Topic Motion In A Circle

circular Motion : Motion in a circular Where

the body maintains a constant distance from a

fixed Point called as center the circle

of .

Examples of Circular Motion

path
N
N
N

N
N

FG
.

¥ Fu

center
radius
 Background knowledge of Uniform circular Motion
directed

.
Fnet Helo always along
at is
a
uity
the
tangent
-

velocity .

velocity
. force is
always directed
towards the center .

:
.
. f I 11

velocity .
The direction of motion

changes

magnitude (speedr)emains
Velocity The
.

Towards the center the same .

centripetal .
Force

The force which is directed towards

the center and is
responsible for
the bodies in a
keeping moving
a circular orbit I path .

The electrostatic force provides

e-
the centripetal
a.

p force .

The force provides

gravitational
⇐
-
-
.
.
aim the
centripetal force .

-
 kinematics Circular Motion
of

*
Radios curvature Gtm ) constant distance between
.

of
circular
me points moving
in a
path and the

fixed called center circle

point as
of the
Arclength
As .

Arc ( Slm ) IO
ii
length Distance travelled
-

no
-

.9A
V
circular
along the
path .

radius

101rad )
Angular Displacement

Tlhhee
iii -

B
The amount swept out
of angle .
.

the radial while

by segment

j
so
a

from point to another

'
moving one
Si
in a circular orbit .

go ,

I
r, ✓z
if Va TV , ⇒ Sz 7 Si 01=02 →
i

'
i
e-

=
constant r .

) Sama .

cowboy
÷ =

Arc S d Radios of ceewatuee Cr )

length
-

s a n
'
s E' lcousiaut ) r

S O A
→
=
-

radios
I to .

( linear distance ) Arc Angular displacement

length
- =

O Are B
length
sq
⇐
-

= -

radios .

¥
←
§ =
I O
 radial
swept out the
segment
.

A Angle by

Angular Displacement lO) =

Arleigh
Radius Cv )
( Radians ) g =
s
( meter )
J C meter ) '

KB
3cm

Radian It
I : is the
angle o=

subs tended at the center

of quad
. -

A
circle
the
by length
an arc 3cm

which is equal in
length to

the radius of curvature of the

circle .

For One
complete Rotation
Are Chicle
length circumference of
-

S =
2 a- r radians .

I
=
TO
f2A radians
O = 2A radians .

Also 0 = 360°

⇒ 360° =
2A radians .

188 = a radians .

To convert In In Radians
Angle degrees Into
Angle

In radians in
I
Angle =
Angle degrees x

1800

To convert In Radians Into In

Angle Angle Degree
.

Angle in
degree Angle In Radians x
180°F
=
4-
Angular Velocity ( owl rads D
-

radial
The
angle swept out
by the segment peer unit time .

Its
angular displacement

displacement
of change of
.

Angular Velocity l radians )

Angular
=

time taken . C seconds ) .

w =
O or do

I I
blue linear And The
Relationship the
Angelou Velocity

T
w = I w

Jw
-

,
t 902
go ,

O
: s r ⇒ D= I
-

r if 02 > Oi

a > a
w =
=

w
I V r cow
⇒
=
=

Ems¥
'
"
m
*

→ tradios
linear Velocity Angular
or
velocity
.

Instantaneous
velocity
, y or
Tangential velocity .

i -
s = r O

il H r w
.

- =

E
if
iii - w =
or

Practice Question t

180 radian
.

N -
= d- -

radios
Calculate the
angular Velocity of 030 m

when travels
of a cal
tyre the car at 54 km/h .
 centripetal Force And
Centripetal Acceleration

[ FEIN ) Lack ]

'
i -
Ac =
V
Ac
-

Sii
r
( mist ⇐

Fc
Ac →
centripetal acceleration
U → linear
velocity ( m/s )
r → radios .

( m )
9 =

'
)
r
(
rt
'
- Ac =
rw co
=
- -

r x s = VO

Ac =
V Wh

co =
I
( )
Angular rads "
t
w →
Velocity .

✓ = VW

iii - Fc = mac

Fc
'

MTV
=

Fe
'
N -
= rn r w
 ";)
-
-
-

pw
C Tls
, ,

Period ) ,
i

Time produce
i
taken to one
'

. I

.
round
'
r
.
trip
.

-
'
.

Period Total Time taken

-
-
-
-

No of cycles I round trip .

T = I
n

In one round trip

complete
arc -

length =
circumference
S =
2A *

w E II
t

t → T

O radians
2y
2A
Sy
= = =

W
2.AT
=

Period

T = ZI co →
Angela Velocity
w
Angelou frequency

[ Hitz )
Frequency
No
of pee unit time
cycles
.

frequency = No .

of cycles lroond -

trip
Total time taken

f- = in Also T =
I
t n

f- =

÷ f =
w_
2 A
W = 2A f
Revolutions lratahons per minute da Radiaue Per See .

I revolution → 2A radians .

I minute → 60 seconds .

n rev
per men = n x 2A red per nun .

n
=
x 2T rad per see .

Go

AS
W/ rads
- '
= RPM W ( ) x 2A

60

TAO
Important formulas Ee
Equations .

=
⇒ A @ = AS .

: co =

V at

AO for 1 rotation
A
w =
or

At E AO = 2A rad ,
It =T

⇒ w = 2A

V = V W or w =L T

✓ T = 29

'

a = V
'
Fc = MV
-
f = ou
C

-y ✓ 2A

'

Fc
"
VW MV W
Ac = = w = 2A f-
Deriving Expression for centripetal force And

centripetal Acceleration .

\
.

My
AO → Small

Is
→
straight line r:

Similar
Two
Triangles . -1%1
Hat = V

IS = II B A .
' IV
'

V VA % =

v r

IV =
VA AS "
AO
=
VB VA
AO
-

✓
O

:lVH=lVBl

Y
=
-

IVids
.

AI
.
.
.

AV =
.
n
: a =

i t
-

- -

t .

✓ v rw
Ac =
-

✓
⇒ ac =C
Ac =
I r

HI
ace =

X
'

: Fe = mac
-

Ac = rw

F = mm
c -

'
V = M TW
c
 w

E
.

f
#f
Arc )
length C s

)
W
Radius of curvature .
Porto
r
Angular Displacement s
.

r
O = I
⇒ s ro
f
r
-
-

Radian ⇒Arc length radius

'
=
-

22
rads D radian
Angular C w/
=aa¥→
-

Velocity co

→ second .

✓ r
÷ Yoji
= r w -
-

centripetal Acceleration ( ac ) it
V
rxco

Fc=m
=

Ac
V÷
=
⇒

Fc
" '
Ac = Vw ↳ = Mirco

Period ( Tls )
00

for I
complete cycle
time → period =
T

displacement 2A rad
Angular =

Angular velocity AT If
=
=

co T I T 2 I
241
= =
⇒
w w

)
Frequency HIHE
f = b- →
2A • =
⇒

↳
-

Angela frequency
-
 Practice Questions

circular Linear
A car is
travelling along a path with
"

Speed angular speed rads

"
15ms and 0.36 .

what is the radios curvature track

of of
.

z -
The drum
of a
spin dryer Has a radius
of 20cm

and rotates at Goo revolutions per minute .

(a) show that the

angular speed of the drum is

rad
'
about 63 5
.

Calculate point drum

(b) for a on the
edge of the

i - Its Linear speed

ii Its acceleration towards the center the drum

of
- .
 A train round circular track
3
toy moves of
a
.

diameter revolution
atom ,
completing one in ios -

calculate for this tram .

(a) the linear speed

(b) the
angular speed

(c) the centripetal acceleration

A stone attached
string
to is in
mooing
4- a a

horizontal circle radios 90cm The Stone has mass

of
.

65 and completes one revolution in 0.70 s

Kg .

Calculate the
keeping the
tension
string stone
in
,

in its circular path -

 NASA used
centrifuge
's
5 20 G for
testing
is
space
-
-

and acceleration
equipment
the
effect of
on humans -

The
centrifuge consists
of
an arm
of length
178M
,

rotating at constant speed and

producing
acceleration
an

equal to 20 times the acceleration

of
calculate
free -

fall .

(a) the
angular speed required to produce
centripetal
a

acceleration
of

20g
(b) the rate rotation the
of of
arm -

orbit
6- A satellite orbits the Earth Zoo km above its

The acceleration towards the

surface .
centre of
the Eceeth is 9.2ms
"
.
The radius of
Earth is

6400 km - Calculate :

(a) The linear speed of Satellite

(b) Time to
complete one -
 :
Fy F Sino ↳ F Cos O

yo ]
.

Fx =
Fx Cos O

F x SMO

} changes the
magnitude of
a speed V

Changes the direction without

} changing
the
magnitude .

changing the direction

f-
.

F- SMO

F Caso the
magnitude of
→
changing
;o
speed
 Horizontal circles

①
I
circular path
If the
body mores in -1

maintaining constant distance from the center

such that their altitudes I height ) remain

the same .

CONICAL PENDULUM

T SMO → Fc

Fc =
T Sano

speed
it
it the → constant

miff
then the radios r →
constant =
T SMO

E from the vertical O → constant

Angle
.

'
my W =3 SMO

and the 0
if the speed increases ,
r → increases

also increases .

Ill
O
:
.

O balances 7"

! hi
T cos the Caso :
"
:

weight
II. , case

T Cos O Tsin a
rug
'
=
Practice Question
Corso O

V i
Show that the speed of the

!
' '
:

body of mass m
,
moving
in
:

horizontal circle V is
of
m

Tano

by
"

given
y
-
=
jgrtano
=
-

mg

T SMO = Fc T Cobo = al

SMO O
m ②
T =
T cos
① mg
= -
-

Divide Eq ① by Eq ②

Fmg
SMO mvyr
÷
T =

tan O = I
gr
'
tano U
gr
=

✓ =

Jgr tano
 Load

:
.

"
IN
.

Fc = NEMO

R IN
R
=

¥
2 Sin O -

O
)
 Banked
Turning of Aeroplanes

When at constant
Hyung
height , lift force is
equal
to the and
weight is

directed upwards
vertically
.

In order to
of circular
move in an ale
length
-

'
radios to be
'

path , of r
,
the lift force L has
'
directed
'
at certain O from the vertical
angle .

The Vertical Ly balances

component of lift force
Are
weight
Ly = IN
Leeds @ → eat
I cos O
mg
=

LSMO
Fc ←

The horizontal
component of Lift force La

provides the
centripetal force .

Lx = Fc
'
I SMO = MV or Mirco
'

'
V

Ly balances IN
the
weight ⇒ Cy =

Leos O
mg
= .
 Practice Question

tilted
he
given diagram shows an aircraft flying
order to in
horizontal radios
fly Circle
in a r
of .

This has
aircraft a
weight
of 3.92×105 N and a constant
I
250
Speed Of
35
-

ms .

.
in the vertical plane ,
two c

forces
acting
on the aircraft
eggs
are the lift force L
acting
at 350 to the vertical and the

w
weight .

calculate :

(a) The
magnitude of
the Lift force L .

(b) The acceleration aircraft towards the center

of of
circle .

(c) The radius

of the circular path .fr )
Show that the
velocity of
the
aeroplane is
given by
the
expression , -

y =
jgrtano go
spring Moving
v, > vav ,

¥ takin
I

> '
,
, .

I ,

when starts
When not
moving
the
spring
Motion
speed V = o

moving
in circular

Tension T = O
Tension → T = kN

Fc = O T = Fc
'
O
Ac =
Kk = MV

If the speed increases

,
Kk =
.mI
the extension re also Ltu

MCheas.es .
 Q . /0 /P /Q

1 (a) (i) Define the radian.

..................................................................................................................................

..................................................................................................................................

.............................................................................................................................. [2]

(ii) A small mass is attached to a string. The mass is rotating about a fixed point P at
constant speed, as shown in Fig. 1.1.

mass rotating
at constant speed

P
Fig. 1.1

Explain what is meant by the angular speed about point P of the mass.
..................................................................................................................................

..................................................................................................................................

.............................................................................................................................. [2]

page 16
(b) A horizontal flat plate is free to rotate about a vertical axis through its centre, as shown
in Fig. 1.2.

plate

M
d

Fig. 1.2

A small mass M is placed on the plate, a distance d from the axis of rotation.
The speed of rotation of the plate is gradually increased from zero until the mass is
seen to slide off the plate.

The maximum frictional force F between the plate and the mass is given by the
expression

F = 0.72W,

where W is the weight of the mass M.

The distance d is 35 cm.

Determine the maximum number of revolutions of the plate per minute for the mass M to
remain on the plate. Explain your working.

number = ........................................... [5]

(c) The plate in (b) is covered, when stationary, with mud.

Suggest and explain whether mud near the edge of the plate or near the centre will first
leave the plate as the angular speed of the plate is slowly increased.

..........................................................................................................................................

..........................................................................................................................................

...................................................................................................................................... [2]

page 17
 6

2 A large bowl is made from part of a hollow sphere.

A small spherical ball is placed inside the bowl and is given a horizontal speed. The ball follows a
horizontal circular path of constant radius, as shown in Fig. 2.1.

ball

14 cm

Fig. 2.1

The forces acting on the ball are its weight W and the normal reaction force R of the bowl on the
ball, as shown in Fig. 2.2.

wall of R
ball
bowl

Fig. 2.2

The normal reaction force R is at an angle to the horizontal.

(a) (i) By resolving the reaction force R into two perpendicular components, show that the
resultant force F acting on the ball is given by the expression

W = F tan .

[2]
© UCLES 2014 9702/41/O/N/14
 7

(ii) State the significance of the force F for the motion of the ball in the bowl.

...........................................................................................................................................

...................................................................................................................................... [1]

(b) The ball moves in a circular path of radius 14 cm. For this radius, the angle is 28°.

Calculate the speed of the ball.

speed = ............................................... m s−1 [3]

© UCLES 2014 9702/41/O/N/14 [Turn over

 vertical circles Horizontal circles .

÷
:

i
:
'

Hitude GPE remains the

of

•
Attitude of GPE
changes same
.

force
④
The size resultant The resultant force
..

of site
of
will keep on
changes He) remains the same

speed all The Lv)

•
The at
points .
speed remains the

same
vary
.
.

④
The site
of centripetal and The size
centripetal
.
of
Tangential acceleration both acceleration remains the

same
vary
. .

For Bodies Vertical Circular Motion

Moving m A
o

Resultant force =
centripetal force

of forces aching of
)
some forces F'
( ) (
som -

aching =

towards the center from the water

away
 "

Fc T w Castro 3
Ncos @ provides the
-

Fc T WH )
I ; deceleration

-1¥
-

=
.

Fc =Ttw
¥ .

0=270 TO z Fc = T -
IN Caso
>

T
Cos 2700=0 i . . .

,
to 8
Fc= T .
a 0=900
p
a 900=0
.

cos
.
.

'

¢ . ooh
.
u d Fc = T
% w ? ⇒
*
No I i
"

g
Fc = T -
IN Caso WSM

Fc = T -
IN

T = Fc t IN Cos @

" "" "

"

"
"
"

⇐ w .

i is
! ;
o

;
.

TI Fc IN
Fc -1 IN Caso
-

T =

0=180 Fc = T -
ING ) = T TIN

Fc = T -
Noosa

0=900
0=270 Fc T
T
J
=

Fc
-
-

-
piety

sin
or a
1-

INCOSO
f
IN
0=0 Causes deceleration
,

IN
Fc = T -
IN

T = Fct IN
 Vertical Circular Motion → Non -

Uniform circular

Motion

Altitude GPE
of changes
.

.
Magnitude of the forces acting
rang continuously
.

As the
.

body goes up ,
its speed will decrease
and the speed increases

as the body comes

y
2

down .

.
Both centripetal and
tangential acceleration
vary
continuously
.
1

For oh Vertical cerudae Motion

a
body moving a .

Fe Sam

facing :⇐
of forces Sum
forces
of
www.eu.nl
=

, I ,

fret towards the centre L

m → mass
of object
Tension in
T →
string
.

Fc →
centripetal force

weight ( neg )
w → .
 or
(a) lathe the
Object
is at the bottom
,

circle
of :

•
Tension overcomes I balances the
weight
and provides Fc

Fc = Ti - IN

Greats
→ T, =
Fc t IN
tension

Tz
→
I , mg
.

WW2
T

B - At Mid -

pom In level with the v

centre circle
of
.

weight causes deceleration •

r Tension provides Fc centre

Fc = = MI IN
✓

At the the circle

lap of
c-

Fc = Ts t W Both Tension h

provide Fc

173
=
Fc - W
weight
I,
Minimum
value Ts =
Mfd -

mg
tension
of
f- c
=
Ts t W
-

To the circular
moving motion
in
keep mass

Fc 7, W

speed the
highest point
m
1mg
at
y
must be than
'
greater
✓ F particular
gr equal to a value
called
-

Minimum
I 7
Jgr of speed
as
,

safe speed -
 Fe = Tv
.

weight cause acceleration

.

IN IN

since since it acts

mg mg ,

causes acceleration opposite to the

direction of
.

velocity produces
,

u deceleration
O

÷
-
mgcoso

mg and
ng

Fa COO
mg
=

y Fc t CoA
=

mg
I

Fat W

⇐ ,

Fc
-

-
- w
 cation A In A circular Track
of Caeeiage
.

with
in contact
→
If
track
-

RYO the

RY
Rs
Rz Rz
Fc = IN t R → Fc J IN

FcIN
m
Mm Value of >
mg
=
-
,

←
contact farce
hair
✓
2
.

g- 7g safe
speed
f
.

2
>
✓
gr
V >
jgr

Fc = R .
.

Fc = R 2

And the

IN
cat provides a
causes

acceleration .
deceleration .

Fc = R -
IN

R = Fc t IN
"
R mu
mg
=
t

g-
,
Max . value
of contact force .
Motion Of Bucket filled with water In
Morong circle .

Rt IN =
Fc

Fc 7' lat

>
MI mg
2

✓ q gr

:
IN R Fc
-

t =

Practice Question
Unni =
JGT
✓ = I m

ruin = ?
= 198171
'
T ? 3. I
-

=
= Ms

w ZI
-

=
.

HI ohhhh ynrco
'

Fc = =

' racer
✓ =

T
Vf2I= 1×21
=

2ft
✓ V
=
⇒
3 .
I
T = 2. O S .
 Practice Question
"
6. O MS

YE
¥§
%
Ba
m
**"
&
E
%
" '" Me

ma
" go .
µ
"•

attached
An
object of
mass 0.1
kg is to
.

Ghq
www.t
cord and in vertical circle
swing
a a .
. o.sn .

radios At the the

3:

Of 0.50 m
top of
.

Be.

speed
'
circle the The
Object is 6.0ms or
-

of
-

ma
, .

"

Find the tension at circle

i. the
top of the -

ii - Calculate the maximum tension needed to

more the object .

The Ball The R
Rolling Over
Loop

R = Fc

IN = Fc t R

W 7 Fc R # O In order for the ball to

{ I remain , in contact with the

rug 4 MI ground ,
R > o

> V safe speed

'
v Maximum
gr
→

✓ a
Jgr

e*
! IN -
R = Fc

:
Fi

Rtd = Fc
3

'
'
He

R -
IN = Fc
 Ball Over the
Rolling loop
* .

W Fct R
Fc R
=
IN
managing
=
-
⇒

for R to imam
:

Fe C lat a

Lyng
managing

V V

'
✓ C imma
gr :

Max .

safe → V a
Jgr At the
biggest point ,
part of
and
speed Weight provides Fc the

rest is balanced contact

by
force .

Roller -
Coaster Ride

R
Fc = w -

Fc = wt R

R W

Fc = R -
W
 4

Section A For
Examiner’s
Answer all the questions in the spaces provided. Use

1 (a) Define the radian.

..........................................................................................................................................

..........................................................................................................................................

...................................................................................................................................... [2]

(b) A stone of weight 3.0 N is fixed, using glue, to one end P of a rigid rod CP, as shown
in Fig. 1.1.

glue

85 cm P

stone,
C weight 3.0 N

Fig. 1.1

The rod is rotated about end C so that the stone moves in a vertical circle of
radius 85 cm.
The angular speed of the rod and stone is gradually increased from zero until the glue
snaps. The glue fixing the stone snaps when the tension in it is 18 N.

For the position of the stone at which the glue snaps,

(i) on the dotted circle of Fig. 1.1, mark with the letter S the position of the stone, [1]

(ii) calculate the angular speed of the stone.

angular speed = ................................... rad s–1 [4]

© UCLES 2010 9702/41/M/J/10