NOTA TUTORAN SP015 CHAPTER 6 20_21

TOPIC 6: CIRCULAR MOTION Motion characteristics for circular motion
6.1 Uniform circular motion
6.2 Centripetal force

6.1 Uniform circular motion
Learning Outcome:

a) Describe uniform circular
motion

b) Convert units between degrees,
radian, and revolution or rotation..

What is circular motion? Angular velocity , ω

▪ Circular motion is a motion which occurs when bodies ▪ Rate of change of angular displacement
rotate in circular path

▪ Unit ω : rad s-1

▪ Other units : rps or rpm

▪ Vector quantity and its direction is perpendicular to the
plane of motion (right hand rule).

The relationship between linear velocity, v, and
angular velocity, ω

s = r

▪ Uniform circular motion is the motion of an object in a Divide both sides by t:
circle with a constant/uniform speed.
▪ magnitude of its linear velocity (speed) remains s = r v = r
constant. t t
▪ direction of its velocity changes continually.
▪ The direction of linear velocity, v at every point along
Comparison between Linear Motion and Circular
Motion the circular path is tangent to the point.

▪ The direction of the angular velocity, ω depends on
the rotation of the object (clockwise or
counterclockwise ).

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Linear velocity, v Example 6.1.3
can be written in terms of period, T and frequency, f :
The diameter of a tire is 64.8 cm. A tack is embedded in the
v = 2 r = 2 r f tread of the right rear tire. What is the magnitude and
T direction of the tack's angular velocity vector if the vehicle
is traveling at 10.0 km/h?
Angular velocity, ω
can be written in terms of period, T and frequency, f :

 = 2 = 2f
T

Example 6.1.1 Solution 6.1.3
An object undergoes circular motion with uniform angular v = 10(1000) = 2.78 ms−1
speed 100 rpm.
Calculate : 60(60)
(a) the period, T
(b) the frequency of revolution, f.

Solution 6.1.1 6.2 Centripetal force
Given : ω = 100 rpm Learning Outcomes:
a) Define centripetal acceleration.
Convert to rad s-1 : b) Solve problems related tocentripetal force for uniform

ω = 100 rev = 100(2 ) = 10.47 rads−1 circular motion cases: horizontal circular motion,
1 min 60 vertical circular motion and conical pendulum (exclude
banked curve).
(a) T = 2 = 2 = 0.60 s
ω 10.47 Introduction

(b) f = 1 = 1 = 1.67 Hz ▪ When an object moving in a circle of radius, r , at a
constant speed, v, the direction of the object changes.
T 0.60 Thus it has an acceleration called the centripetal
acceleration, ac
Example 6.1.2
▪ ac is defined as the acceleration of an object moving in
An object travels around the circumference of a circle of circular path and it directed towards the center of the
radius 6 m at a rate of 30 rev/min. Calculate: circle.

(a) its angular speed in rad/s. ▪ Direction of ac – graphically

(b) its linear speed around the circle.

Solution 6.1.2 Δv
r=6m Δt
ω = 30 rev/min

a =

FIGURE 6.2.1

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▪ Magnitude of ac ✓ As a car makes a turn, the force of friction
acting upon the turned wheels of the car
▪ Since v = r ω provides the centripetal force required for
circular motion.
a = rω2
c FIGURE 6.2.3

a = vω ▪ Magnitude of Fc:
c

Example 6.2.1 ▪ since v = r ω , thus
Calculate the centripetal acceleration of a car traveling on a
circular race track of 1000 m radius at a speed of F = mrω2 = mvω
180 km h-1. c
Solution 6.2.1
Given : r = 1000 m ▪ Direction of Fc - towards the centre of the
circle and same direction as the centripetal
v = 180 km h-1 acceleration, ac

Convert to m s-1 FIGURE 6.2.4
v = 180(1000) = 50 ms−1
60(60) ▪ Fc is perpendicular to the direction v, so it
does no work on the object.
hence

ac = v2 = 502 = 2.5 ms−2
r 1000

Centripetal Force, Fc Circular Motion In Horizontal Plane (Uniform
Circular Motion)
▪ Fc is defined as the net force required to keep an
object of mass, m moving at a speed v on a Case 1 : object moves in a horizontal circle with
circular path of radius, r. steady speed.

▪ Examples: T

✓ As the moon orbits the Earth, the force of
gravity acting upon the moon provides the
centripetal force required for circular
motion

mg

FIGURE 6.2.5

FIGURE 6.2.2 3

▪ Two forces acting on the object : Example 6.2.3

✓ The force of gravity ( weight i.e. mg ) A car travels around a flat curve of radius r = 50m. The
✓ The tension in the string - is the only
coefficient of the static friction between the tires & the road
component in the radial direction that
is μ = 0.75. Calculate the maximum speed at which the car
provided the centripetal force.
s R

can travel without skidding.

▪ Applying Newton’s 2nd Law: Solution 6.2.3
Given : μs = 0.75 , r = 50m
Fnet = mac = mv2
r T = mv2 fs supplies the centripetal force f
r
fs = mv2 s
r
mg

Example 6.2.2 But fs= µsR

A 0.25 kg rock attached to a string is whirled in a horizontal  m g = mv2
circle at a constant speed of 10.0 ms-1. The length of the r
string is 1.0 m. Neglecting the effects of gravity, find the
tension in the string. v= rg

Solution 6.2.2 = (0.75) (50) (9.81)

T = Fc = 19.18 m s−1

T = mv 2 = 0.25(10)2 = 25 N Example 6.2.4
r 1.0
A 1200 kg car with a velocity of 8.0 m/s travels around a

Case 2 : Motion of car round a curve : Flat curve flat curve of radius r = 9.0m.
road
a) Calculate the horizontal force must the pavement exert
▪ If the coefficient of static friction between the tires
& the road is μ then, f = µR on the tires to hold the car in the circular path ?
sR
b) What coefficient of friction must exist for the car not to
Vertical comp
slip ? R

Solution 6.2.4

a)

ΣFy = 0 fs = Fc = mv2 = 1200(8.0)2 = 8533N f
R − mg = 0 r
9.0 s
R = mg … ( 1 )
f b) fs = mv2 mg

s r

mg R = mg

Horizontal comp

fs = Fc = mv 2  R = mv2
r r

μR = mv 2 …( 2 )  m g = mv2
r
r
v2 (8.0)2
(1) into (2)  = rg = (9.0) (9.81) = 0.72

 mg = mv2
r

∴ The maximum velocity without slipping on the road is

v = μr g

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Case 3 : Conical Pendulum Component-y : Component-x :
Tsin θ = mv 2
A conical pendulum moving in uniform circular motion T cos 30° = mg
with speed v : T = 0.15(9.81) r

Component – x cos 30 v = r Tsin θ
m
Fx = mv 2 T = 1.70 N
r v = (Lsin θ)Tsin θ
m
Tsin  = mv 2 FIGURE 6.2.6
r = 1.84 m s−1

Component – y

Fy = 0 Circular Motion In Vertical Plane
Tcos  = mg
Case 1 : A ball is attached to a string & moves in a
vertical circle.

…Tsin  = mv2 … (1) At the TOP:
r
r = L sin θ FIGURE 6.2.8
Tcos  = mg …(…(21)
) r

)(2(1) : (mv 2 ) FIGURE 6.2.7 At the top of the circle ( point A ) : both T & mg are
Tsin  r directed downwards
=
(2) Tcos  mg
Fnet = mac

tan  = v 2 T + mg = mv2 (T is minimum)
rg r

v = r g tan  = Lg sin  tan  T = mv2 − mg
r

At the BOTTOM:

Example 6.2.5

A 0.15 kg ball attached to a string which is 1.2 m in length
moves in a horizontal circle. The string makes an angle of
30° with the vertical. Find the tension in the string & the
speed of the ball.

Solution 6.2.5

FIGURE 6.2.9

r At the bottom of the circle (point B) :
T & mg point in opposite direction
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Fnet = mac (T is maximum ) Example 6.2.7

T − mg = mv2 A 2 kg ball is tied to the end of a 80 cm length of string.
r The ball is then is whirled in a vertical circle and has a
velocity of 5 m/s at the top of the circle.
T = mv2 + mg
r a) What is the tension in the string at that instant ?
b) What is the minimum speed at the top necessary to
Circular motion is possible as long as the cord remain taut, maintain circular motion ?
thus there is a critical (minimum) speed to be maintained.
Solution 6.2.7
T + mg = mv 2
r m = 2 kg , r = 80 cm , v = 5 m/s at the top.

a) Top

T = mv2 − mg
r

T = 42.9 N

If the rope is sagging, T = 0 , thus : b) T = mv2 − mg
r
mg = mv 2
r For v , T = 0
min
v min = r g
mg = mv 2
r

vmin = r g

= (0.8)(9.81) = 2.8 m/s

Example 6.2.6 Example 6.2.8
A 1.2 kg rock is tied to the end of a 90 cm length of string.
The rock is then is whirled in a vertical circle at a constant A rope is attached to a bucket of water and the bucket is
speed of 8 m/s. What are the tensions in the string at the top then rotated in a vertical circle of 0.70 m radius. Calculate
and bottom of the circle ? the minimum speed of the bucket of water such that the
Solution 6.2.6 water will not spill out.
m = 1.2 kg , r = 90 cm , v = 8 m/s
Solution 6.2.8
Top
Fnet = ma
T + mg = mv2 c
r
T + mg = mv 2
T = mv2 − mg r
r
The water will not spill out if the T=0, thus :
T = 73.6
N mg = mv 2 v min = r g
r vmin = (0.7) 9.81 = 2.62 m s−1
Bottom

T − mg = mv2
r

T = mv2 + mg
r

T = 97.1 N

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Case 2 : Roller coaster on a circular track / Ferris wheel Example 6.2.10
Top of the circle :

Fnet = ma v
c

mg R + mg = mv 2
r

R R = mv 2 − mg vFigure 6.2.10
r

Bottom of the 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
R 6.2.10. If the time taken to makes one rotation is 10 s and
mg the mass of the rider is 60 kg, Calculate the normal force
exerted on the rider

Fnet = ma a. at the top of the circle,
c

R − mg = mv 2 b. at the bottom of the circle.
r
(Given g = 9.81 m s-2)

R = mv 2 + mg Solution 6.2.10
r
m = 60 kg; r = 8 m; T = 10 s

a. The constant speed of the rider is

A minimum velocity (when R = 0) is required in order to v = 2r v = 2π(8)
keep a roller coaster car on a circular track. T
10

0 + mg = mv 2 v = 5.03 m s−1
r
v min = r g The free body diagram of the rider at the top of the circle :
N

Example 6.2.9 ac Fnet = mv2
What minimum speed must a roller coaster be traveling mg r
when upside down at the top of a circle (refer to the figure)
if the passengers are not to fall out ? Assume r = 8.0 m. mg − Nt = mv2
Solution 6.2.9 r
r = 8.0 m
(60)(9.81) − Nt = (60)(5.03)2
Top
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Nt = 399 N

b. The free body diagram of the rider at the bottom of the

circle :

mg Fnet = mv2
r

R Nb − mg = mv2
r
Fnet = ma N
c
(60)(5.03)2
R + mg = mv 2 Nb − (60)(9.81) =
r 8
ac
For vmin , R = 0
Nb = 778 N

v min = r g mg
Caution :
vmin = (8.0)(9.81) = 8.9 m/s 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.

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