Download Examples of circular motion effects

Fizz sicks Unit 3 2015
6. Circular Motion
1 of 25
6. Circular motion
Study Design
• analyse the uniform circular motion of an object moving in a horizontal plane (Fnet = mv2/R) such
as a vehicle moving around a circular road; a vehicle moving around a banked track; an object on
the end of a string;
• apply Newton’s second law to circular motion in a vertical plane; consider forces at the highest
and lowest positions only; (except in 2013, when you were required to know the direction of the
acceleration at all points in the vertical circle).
In Physics, so far, whenever there has been a non-zero resultant force, then there has been
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 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.
Fizz sicks Unit 3 2015
6. Circular Motion
2 of 25
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.
Fizz sicks Unit 3 2015
6. Circular Motion
3 of 25
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 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.
Fizz sicks Unit 3 2015
6. Circular Motion
4 of 25
Forces of Incorrect Magnitude (not on the course, but interesting)
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 vertical circle motion
Consider a car passing through a vertical loop. The force towards the centre of the circle is given
by F + mg =
And F - mg =
mv 2
at the top.
r
mv 2
at the bottom.
r
Between these points the car’s speed is affected by gravity. It slows
downs on the upward section and speeds up on the downhill
section.
F
mg
F
Except at the top and bottom of the loop, the force due to gravity
provides a component of acceleration in the direction of the car’s
motion.
The car will be travelling faster at the bottom hence it has more KE
and less PE than at the top.
mg
mv 2
mv 2
= ma at the top, and F - mg =
= ma at the bottom.
r
r
The force ‘F’ will differ from top to bottom, as this is not a case of uniform circular motion.
Using ΣF = ma gives F + mg =
Direction of the acceleration in vertical circles
The direction the acceleration in vertical circles always remains Radially Inwards. So the
acceleration (net force) will point towards the centre from any point on the vertical circle.
Examples of circular motion effects
Person on a swing
Fizz sicks Unit 3 2015
6. Circular Motion
5 of 25
At the bottom of the swing, the forces on the person are the reaction from the seat and the weight
mv 2
. In this case R > mg, so the person ‘feels’ heavier.
r
Person in a car going over a hump.
R
force. R - mg =
v
mg
mg – R =
mv 2
r
R = mg -
mv 2
r
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
distance 2πr
speed =
=
time
T
mv 2
F=
, but
r
This result can be used to write an equation for the force in terms
2πr
of the period.
v=
, so
T
Equations for the magnitude of the acceleration can be obtained
4π2r m
F=
by dividing the force equations by the mass m.
T2
The diagram below shows a section of a roller coaster track, with a roller coaster car travelling at
constant speed from the left.
Example 1 2001 Question 10
(2 marks)
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?
Fizz sicks Unit 3 2015
6. Circular Motion
6 of 25
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.
Example 2 2000 Question 15
(3 marks)
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.
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.
Example 3 1999 Question 14
(2 marks)
Calculate the speed of the car.
Fizz sicks Unit 3 2015
6. Circular Motion
7 of 25
Example 4 1999 Question 15
(4 marks)
Explain:
• why the car needs a horizontal force to turn the corner
• where this force comes from.
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.
Example 5 1998 Question 13
(1 mark)
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.
Example 6 1998 Question 14
(1 mark)
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.
Example 7 1998 Question 15
(2 marks)
Calculate the speed of the cart at the point Y.
Example 8 1998 Question 16
(2 marks)
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.
Fizz sicks Unit 3 2015
6. Circular Motion
8 of 25
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.
Example 9 1997 Question 10
(2 marks)
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.
A mass of 0.10 kg is attached to a light string of length 3.5 m and caused to move in a horizontal
circle at constant speed. The situation is pictured below.
The string makes an angle of 450 to the ground and the acceleration of gravity is 10 ms-2.
Example 10 1985 Question 25
What is the tension in the string?
(1 mark)
Example 11 1985 Question 26
(1 mark)
At what speed is the mass moving.
Fizz sicks Unit 3 2015
6. Circular Motion
9 of 25
A light spring has a force-length relationship as shown below.
Example 12 1984 Question 15
What is the spring constant?
(1 mark)
Example 13 1984 Question 16
(1 mark)
How much work is necessary to change the length of the spring from 0.75 m to 1.0 m?
This spring is now placed on a horizontal frictionless surface as shown below. One end is fixed and
can pivot about the point A; the other end has a mass of 2.0 kg attached to it. The mass is set
moving at a constant speed in a circle of radius 1.0 m.
Example 14 1984 Question 17
What is the speed of the mass?
(1 mark)
Fizz sicks Unit 3 2015
6. Circular Motion
10 of 25
Two masses, of 1.0 kg and 5.0 kg, are attached to a light rod and suspended from a string as
shown.
When the rod is horizontal, ie. The masses are balanced, the length d is 0.4 m.
Example 15 1982 Question 8
What is the tension in the string?
(1 mark)
The horizontal rod is now set into motion so that both masses rotate about the vertical string with a
frequency of 2 revolutions per second.
Example 16 1982 Question 9
What is the speed of mass X?
(1 mark)
Example 17 1982 Question 10
(1 mark)
acceleration of Y towards string
What is the value of the ratio
acceleration of X towards string
Fizz sicks Unit 3 2015
6. Circular Motion
11 of 25
P is a frictionless puck travelling in a horizontal circle at constant speed at the end of a string. Q is
a car travelling at constant speed on a horizontal circular track.
When they are in the position shown in the diagram, which of the vectors A, B, C, D or E best
represents:
Example 18 1981 Question 17
i
the acceleration of P?
ii
the resultant force on P?
Example 19 1981 Question 18
i
the acceleration of Q?
ii
(1 mark)
(1 mark)
the resultant force on Q?
An object is travelling in a horizontal circle of radius 0.20 metre. A student takes a multi-flash
photograph of this object, using a flash which fires every 0.125 second. The following diagram
represents this photograph.
Example 20 1980 Question 10
What is the speed of the object?
(1 mark)
Example 21 1980 Question 11
(1 mark)
What is the acceleration of the object towards the centre of the circle?
Fizz sicks Unit 3 2015
6. Circular Motion
12 of 25
A mass of 0.5 kg, tied to the end of a string of length 0.50 m, is travelling in a circular path at
uniform speed on a horizontal frictionless table.
Example 22 1979 Question 12
(1 mark)
What is
(i)
the magnitude,
(ii)
the direction, of the force of the table on the mass?
Example 23 1979 Question 13
(1 mark)
While the mass is travelling at a uniform speed
A. its kinetic energy and momentum are both constant
B. its kinetic energy is constant but its momentum is changing
C. its kinetic energy is changing but its momentum is constant
D. its kinetic energy and momentum are both changing.
Example 24 1979 Question 14
(1 mark)
If the string breaks when stretched with a force just greater than 4.0 x 102 N, what is the maximum
possible speed it can have?
Fizz sicks Unit 3 2015
6. Circular Motion
13 of 25
A wooden horse and a wooden elephant are fixed to a merry-go-round platform. Each travels in a
horizontal circular path at constant speed. The horse has a mass of 75 kg and is travelling in a
circle of radius 2.0 m. The elephant has a mass of 150 kg and travels in a circle of radius 6.0 m.
What is the value of each of the following ratios?
Example 25 1977 Question 5
(1 mark)
frequency of revolution of the elephant
frequency of revolution of the horse
Example 26 1977 Question 6
speed of the elephant
(1 mark)
speed of the horse
Example 27 1977 Question 7
net force on the elephant
(1 mark)
net force on the horse
Example 28 1977 Question 8
(1 mark)
kinetic energy of the elephant
kinetic energy of the horse
Fizz sicks Unit 3 2015
6. Circular Motion
14 of 25
A 0.10 kg mass is attached 0.15 m from the centre of a turntable. The turntable rotates at 0.5
revolutions per second. Light shines horizontally at the turntable and the shadow of the mass
moves in a horizontal straight line on a screen. X and Y represent the extreme positions of the
shadow.
Example 29 1976 Question 30
(1 mark)
What is the magnitude of the net force acting on the mass?
Example 30 1976 Question 31
(1 mark)
How much work is done on the mass during one complete revolution?
Example 31 1976 Question 32
(1 mark)
What is the magnitude of the acceleration at X?
Example 32 1976 Question 33
(1 mark)
What is the speed of the shadow at the mid-point of its path?
Fizz sicks Unit 3 2015
6. Circular Motion
15 of 25
Two spheres of equal mass are attached to a string of length 1.00 m as in the figure. The string
and the spheres are whirled in a horizontal circle about O at a constant rate.
Example 33 1972 Question 15
(1 mark)
speed of sphere 2
What is the ratio
?
speed of sphere 1
Example 34 1972 Question 16
(1 mark)
acceleration of sphere 2
What is the ratio
?
acceleration of sphere 1
Example 35 1972 Question 17
(1 mark)
tension in string between spheres 1 and 2
What is the ratio
?
tension in string between spheres 1 and O
Fizz sicks Unit 3 2015
6. Circular Motion
16 of 25
A small car travels in a circle of radius 10.0 m at a constant speed. The first figure shows the car
from above and the second figure shows the car from behind.
Example 36 2003 Question 11
(2 marks)
In the position X, shown above, which of the arrows (A. – I.) best shows the direction of the net
force on the car?
Example 37 2003 Question 12
(2 marks)
On the second figure, draw arrows to show all the separate forces acting on the car at the
position X, ignoring air resistance. (You must show both the direction and point of application of
each separate force.)
Fizz sicks Unit 3 2015
6. Circular Motion
17 of 25
Example 38 2003 Question 13
(2 marks)
In the position X, shown in the second figure, which of the arrows (A. – H.) best shows the
direction of the force that the tyre exerts on the road.
Non-uniform circular motion
A ’loop the loop’ toy is designed as shown in the diagram below. A toy car of mass m can be
placed on the track at various heights above the ground so that, when released, it enters the loop
which has a radius r. If the height is large enough, the car will complete the loop without falling
away from the track.
In the questions below ignore the effects of friction.
Example 39 1985 Question 31
(1 mark)
Write an expression, in terms of the symbols g and r, for the minimum speed, v, that the car can
have at the point P if it is not to lose contact with the track.
Example 40 1985 Question 32
(1 mark)
Which of the expressions (A – E) below gives the minimum height from which the car can be
released if it is to just loop the loop?
A.
9r
4
B.
5r
2
C.
2r
D.
r
E.
r
2
Fizz sicks Unit 3 2015
6. Circular Motion
18 of 25
A car travels at increasing speed around a circular section of horizontal road of radius 50 m. When
the car is at P, it is travelling south with a speed of 10 ms-1. Its speed is increasing at the rate of
2.0 ms-2.
Example 41 1983 Question 18
(1 mark)
Which arrow in the key best indicates the direction of the car’s total acceleration at the instant the
car is a P?
Example 42 1983 Question 19
(1 mark)
What is the magnitude of the component of the car’s acceleration in the westerly direction at the
instant the car is at P?
Example 43 1983 Question 20
(1 mark)
What is the magnitude of the acceleration of the car at the instant the car is at P?
A pendulum bob suspended from P is released from X, and swings through midpoint O to Y.
Example 44 1980 Question 23
(1 mark)
When it is passing through O, the tension in the string is
A.
B.
C.
D.
E.
zero
greater than the weight of the pendulum bob
equal to the weight of the pendulum bob
slightly less than the weight of the pendulum bob
about half the weight of the pendulum bob
Fizz sicks Unit 3 2015
6. Circular Motion
19 of 25
Example 45 1980 Question 24
(1 mark)
The pendulum bob is now released from X’, such that the string makes an angle 2 with the
vertical. The tension in the string as the bob passes through O will now be
A.
B.
C.
D.
E.
double what it was before
greater than what it was before, but not double
the same as it was before
slightly less that it was before
half of what it was before
Fizz sicks Unit 3 2015
6. Circular Motion
Solutions
Example 1 2001 Question 10, 35%
At the lowest point, the roller coaster is in the
bottom part of a section of circular motion.
This means that the force causing this
acceleration is directly up. If we now take
friction and air resistance Opposing forces
into consideration, (these
will both be opposing the
Central
motion) then the net force is
force
given by the sum of the two
vectors.
E
(ANS)
Examiners comment Question 10
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).
Example 2 2000 Question 15, 52%
The net force is the centripetal force which is
calculated by
v2
F=m
r
242
F = 50×
12
The magnitude of the net force is
F = 2.4 ×103 N
The direction of the net force is always towards
the centre of the radius.
C
(ANS)
Example 3 1999 Question 14, 75%
ΣF = 6400 N =
=
v2 =
mv 2
r
800 × v 2
36.0
6400 × 36
800
= 288
 v = 17 m/s
(ANS)
20 of 25
Example 4 1999 Question 15, 48%
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
This force is given by ΣF =
.
r
This force must be supplied by the friction
between the car tyres and the road.
Example 5 1998 Question 13, 43%
end of
ride
X
m
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.
Fizz sicks Unit 3 2015
6. Circular Motion
Example 6 1998 Question 14, 65%
21 of 25
Example 10 1985 Question 25
There are two forces acting on the 0.01 kg
mass, the tension in the string and the weight.
Tension
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.
Example 7 1998 Question 15, 25%
The total energy must be constant.
 the sum of PE and KE = constant.
At X TE (total energy)
= mgh + ½ mv2
= 500 × 10 × 7.8 + ½ × 500 × 102
= 64 000 J
At Y TE
64 000 = ½ mv2
= ½ × 500 × v2
63220
v2=
250
v
= 16 m/s
(ANS)
Example 8 1998 Question 16, 50%
F = ma
mv 2
=
r
500×15.92
=
8.0
= 15805 N
(ANS)
Example 9 1997 Question 10, 70%
F = ma
mv 2
=
r
1200× v 2
2800 =
90
2800
× 90
 v2 =
1200
2
 v = 210
 v = 14.5 m/s
(ANS)
Weight
Tsin450 = mg
0.01×10
T=
sin450
 T = 1.4 N (ANS)
Example 11 1985 Question 26
The net force acting is radially inwards. Since
the angle is 450, it will have the same
magnitude as the weight. (Because the
horizontal and vertical components of the
tension are the same.)
mv 2
 mg =
r
To find r, use r = 3.5cos450
= 2.47
0.01 v 2
So
 0.01 x 10 =
2.47
2
 v = 10 x 2.47
 v = 5.0 ms-1
(ANS)
Example 12 1984 Question 15
The spring constant is the gradient of the forceextension graph.
50 - 0
k=
1.0 - 0.5
k = 100 Nm-1
(ANS)
Example 13 1984 Question 16
The work done is the area under the forceextension graph.
When the length of the spring was 0.75 m the
force was 25 N.
The area required is the area from 0.75 m to
1.0 m.
This trapezium has an area given by
WD = ½ (25 + 50) x (1.0 - 0.75)
= 9.4 J
(ANS)
Fizz sicks Unit 3 2015
6. Circular Motion
Example 14 1984 Question 17
When the spring is extended to 1.0 m, the force
acting on it is 50 N
mv 2
Using F =
gives
r
2.0  v 2
50 =
1.0
 v2 = 25
 v = 5.0 ms-1
(ANS)
Example 15 1982 Question 8
The net force acting down is the combined
weight of both masses. This is opposed by the
tension in the string.
 T = (1.0 + 5.0) x 10
= 60 N
(ANS)
Example 16 1982 Question 9
The speed is given by distance travelled over
time taken. As the frequency is 2 revolutions
per second the period = 0.5 sec.
2 r
s=
, where r = 2.0 m
0.5
2 × 2
s=
0.5
 s = 25 ms-1
(ANS)
Example 17 1982 Question 10
X is exerting the same force on Y as Y is
exerting on X.
 myay = mxax
 5ay = 1ax
ay

= 0.2
(ANS)
ax
Example 18 1981 Question 17
i
P is moving in circular motion.
 net acceleration is radially inwards
A
(ANS)
ii
P is moving in circular motion.
 net force is radially inwards
A
(ANS)
22 of 25
Example 19 1981 Question 18
i
Q is moving in circular motion.
 net acceleration is radially inwards
A
(ANS)
ii
Q is moving in circular motion.
 net force is radially inwards
A
(ANS)
Example 20 1980 Question 10
There are 4 time intervals while the object
travels through a quarter circle.
 It takes 4 x 0.125 = 0.5 sec to travel a
quarter circle.
To complete the entire circle it will take 2 sec.
distance
 speed =
time
=
2× π × 0.20
2
= 0.63 ms-1
(ANS)
Example 21 1980 Question 11
v2
a=
r
a=
0.6284 2
0.2
 a = 1.97
 a = 2.0 ms-2
(ANS)
Example 22 1979 Question 12
(i)
The table is a frictionless surface,
so the only force of the table on the
mass can be vertical.
 The normal reaction has the
same magnitude as the weight.
 Fnormal = mg
= 0.50 x 10
=5N
(ANS)
(ii)
The direction is UP
(ANS)
Example 23 1979 Question 13
Kinetic energy is not a vector, therefore its
direction is not required.
Momentum is a vector, therefore you need a
direction to specify it completely.
Since the mass is moving in circular motion, its
direction is constantly changing.
 KE is constant and “P” is changing
 B (ANS)
Fizz sicks Unit 3 2015
6. Circular Motion
Example 24 1979 Question 14
mv 2
Use F =
r
0.5  v 2
 400 =
0.5
2
 v = 400
 v = 20 ms-1
(ANS)
Example 25 1977 Question 5
Both the horse and the elephant will do the
same number of revolutions every second.
1
(ANS)
Example 26 1977 Question 6
We don’t know how long it takes to complete a
revolution, but we know that in the time it takes
to complete one revolution the distance
travelled by the elephant will be 3 times the
distance travelled by the horse.
This comes from distance travelled = 2 π r.
r for the elephant is 6, compared with 2 for the
horse.
speed of the elephant

=3
(ANS)
speed of the horse
Example 27 1977 Question 7
me v e2
net force on the elephant
re

=
mh v h2
net force on the horse
rh
me v 2erh
mh v h2re
150 2 2
=
3 
75
6
=6
=
Example 28 1977 Question 8

kinetic energy of the elephant
kinetic energy of the horse
me v e2
mh v h2
150 2
=
3
75
= 18
(ANS)
=
1
m e v 2e
2
=
1
mh v h2
2
23 of 25
Example 29 1976 Question 30
mv 2
Net force =
r
Since the turntable rotates at 0.5 revolutions
per second, (i.e it takes 2 seconds to complete
2πr
a revolution) and v =
T
2  π  0.15
v=
2
 v = 0.4712 ms-1
0.10  0.47122
0.15
 F = 0.15 N
(ANS)
F=
Example 30 1976 Question 31
After one complete revolution the displacement
is zero.
 WD = 0 J
(ANS)
Example 31 1976 Question 32
f
a=
m
0.15
a=
0.10
 a = 1.5 ms-2
(ANS)
Example 32 1976 Question 33
mv 2
F=
r
0.10  v 2
 0.15 =
0.15
0.15×0.15
 v2 =
0.10
 v = 0.47 ms-1
(ANS)
Example 33 1972 Question 15
Sphere 2 travels twice as far as sphere 1 in the
same time.
speed of sphere 2

=2
(ANS)
speed of sphere 1
Fizz sicks Unit 3 2015
6. Circular Motion
Example 34 1972 Question 16
v2
a=
r
v 22
acceleration of sphere 2
r

= 22
v1
acceleration of sphere 1
r1
=
There are two forces that the road exerts on
the tyre, the normal reaction and the frictional
force. These two forces will sum to be ‘H’.
 the answer is the opposite to ‘H’ which is “G”
v 22r1
v 12r2
= 22 x
=2
24 of 25
1
2
(ANS)
Example 35 1972 Question 17
The force acting is the tension in the string.
The tension between 1 and 2, acts inwardly on
sphere 2 and outwardly on sphere 1.
The tension between 1 and O must have a net
acting radially inward.
To get sphere 2 to accelerate, let the tension
be T2. Therefore the net tension acting on
sphere 1 needs to be ½T2. (from Q16)
Therefore the tension between 1 and O, must
be 1.5T2, therefore
tension in string between spheres 1 and 2
2
=
3
tension in string between spheres 1 and O
 0.67
(ANS)
Example 36 2003 Question 11, 75%
For circular motion, the net force is always
‘radially inwards’.
 C (ANS)
Example 37 2003 Question 12, 30%
Points to note: Normal reaction is a contact
force, so it must start at the tyre-road interface
Weight force must act from the centre of mass
Friction forces act radially inwards from the
tyre-road interface
The normal reaction should be the same length
as the weight force.
Example 38 2003 Question 11, 21%
The force that the ‘tyre exerts on the road’ is
opposite to the force the ‘road exerts on the
tyre’.
Non-uniform circular motion
Example 39 1985 Question 31
At the top of the track, for the normal reaction
mv 2
to be ‘just 0’, then mg =
r
2
 gr = v
 v = gr
(ANS)
Example 40 1985 Question 32
The total energy at point P (at the top) is the
sum of the KE and the PEP.
The KEP = ½mv2
= ½ x m x gr (from Q31)
PEP = mgh
= m x g x 2r
 At P the Energy = ½mgr + 2mgr
= 2½mgr.
This energy comes from the initial PE.
 mgh = 2.5mgr
 h = 2.5r
B
(ANS)
Example 41 1983 Question 18
When the car is at P, it has two components to
its acceleration. One component is radially
inwards, and the other is 2.0 ms-2 in the
direction of motion.
 the net acceleration is between these two.
H
(ANS)
Example 42 1983 Question 19
v2
A=
r
102
=
50
 a = 2 ms-2
(ANS)
Example 43 1983 Question 20
The net acceleration in the direction H is the
vector sum of 2.0 radially inwards (westerly)
and 2.0 southerly.
 a = 2.8 ms2
(ANS)
Fizz sicks Unit 3 2015
6. Circular Motion
Example 44 1980 Question 23
When the bob passes through the point O, it
has a net force acting radially inwards. In this
case that means upwards. Therefore the
tension needs to be greater than the weight, so
that the net force is radially upwards
B
(ANS)
Example 45 1980 Question 24
If the bob is released from X’, it will have
greater PE, so it will have more KE at the point
O.
The increase in KE will not be double, because
the increase in PE isn’t double.
Since the bob is travelling faster, it will require
a greater force to keep in in circular motion.
B
(ANS)
25 of 25