Introduction to Unit 1
Advanced Higher Physics
Unit 1
Saturday, 20 May 2006
Mechanics
Mechanics Unit 1 Summary
►
Linear Kinematics
►
 1D & 2D kinematics
►





Relativistic Motion
 Relativistic Mass
 Relativistic Energy
 Other Relativistic Effects
►
Angular Kinematics
 Angular Notation
 Kinematic Relationships
 Tangential Speed
►
►
Centripetal acceleration / force
Torque
Newtons Laws
Angular Momentum
Angular Kinetic Energy
Newtons Law of Gravitation
Gravitational force & fields
Gravitational Potential & Energy
Orbital Mechanics
Escape Velocity
Simple Harmonic Motion
 Analysis of SHM
 Energy in SHM
 Damping
Rotational Dynamics





Gravity
►
Wave particle duality
 Theory
 Examples of Duality
►
Quantum Mechanics
 Atomic Models
 Quantum Mechanics Theory
Friday, 18 March 2016
1D Mechanics
At higher we dealt exclusively with situations where the acceleration of an
object was constant. This is not always the case. One common
example is in friction where acceleration is dependant on velocity.
If acceleration is a function of time (eg a=0.4t), then the standard
equations of motion DO NOT APPLY. Instead we must use a calculus
method.
In general (and by definition)
dv
a
dt
v   a dt
ds
v
dt
s   v dt
Friday, 18 March 2016
1 D Mechanics
Worked Example
A 4kg object, initially at rest is subject to a force which
varies with time according to the equation F=0.80t.
►
1. Calculate an expression for the acceleration of this object
2. Now, by calculus find expressions for v and s in terms of
time
3. Find the displacement of the object after 12 seconds
4. Find the velocity of the object after 18 seconds
Friday, 18 March 2016
Do you get it – Try this.
A 14kg object is initially moving at 6ms-1. From this moment it
is subject to a force in Newtons according to the expression
F=18-0.4t.
1. Formulate expressions for the acceleration, velocity and
displacement of the object.
2. At what time is there no force on the object and what is its velocity
at that time?
3. How long after the start of the experiment does the object come to
rest?
4. How long after the start of the experiment does the object return to
its initial position and what is its velocity at this time?
Friday, 18 March 2016
Here’s why its useful
In an experiment we tracked the position of an object in one
dimension and plotted the data in the graph below.
Use your maths
skills to get the
equation of this
graph, and use it
to form an
equation for the
acceleration of the
object. If we
know the mass of
the object 1.35kg,
we can calculate
the force on the
object
d/m
60
50
40
30
20
10
0
0
5
10
15
t/s
20
Friday, 18 March 2016
Rigorous proof of higher formulae
Using your knowledge of calculus:
Considering an object with initial speed ‘u’ at t=0 and
a constant acceleration of ‘a’, and knowing that
a=dv/dt, prove that a=(v-u)/t.
Now considering the same object and knowing that
v=ds/dt, prove that s=ut+½at2
For both proofs use either
► definite integrals,
► or indefinite integrals and evaluate the constant of
integration.
Friday, 18 March 2016
Relativity
Advanced Higher Physics
Unit 1
Saturday, 20 May 2006
Mechanics
A challenge to start with
► How
fast are we moving???
► Data
Diameter of earth = 12000 km
Radius of our orbit of Sun = 940 million kilometres
Assume you are standing on the equator and that the
Earth and the orbital path are perfectly circular.
► Velocity
due to earth rotation = 436 m/s
► Velocity due to solar orbit = 187,284 m/s
Friday, 18 March 2016
Relativistic mass
Although the concept of relativistic mass is odd, it is quite
simple. As the velocity of an object relative to an observer
increases, its mass will also increase. These effects are
very small until we start to go faster than 0.1c (10% the
speed of light).
The relativistic mass of an object with rest mass mo moving at
a speed v relative to a stationary observer is given by:
m
mo
1 v
2
c
2
Friday, 18 March 2016
Relativistic Mass Questions
►
Find the relativistic mass of each of the following
An electron moving at 2.3x108 ms-1
2. A neutron moving at 1.3x108 ms-1
3. A 700kg space probe moving at 8x107 ms-1
4. A 3kg meteorite accelerated to 9x106 ms-1
1.
If an object has a rest mass of 6kg, but a
relativistic mass of 6.7kg, calculate its velocity
6. An object has a rest mass of 24x10-5 kg. Its
relativistic mass is 12% greater. Calculate its
velocity.
5.
Friday, 18 March 2016
Change in mass as velocity changes
► Using
your knowledge of relativistic mass, plot a
graph of how the mass of an object changes as its
velocity increases from rest to 97%, if the object
has a rest mass of 1kg.
► (calculate
m at 10% intervals,
and then 95% and 97%)
Friday, 18 March 2016
Relativistic Energy
The most important concept in relativity is that the mass of an object is a
measure of its total energy. The Newtonian concept of kinetic energy
(½mv2) begins to fail at relativistic velocities (>0.1c) as the mass is no
longer constant as the object was accelerated. Instead we use Einstein’s
equation as follows:
The Rest Energy of an object is calculated using its rest mass (Eo=moc2),
while the Relativistic Energy is calculated using the relativistic mass
(E=mc2). The kinetic energy of the object is the difference between its
relativistic energy and rest energy.
E = Ek + Eo
Ek = E – Eo
Ek = mc2 - moc2
Friday, 18 March 2016
Relativistic Energy
► The
graph below shows how the Newtonian and
Relativistic definitions of kinetic energy differ and
how the difference between the theories increases
dramatically as the speed of light is approached.
Use Excel to plot a graph of the kinetic energy
of a 1kg object as its velocity is
Increased from rest to 0.85c.
Plot both the kinetic energy calculated by
Newtonian theory and the relativistic theory
and note at what velocity
Newtonian mechanics begins to break down
Plot another graph showing the
correlation between Newtonian
mechanics and relativistic
mechanics at velocities <0.1c
Presentation is important here, so please
play about with scales, etc to get
the graph looking appropriate.
Friday, 18 March 2016
Relativistic Energy Problems
Accelerating particles.
Particle accelerators are used to accelerate protons, electrons
and even small atomic nuclei to very high velocities for
experimental work. However, as the velocities reached are
relativistic (>0.1c), the effects of relativity must be
accounted for.
We can assume that the charge of a particle does not change
as it accelerates, not does the voltage of the field
accelerating it.
Equating the work done on the charge to the relativistic kinetic
energy, we can calculate the velocity of the particle.
►
Combine the formulae for work done on a charge in an
electric field with the formula for relativistic kinetic energy
and rearrange this formula for voltage.
►
Also rearrange the formula for velocity (quite tricky).
Friday, 18 March 2016
Now you can work these out….
1.
2.
3.
4.
5.
6.
Calculate the voltage of field required to accelerate
an electron to 8x107 ms-1. (19.2kV)
Calculate the velocity a proton would reach in a
field of 30kV.
(2.40x106 ms-1)
Calculate the velocity an alpha particle would reach
in the same field.
(1.69x106 ms-1)
Calculate the voltage of field required to accelerate
an alpha particle to 0.8c.
(1.26x109 V)
Calculate the voltage required to accelerate a
deuterium (21H) nucleus to the same speed.
From your knowledge of work done, calculate the
average force on the deuterium nucleus in Q5 if it
accelerates over a distance of 800m. (2.58x10-13N)
Friday, 18 March 2016
Other Relativistic Effects
Time Dilation
In AH we do not need to analyse this effect mathematically as it
is quite complex and has effects which are difficult to
understand. One of the side effects of relativity is that when
two objects are in motion relative to each other, time does
not run at a constant rate (it runs faster for the “stationary”
observer.
If we set two clocks to run at identical times, then place one on
a ship at a velocity close to the speed of light and fly it in a
circle, the clock which has been moving will run slower than
the stationary one.
Essentially, as velocity increases, time slows down.
►
Length contraction
In addition to this, at relativistic velocities, the length of an
object in the direction of travel will decrease as the object
approaches the speed of light.
►
Friday, 18 March 2016
Post Summer Brain Warm-up
Advanced Higher Physics
Unit 1
Saturday, 20 May 2006
Mechanics
Mini testy type thingamajig….
1.
Write down the correct units for each of the following
(a) Momentum
(b) Charge
(c) Pressure
(d) Acceleration
(e) Capacitance
(f) Weight
(g) Relativistic Mass
2.
From the expression a = dv/dt , derive the equation of motion
v = u + at
3.
Bobby (a 67kg ice skater) was skating along at 14ms-1, minding his
own business, when he collided with a stationary 39kg kangaroo which
was lost. The pair become entangled. What was their velocity just
after the collision?
4.
A small elephant (137kg) is accelerated in space to a velocity 1.5 times
greater than “the velocity of light in a material with a refractive index
of 1.65”. Calculate the relativistic mass of the elephant at this
velocity.
5.
If Mr Colquhoun had a rest mass of 87kg before the holidays, but lost
7kg of mass between then and now, to what speed would he have to
be accelerated such that his relativistic mass equalled his rest mass
before summer?
Friday, 18 March 2016
Mini testy type thingymajig….
1.
Write down the correct units for each of the following
(a) kgms-1 or Ns
(b) C
(c) Pa or Nm-2
(e) F
(f) N
(g) kg
2.
Derivation
3.
8.85 ms-1
(d) ms-2
dv
a
dt
t
v
0
u
 a dt   dv
4.
5.
329 kg
1.18 x 108 ms-1
at t0  vuv
at  0  v  u
v  u  at
Friday, 18 March 2016
Some Quick Revision…post summer
1.
The motion of an object is the be analysed and a student
measures its displacement. If its motion is described by the
equation s=0.3t3-0.6t2+1.3t, calculate:
(a) Its velocity after 5 seconds (17.8m/s)
(b) Its acceleration after 8 seconds (13.2 m/s2)
2.
Find the relativistic mass of a 1.6 tonne African elephant
moving with a velocity of 1.3x108 ms-1.
3.
By calculating the difference between the rest energy and
the relativistic energy, determine the kinetic energy of an
87kg human being moving with a velocity of 60,000 kms-1.
Friday, 18 March 2016
Rotational Motion
Advanced Higher Physics
Unit 1
Saturday, 20 May 2006
Mechanics
Angular Terminology
Rotational motion works in much the same way as
motion in a straight line if we analyse it in the
correct way. The following symbols will be used
throughout this section.
θ ω0 ω  -
Angular displacement
Initial angular velocity
Final angular velocity
Angular acceleration
-
rad
rad s-1
rad s-1
rad s-2
Terminology aside, the basic equations of motion for
angular kinematics can be derived in exactly the
same way as they were for linear kinematics.
Friday, 18 March 2016
Deriving the angular equations of motion
►
Starting from
d

dt
►
Derive the angular equations of motion for an object
accelerating from ‘ω0’ to ‘ω’ with a constant angular
acceleration ‘’ in time ‘t’. In performing this motion, the
object passes through angular displacement θ.
Friday, 18 March 2016
Working with angular velocities
When using angular velocities in calculations, they
must always be measured in rad s-1. Figures for
angular velocity are often given in different ways.
Convert each of the following to radians per
second.
1.
2.
3.
4.
5.
3000 rpm
24 rpm
1 revolution per day
1 revolution in 28 days
50 revolutions per second
1 . 314 rad/s
2. 2.51 rad/s
3. 7.3x10-5 rad/s
4. 2.6x10-6 rad/s
5. 314 rad/s
Friday, 18 March 2016
Centripetal acceleration
In order for an object to move in anything other than
a straight line, it must accelerate. In the case of
pure circular motion, this acceleration is always
towards the centre of the circle. This is called
centripetal acceleration and is entirely different from
angular acceleration.
Two formulae for centripetal acceleration can be
derived from your knowledge of circular geometry
and vector addition and subtraction.
This derivation can be found in your SCHOLAR
textbooks.
Friday, 18 March 2016
Centripetal acceleration
We consider a point moving in a circle with radius r and with constant
angular velocity ω moving though an angle Δθ from point A to point B as
shown here.
In order to find the centripetal
acceleration we must
determine the change in
ω
velocity of the point in
travelling from A to B. Its
angular velocity will not have
changed, nor will the
vb
magnitude of its linear velocity,
but the direction of the
Δθ
tangential velocity has
r
changed. This change in
B
velocity can be expressed as
A
va
Δv = vb - va
Friday, 18 March 2016
Centripetal acceleration
Δv = vb - va
This is a vector sum, and is carried out as follows
-va
Δv
ω
vb
vb
r
Δθ
B
A
va
Now, by some clever maths, we can
show that as we make Δθ very
small, the change in velocity
becomes perpendicular to va. In
other words the acceleration of
the point is towards the centre of
the circle.
Centripetal acceleration is
written as

a
Friday, 18 March 2016
Centripetal acceleration
-va
Δv
Δθ
vb
With this right angled triangle we can say that
Δv = v Δθ , which is
true for small angles measured in radians (θ = tan θ for small angles)
We can define the centripetal acceleration as
Then substitute in for Δv = v Δθ
Then in the limit as Δt tends to zero we get
the differential
And finally since v=rω, ω=v/r giving
v
t
v
a 
t
d
a  v
dt
a  v
a 
v2
a  v 
r
Friday, 18 March 2016
Centripetal Force
►
►
Firstly, a myth to dispel.
You are in a car driving round a circular track anticlockwise.
There is a force acting on you, but in which direction does it
act?




►
►
Forward?
Backward?
Left?
Right?
Centrifugal force does not exist, it is a reaction force
experienced as a result of centripetal force which acts
towards the centre of the circle.
In the same way that when standing on the floor you are
pushed up by the floor, when travelling in a circle you feel
like you are being pushed out from the centre – this is not
the case!!
Friday, 18 March 2016
Friday, 18 March 2016
Centripetal Force
►
The basic concept of centripetal force is very simple to
understand. We have proved the expression for centripetal
acceleration.
►
If we consider the case of a point mass following a circular
path, a force must exist to give it this centripetal acceleration.
This force must act towards the centre of the circle and must
obey Newton’s 2nd law.
F  ma
2
mv
2
F  mr 
r
Friday, 18 March 2016
Applications of centripetal force
► Confirming
the formula experimentally
► Motion
in a horizontal circle
► Motion
in a vertical circle
► Conical
pendulum
► Cars
cornering
► Banked
corners
Friday, 18 March 2016
Confirming the formula
Mass experiencing
centripetal force
m1
F  mr
2
Mass used to counter
centripetal force
m2
In this experiment we slowly increase the angular velocity of the turntable until the
centripetal force on m1 exceeds the weight of m2. When this occurs m1 will move.
When m1 moves, maintain the rotational velocity and measure it (time a number of
rotations). Since the mass and initial radius of m1 , the weight of m2 and the angular
velocity are known, the formula can be checked.
Perform the experiment twice at each of two radial positions. (4 times in total)
Friday, 18 March 2016
Motion in a Horizontal Circle
►
1.
2.
3.
Try the following questions
A young girl of mass 37kg is on a roundabout at a radius of
1.6m. Her angular velocity is constant at 2.4 rad s-1.
Calculate her tangential velocity, her centripetal
acceleration and the centripetal force acting on her.
In a high g simulator a 300kg pod is rotated at high speed
around a central pivot. The radius of the arc is 6m. If the
tension force safety limit of the arm connecting the pod to
the pivot is 20kN, Calculate the maximum angular velocity
at which the simulator can safely be used. Calculate the
corresponding tangential velocity at maximum safe limits.
A 850kg car drives round a corner at 40.0 km h-1 . If the
corner has a radius of 8.00m, calculate the centripetal
force required to negotiate the corner safely.
Friday, 18 March 2016
Motion in a Vertical Circle
► Consider
an object being whirled vertically on a
piece of string. The forces acting on it will change
depending on the position of the object as shown
below.
In each position, we can state
that the sum of the forces (T
and mg) must equal the
centripetal force (mrω2).
a⊥
T
a⊥
mg
T
mg
a⊥
T
mg
T
a⊥
mg
Form an equation for the tension
in the string when the mass m
is at
(a) the top
(b) the bottom
(c) the horizontal position
Friday, 18 March 2016
Conical Pendulum
The period of a conical pendulum, similarly to a standard
pendulum, can be proven to be independent of the mass on
the end as long as the connection (eg string) is of negligible
mass.
Φ
Φ
T
T cos Φ
l
T sin Φ
ω
mg
mg
a⊥
a⊥
Use the above free body diagrams and with
the aid of your scholar books derive
expressions for
(a) the tension in the string
(b) the angular velocity of the bob
(c) the period of the bob
Friday, 18 March 2016
Cars cornering on the flat
We can assume that a car, or other object driving round a
corner is travelling in a horizontal circle with a constant
angular velocity. As a result we know the only unbalanced
force on the car is the centripetal force.
On a flat road, this force must be provided by the friction of the
tyres. The weight of the car is exactly balanced by the
reaction force from the road.
This means that at a given velocity, there is a limit to how tight
a corner a car can go round before skidding.
Reaction
Force
Friction
from tyres
Weight
Example
A 700 kg car travelling at 50kmh-1 is
fitted with tyres providing a
maximum frictional force of 4500N.
What is the smallest radius of
corner it can go round at this
velocity?
Friday, 18 March 2016
Banked corners
On a banked corner, similar to a conical pendulum, a
component of the reaction force from the road provides part
of the centripetal force required.
The car is still assumed to be travelling in a horizontal circle, so
the weight and the vertical component of the reaction force
must be balanced.
Reaction
Force
Weight
Φ
Example
For a 600kg car, calculate the angle of
banking required such that a car
travelling at 50 kmh-1 need produce
no friction from the tyres (ie,
horizontal component of the
reaction force equals the centripetal
force required), for a corner with a
radius of 36m.
Friday, 18 March 2016
Rotational Dynamics
Torque & Moment
Advanced Higher Physics
Unit 1
Saturday, 20 May 2006
Mechanics
Torque & Moment
Like the equations of motion, there are commonalities between
linear and angular dynamics (the study of forces and
motion).
Torque and moment are in essence the same thing. They are a
force acting on a mass which causes it to rotate or turn.
Technically, a moment is defined as:
The tangential component of a force multiplied by its radius
from the point of rotation. It is measured in Nm.
If the force is perpendicular
to the radius
If the force is not perpendicular
to the radius
F sin Φ
F
Φ
r
T = Fr
F
F cos Φ
r
T = Fr sin Φ
Friday, 18 March 2016
Balancing torques
Balanced torques work in exactly the same way as balanced forces. For an
object which is stationary or rotating, the torques must be balanced.
For each of the following, draw out the diagram, and find the value of the
missing force or radius.
F
6N
0.6m
0.3m
1.5m
160N
96N
7m
r
20N
0.9m
90N
4m
85N
0.9m
F
3N
F
3m 5m
12N
r
40N
1.3m
1.9m
60°
F
Friday, 18 March 2016
Basic Structural Mechanics
This principle can be applied to any object or
structure, or section of a structure
Mass=200kg
Φ
Φ
Mass=300kg
3m
5m
mg
2.5m
mg
3m
Friday, 18 March 2016
Torque & Acceleration
The aim of this experiment is to prove the linear
relationship between torque and angular
acceleration.
r
m2
Mass used to
create force
Friday, 18 March 2016
Torque & Angular Acceleration
To derive the angular equivalent of Newton’s 2nd Law (F=ma)
we analyse a point mass ‘m’ moving in a circle with an
angular acceleration ‘’. We already know that:
F = ma
T = Fr
a = r
And all of these laws must hold for our object, so:
F=ma
F=m r 
I is the moment of inertia
measured in kgm2
T/r = m r 
T = I
T = mr2 
The quantity ‘mr2’ is a property of the object which is rotating
and is called the moment of inertia.
For a rotating point mass the moment of inertia is given by:
I = mr2
Friday, 18 March 2016
Finding Compound Moments of Inertia
►
►
►
When calculating the moment of
inertia of an object, the back of
the data book provides the
following formulae for different
shapes of object.
Note that it is their distribution of
mass around the axis of rotation
that matters, so a doughnut
shape uses the same formula as
a point mass as all the mass is at
a single radius.
We can also add or subtract
shapes from each other to allow
us to analyse more complex
shapes.
Point mass I  mr 2
I  1 ml 2
12
Rod about centre
Rod about end
I  1 ml 2
3
Disc about centre
Sphere about centre
I  1 mr 2
2
I  2 mr 2
5
Friday, 18 March 2016
Finding Compound Moments of Inertia
►
Find the moment of inertia for each shape
(b)
(a)
3.9kg
Solid sphere
0.36m
1.9kg
0.28m
(c) Metre stick about 1 end
(d) A bouncy ball (rolling)
0.83m
(e) Plastic pipe
i. Around 1 end
ii. Around centre
iii. Around lengthways central axis
5.2kg
0.08m
Friday, 18 March 2016
Angular Momentum
Again, angular momentum works exactly the same as linear
momentum in that momentum is always conserved. This
gives rise to the formula:
Angular Momentum
(kgm2s-1)
L  I
Angular velocity
(rad s-1)
Moment of Inertia
(kg m2)
The same principles apply to the angular system in that
momentum is always conserved, and that in an elastic
collision, the rotational kinetic energy is also conserved.
Ekrot  I
1
2
2
The only additional consideration is that changes in angular
velocity can be brought about by changes in moment of
inertia.
Friday, 18 March 2016
Try these
1.
Calculate the angular momentum of
a) A 12kg disc of radius 0.85m rotating at 6.5rad s-1
b) A 2.6m, 4.3kg rod rotating about the middle at 8.2rad s-1
2.
A ballet dancer is trying to achieve a faster
spinning speed. She starts spinning with her arms
out, giving her an angular velocity of 12.3 rad s-1.
In this position her moment of inertia 15kgm2. She
then draws her arms in which reduces her moment
of inertia by 20%. Calculate her new angular
velocity. Calculate whether her kinetic energy has
increased, decreased or stayed the same.
Friday, 18 March 2016
Outcome 3 – Measuring I
Using the air table fitted with a central pulley, determine the
moment of inertia of the disc and any associated masses.
Determine the torque applied to the disc and its angular
acceleration. Repeat the experiment for a number of
different applied torques. Use this data to determine the
moment of inertia of the disc by graphical method.
It is recommended that the measurement of time and angular
displacement is used to determine the angular acceleration of
the disc, rather than an attempt to measure the
instantaneous velocity of the disc.
A full report of the experiment should be produced using the
guidelines including an analysis of errors and a full evaluation
of the experiment.
Friday, 18 March 2016
Gravitation
The Laws of Attraction
Advanced Higher Physics
Unit 1
Saturday, 20 May 2006
Mechanics
Newtons Law of Gravitation
We are used to thinking of gravity as the source of weight
always acting downwards. In fact there is a force of
gravitational attraction between any two objects in the
universe. Usually, for these forces to be significant, the
masses must be large and the distances reasonable.
The force of attraction (which acts on both objects) is given by
the formula:
Where:
Gm1m2
F
2
r
F = Force of attraction (N)
m1= mass of object 1 (kg)
m2= mass of object 2 (kg)
r = the distance between the objects (m)
G = Gravitational constant
= 6.67x10-11 N m2 kg-2
Friday, 18 March 2016
Interesting ideas
Zero gravity points
It stands to reason that there is some location between the
moon and earth where the moon’s gravity cancels out the
earths leading to a point of “zero gravity”.
Find this point between the earth and the moon
Friday, 18 March 2016
Orbital Mechanics
By equating the gravitational force to the formula for
centripetal force we can perform detailed analysis on objects
in circular orbits of planets or moons.
The orbital period of an object
is only dependant on the mass of
the planet / star around which it is
orbiting, and the radius of the orbit.
Gm1m2
2
F

m
r

1
2
r
Gm2  r 3 2
For example, using the following data, calculate the radius and
therefore, the altitude of the orbit of the moon above the
earth (HINT: first calculate ω from orbital period):
Mass of Earth
= 6.0x1024kg
Mass of Moon
= 7.3x1022kg
Radius of Earth
= 6.4x106m
Answer = 3.84x108m
Lunar Orbital Period
= 28 days
Friday, 18 March 2016
Orbital Mechanics
When an object is in a circular orbit around a planet, the only
force acting on it is the gravitational pull of the planet. This
force must therefore be equal to the centripetal force on the
object.
By equating these two formulae, derive expressions for:
► The tangential velocity of the satellite at a known orbital
radius around a given planet.
► The period of a satellite given the radius of orbit around a
known planet
► The required radius of orbit to achieve a given period of
rotation.
►
Use the last formula to calculate the radius of orbit and
therefore the altitude above earth of a geostationary satellite
Friday, 18 March 2016
Design a solar System
►
►
►
►
You have been tasked with the
design of a single planet within
a solar system, based around
a star slightly larger than our
sun. The star has a mass of
2.5x1030kg and a radius of
7.1x108m.
You must form a “Fact sheet”
for your planet in the following
format:
You should also supply a
calculations sheet stating what
assumptions you made and
showing all your calculations.
Fact sheet should be a single
A4 sheet.
Planet Name
Radius of planet
Mass of Planet
Radius of orbit
Length of 1 “year”
Length of one “day”
Gravity at Surface
Height of Geostationary Orbit
Moons (and associated
geometrical and orbital details)
Any other information
Friday, 18 March 2016
Solar System Design Tips
►
►
All the calculations you’ll be
using are based on a few
formulae which can be
rearranged to calculate the
information you need.
When trying to describe
time, specify whether you
are referring to a day on
earth, or a day on your
planet/moon.
Gm1m2
F
r2
F  mr 2
2

T
Friday, 18 March 2016
Weight on planets
When dealing with a relatively
small object on a planet we
consider the mass and
radius of the planet to work
out what we have
previously used as ‘g’.
So for an object on earth
(mass=5.97x1024 kg)
(Radius=6.38x106 m), we
can say the gravitational
field strength(g) is:
Gm1m2
r2
Gmplanetmobject
Weight 
rplanet 2
F
 Gmplanet 
mobject
Weight  
2 
 r

 planet 
So we can say
 Gmplanet 

g 
2
 r


planet


Friday, 18 March 2016
Cavendish-Boys Experiment
First carried out in the 18th century, this experiment can be used to
determine a reasonably accurate value for G. It uses a torsion balance
which measures the torque due to the gravitational attraction between
the large and smaller masses. A mirror mounted on the torsion wire can
be used to determine the angular displacement of the wire and hence the
torque. (T=kθ, where k is a constant)
Torsion
wire
m1
m2
m1
m2
Torsion
wire
m1
m2
m1
m2
Friday, 18 March 2016
Cavendish-Boys Experiment
m2
m1
Torsion
wire
m1
Length
m2
Since all values except G can be
measured to a reasonable degree
of accuracy in this experiment, it
can be used to determine a value
for G.
Torque from grav forces  torque from balance
2 Fr  k
Fl  k
mm l
G 1 2 2   k
 r 
Friday, 18 March 2016
Gravitational Potential
At SG and higher Ep=mgh, however this assumes that g is constant which for
an object gaining significant height, we know it is not true.
The gravitational potential (V measured in J kg-1) at a point is defined as:
the work done by external forces in moving a unit mass from infinity to
that point.
This will always be a negative value and is calculated by integrating the work
done formula (since F varies with radius) from infinity to r.
r
F
V 
dr
m2

r
Gm1
V   2 dr
r

Perform this integration to
get an expression for V
 Gm
V
r
Friday, 18 March 2016
Gravitational Potential Energy
The actual potential energy of a specific object (m2) due to its
position in the gravitational field of another object(m1) is
simply the gravitational potential at that point multiplied by
the mass of the object in the field(m2).
 Gm1m2
Potential Energy  Vm2 
r
Friday, 18 March 2016
Drawing Gravitational fields
The gravitational field strength (g) can be shown
diagrammatically by drawing lines around a picture
of the object.
► The lines show the direction of the force acting on a
body placed in the field
► The closer together the lines are the stronger the
gravitational field.
► Use
your SCHOLAR books to sketch the gravitational
field around
 A single point mass
 Two identical point masses in proximity
Friday, 18 March 2016
Now try these
►
►
1.
2.
On page 92(ish) of SCHOLAR books try Quiz 1 and Quiz 2
(Page 78 & 83)
Extension (use planetary data from scholar book)
A 1200kg satellite moves from a circular orbit around the
earth at an altitude of 13,000km above the surface into a
higher circular orbit. If its gravitational potential energy is
increased by 5GJ, determine the new altitude of the
satellite.
(17.9x106m)
How much energy would need to be given to a 5kg object
on the surface of the earth in order for it to reach infinity
(ie never stop). What velocity would it need to be moving
at to have this much kinetic energy (non-relativistic).
(-312MJ , 11.17x103m/s)
Friday, 18 March 2016
Escape velocity
Theoretically, ignoring the effect of air friction, if an object is
launched vertically with a high enough velocity, it will
continue on indefinitely rather than fall back to the
surface. This is when the kinetic energy it has at launch
plus its gravitational potential energy (negative
remember!) equals zero.
For escape velocity on the surface of a planet
Ek  E p  0
  Gm1m2 
1
2
0
m2 vesc  
 r

2
 planet 
now rearrange this for an equation for v esc
Using data in your scholar books find the escape velocity from the moon and the earth
Friday, 18 March 2016
Black Holes
►A
black hole is essentially a massive star which is
extremely dense. The result is that it has a
relatively small diameter and a mass many times
greater than our sun.
► If the escape velocity from the surface of the star is
greater than the speed of light, then no light will be
emitted by the star and so to an external observer,
it will appear black.
► Black holes also have the tendency to “suck in”
other objects. This increases their mass further.
Friday, 18 March 2016
Planetary Data
Object
Name
Mass (kg)
Radius (m) Orbital
Radius (m)
Planet
Mercury
3.30x1023
2.44x106
57.9x109
Planet
Venus
4.87x1024
6.05x106
108x109
Planet
Mars
6.42x1023
3.40x106
228x109
Planet
Jupiter
1.90x1027
71.5x106
778x109
Jupiter
Moon
Io
8.94x1022
1.82x106
422x107
Star
Sol
1.99x1030
695x106
N/A
Friday, 18 March 2016
Simple Harmonic Motion
Advanced Higher Physics
Unit 1
Saturday, 20 May 2006
Mechanics
Springs and spring constants
Aim
To determine the spring constant of a given
spring.
Method
The change in displacement of a spring is
measured with a ruler as known masses are
hung from it.
Results
Mass (kg)
Weight (N)
∆ Displacement (m)
Friday, 18 March 2016
Springs and spring constants
Analysis
Plot a graph of force against
displacement. Find the gradient of
this graph (dF/ds). This is the
spring constant.
F
∆s
Mass (kg)
Weight (N)
∆ Displacement (m)
Friday, 18 March 2016
Wobbly things
Anything wobbling or vibrating is usually undergoing simple
harmonic motion. Take this mass attached at both sides to
rubber bands.
positive
Think about the forces acting on the box.
Where would the following occur:
(a) Maximum positive acceleration?
(b) Maximum negative acceleration?
(c) Maximum positive velocity?
(d) Maximum negative velocity?
Friday, 18 March 2016
Intro to simple harmonic motion
Many objects exhibit simple harmonic motion, which is simply
an oscillation of an object which is disturbed from its rest
position. If we examine a spring and mass in a horizontal
frictionless arrangement as shown we can analyse this type
of motion.
Equilibrium position
Position y=0
Force = 0
Spring Compressed
Position y=negative
Force = positive
Spring Extended
Position y=positive
Force = negative
Friday, 18 March 2016
SHM Maths
y
F   ky
ma   ky
If we assume the object is moved to a positive y
position and then released we can say the force
acting on it and causing it to accelerate is:
This differential equation is specific to the spring
scenario, but represents a more general
equation for SHM.
The constant ω is called the angular frequency,
but is simply a constant for any given situation
relating to the frequency of the oscillation.
The mathematical solution of this differential
equation is beyond the scope of advanced
higher.
 ky
a
m
d 2 y  ky

2
dt
m
k

m
d2 y
2



y
2
dt
Friday, 18 March 2016
Solutions to the SHM Equation
There are two solutions (really only one) to this equation. Both
satisfy the equation, but are only in fact different in their
initial conditions as we will see later.
y  a sin t
y  a cos t
dy
 a cos t
dt
d2 y
2


a

sin t
2
dt
d2 y
2



y
2
dt
dy
  a sin t
dt
d2 y
2


a

cos t
2
dt
d2 y
2
  y
2
dt
Friday, 18 March 2016
Velocity in SHM
►
►
Assuming the initial conditions for an object given an initial
velocity at zero displacement, we can say:
y  a sin t
v  a cos t
y
sin t 
a
v
cos t 
a
Starting with the mathematical relationship
sin 2   cos 2   1
sin 2 (t )  cos 2 (t )  1
►
Show that the velocity of the object is given by the
expression
v   a 2  y 2
Friday, 18 March 2016
Energy in SHM
So far we know that a mass oscillating on a horizontal
spring on a frictionless surface will have a
displacement of the form y = a sin ωt. We also
know that the velocity of the mass is given by
v=±ω√(a2 - y2).
First on a scrap bit of paper, then in you jotter….
Substitute the velocity equation above into the
kinetic energy equation
Look at the Energy in SHM section of your scholar
book and note the proof for the potential energy
stored in the spring.
Finally show that the total energy is given by
Ep+Ek= ½mω2a2
Friday, 18 March 2016
Scholar graph sketching exercise
At the the end of the energy in SHM section there is a
scholar graphing exercise. Sketch both these
graphs by hand and stick in your jotter.
You may draw both graphs on the same set of axes
assuming you clearly identify which is which.
Friday, 18 March 2016
Vertically oscillating masses
►
Design and write up an experiment to determine the spring
constant of a spring using a vertically suspended mass and
measuring the angular frequency and mass to determine k.
Carry out the experiment, determine a value for k and then
check this value for your spring as we did in the first
experiment by calculating the ratio of the weight to the
extension of the spring.
►
In order to do this you will need to derive the expression for
angular frequency of a vertically suspended oscillating mass
on a spring.
Take five values for time for 10 oscillations. Find
approximate random error, then find k and the error in k.
Good luck
►
►
Friday, 18 March 2016
Build a clock
► Read
your scholar textbook section on simple
pendulums in SHM. Use the information to find the
length of pendulum on earth (g=9.8Nkg-1) which
would pass the central point once per second
(period=2s).
► Build
this pendulum and test it for accuracy.
Friday, 18 March 2016
Damping
When most objects undergo simple harmonic motion
there is some friction in the system. This causes the
oscillation to decrease in amplitude, but usually not
to change in frequency.
Sometimes damping is to be minimised when an
oscillation is desired, however often vibrations are to
be minimised and so additional damping is required
such as in shock absorbers and suspension systems.
Critically damped systems are damped such that
when disturbed, they return to their equilibrium
position in the minimum possible time.
Friday, 18 March 2016
Wave Particle Duality
Advanced Higher Physics
Unit 1
Saturday, 20 May 2006
Mechanics
Schizophrenic Waves
Discuss
What is light????
► Think







Wave
about:
Diffraction
Photoelectric effect
Absorption and emission spectra
Reflection
Refraction
Photons
Waves
OR
Particle
What do you think????
Friday, 18 March 2016
Wave Particle Duality
Under certain conditions waves can behave like particles and
particles can behave like waves.
We know from higher that:
f=v/λ for light
E=hf for a photon
E=mc2 for a particle
The theory of wave-particle duality combines these three ideas
and applies them all to waves and particles.
For this you must imagine light as a particle moving at the
speed of light.
Form an equation for the momentum (mv) of the photon as a
function of its wavelength.
Friday, 18 March 2016
de Broglie Wavelength
►
The de Broglie wavelength is given by the equation
Wavelength (m)
h

p
Plank’s Constant
6.63x10-34 Js
Momentum (kgms-1)
►
1.
2.
3.
4.
Find the wavelength of
A neutron moving at 10% of the speed of light
An electron moving at 6x105 ms-1
A golf ball moving at 5 ms-1
A person at full sprint (estimate values)
Friday, 18 March 2016
de Broglie Wavelength
The wavelengths of larger objects are so small that no
wave phenomena are observable.
As a result of this, the concept of de Broglie
wavelength is only applied to particles of sub atomic
or at most, atomic scales.
Friday, 18 March 2016
Compton Scattering
In 1923, Arthur Compton provided more evidence for
wave particle duality by showing that X-rays (high
frequency EM waves) can behave like particles.
thin
graphite
sheet
Friday, 18 March 2016
Compton Scattering
If he thought of the X-rays as particles (photons),
which collided with the electrons in the graphite,
and analysed the momentum of the collision.
electron
Some momentum
transferred to electron
Less momentum
so lower frequency
thin
graphite
sheet
Friday, 18 March 2016
Wave Particle Duality
Atomic Models
Advanced Higher Physics
Unit 1
Saturday, 20 May 2006
Mechanics
In the beginning
► Earth
► Water
► Air
► Fire
Friday, 18 March 2016
Then came the chemists…..(in 1869)
Friday, 18 March 2016
But what about atoms
The concepts of positive and negatives were known about first.
So Scientists made an assumptions. Large positive lump with
little negative bits stuck in it. The plum pudding model.
Originally suggested by J.J Thomson (the bloke who
discovered electrons) in 1904
Friday, 18 March 2016
Then came Rutherford (in 1911)
Rutherford scattering suggested a different model, mostly made
up of empty space with a small dense nucleus which was
positively charged, orbited by much smaller electrons, which
were negatively charged.
Thin gold
foil
Experiment carried
out in a vacuum
 source
large
deflection by
nucleus
Small
deflection by
electrons
Zinc sulphide screen
and eyepiece used to
detect alpha particles
Friday, 18 March 2016
But there were still questions
► The
Rutherford model didn’t explain all the
interesting work being done on absorption and
emission spectra at the time. It was all very clever,
but nobody knew how it all worked. There was a
missing piece of the puzzle.
Friday, 18 March 2016
Neils Bohr (1913) - Denmark
Mr Bohr cracked the basis for quantum theory in 1913 when he
modified Rutherford’s model to explain absorption and
emission spectra and opening that whole can of worms that
is “quantum mechanics”.
Bohr proposed that there were distinct
values at which the amount of energy
represented a wavelength that would
exactly fit into the orbit. After
messing with some complicated maths
he discovered that for a stable orbit
the angular momentum was a multiple
of h/2π
ie.
L = nh/2π
mvr = nh/2π
Friday, 18 March 2016