Essential idea: A force applied perpendicular to its displacement can result in circular motion.
6.1 Circular motion
Nature of Science
Observable universe: Observations and subsequent deductions led to the realization that the force must act
radially inwards in all cases of circular motion.
Understandings:

Period, frequency, angular displacement
and angular velocity

Centripetal force

Centripetal acceleration
Applications and skills:

Identifying the forces providing the
centripetal forces such as tension, friction,
gravitational, electrical, or magnetic

Solving problems involving centripetal force,
centripetal acceleration, period, frequency,
angular displacement, linear speed and
angular velocity

Qualitatively and quantitatively describing
examples of circular motion including cases
of vertical and horizontal circular motion
Theory of knowledge
Foucault’s pendulum gives a simple observable proof of the rotation of the Earth, which is largely
unobservable. How can we have knowledge of things that are unobservable?
Your opinion?
What force is responsible for the moon to continue
in orbit around the earth?
What force is responsible for a car to round a curve
without sliding off the road?
Angular displacement
Angular velocity
Centripetal acceleration
1.
A particle P is moving in a circle with uniform speed. Which one of the following diagrams correctly shows
the direction of the acceleration a and velocity v of the particle at one instant of time?
a
A.
a
B.
v
v
P
P
v
C.
D.
a
P
a
v
P
(1)
2.
A point mass is moving in a horizontal circle with a velocity of constant magnitude v. At one particular time,
the mass is at P. A short time later, the mass is at Q, as shown below.
v
Q
v
P
Which vector diagram correctly shows the change in velocity Δv of the mass during this time?
A.
B.
v
v
v
v
v
v
C.
D.
v
v
v
v
v
v
(1)
3.
The centripetal force that causes a car to go round a bend in the road is provided by
A.
the force produced by the car engine acting on the wheels.
B.
the friction between the tyres and the road.
C.
the weight of the car.
D.
the force exerted by the driver on the steering wheel.
(1)
4.
The centripetal force F acting on a particle of mass m that is travelling with linear speed v along the arc of a
circle of radius r is given by
A.
F=
v2
.
mr
B.
F = mv2r.
C.
F = mr2v.
D.
F=
mv 2
r
.
(1)
5.
A brick is placed on the surface of a flat horizontal disc as shown in the diagram below. The disc is rotating at
constant speed about a vertical axis through its centre. The brick does not move relative to the disc.
Which of the diagrams below correctly represents the horizontal force or forces acting on the brick?
(1)
6.
A satellite orbits the Earth at constant speed as shown below.
satellite
Earth
(a)
Draw on the diagram
(i)
an arrow labelled F to show the direction of the gravitational force of the Earth on the satellite.
(ii)
an arrow labelled V to show the direction of the velocity of the satellite.
(2)
(b)
Although the speed of the satellite is constant, it is accelerating. Explain why it is accelerating.
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.....................................................................................................................................
.....................................................................................................................................
(2)
(c)
Discuss whether or not the gravitational force does work on the satellite.
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(3)
(Total 7 marks)
7.
This question is about circular motion.
A linear spring of negligible mass requires a force of 18.0 N to cause its length to increase by 1.0 cm.
A sphere of mass 75.0 g is attached to one end of the spring. The distance between the centre of the sphere M
and the other end P of the unstretched spring is 25.0 cm, as shown below.
P
M
25.0 cm
The sphere is rotated at constant speed in a horizontal circle with centre P. The distance PM increases to 26.5
cm.
(a)
Explain why the spring increases in length when the sphere is moving in a circle.
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.....................................................................................................................................
(2)
(b)
Determine the speed of the sphere.
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(4)
(Total 6 marks)
8.
This question is about the kinematics and dynamics of circular motion.
(a)
A car goes round a curve in a road at constant speed. Explain why, although its speed is constant, it is
accelerating.
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(2)
In the diagram below, a marble (small glass sphere) rolls down a track, the bottom part of which has been bent
into a loop. The end A of the track, from which the marble is released, is at a height of 0.80 m above the
ground. Point B is the lowest point and point C the highest point of the loop. The diameter of the loop is 0.35
m.
A
marble
C
0.80 m
0.35 m
ground
B
The mass of the marble is 0.050 kg. Friction forces and any gain in kinetic energy due to the rotating of the
marble can be ignored. The acceleration due to gravity, g = 10 ms–2.
Consider the marble when it is at point C.
(b)
(i)
On the diagram opposite, draw an arrow to show the direction of the resultant force acting on the
marble.
(1)
(ii)
State the names of the two forces acting on the marble.
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...........................................................................................................................
(2)
(iii)
Deduce that the speed of the marble is 3.0 ms–1.
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(3)
(iv)
Determine the resultant force acting on the marble and hence determine the reaction force of the
track on the marble.
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(4)
(Total 12 marks)
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Essential idea: A force applied perpendicular to its displacement can result in circular motion.
6.1 Circular motion
Nature of Science
Laws: Newton’s law of gravitation and the laws of mechanics are the foundation for deterministic classical
physics. These can be used to make predictions but do not explain why the observed phenomena exist.
Understandings:

Newton’s law of gravitation

Gravitational field strength
Applications and skills:

Describing the relationship between
gravitational force and centripetal force

Applying Newton’s law of gravitation to the
motion of an object in circular orbit around
a point mass

Solving problems involving gravitational
force, gravitational field strength, orbital
speed and orbital period

Determining the resultant gravitational field
strength due to two bodies
Theory of knowledge
The laws of mechanics along with the law of gravitation create the deterministic nature of classical
physics. Are classical physics and modern physics compatible? Do other areas of knowledge also have a
similar division between classical and modern in their historical development?
Your opinion?
Law of Universal Gravitation and Field Strength
Law of Universal Gravitation:
Every particle in the universe attracts every other particle in the universe with a force (F) that is proportional to
the product of their masses ( m1 and m2 ) and inversely proportional to the square of the distance (r) between
them. This force acts along the line joining the two particles.
F G
m1 m2
r2
where G  6.67 1011 N m 2 kg 2
*This is only valid if we assume the following: the two masses are p ________ m___________ and the
s_____________ of the two masses is much, much g_____________ than their r____________
**Please note that the mass in this formula determines the gravitational force between the two bodies and is
thus referred to as gravitational mass. This is conceptually different than inertial mass which represents a
resistance to force, or inertia (recall F = ma). So far, all experiments indicate that these two types of masses are
equal for any body!
Attraction between two people:
A 50 kg person and a 75 kg person are sitting on a bench such that their centers are about 50 cm apart. Find
the gravitational force each exerts on the other.
Spacecraft in orbit:
What is the force of gravity acting on a 2000 kg spacecraft when it orbits two Earth radii from the Earth’s centre.
The earth’s radius is about 6380 km, and its mass is
Note: The spacecraft is twice as far from the earth’s center as when it is on the surface of the earth. You
should notice that the force of gravity decrease as the square of the distance. So if you double the distance, the
force should be one-fourth as big. This means the force of gravity on the spacecraft in orbit will be ¼ of its
weight on earth!
Gravitational Field Strength
We can express the acceleration due to gravity on earth (or any planet) in terms of the universal gravitation law.
We can represent this as gravitational field strength. This refers to the gravitational force per unit mass at a
particular location.
On the earth’s surface we can write, the gravitational force exerted on a mass m , a distance r away from the
earth (mass me ) is given by:
m me
F G 2
r
Using Newton’s second law, we can determine the acceleration, g , of the mass m .
g
g G
G
F
m
so,
g
m me
re 2
m
me
Gravitational field strength on earth's surface
re 2
In general, at any distance r from a body of mass M the gravitational field strength is:
g G
M
r2
Gravitational Field Strength on Mount Everest:
Estimate the effective value of g on the top of Mt. Everest, 8848 m above the earth’s surface. That is, what is
the acceleration due to gravity of objects allowed to fall freely at this altitude?
We must add the 8848 m to the radius of the earth for our calculation.
1.
Newton’s law of gravitation for the force F between two point objects of masses M and m, separated by a
distance d may be written as
Fd2  Mm.
The expression may also be used for the force of attraction between the Sun and the Earth, although they are
not point masses. This is because
A.
the gravitational constant G is not involved in the expression.
B.
the force between the Sun and the Earth is very large.
C.
the separation of the Sun and the Earth is much greater than their radii.
D.
the mass of the Earth is much less than the mass of the Sun.
(1)
2.
In Newton’s universal law of gravitation the masses are assumed to be
A.
extended masses.
B.
masses of planets.
C.
point masses.
D.
spherical masses.
(1)
3.
Gravitational field strength at a point may be defined as
A.
the force on a small mass placed at the point.
B.
the force per unit mass on a small mass placed at the point.
C.
the work done to move unit mass from infinity to the point.
D.
the work done per unit mass to move a small mass from infinity to the point.
(1)
4.
The mass of Mars is approximately 0.1 times the mass of Earth and its diameter is approximately 0.5 times
that of Earth.
What is the approximate gravitational field strength on the surface of Mars?
A.
2N kg–1
B.
4N kg–1
C.
25N kg–1
D.
50N kg–1
(1)
5.
The acceleration of free fall of a small sphere of mass 5.0 × 10–3 kg when close to the surface of Jupiter is 25
ms–2. The gravitational field strength at the surface of Jupiter is
A.
2.0 × 10–4 N kg–1.
B.
1.3 × 10–1 N kg–1.
C.
25 N kg–1.
D.
5.0 × 103 N kg–1.
(1)
6.
The Earth is distance RM from the Moon and distance RS from the Sun. The ratio
gravitatio nal field strength at the Earth due to the Moon
gravitatio nal field strength at the Earth due to the Sun
is proportional to which of the following?
A.
B.
C.
D.
RM
2
RS
2
RM
RS
RS
2
RM
2
RS
RM
(1)
7.
Planet X has radius R and mass M. Planet Y has radius 2R and mass 8M.
Which one of the following is the correct value of the ratio
gravitational field strength at surface of planet X
?
gravitational field strength at surface of planet Y
A.
4
B.
2
C.
1
2
D.
1
4
(1)
8.
This question is about gravitational fields.
(a)
Define gravitational field strength.
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...................................................................................................................................
...................................................................................................................................
(2)
(b)
The gravitational field strength at the surface of Jupiter is 25 N kg–1 and the radius of Jupiter is 7.1 
107 m.
(i)
Derive an expression for the gravitational field strength at the surface of a planet in terms of its
mass M, its radius R and the gravitational constant G.
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(2)
(ii)
Use your expression in (b)(i) above to estimate the mass of Jupiter.
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(2)
(Total 6 marks)
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Answers:
1.
C
[1]
2.
C
[1]
3.
B
[1]
4.
D
[1]
5.
D
[1]
6.
(a)
(b)
(c)
(i)
from satellite towards centre of Earth;
(ii)
tangent to circle at satellite in correct direction;
(no labels, [1 max])
2
direction of motion is changing / force acts on satellite;
and changing direction means changing velocity / any further detail;
2
work done is product of force and distance moved in direction of force;
force is always normal to direction of motion;
hence no work done;
(accept argument based on changes in Ek and Ep)
3
[7]
7.
(a)
for circular motion, force required towards centre of circle / centripetal force;
this provided as a result of extension of the spring;
2
Do not give credit where candidate implies that the spring is pulled outwards
by a force.
(b)
force produced by spring = 1.5 × 18 = 27 N;
mv 2
use of F =
;
r
(0.075  v 2 )
27 =
;
0.265
v = 9.77 m s–1;
4
[6]
8.
(a)
(b)
Look for an answer on the following lines.
the direction of the car is changing;
hence the velocity of the car is changing;
or
since the direction of the car is changing;
a force must be acting on it, hence it is accelerating;
(i)
arrow pointing vertically downwards;
(ii)
weight;
2 max
1
Do not penalize the candidate if they state “gravity”.
normal reaction;
2
Do not penalize the candidate if they state “push of the track on the marble”.
(iii)
loss in PE = 0.05 × 10 × (0.8 – 0.35);
= gain in KE = 12 mv2;
to give v = 3.0 m s–1;
or
use of v = 2 gh to give v = 4.0 m s–1 at point B;
and then use of v2 – u2 = 2gh with v = 4.0 m s–1 and h = 0.35 m;
to get u = 3.0 m s–1;
3 max
Do not penalize the candidate if g = 9.8 m s–2 is used.
(iv)
recognize that resultant force =
(0.05  9.0)
= 2.6 N;
0.175
mv 2
N=
– mg;
r
= 2.6 – 0.5 = 2.1 N;
mv 2
;
r
=
4
[12]
Solutions – Gravitational Fields
1.
C
[1]
2.
C
[1]
3.
B
[1]
4.
B
[1]
5.
C
[1]
6.
C
[1]
7.
C
[1]
8.
(a)
(b)
the force exerted per unit mass;
on a point (small) mass;
(i)
use of g 
Mm
F
and F  G 2 ;
m
R
combine to get g  G
(ii)
2
M
;
R2
2
gR 2
M
;
G
substitute to get M = 1.9  1027 kg;
2
[6]