2.
Observing circular motion.
Where do you observe things moving with circular
motion?
• The hammer swung by a hammer thrower
• Clothes being dried in a spin drier
• Chemicals being separated in a centrifuge
• Cornering in a car or on a bike
• A stone being whirled round on a string
• A plane looping the loop
• A DVD, CD or record spinning on its turntable
• Satellites moving in orbits around the Earth
• A planet orbiting the Sun (almost circular orbit for many)
• Many fairground rides
• An electron in orbit about a nucleus
Any motion in a curve is an example of circular motion.
3.
Linear velocity.
The instantaneous linear velocity at a point in the
circle is usually given the letter v and measured in
metres per second (m s-1).
To calculate the speed of a body moving in a
circular path, you need to know the distance
moved and the time taken i.e.
Speed = circumference of circle
time to complete one rotation
Since the circumference of a circle = 2πr
ν = 2πr
t
4.
Questions.
1. A Child’s roundabout has a diameter of 2.5 m,
it revolves at 45 rpm.
Calculate the linear velocity of its outside edge.
2. A satellite orbits the earth every 90 min, its
speed is approximately 7.8 Km s-1
Calculate its height above the earth
(radius of the earth is 6400 km)
3. A bucket is being swung on the end of a piece
of rope, it is travelling at 6m/s, the rope is 1.5m
long, what is the time period of rotation?
5.
Degrees and radians
The radian is a more r
‘natural’ unit for
measuring angles. r
θ=1
One radian (or rad for r
rad
short) is defined as
the angle subtended
at the centre of a
circle of radius r by an
arc of length r.
6.
Thus the complete circumference 2πr subtends an
angle of 2πr/r radians.
Thus in a complete circle of 360 degrees there are
2π radians.
Therefore 1 radian = 360° /2π = 57.3°
7.
Use a calculator to complete the table of q in
degrees and radians, sin θ, cos θ, and tan θ
when θ has values in degrees shown in the table:
Sin θ cos θ tan θ
θ degree θ in radians
0.01 1.75E-4 1.75E-4 0.99 1.75E-4
0.1 1.75E-3 1.75E-3 0.99 1.75E-3
0.5 8.7E-3 8.7E-3 0.99 8.7E-3
1 0.017 0.017 0.99 0.017
5 0.087 0.087 0.99 0.087
10 0.175 0.175 0.99 0.175
20 0.35 0.34 0.94 0.36
70 1.22 0.94 0.34 2.75
When θ (in radians) is small what are suitable
approximations for: sin θ, tan θ, cos θ?
8.
Angular velocity.
The angular velocity is the angle through which
the radius to this point on the circle turns in one
second. This is usually given the letter ω (Greek
omega) and is measured in radians per second
(rad s-1)
So if a body completes 4 revolutions every
second it has an angular velocity of?
9.
Time period for one rotation (T) = distance/velocity
= 2πr/v
= 2π/ω
Therefore linear and angular velocity are related by
the formula:
Linear velocity = radius of circle × angular velocity
v = rω
The angular frequency is the number of revolutions
per second
ω = θ/t
10.
A complete revolution of a circle = 2π rad
ω = 2π/t
So
Since ƒ = 1/T
Angular velocity ω = 2πƒ
11.
Questions
1. Calculate the angular velocity of an engine
operating at 2500 rpm.
2. A old fashioned record player, playing an
album. (33 rpm)
3. A satellite in geostationary orbit.
4. A satellite in low polar orbit (orbit time 90min).
5. An ultra sonic motor operating a 500 kHz
6. A motor has an angular velocity of 23 rad/s
what is its time period of rotation?
12.
Centripetal force
When we secure an object to the end of a piece
of string and swing it the tension induced on the
string is the centripetal force. This is the force
that keeps the object orbiting around the same
point.
13.
When an object is in a regular orbit the
apparent forces due to the momentum of
the object are equal to the centripetal
force.
Some people refer to this apparent
outwards force centrifugal force, because
this is due to the momentum and is not
really a force we do not use the term.
14.
Acceleration towards the centre of a circle is the
centripetal acceleration.
An object can only
accelerate if a resultant
force is acting on it.
– the centripetal force –
otherwise the object would
fly off at a tangent to the
circle (Newton’s first law of
motion).
15.
In the previous example, the moving object is
acted on by the force towards the centre of the
circle, but it does not get any closer.
The centripetal force needed to make an object
follow a circular path depends on.
(i) the mass, m of the object
(ii) the speed, ν
(iii) the radius, r of the circle
Centripetal force, F= mν2
r
17.
When an object moves in a circle, its velocity is
at a tangent to the circle. Its velocity is changing
since the direction is changing ∴ the body is
accelerating.
∆ν shows how the velocity vector changes. The
arrow shows the direction of the change in
velocity ∴ the direction of the acceleration.
18.
Since F = ma; a = F/m
a = ν2
r
ν = rω
Now
so centripetal acceleration can be calculated
using angular velocity:
a = rω 2
And centripetal force = ma
= mrω2
19.
Circular motion.
A satellite is orbiting the earth in a low earth orbit
of 60 Km and takes 120 mins to complete one
complete orbit.
Calculate the following:
Linear velocity.
Angular velocity.
Centripetal acceleration.
Radius of the earth is 6400 Km.
The satellite has a mass of 900 kg calculate the
centripetal force.
20.
Gravitation and Gravity.
When we see pictures or film of astronauts
floating we often think that they are experiencing
no gravity at all.
In fact they are being affected by the gravity of
all the things around them, to varying degrees.
Newton understood that gravity was a universal
force of attraction that affected everything, the
gravity that causes us to fall towards the ground
is the same force that keeps the moon in orbit.
21.
Newton’s law of universal gravitation states:
“Every particle in the universe attracts every
other with a force which is proportional to the
product of their masses and inversely
proportional to the square of their separation.”
From this statement we can work out that:
F ∝ m1m2 F ∝ 1/r2
and
22.
When we introduce a constant, the universal
gravitational constant, G, we get the equation:
Gm1m2
F =− 2
r
Where F is the force of gravitational attraction
between the two bodies of mass m1 and m2.
And G is the universal gravitational constant,
which is equal to 6.67 x 10-11 N m2 kg-2.
23.
Force on m2
m2
m1
Displacement of m2
Force and displacement are vector quantities.
They are in opposite directions to each other;
this introduces the –ve sign into the equation.
24.
Earth and moon.
The Earth has a mass (ME) of 5.98 x 1024 kg
The moon has a mass (MM) of 7.35 X 1022 kg.
The radius (R) of the earth is 6.38 x 106 m
The centres of gravity of moon and earth are
separated by 3.84 x 108 m.
Calculate the Force of attraction an 80 kg man
experiences at the earths surface.
Calculate the force of attraction between the moon
and the earth.
25.
Calculate the force a 100kg satellite will
experience at, 100, 200, 300, 400, 500, 600,
700, 800, 900 and 1000 Km above the earths
surface.
Plot a graph of your results.
What does the shape of the graph show?
26.
Questions
Calculate the force of attraction between two members of
the class when they are standing 2 meters apart.
– Would you feel the force?
A canoeist with a mass of 60 kg paddles past a super
tanker weighing 650,000 tonnes at a distance of 4m.
– Calculate the force expereinced.
– Would the force be felt?
A satellite orbits the earth at 700 km, its orbit is changed
so that it now orbits at a height of 500 km.
– Calculate the factor by which the force will increase.
– What will happen to the satellites speed?
27.
The universal constant of gravitation.
If we rearrange Newton’s formula shows the
significance of G.
2
Fr
G=−
m1m2
Suppose the separation of the masses is 1
metre.
The masses m1 and m2 are both 1 kg
If this is the case then G = -F.
28.
The universal gravitational constant is the
attractive force between two 1 kg masses
placed 1m apart, this force is extremely
small, hence the value of the gravitational
constant is small.
29.
Gravitational field strength
All masses produce a gravitational field. The
strength of a gravitational field at a point is
defined as the force that would be exerted on a
unit mass placed at that point.
Gravitational field strength = force per unit mass
F
g= (Linear field)
m
The units for gravitational field strength are N kg-1.
30.
We can use Newton’s law of gravitation to find
an expression for g at the surface of the earth as
the force of gravity on an mass on the surface of
the Earth is given by:
GM E m
F =− 2
r
F GM E
g= =− 2 Radial
so field
m r
The second equation gives the field strength for
a radial gravitational field, such as that produced
by a spherical mass such as a planet or moon.
31.
Calculate g for the Sun, Earth and the Moon.
The Sun’s mass is 1.989 x 1030 kg
r = 695 500 km
The Earth’s mass is 5.978 x 1024 kg
r = 6400 km
The Moon’s is 7.35 X 1022 kg.
r = 1700 km
32.
The neutral point
While travelling between two celestial bodies
you will pass through a point where the forces
due to each body are the same, this is called the
neutral point.
GM M GM E
gm = g E ∴ =
2 2
r R
2
R ME
=
2
r MM
33.
Earth-moon neutral point.
When the Apollo missions went to the moon they
passes through the neutral point when they
where 90% of the way to the moon.
The mean separation between the earth and the
sun is 1.496 x 1011m.
Work out how far from the earth the Earth – sun
neutral point is.
34.
Gravitational potential
• If an object is falling back towards Earth from a
neutral point it will gain a velocity as it loses
gravitational potential energy
1
mg∆h = mv 2
2
v = 2 g∆h
• This works well for a small value of change in
height but for large values we need to take
account of the change in gravitational field
35.
Work done in moving a mass to
infinity
• The work to move a mass from distance R to
infinity is:-
GM E m
W=
R
36.
Gravitational potential
Gravitational potential is always negative.
The potential at a point is the amount of energy needed to
move 1 kg from infinity to that point.
A distant object has zero PE
W GM E
V =− =−
m r
V = -GM/r (In some texts U = -GM/r)
37.
0
Potential
J/kg
planet Max
Distance from Centre / m
38.
As the potential is the energy needed to move
1Kg from infinity to the point in question.
We can calculate the potential energy of any
object at that point by simply multiplying the
potential by the mass of the object.
Ep = mV
39.
Gravitational potential (additional notes)
Gravitational potential is a field, this is because it
is defined at every point in space, unlike
gravitational field strength it is scalar rather than
vector.
If a mass is positioned at one point where the
potential is V1 and is moved to an other point
with a potential of V2 then the work done is equal
to: W = m(V1-V2)
W = m∆V
40.
Escape velocity.
If a mass at the earths surface can be given
enough Kinetic Energy equal to its gravitational
potential energy (= mV), it will escape
completely from the Earth’s gravitational field
(and end up at infinity). This velocity is called the
escape velocity. The escape velocity is not
dependant on the mass.
41.
We can derive the formula for determining the escape
velocity using the formulas for Gravitational potential energy
and kinetic energy.
1 2 GM E m
mv =
2 R
2GM E
v=
R
42.
Calculate the escape velocities for the Moon,
Earth and Jupiter.
Body Mass (kg) diameter (km)
Moon 7.35×1022 3476
Earth 5.97×1024 12,104
Jupiter 1.89×1027 142,800
43.
Geostaionary Satellites
Geostationary satellites travel with a speed of
3068 m/s
Calculate the radius of their orbit.
The radius of the earth is 6400 km, what is the
altitude of the satellite.
Calculate the gravitational force acting on a
150Kg satellite.
44.
International space station
The ISS is a low earth orbit satellite, it orbits the
planet 16 times a day.
Its average altitude above the earth is 333.3 km
and Its mass is 233,000 km.
Calculate the velocity of the ISS
Calculate the gravitational force acting on the
ISS.
45.
Mercury and Venus.
Planet Mass Mean R T of orbit
Mercury 3.3022×1023 kg 57.9 x 106 m 88 days
Venus 4.8685×1024 kg. 108.1 x 106 m 224.7 days
• Calculate the gravitational forces between the planets and
the Sun.
• The mass of the sun is 2 x 1030 kg.
• Calculate the velocities of the planets orbits.
• Using the formulas F=ma and F = -GMm/r2
• We have already calculated that the time period of the orbit
squared is proportional to the radius of the orbit cubed. (T2
∝ R3)
• Use this relationship to calculate the radius of Jupiter’s
orbit; Jupiter orbits the sun every 11.86 years.
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