smooth horizontal table. A
small particle of mass m travels at speed v0 on
the table at 90 to the rod. It collides with the
end of the rod and sticks to it. Calculate the
speed of the centre of mass of the combined rod
and particle after the collision, and find the
new position of the centre of mass,

• a) Linear momentum before linear momentum after

83
b) New position of the centre of mass moments
about it sum to zero
84
b) New position of the centre of mass moments
about it sum to zero
85
Example 10 (Tripos 2000) and the angular speed
of rotation about the centre of mass.
The principle of the conservation of angular
momentum may be applied about the centre of mass
?
86
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87
The Gyroscope
MO 164 TM 339

• A gyroscope is heavy flywheel which is rapidly
  rotating about an axis. It is usually mounted so
  that it can be turned about either of the two
  other orthogonal axes

L I?
88

• Whilst the flywheel is stationary, there is
  nothing unexpected about the gyroscope, so that a
  moment applied about the y or z axes produces
  rotation about the y or z axes.
• When the flywheel is rapidly rotating about the x
  axis, a moment about y produces slow rotation
  (called precession) about z and vice-versa. This
  is apparently counter to expectations, but is
  readily understood, either in terms of angular
  momentum, or in terms of the forces acting on
  particles in the flywheel.

89
A precessing gyroscope

• The gyroscope is precessing about the
• z axis. Look down from on top along z

O
Consider the motion of a particle (blue square)
which is just coming up towards you at A, moving
over the top at B, and disappearing at C
C
z, O
?
B
x, L
The blue particle moves from A to B to C. At B,
it is moving in a curved path. Therefore it must
feel a force to the left
A
t
t?dt
tdt
y
90

• Looking from the side along y

z, ?
F
y
x, L
F
91
Precession using vectors

• Apply a couple of moment G to a rotating flywheel

In time dt the moment, G, of the couple exerts a
change in angular momentum of Gdt, so that dL
Gdt. Note that this change is directed along G.
If G is perpendicular to L (as here) then the
change in angular momentum is also perpendicular
to the angular momentum. L then keeps constant
magnitude, but is constantly changing direction
in the plane of G and L, so precession O takes
place.
92

• Hence, there is precession about an axis
  perpendicular to both L and G. The rate of
  precession, O rad s?1, is easily calculated. In a
  small time, dt, it precesses through small angle
  Odt.

In this triangle, we can see that, as dt is
small, we can write G dt Odt L ?
G O L
In fact, we can write this in vector form as
follows
93

• Hence, there is precession about an axis
  perpendicular to both L and G. The rate of
  precession, O rad s?1, is easily calculated. In a
  small time, dt, it precesses through small angle
  Odt.

In this triangle, we can see that, as dt is
small, we can write G dt Odt L ?
G O L
In fact, we can write this in vector form as
follows To see this, consider a gyroscope at
an angle to the horizontal plane
94

• In the case where the gyroscope is at an angle,
  first resolve the angular momentum into vertical
  and horizontal components

LV L cos(? ) is constant. LH L sin(?) is
affected by the couple and is therefore changing
direction but not magnitude. ? G O L
sin(? ) G is perpendicular to the plane
containing O and L, so we may write
G O ? L I O ? ?
95
Gyroscope examples

• (i) Luni-solar precession

The effect of the unequal pull of gravity from
the Moon and Sun on the non-spherical Earth
applies a moment which causes the N-S axis to
precess with a period of about 26,000 y. We see
this a slow change in the positions of the stars
with time.
L
N
G
Moon, Sun
S
96

• (ii) Atomic precession

An atom has an angular momentum and a magnetic
moment. The magnetic moment subjects the atom to
a couple when a magnetic field is applied which
results in small changes in the energies of its
electronic states. This results in the splitting
of spectral lines the Zeeman effect.
B
L, m
97
Einsteins theory ofSpecial Relativity
MO 193-227 TM 1319-1356
98
Frames of reference

• A frame of reference is just a set of axes which
  we can use to define points (or events). We are
  all familiar with Cartesian frames (x,y,z), but
  there are others commonly used.
• We need a frame of reference in which to define
  positions, velocities, and accelerations. For
  example, a position vector r might have the
  coordinates (2,4,7) in one Cartesian frame. In
  another, the same vector might be (6,5,?11). A
  reference frame helps us to be specific about our
  measurements.

99

• We use this concept of a frame of reference
  widely in Physics. The laboratory frame, for
  example, is often the one in which you, the
  observer, are situated. You must imagine a set of
  axes fixed to the floor, and you are standing
  stationary at the origin. It is often helpful,
  however, to change your viewpoint to another
  frame, say one which is moving at a steady speed
  through the laboratory frame parallel to the x
  axis. For example, you can imagine standing on
  the platform of a station (the laboratory
  frame) watching someone run past you at 10 mph.
  Coming into the station is a train,

100

• also moving at 10 mph parallel to the runner. If
  you were to transform your point of view into the
  trains frame (the moving frame) you would see
  the runner apparently running on the spot, not
  making any progress at all relative to the train.
  We make wide use of the concept of a frame of
  reference in the theory of special relativity

101

• Example 1 a particle, P, is situated at position
  vector (5,3) in a two-dimensional coordinate
  frame S. What are its coordinates measured in
  frame S which has the same origin as S but is
  rotated clockwise by 30 degrees relative to S?

102

• Example 1 a particle, P, is situated at position
  vector (5,3) in a two-dimensional coordinate
  frame S. What are its coordinates measured in
  frame S which has the same origin as S but is
  rotated clockwise by 30 degrees relative to S?

103

• Before coming on the the development of
  Einsteins Special Theory of Relativity, it will
  be useful to examine in general terms some
  assumptions we have been making about the frames
  of reference in which we have been thinking about
  physics. Thus
• (a) A frame of reference is just a set of
  calibrated axes or coordinate system against
  which we can measure positions, velocities,
  accelerations, and times. Typically, we might

104

• say that an event occurs at a particular
  instant in space and time defined in one
  coordinate system, S, by (x,y,z,t). The same
  event, seen in another coordinate system, S?, is
  defined by the coordinates (x?,y?,z?,t? ).
• (b) The geometry of the space in which we are
  working obeys the axioms derived from the
  postulates of Euclid in this space, two parallel
  lines meet at infinity, the sum of the angles in
  a triangle is 180 degrees, etc. We refer to this
  as Euclidean space.
• (c) The distance between two events in space seen
  in one frame is the same when viewed in any other
  frame.

105

• Thus if events A and B occur in Cartesian frame
  S at positions (xA,yA,zA,tA) and (xB,yB,zB,tB),
  then the space interval, ?sAB, between the two
  events viewed in S is given by Pythagoras
  Theorem.

106

• Thus ?s2AB (xB? xA)2 (yB? yA)2 (zB? zA)2.
• Now view the same two events from the point of
  view of another frame, S?.

y
y?
?y?
B
A
?s?AB
S?
x
S
?x?
x?
107

• The two events occur in S? at coordinates of
  (x?A,y?A,z?A,t?A) and (x?B,y?B,z?B,t?B). Again
  the space interval, ?s? AB , between the two
  events, viewed in S?, is given by Pythagoras
  Theorem, i.e.
• ?s? 2AB (x?B? x?A)2 (y?B? y?A)2 (z?B?
  z?A)2.
• Our assumption is that ?s2AB ?s? 2AB .

108
y?
?y?
B
?y
A
?x
S?
S
?x?
x?
Our assumption is that ?s2AB ?s? 2AB .
109

• (d) The time interval is the same in any frame.
  Thus ?tAB (tB? tA) ?t?AB (t?B? t?A). In
  fact we have a strong notion that time and space
  are absolute quantities. We think that we can
  define a point in absolute space and absolute
  time, and that space and time are the same for
  everyone, no matter how they are moving with
  respect to each other. These ideas obviously work
  very well in everyday life, but need closer
  examination.
• (e) We can express the transformation between
  coordinates seen in one inertial Cartesian

110

• frame and those in another using the Galilean
  transformation. Suppose that frame S' is moving
  along the positive x axis of frame S at constant
  speed v, and their origins coincide at t t?
  0.
• An event, A, occurs in S at (xA,yA,zA,tA), and
  in S? at (xA?,yA?,zA?,tA? ).

MO 173 TM 1322
y?
x?
S?
111

• We see xA? xA? vtA, which is the only
  coordinate affected, so the Galilean
  transformation is
• x? x ? vt
• y? y
• z? z
• t? t
• This is the transformation which applies to all
  the Newtonian Physics youve done so far. As we
  shall see, it only works for transformations
  between frames in which v ltlt c, the speed of
  light.

112

• (f) Note that the quantities x, x?, etc. are
  really intervals between the two events (i) the
  origins coinciding with each other, and (ii)
  event A. Even here, we are expressing the
  transformation between space and time intervals,
  not between absolute space and absolute time
  positions. We could therefore equally well write
  
• ?x? ?x ? v?t and equally ?x ?x
  v?t
• ?y? ?y ?y ?y
• ?z? ?z ?z ?z
• ?t? ?t ?t ?t
• where ? denotes the interval between events.

113
Example 2 an observer in a high-speed train,
travelling at 575 km h?1 (currently the speed
record held by the TGV) measures the time between
his passing two signals as precisely 3 s. What is
the distance between the two signals measured by
a second observer on the track using a tape
measure?
The observer in the train is present at both
events, so he measures a space interval of
zero. The Galilean transformation gives
?xAB
A
B
S (track frame)
v
A,B
?xAB 0
?tAB 3
S (train frame)
114
Example 2 an observer in a high-speed train,
travelling at 575 km h?1 (currently the speed
record held by the TGV) measures the time between
his passing two signals as precisely 3 s. What is
the distance between the two signals measured by
a second observer on the track using a tape
measure?
The observer in the train is present at both
events, so he measures a space interval of
zero. The Galilean transformation gives
?xAB
A
B
S (track frame)
v
A,B
?xAB 0
?tAB 3
S (train frame)
115
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116
Problems with classical physics

• Nineteenth-century physicists sought mechanical
  explanations for everything. Even Maxwells
  electromagnetic theory was based on mechanical
  interactions. In particular, it was thought that
  light waves, like all other known waves, must
  travel in a medium. Thus water waves travel on
  the surface of water sound waves travel through
  the air waves on my mothers washing line
  travelled down the line (waves on a string). What
  about light, radio etc.? The work of Huygens,
  Young,

117
Problems with classical physics
1D wave equation for waves on a string, or water
waves, or sound waves

• Nineteenth-century physicists sought mechanical
  explanations for everything. Even Maxwells
  electromagnetic theory was based on mechanical
  interactions. In particular, it was thought that
  light waves, like all other known waves, must
  travel in a medium. Thus water waves travel on
  the surface of water sound waves travel through
  the air waves on my mothers washing line
  travelled down the line (waves on a string). What
  about light, radio etc.? The work of Huygens,
  Young,

represents the amplitude of the wave, that is
the displacement of a point in the medium, and c
is the speed of the wave. For example, for waves
on a string
James Clerk Maxwell showed that, for
electromagnetic waves (light, radio, x-ray etc.)
the equation was
118
Problems with classical physics

• Nineteenth-century physicists sought mechanical
  explanations for everything. Even Maxwells
  electromagnetic theory was based on mechanical
  interactions. In particular, it was thought that
  light waves, like all other known waves, must
  travel in a medium. Thus water waves travel on
  the surface of water sound waves travel through
  the air waves on my mothers washing line
  travelled down the line (waves on a string). What
  about light, radio etc.? The work of Huygens,
  Young,

119

• and Fresnel seemed to account for the properties
  of light in terms of waves. Therefore, (it was
  thought) there must be a corresponding medium for
  them? They made the hypothesis that there was
  indeed such a medium, and they called it the
  luminiferous aether. Thus, light travelled at
  3?108 m s?1 through this medium which was
  all-pervading. This hypothesis could be tested by
  looking for effects caused by motion through the
  aether.One such piece of evidence was supplied by
  Bradley in 1725 who observed stellar aberration.

120
Stellar aberration

• Imagine you are in vertically-falling rain. Now
  get on your bike the rain appears to come down
  at you from an angle to the vertical.

• ? v/c ? 10?4 radians
• This was easily measured by Bradley, and appeared
  to show evidence for a stationary aether.

c
?
aether wind
v
Earths orbital velocity through the stationary
aether
121
The Michelson-Morley experiment

• One of the most-famous attempts to measure
  motion through the aether was the
  Michelson-Morley experiment. The Earth moves at
  about 30 km s?1 in its orbit around the Sun,
  which is an appreciable fraction of the speed of
  light, 300,000 km s?1. M M set up an optical
  interferometer which would have been easily
  sensitive enough to detect this motion. The
  principle was that a coherent light beam was
  divided into two parts, and each part sent along
  perpendicular paths as follows

122
This is a simplified diagram of a MM
interferometer. You will meet this again in more
detail next year.
123

• A beam of light from a coherent light source, S,
  is split by the half-silvered mirror, A, into two
  beams, one travelling towards B, and the other
  towards C. The beams are reflected by the
  fully-silvered mirrors B and C. That from B
  passes through A to T, and that from C is
  reflected at A towards T. The two beams combine
  in the telescope, T, and interfere to produce an
  interference pattern, which is measured by the
  observer O.
• Let us suppose that the apparatus is moving
  through a stationary aether from right to left,
  at speed v, so that there is an aether wind

124

• blowing from left to right parallel to AC. The
  two light paths ABA and ACA are then not
  identical. We can see this as follows. First, the
  path ACA
• AC is with the flow, so tAC d/(cv)
• CA is against the flow, so tCA d/(c?v).
• tACA d/(cv) d/(c?v).
• Now consider the path ABA
• Both AB and BA are across the wind. The light
  gets blown to the right, so the path from A to
  B is slightly against the flow and takes longer.

125
v
aether wind
B
v
c
A

• The speed from B to A is the same as that from A
  to B, so the total time of flight is

126

• ? the time difference between ACA and ABA is
• Now, if x is much less than 1 we can use the
  first two terms of the binomial expansion of

127

• (1x)n ? 1nx, so that
• With the apparatus used by MM, this corresponded
  to a shift in the fringe pattern of about half a
  fringe, very easily seen if it existed.

128

• However, no shift was ever seen, despite the
  experiment being repeated with first AC then AB
  parallel to the Earths orbital velocity, and at
  six month intervals (just to check that the
  aether wasnt coincidentally moving at the same
  velocity as the Earth when the experiment was
  first done).
• Although this experiment is often cited as
  evidence that the aether does not exist, Einstein
  was probably not aware of it when he formulated
  his theory of special relativity.

129
Classical electromagnetism

• Newtonian physics appears to operate in accord
  with the Galilean transformation, i.e. we can
  transform the (x,y,z,t) coordinates of events
  seen in one inertial frame into those seen in
  another using this transformation.
• However, Einstein was aware that Maxwells laws
  of classical electromagnetism (which you will
  come to next year) did not transform in the same
  way. In particular, the speed of radio or light
  waves was predicted to be given
• by the expression , where
  e0 and µ0

130

• are constants associated with the vacuum. This
  seems to say that the speed of light in a vacuum
  is independent of the motion of the source or the
  observer, since there is no meaning to motion
  relative to a vacuum. No need for an aether it
  just doesnt exist. He postulated that this was
  not some quirk of electromagnetism, but that its
  consequences applied to the whole of physics. He
  formulated the theory of relativity special
  relativity (1905) applying to un-accelerated
  frames, and general relativity (1916) which is
  about gravity. Here we do the special theory.

131
Einsteins postulates
MO 194 TM 1321

• Einsteins ideas about electromagnetism and the
  nature of physical laws may be summarised in his
  two postulates
• All of the laws of physics are the same in every
  inertial (un-accelerated) frame.
• The speed of light in a vacuum is the same for
  all observers.
• The first of these postulates is quite consistent
  with the Newtonian mechanics that you have
  already met in school. You made the assumption
  that

132

• Newtons laws were true and obeyed in any
  inertial frame, and were consistent with the
  Galilean transformation. However, the new thing
  is that all the laws of Physics, including
  electromagnetism and anything else that you can
  mention, are the same in every inertial frame.
• Turning that around, the first postulate states
  that, if you are in an inertial frame, there is
  no internal experiment which you can do which can
  distinguish between that frame and any other
  inertial frame all are equivalent. For example,
  there is no such thing as an absolute rest frame.

133

• When I am at rest in my inertial frame, that
  state of rest is the same as any other in any
  inertial frame no matter how fast it is moving
  relative to mine ? a revolutionary concept for
  people seeking an all-pervading aether.
• The second postulate really follows on from the
  first. If there is no internal experiment I can
  do to tell which particular inertial frame I am
  in, then the speed of light ? in a vacuum ? must
  be the same for me as for anyone else. This has
  far-reaching implications for space and time.

134
What is time?

• Time is a notion or a concept. We know that it is
  not a substance. We know that it always moves in
  one direction, i.e. time passes or time
  advances. We assume that its rate of flow is
  uniform, and that it is universal.
• We measure time in terms of intervals, that is we
  identify events and then we measure how many
  ticks (assumed equally spaced) there are
  between the events using a machine a clock
  that has been designed to produce ticks at as
  uniform a rate as possible.
• Time is the most-accurately measured of all
  physical quantities by many orders of magnitude.

135
How is time measured?

• Time intervals are measured using processes which
  are assumed to be exactly periodic the swinging
  of a pendulum the oscillation of an escapement
  mechanism the rotation of the Earth the orbit
  of the Earth around the Sun the vibration of an
  excited atom.
• Since 1967, we have used atomic clocks to measure
  time. The SI second is defined to be exactly
  9,192,631,770 cycles of vibration in an atomic
  clock controlled by one of the characteristic
  frequencies of caesium 133.

136

• Notes
• Atomic time is now independent of astronomy.
• We keep times consistent with astronomy by
  inserting leap seconds up to twice per year at
  midnight on June 30th or December 31st.
• The pulses received from highly regular ms
  pulsars may supersede atomic clocks in the
  future.
• The time scale disseminated by radio (UTC) e.g.
  the time pips on the (analogue) BBC is an
  average over many atomic clocks in many countries.

137
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138
Time dilation
MO 195 TM 1324

• One of the consequences of Einsteins postulates
  is that identical clocks appear to run at
  different rates depending on their relative
  motion. To see how this comes about, we will
  consider a special kind of clock. Its pendulum
  is a photon reflected back and forth between
  mirrors. Since the speed of light is directly
  involved with the mechanism of this clock, we can
  see quite easily how Einsteins postulates,
  especially the second one, affect the performance
  of the clock. His first postulate tells us that
  all clocks, whatever the mechanism, must be
  affected in

139

• the same way (as otherwise we would be able to
  measure the speed of our inertial frame
  relative to absolute rest by comparing this clock
  with one having a different mechanism)

mirrors
photon
clock face
140
(No Transcript)
141

• First view the situation in the upper clocks
  rest frame. Let events A, B, C be a photon
  leaving the base mirror, arriving at the upper
  mirror, and arriving back at the base mirror
  respectively. In the clocks rest frame, S?, the
  photon takes a time 2h/c to travel from A to B
  and back to C (where c is the speed of light).
  Let this time interval be . Then

B
h
A
C
rest frame S?
back
142

• Now lets look at the same events as seen in the
  lower frame, S. The time interval between events
  A and C in this frame is ?tAC. Remember that the
  photon still travels at speed c, but has further
  to go, so takes longer

B
h
A
D
C
v
143

• In the triangle ABD, by Pythagoras theorem, we
  have
• But in the moving clocks rest-frame, S?, we have
• So substituting for h gives us

144

• Rearranging, we find
• where
  .
• Note that ? is greater than one for all speeds
  such that 0 ? v ? c, and is undefined for speeds
  of c or greater. This equation therefore shows us
  that the time interval between events A and C is
  shortest in the rest frame of the clock, and that
  the time interval measured between the same two
  events viewed in a frame in which that clock is
  moving is longer.

145

• Notes
• We are forced to conclude that the rate of the
  passage of time depends on relative motion.
• The shortest time interval between two events is
  measured by a clock which is present at both
  events.
• Such a time interval is called a proper time
  interval, and such a clock measures proper time.
• The time interval measured in another moving
  frame, using two clocks each of which is at only
  one of the events, is always longer.

146

• (e) This is sometimes summarised in the statement
  moving clocks run slow. Be careful with this
  statement as it can cause confusion. Always ask
  Which clock was present at both events? (It
  measures the shortest time interval.)
• (f) This true for every kind of clock, not just
  light-clocks (remember Einstein P1).
• (g) The effect is tiny in every day life 70 mph
  for 6 years causes a 1 ?s shift
• (h) The effect is larger for space travellers at
  four-fifths of c, the shift is from 3 s to 5 s

147
Two quotations on the relativity of
time Einstein 1905 Thence w