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2002 Physics

Chapter 4
Circular motion
CHAPTER 4
circular motion: uniform circular motion using
a =
Page 30
CIRCULAR MOTION
v2
42r
=
, comparison with non-uniform circular motion using conservation
r
T2
of energy in a uniform gravitational field;
When a mass is moving in a horizontal circle with constant speed. At every instant the velocity is
changing, because the direction is constantly changing. Since the velocity is changing there must be an
acceleration, which means that there must be a net force acting.
In Physics, so far, whenever there has been a non-zero resultant force, then there has been an
acceleration in the direction of the net force, which has produced a change in speed. We now need to
consider the case when the velocity changes but the speed doesn’t. This occurs when the net force is
acting at right angles to the direction of motion.
This is circular motion, where the direction of the object is constantly changing, therefore the velocity is
changing because ‘v’ – ‘u’ is non-zero.
Consider the following 5 cases.
Case 1
F
the object will speed up, but there will not be a change of direction
v
Case 2
F
the object will slow down, but there will not be a change of direction
v
Case 3
F
the object will speed up, with a change of direction
v
Case 4
F
the object slows down, with a change of direction
v
Case 5
F
No change of speed, but a change of direction
v
In all the cases above, there is an acceleration, because there is a net force acting. But this acceleration
does not always result in a change of speed.
2002 Physics
Chapter 4
Circular motion
Page 31
UNIFORM CIRCULAR MOTION
When a constant force is applied at right angles to the direction of motion, then since the force does not
have a component parallel to the velocity, the velocity will not either increase or decrease.
Uniform circular motion results when a resultant force of constant magnitude acts normal to the motion
of the body.
vi
mass ‘m’
Consider a mass ‘m’ with a constant speed in uniform circular motion. In the position shown above, the
instantaneous velocity and the resultant force are shown as vectors.
If we look at this mass a short time later
vf
mass ‘m’
Then the change in velocity is given by vf - vi =
-vi
vf
v
The resultant force must always act radially inwardly.
Carousel Application
If you were to think about swinging a mass on the end of a piece of string in a horizontal circle, then the
force being applied by the string can only be inwards (can you push with a piece of string?).
What would happen to the mass if the string broke? It would move away tangentially, the string stops
this, so it must be providing an inward force.
2002 Physics
Chapter 4
Circular motion
Page 32
When a body travels in a circle, the force is always towards the
centre, while the velocity is tangential. The force is always at
right angles to the velocity. The force is usually provided by a
cord.
If the cord breaks the force is removed and
the circular motion ceases. The object will fly
off in the direction it was travelling at the
time of the break. If this was originally an object being
swung in a vertical circle rather than a horizontal one,
the object would move off into projectile motion. You
would have to analyse the vertical and horizontal components
of the motion separately from the point where the cord
broke.
From Newton’s laws
Fc = mac
2
2
4 r
mv
=m
 F = ma =
2
r
T
FORMULAE
A mass moving with a uniform speed in a circular path of
radius ‘r’ and with a period ‘T’ has an average speed given by
distance travelled
2r
=
time taken
T
where Fc = central force and ac = central acceleration
The vector form of this last equation is
a = -
2
4 r
2
T
(the negative sign is needed because the acceleration and radius are in the opposite directions.)
WORK DONE
There is no work done by the centripetal force, since work done = force x distance moved in the
direction of the force. In circular motion the motion is always at right angles to the force. If there was
work being done, then there would be a change in KE, from WD =  KE, but the speed does not change, so
there can‘t be any work being done.
SOME ASSOCIATED FACTS
1. If the central force is continually increased, the body will spiral in. this gives a displacement in
the direction of the force, work is done and the body speeds up.
2. If the central force is increased and then held constant, a smaller faster circular path results
3. If the force is held constant and the speed increased, the body spirals outwards.
4. reducing speed under constant force gives an inward spiral.
2002 Physics
Chapter 4
Circular motion
Page 33
NON-UNIFORM CIRCULAR MOTION
Consider a car passing through a vertical loop. At the top, the force towards the centre of the circle is
given by
At the top
2
mv
F + mg =
F
r
Here the car speed is affected by gravity.
mg
It slows down on the upward section and speeds
up on the downhill section.
Except at the top and bottom of the loop, the force
of gravity means that there is a component of
the car’s motion.
At the bottom
2
mv
F - mg =
r
F
mg
EXAMPLES OF CIRCULAR MOTION EFFECTS
Person on a swing
At the bottom of the swing, the forces on the person are the reaction from the seat and the weight
2
mv
force. R - mg =
. In this case R > mg, so the person ‘feels’ heavier.
r
Person in a car going over a hump.
v
mg – P =
mv
r
2
P = mg -
mv
r
2
The person feels lighter.
2002 Physics
Chapter 4
Circular motion
Page 34
Examples
2000
Jo is riding on a roller-coaster at a fun fair. Part of the structure is shown in Figure 8.
When Jo is at point X her velocity is 10 m s-1 in a horizontal direction, and at point Y
it is 24 m s-1 in a horizontal direction. At Y the radius of curvature is 12 m.
Figure 8
Question 15
Jo has a mass of 50 kg. What is the magnitude and direction of the net force on Jo
at point Y? Use the key (A–G) below to select the best indication of the direction.
Reproduced by permission of the Victorian Curriculum and Assessment Authority, Victoria, Australia.
Solution Q15
The net force is the centripetal force which is calculated by
242
v2
 F  50 
The magnitude of the net force is F  2.4  10 3 N
r
12
The direction of the net force is always towards the centre of the radius, therefore
the answer is C
Fm
Examiners comment Q15
The motion at the instant of point Y can be treated as uniform circular motion. Thus,
the net force must be directed towards the centre of the circle, that is, upwards in
mv 2
r
resulting in a value of 2400 N. The most common errors were in calculating the weight
the direction of arrow C. The net force can be calculated via the equation F 
mv 2
. Students need to
r
understand that ‘centripetal force’ is another name for ‘net force’; it is not a force in
its own right. Most students were able to choose C as the direction of the net force.
force mg or the weight force less the centripetal force mg -
2002 Physics
Chapter 4
Circular motion
Page 35
1999
Eddie Irvine and his Formula 1 racing car are taking a corner in the Australian Grand
Prix. A camera views the racing car head-on at point X on the bend where it is
travelling at constant speed. At this point the radius of curvature is 36.0 m. The total
mass of car and driver is 800 kg.
1999 Question 13
On Figure 7, showing the camera’s view of the racing car, draw an arrow to
represent the direction of the net force acting on the racing car at this instant.
Reproduced by permission of the Victorian Curriculum and Assessment Authority, Victoria, Australia.
Solutions Q13
The car is moving in circular motion, so the net force must be radially inwards. In this
case that is to the right.
Examiners comment Q13
For circular motion at constant speed the acceleration, and hence the net force, must
be directed towards the centre of the circle. Thus, an arrow in the horizontal
direction, and to the right, needed to be drawn on the camera view of Eddie. The point
of application of this force was not considered as essential when marking this question.
The average mark of 1.1/2 indicated that about 55% of candidates understood the
concept of uniform circular motion in these contexts.
1999 Question 14
Calculate the speed of the car.
Reproduced by permission of the Victorian Curriculum and Assessment Authority, Victoria, Australia.
2002 Physics
Chapter 4
Circular motion
800  v2
36.0
 v2 =
Page 36
Solution Q14
F = 6400 N =
mv 2
r
=
6400  36
= 288
800
 v = 16.97 m/s
Examiners comment Q14
Application of the uniform circular motion equations resulted in a speed of 17 m s -1 for
Eddie. The average score of 1.5/2 indicated that the majority of candidates were quite
confident in applying these equations. The major point of concern here was in noting the
number of students who experience difficulty with the algebraic manipulation of
expressions or in using a calculator to determine numerical values.
1999 Question 15
Explain:
• why the car needs a horizontal force to turn the corner
• where this force comes from.
Reproduced by permission of the Victorian Curriculum and Assessment Authority, Victoria, Australia.
Solution Q15
The car is moving in a circular path. For uniform circular motion there is a net force
acting. This force must act horizontally, and radially inwards.
mv 2
.
r
This force must be supplied by the friction between the car tyres and the road.
This force is given by F =
Examiners comment Q15
Students were expected to include in their explanation these points:

Uniform circular motion requires a net force acting towards the centre of the
circle.

This force must be provided from the contact with the track in both cases. For
Eddie it is the friction force between tyres and track that accelerates the car
inwards.
The average score for this question was 1.9/4, indicating only a moderate understanding
of this concept. It was disappointing to note that many students omitted to relate the
question to circular motion and addressed the question from the perspective that Eddie
would skid off the track without some force being applied. While this approach was
quite correct, these students often failed to fully address the nature of the force
required. But, by far the most difficult aspect of this question, was in describing the
origin of the force.