CHAPTER 6 CIRCULAR MOTION

Chapter --- Circular Motion

6.1 Uniform Circular Motion [1 hour]
6.2 Centripetal Force [2 hours]

Chapter --- Circular Motion

OVERVIEW: Circular Motion

Uniform Application/ Example
circular
motion

Characteristics Horizontal Vertical Conical
of circular circular circular pendulum
motion motion motion

Chapter --- Circular Motion

• Planets rotates • Satellite revolves
about its axis about the earth

Circular motion is motion in a circular path/ motion
which occurs when bodies rotate around something

Chapter --- Circular Motion

Chapter --- Circular Motion

Development

Circular Motion
A motion in a
circle/ circular
path

Chapter --- Circular Motion

6.1 UNIFORM CIRCULAR
MOTION

Describe uniform circular motion in terms of
change in the direction of velocity but not
magnitude

Chapter --- Circular Motion

Circular motion Uniform the motion of a body
Non-Uniform traveling a circular path

at constant speed

an object moving in a
circular path has a
varying speed

6.1: UNIFORM CIRCULAR MOTION

Chapter --- Circular Motion

Development

Circular Motion Uniform circular
motion
A motion in a
circle/ circular the motion of an
path object traveling at a
constant (uniform)
speed on a circular
path

6.1: UNIFORM CIRCULAR MOTION

Chapter --- Circular Motion

Characteristics of Circular Motion

Linear distance, s

θs Consider an object which
does move with uniform
O circular motion as shown.
The length of a circular arc
(path), s is given by

Angular displacement,  s  rθ

• The angle undergone by an where
object from a fixed reference
point  = angular displacement/

• Unit: radian (rad) = 180° angle turned through
2 = 360° r = radius of the circular path

1 revolution = 360o = 2 rad

6.1: UNIFORM CIRCULAR MOTION

Chapter --- Circular Motion  
v v
Linear (tangential) velocity, v
• It is directed tangentially to r r

the circular path and

always perpendicular to the O
radius of the circular path

v s r 
t v
Angular velocity, 
• In uniform circular motion,
the magnitude of the linear • Is defined as rate of change
velocity (speed) of an
object is constant but the of angular displacement
direction is continually
changing  
t

• Unit: rad s-1/ rpm

6.1: UNIFORM CIRCULAR MOTION

Chapter --- Circular Motion

Development

Circular Motion Uniform circular Uniform circular motion
motion
A motion in a The motion of an
circle/ circular path the motion of an object in a circle with a
object traveling at a constant speed but the
constant (uniform) direction of the
speed on a circular velocity is continually
path changing with time

6.1: UNIFORM CIRCULAR MOTION

Chapter --- Circular Motion

6.2 CENTRIPETAL FORCE

(a) Define and use centripetal acceleration,

ac  v2
r
(b) Define and solve problems on centripetal

force,

Fc  mv 2
r

Chapter --- Circular Motion

Centripetal Acceleration, ac
v
• In 6.1 we’ve define uniform 
circular motion  The motion of v
an object in a circle with a
rr

constant speed but the direction O

of the velocity is continually r
v
changing with time
• The direction of this velocity

which is continually changing

with time  will produce

centripetal acceleration, ac ac ac
• The direction of this centripetal

(radial) acceleration is always
directed toward the centre of

the circle and perpendicular to v ac
the linear (tangential) velocity

v

6.2: CENTRIPETAL FORCE

Chapter --- Circular Motion

The centripetal acceleration, ac  v2 OR ac  r 2  v
r
ac – the acceleration of an
object moving in circular path

whose direction is towards ac : centripetal acceleration

the centre of the circular path v : linear(tangential) velocity
and whose magnitude is
r : radius of circular path
equal to the square of the
ω : angular velocity (angular frequency)
speed divided by the radius

6.2: CENTRIPETAL FORCE

Chapter --- Circular Motion

• We can obtain the alternative expression of
centripetal acceleration is

From ac  v2 and v  2r ac  v2
r T r

 2r 2  r2
ac  T
 v
r

ac  42r  4 2r
T2
T 2

6.2: CENTRIPETAL FORCE

Chapter --- Circular Motion

Example 6.2 Example 6.3

Calculate the centripetal A motorbike moving at a constant

acceleration of a car traveling speed 20.0 m s1 in a circular
on a circular racetrack of 1 000 track of radius 25.0 m. Calculate
m radius at a speed of 180 km a) the centripetal acceleration of
h–1. the motorbike, [16.0 m s-2]
b) the time taken for the
Solution: [2.5 m s-2]
motorbike to complete one
Given
revolution. [7.85 s]
r = 1 000 m, v = 180 km h–1

STEP Solution:  v2
a) Equation r
1. Convert v, from km h-1 to ac

2. m s-1 ac  v2 b) Equation ac  42r
Equation r T2

6.2: CENTRIPETAL FORCE

Chapter --- Circular Motion

Example 6.4
A car initially travelling eastward turns
north by travelling in a circular path at
uniform speed as shown in diagram.
The length of the arc ABC is 235 m and
the car completes the turn in 36.0 s.
Determine
a. the acceleration when the car is at B located at an angle of

35.0,
b. the car’s speed,
c. its average acceleration during the 36.0 s interval.

6.2: CENTRIPETAL FORCE

Chapter --- Circular Motion

Solution: sABC  235 m, t  36.0 s

a. The period of the car is given by T  4t  436.0

T  144 s

The radius of the circular path is s ABC  rθ π 
235  r 
2
r  150 m

Therefore the magnitude of the centripetal acceleration is

ac  42r ac 4π2 150
T2 1442

ac  0.286 m s2 (towards point O)

6.2: CENTRIPETAL FORCE

Chapter --- Circular Motion

Solution: sABC  235 m, t  36.0 s

b. From the definition of the speed, thus

v  s  sABC
t t

v  235 v  6.53 m s1
36.0

6.2: CENTRIPETAL FORCE

Chapter --- Circular Motion

Solution: sABC  235 m, t  36.0 s

c. At point A,

aav   vA  6.53 vC
A t 36.0

aav   0.181 m s2 aC
A vA

At point C aA

aav   vC  6.53
t 36.0
C

aav   0.181 m s2
C

6.2: CENTRIPETAL FORCE

Chapter --- Circular Motion

Solution: sABC  235 m, t  36.0 s

Therefore the magnitude of the average acceleration is

   aav  2  2
aav x aav y

aav   0.1812  0.1812

aav  0.256 m s 2
tan 1  aav y 
θ 
aav 
x y
aav
and θ  tan1 0.181 
  0.181 45
θ  45
x

6.2: CENTRIPETAL FORCE

Chapter --- Circular Motion

Example 6.5

A boy whirls a marble in a horizontal circle of radius 2.00 m and
at height 1.65 m above the ground. The string breaks and the
marble flies off horizontally and strikes the ground after traveling
a horizontal distance of 13.0 m. Calculate

a. the speed of the marble in the circular path,

b. the centripetal acceleration of the marble while in the motion.

Solution: r =2.00 m 
u
1.65 m 
u 1.65 m

Before 13.0 m
After

6.2: CENTRIPETAL FORCE

Chapter --- Circular Motion

Solution:

a. From the diagram : ux  u; uy  0
sx  13.0 m ; sy  1.65 m

The time taken for the marble to strike the ground is
1 gt 2
 sy  u yt  2 2
1.65  0  1
9.81t
2
t  0.580 s

The initial speed of the marble after the string breaks is equal to

the tangential speed of the marble in the horizontal circle.

Therefore sx  uux0t .580
13.0 

u  22.4 m s1

6.2: CENTRIPETAL FORCE

Chapter --- Circular Motion

Solution:
b. From the definition of the centripetal acceleration, thus

ac  v2  u2
r r

ac  22.42

2.00

ac  251 m s2

(towards the centre of the horizontal circle)

6.2: CENTRIPETAL FORCE

Chapter --- Circular Motion

1.

11.0 m s1 5.0 m 5.0 m

A centre B 11.0 m s1

Figure 6.7

A particle moves in a semicircular path AB of radius 5.0 m with
constant speed of 11.0 m s-1 as shown in Figure 6.7. Calculate

a. the time taken to travel from A to B,

b. the average velocity,

c. the average acceleration.

ANS.: 1.43 s; 6.99 m s1 (to the right); 15.4 m s2 (downward)

6.2: CENTRIPETAL FORCE

Chapter --- Circular Motion

2. The astronaut orbiting the Earth is preparing to dock with
Westar VI satellite. The satellite is in a circular orbit 600 km
above the Earth’s surface, where the free fall acceleration is
8.21 m s2. Take the radius of the Earth as 6400 km.
Determine
a. the speed of the satellite,
b. the time interval required to complete one orbit around
the Earth.
ANS.: 7581 m s1; 5802 s

3. The radius of the mercury’s circular orbit around the sun is
5.79  107 km and the mercury travels around this orbit in
88.0 days. Calculate
a. the linear speed of the mercury,
b. the radial acceleration of the mercury.
ANS.: 4.79  104 m s1; 3.96  102 m s2 (towards the sun)

6.2: CENTRIPETAL FORCE

Chapter --- Circular Motion

Centripetal Force, Fc

In order for an object to move Without a centripetal
at constant speed in a circle, force, an object in motion
an outside force must continues along a straight-
constantly turn the object line path.
toward the
center of circular path Newton’s
1st Law

6.2: CENTRIPETAL FORCE

Chapter --- Circular Motion

Equation of Centripetal Force

• From Newton’s second law of motion, a force must

be associated with the centripetal acceleration. This

force is known as the centripetal force and is given

 by 
F   
 a  ac and F  Fc
Fnett  ma where  v2
Fc  mac and ac  r  r2  v

Fc  mv 2  mr 2  mv
r

where Fc : centripetal force

6.2: CENTRIPETAL FORCE

Chapter --- Circular Motion

• The centripetal force is • Its direction is in the

defined as a force same direction of the

acting on a body centripetal acceleration

causing it to move in a as shown in figure

circular path of below.

magnitude  
ac v
mv 2  
Fc  r v Fc 

 Fc
ac 
and it always directed 
towards the centre of Fc ac
the circular path. 
v

6.2: CENTRIPETAL FORCE

Chapter --- Circular Motion
Where does the centripetal force come from ?

As a car makes a turn, the As a bucket of water is tied As the moon orbits
force of friction acting to a string and spun in a the Earth, the force
upon the turned wheels of circle, the tension force of gravity acting
the car provides acting upon the bucket upon the moon
centripetal force required provides the centripetal provides the
for circular motion. force required for circular centripetal force
required for circular
Fc  f motion. Fc  T motion

mv 2  k N mv2  T
r r

6.2: CENTRIPETAL FORCE

Chapter --- Circular Motion

What happens if the centripetal force suddenly stops acting on
the object?

6.2: CENTRIPETAL FORCE

Chapter --- Circular Motion

Uniform circular motion Horizontal Circular Object revolving in
Motion a horizontal circle

Flat curve

Conical Pendulum

Vertical Circular Ferris Wheel
Motion
Object revolving in
6.2: CENTRIPETAL FORCE a vertical circle

Chapter --- Circular Motion

Example 6.6: Motion in a horizontal circle

A ball of mass 150 g is attached to one end of a string 1.10 m
long. The ball makes 2.00 revolution per second in a horizontal
circle.

a. Sketch the free body diagram for the ball.

b. Determine

i. the centripetal acceleration of the ball,

ii. the magnitude of the tension in the string.

Solution: m  0.150 kg; l  r  1.10 m; f  2.00 Hz

a. The free body diagram for the ball :

ac 
T
r

mg

6.2: CENTRIPETAL FORCE

Chapter --- Circular Motion

Solution: m  0.150 kg; l  r 1.10 m; f  2.00 Hz

b. i. The linear speed of the ball is given by

v  2r  2rf v  13.8 m s1
T

v  21.102.00

Therefore the centripetal acceleration is

ac  v2 ac  13.82 (towards the centre
r of the horizontal
1.10
ac  173 m s2 circle)

ii. From the diagram in (a), the centripetal force enables the

ball to move in a circle is provided by the tension in the string.

Hence Fx  Fc  mac T  0.150173

T  mac T  26.0 N

6.2: CENTRIPETAL FORCE

Chapter --- Circular Motion

Example 6.7: Motion rounds a curve on a flat (unbanked) track
(for car, motorcycle, bicycle, etc…)

A car of mass 2000 kg rounds a circular turn of radius 20 m. The
road is flat and the coefficient of friction between tires and the
road is 0.70.

a. Sketch a free body diagram of the car.

b. Determine the maximum car’s speed without skidding.

Solution: m  2000 kg; r  20 m; μ  0.70  
ac N
a. The free body diagram of the car :

Centre of 
circle f


mg

6.2: CENTRIPETAL FORCE

Chapter --- Circular Motion

6.2: CENTRIPETAL FORCE

Chapter --- Circular Motion

Solution: m  2000 kg; r  20 m; μ  0.70

b. From the diagram in (a),

y-component : Fy  0 N  mg

x-component : The centripetal force is provided by the

frictional force between the wheel (4 tyres) and the road.

Therefore  Fx  mv 2

f r 2
mv

r 2
mv
μmg 
r

v  μrg

v  0.70209.81 v  11.7 m s1

6.2: CENTRIPETAL FORCE

Chapter --- Circular Motion

acHorizontal circular motion Motion rounds a curve on a
Tr flat track (for car , etc)


mg



ac
T


mg

T  mv2 fs  mv2
r r

6.2: CENTRIPETAL FORCE

Chapter --- Circular Motion

Example 6.8

Conical Pendulum
Figure below shows a conical
pendulum with a bob of mass 80.0 kg
on a 10.0 m long string making an
angle of 5.00 to the vertical.
a. Sketch a free body diagram of the

bob.
b. Determine

i. the tension in the string,
ii. the speed and the period of

the bob,
iii. the radial acceleration of the

bob. (Use g =9.81 m s2)

6.2: CENTRIPETAL FORCE

Chapter --- Circular Motion

Solution: m  80.0 kg; l  10.0 m; θ  5.00

a. The free body diagram of the bob :

  θ T cosθ
ac T

T sin θ


mg

b. i. From the diagram,  Fy  0

T cosθ  mg

T cos5.00  80.09.81 T  788 N

6.2: CENTRIPETAL FORCE

Chapter --- Circular Motion

Solution: m  80.0 kg; l  10.0 m; θ  5.00

b. ii. The centripetal force is contributed
by the horizontal component of the

sin θ  r tension. Fx  Fmc v 2
l
l T sin θ  r
r  l sin θ mv
2

T sin θ  l sin θ

r v  Tl sin2 θ
m

 v  78810.0 sin 5.00 2
80.0
v  0.865 m s1

6.2: CENTRIPETAL FORCE

Chapter --- Circular Motion

Solution: m  80.0 kg; l  10.0 m; θ  5.00

b. ii. and the period of the bob is given by

v  2r  0.865  2 10.0sin 5.00
T T
T  6.33 s
v  2l sin θ

T

iii. From the definition of the radial acceleration, hence

ar  v2 ar  v2
r l sin θ

ar  0.8652 (towards the
centre of the
10.0sin 5.00 horizontal circle)

ar  0.859 m s2

6.2: CENTRIPETAL FORCE

Chapter --- Circular Motion

Example 6.9

Motion in a vertical circle 3.0 m s1 A
E
A sphere of mass 5.0 kg is tied to 3.0 m s1
an inelastic string. It moves in a D
vertical circle of radius 55 cm at a
constant speed of 3.0 m s1 as 3.0 m s1
shown in Figure 6.13. By the aid of
the free body diagram, determine
the tension in the string at points
A, D and E.

(Given g = 9.81 m s-2)

6.2: CENTRIPETAL FORCE

Chapter --- Circular Motion

Solution: m  5.0 kg; r  0.55 m; v  3.0 m s1

The free body diagram of the sphere at :

Point A, A  F  mv2 mv 2
r r
 TA  mg 
TA 
TA  5.09.81  5.03.02
 mg
ac 0.55

Point D,   TD  mv 2 TA  32.8 N
ac TD r TD  81.8 N

D TD  5.03.02

 0.55
mg

6.2: CENTRIPETAL FORCE

Chapter --- Circular Motion

Solution : m  5.0 kg; r  0.55 m; v  3.0 m s1

The free body diagram of the sphere at:

Point E,   TE  mg  mv 2
Caution : ac TE r

TE  5.09.81  5.03.02

E 0.55

 TE  131 N
mg

 For vertical uniform circular motion only,

 the normal force or tension is maximum at the bottom of

the circle

 the normal force or tension is minimum at the top of the

circle

6.2: CENTRIPETAL FORCE

Chapter --- Circular Motion B

Example 6.10 
v
Motion in a vertical circle
A small remote control car with 3.00 m
mass 1.20 kg moves at a constant
speed of v = 15.0 m s1 in a vertical 
circle track of radius 3.00 m as v
shown in figure below. Determine
the magnitude of the reaction A
force exerted on the car by the
track at
a. point A,
b. point B.

(Given g = 9.81 m s2)

6.2: CENTRIPETAL FORCE

Chapter --- Circular Motion

Solution: m  1.20 kg; r  3.00 m; v  15.0 m s1

a. The free body diagram of the car at point A :

 
ac NA


mg

 F  mv2 NA  mg  mv 2
r r

N A  1.209.81  1.2015.02

3.00

N A  102 N

6.2: CENTRIPETAL FORCE

Chapter --- Circular Motion

Solution: m  1.20 kg; r  3.00 m; v 15.0 m s1

b. The free body diagram of the car at point B :

 
NB mg
ac

 F  mv2 NB  mg  mv 2
r r

NB  1.209.81  1.2015.02

3.00

NB  78.2 N

6.2: CENTRIPETAL FORCE

Chapter --- Circular Motion v
v
Example 6.11

Motion in a vertical circle
A rider on a Ferris wheel moves

in a vertical circle of radius, r = 8
m at constant speed, v as shown in

Figure 6.12. If the time taken to
makes one rotation is 10 s and the
mass of the rider is 60 kg, Calculate
the normal force exerted on the
rider
a. at the top of the circle,
b. at the bottom of the circle.

(Given g = 9.81 m s-2)

6.2: CENTRIPETAL FORCE

Chapter --- Circular Motion

Solution: m  60 kg; r  8 m; T  10 s

a. The constant speed of the rider is

v  2r v  2π8
T
10
v  5.03 m s1

The free body diagram of the rider at the top of the circle : 
Nt Nt
 F  mv2
 r  
ac mg
ac  mv 2
mg mg  Nt  r

609.81  Nt  605.032

8
Nt  399 N

6.2: CENTRIPETAL FORCE