Physics Unit 3 2009
6. Circular Motion
1 of 11
6. Circular motion
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.
Consider a mass is moving in a horizontal circle with constant speed. Remember that velocity is a
vector, and has a magnitude and a direction. When an object is traveling in circular motion its
speed remains the same, but its velocity is constantly changing because the direction of motion of
the object is changing. Since the velocity is changing the object is accelerating, so there must be
a net force acting.
Physics Unit 3 2009
6. Circular Motion
2 of 11
The Direction of the Force
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.
It is not possible to tie a piece of string to an object
and push it along. The only way to move an object
using a piece of string is to pull it along. The force is
directed along the string. Thus the string indicates
the direction of the force on the object.
Physics Unit 3 2009
6. Circular Motion
3 of 11
Imagine that you have a set of keys tied to a string and you are swinging them around your head.
The string indicates the direction of the force, which is at all times towards the centre of the circle,
i.e. radially inwards.
When a body travels in a circle, the force (and
acceleration) is towards the centre while the velocity is
tangential. The force is always at right angles to the
velocity. The force is sometimes provided by a string or
rope, but it could be supplied by friction (cars going around
corners, or people running on circular paths) or gravity (the
moon and other satellites around the earth).
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 the object were originally
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.
The Size of the Force
Now that we understand the direction of the force necessary to keep an object in circular motion,
we can consider the size of the force. Imagine that you are twirling an object attached to a piece
of string over your head. The string provides the force that keeps the object in a circular orbit.
What are the factors that determine the size of this force, i.e. the tension in the string?
If you swing the object around at a faster rate that the tension force will increase. It is also easy to
imagine that if you had a heavier object on the end of the string, it would require a larger tension
force that a lighter object in the same orbit. The length of the string also affects the size of the
force.
The formula for the magnitude of a net force necessary to keep an object in circular motion is:
F=
mv 2
r
Where F is the magnitude of the force, m is the mass of the object in uniform circular motion, v is
the magnitude of the velocity of the object and r is the radius of the orbit.
Physics Unit 3 2009
6. Circular Motion
4 of 11
Forces of Incorrect Magnitude
Consider what you happen if the force on the object was not the correct magnitude to keep the
object in a circular orbit, or if the magnitude of the force was changed. Think about the following
statements, and see if you can understand them.
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.
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
F + mg =
mv2
Here the car speed is affected by gravity.
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
F - mg =
mv
F
r
mg
F
2
mg
r
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
force. R - mg =
mv
r
2
. In this case R > mg, so the person ‘feels’ heavier.
Physics Unit 3 2009
6. Circular Motion
5 of 11
Person in a car going over a hump.
v
mg – P =
mv
r
2
P = mg -
mv
r
2
The person feels lighter.
Other Formulae
Several other formulae for circular motion can also be derived. The period (T) of an orbit is the
time it takes for an object to complete one full circuit. We can the fact that the speed of the object
is v and the distance the object travels during one orbit is the circumference of the circle, 2r to
derive the formula
speed =
distance 2πr
=
time
T
mv 2
, but
r
2πr
v=
, so
T
4π 2rm
F=
T2
F=
This result can be used to write an equation for the force in terms of the
period.
Equations for the magnitude of the acceleration can be obtained by
dividing the force equations by the mass m.
Physics Unit 3 2009
6. Circular Motion
6 of 11
Examples
2001
The diagram below shows a section of a roller coaster track, with a roller coaster car travelling at constant
speed from the left.
2001 Question 10
Which of the arrows (A–H) best indicates the direction of the net force on the roller coaster car at the lowest
point, assuming that friction and air resistance cannot be neglected?
2000
Jo is riding on a roller-coaster at a fun fair. Part of the structure is shown below.
When Jo is at point X her velocity is 10 ms-1 in a horizontal direction, and at point Y it is 24 ms-1 in a
horizontal direction. At Y the track has a radius of curvature of 12 m.
2000 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.
Physics Unit 3 2009
6. Circular Motion
7 of 11
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.
The magnitude of the horizontal force on the car is 6400 N.
1999 Question 14
Calculate the speed of the car.
1999 Question 15
Explain:
• why the car needs a horizontal force to turn the corner
• where this force comes from.
1998
A cart on a roller coaster rolls down a track as shown below. The upper section of the track is straight and
the lower section forms part of a circle. The effect of friction can be ignored in the following questions.
1998 Question 13
On the sketch of the roller coaster, draw one arrow which shows the direction of the net force on the cart
when it is at the point X.
1998 Question 14
On the sketch of the roller coaster, draw one arrow which shows the direction of the net force on the
cart when it is at the point Y.
The mass of the cart is 500 kg, and at point X it is travelling at a speed of 10 m s-1.
1998 Question 15
Calculate the speed of the cart at the point Y.
1998 Question 16
Calculate the net force acting on the cart at the point Y if the radius of curvature of the track at this point is
8.0 m.
Physics Unit 3 2009
6. Circular Motion
8 of 11
1997
A traffic engineer needs to put a maximum speed-limit sign on a dangerous bend. If a car travels too fast the
sideways frictional force will not be large enough to keep it on the road.
The maximum sideways frictional force that the tyre-road combination can produce without slipping is
estimated to be 2800 N for a car of mass 1200 kg.
1997 Question 10
If the bend is modelled on the arc of a circle of radius 90.0 m as shown above, calculate the maximum speed
that a car can have and remain on the road without slipping.
Physics Unit 3 2009
6. Circular Motion
9 of 11
Solutions
2001 Q10 solution
At the lowest point, the roller coaster is in the bottom part of a section of circular
Opposing forces
motion. This means that the force causing this acceleration is directly up. If we
now take friction and air resistance into consideration, (these will both be
Central
opposing the motion) then the net force is given by the sum of the two vectors.
force
E
Examiners comment Question 10 (0.69/2)
Arrow E best indicates the direction of the net force on the roller coaster car. The motion of the car at this
point can be considered to be made up of both a uniform circular motion component (force towards the
centre of the circle) and a friction component (force in the opposite direction to the motion).
2000 Q15 solution
The net force is the centripetal force which is calculated by
F=m
v2
r
242
12
The magnitude of the net force is
F = 2.4 ×103 N
F = 50 ×
The direction of the net force is always towards the centre of the radius, therefore the answer is C
1999 Q14 solution
17 m/s
F
mv 2
800  v2
=
r
36.0
6400  36
v2 =
800
= 288
v = 16.97 m/s
= 6400 N =
1999 Q15 solution
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
F =
This force is given by
.
r
This force must be supplied by the friction between the car tyres and the road.
Physics Unit 3 2009
6. Circular Motion
10 of 11
1998 Q13 solution
end of ride
X
mg
Y
If the effects of friction are to be ignored, then the only forces acting are the weight force and the normal
reaction from the track.
Net force acting
The cart is restricted to move in one direction,  the acceleration must be in this direction. We know that the
cart is accelerating because it is gaining KE (by losing PE), so its speed is increasing, hence it is
accelerating.
 the net force acting is in the direction of the acceleration, parallel to, and down the slope.
1998 Q14 solution
end of ride
X
Y
At the point Y, the cart is undergoing uniform circular motion. This means that the net force must be acting
radially inwards.
 the net force is vertically upwards.
Physics Unit 3 2009
6. Circular Motion
1998 Q15 solution
16 m/s
The total energy must be constant.  the sum of PE and KE = constant.
At X TE (total energy)
= mgh + ½ mv2
= 500 × 9.8 × 7.8 + ½ × 500 × 102
= 63 220N
At Y TE
63 220 = ½ mv2
= ½ × 500 × v2
63220
v2 =
250
v = 15.9 m/s
1998 Q16 solution
1.6 × 104N
F = ma =
mv 2
500  15.92
=
8.0
r
= 15805 N
1997 Q10 solution
v = 14.5 m/s
F = ma =
mv 2
r
2800 =
1200  v2
90
2800  90
1200
 v2 = 210
 v = 14.5 m/s
 v2 =
11 of 11