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UNIT 6: Circular Motion

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6.1 Uniform Circular Motion { Definition - is motion in a circle (circular arc) at a constant speed. { Uniform circular motion of a particle about the axis of rotation, O can

be described by the angular position ( ) of a reference line with respect to the x-axis as shown below.

rs=

motioncircular theof radius the: length) (arc arc. oflength the:

radianin position angular : where

rs

{ If the particle changes its angular position of the reference line from 1 to 2, the angular displacement is given by

12 =clockwise.ismotion theifnegative is

ise.anticlockw ismotion theif positive is

y

O1 x

2

t1

t2

y

s

O

x

r

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6.1.1 Angular velocity (angular frequency),{ Average angular velocity, av

z Definition is the rate of change of angular displacement.

{ Instantaneous angular velocity, z Definition is the instantaneous rate of change of angular

displacement.

{ Vector quantity.{ The unit of angular velocity is radian per second (rad s-1).{ Notes :

{ For uniform circular motion, the angular velocity is constant therefore the angular displacement in time, t, =t

{ Unit conversion :

{ The direction of angular velocity will be learnt in unit 7.

r

tttav =

= 12

12

dtd

tt =

= 0

limit

o

o

360rad 2180rad

==

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6.1.2 Tangential velocity (linear velocity),{ From the definition of instantaneous linear velocity, we get

{ Vector quantity.{ The unit of the tangential (linear) velocity is m s-1.{ It is directed tangentially to the circular path. { The magnitude of the linear velocity (speed) of an object is constant

in uniform circular motion but the direction is continually changing.{ The linear velocity is difficult to measure but we can measure the

period, T of an object in circular motion.

vr

rsdtdsv == where

==dtd

dtdrv where

rv = The relationship between angular velocity and linear velocity

y

s

O

x

vv

v

r

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{ The period, Tz Definition is the time taken for one complete revolution

(cycle/rotation).z The S.I. unit of the period is second.z If the object makes one revolution (rotation), the angular

displacement, = 2 radian and the time interval, t = T hence

{ The frequency, fz Definition is the number of revolutions (cycles/rotations)

completed in one second.z The S.I. unit of the frequency is hertz (Hz) or s-1.

{ Therefore we can determine the linear velocity by using equationbelow.

t

=

rf2T

r2v ==

T1ff2

T2 === whereor

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6.1.3 Centripetal (radial) acceleration,{ The figure below shows a particle moving with constant speed in a

circular path of radius, r with centre at O. The particle moves from A to B in a time, t.

{ Consider sector OAB of the circle. The arc length AB is given by

{ The velocities of the particle at A and B are v0 and v1 respectively where

rs =rs =

vvv 10 == rr

(1)

rc aarr or

0vr

1vr

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{ Let PQ and PR represent the velocity vectors v0 and v1 respectively, as shown in figure below.

{ Then QR represent the change in velocity vector v of the particle in time interval t. Since the angle between PQ and PR is small hence

{ Equation (1) is equal to equation (2) then

Dividing by time, t, we get

01 vvvrrr =

1vr

0vr

P Q

R

( ) ( ) PQQR = vv =

vv = (2)

vv

rs =

=

tv

v1

ts

r1

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As t 0, we have

{ Definition is defined as the acceleration of an object moving in circular path whose direction is towards the centre of the circular path and whose magnitude is equal to the square of the speed divided by the radius.

va

rv =

=

dtdv

v1

dtds

r1

motion circular of radius the: velocitygential)linear(tan the:

onaccelerati lcentripeta the: where

rv

arva c

2

c =

vra 2c ==

=

t

vv1

ts

r1

0t0t

limitlimit

or

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car

car

car

carca

r

car

{ The direction of centripetal (radial) acceleration is always directed toward the centre of the circle and perpendicular to the linear (tangential) velocity as shown below.

{ Because of

therefore we can obtain the alternative expression of centripetal acceleration is

2

2

c Tr4a =

Tr2v =

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{ Example 1: A sprinter is running at constant speed 9.2 m s-1 in a circular track with centripetal acceleration 3.8 m s-2. Finda. the radius of the circular track.b. the time required by the sprinter to make one revolution.Solution: v = 9.2 m s-1 and ac = 3.8 m s-2

a. Applying equation of centripetal acceleration:

b. the period of the sprinter,

arv2

c =( ) ..

r2983

2

=m322r .=

T

r2v = . s215T =

or 22

c Tr4a =

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{ Example 2: A boy whirls a stone in a horizontal circle of radius 1.5 m and at height 2.0 m above level ground. The string breaks, and the stone flies off horizontally and strikes the ground after traveling a horizontal distance of 10 m. What is the magnitude of the centripetal acceleration of the stone while in circular motion?(HRW.70.51)(Use g = 9.81 m s-2)Solution:

The time taken by the stone to strikes the ground,

0vgt21tvs y0

2y0y == where

0v

2.0 m

1.5 m

Before After

2.0 m

10 m

s640gh2t .==

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Considering the horizontal distance, x = 10 m

The initial speed of the stone after the string breaks is equal to the speed of the stone in horizontal circular motion. Therefore

{ Example 3: (exercise)The astronaut orbiting the Earth is preparing to dock with Westar VI satellite. The satellite is in a circular orbit 600 km above the Earths surface, where the free fall acceleration is 8.21 m s-2. Take the radius of the Earth as 6400 km. Determine the speed of the satellite and the time interval required to complete one orbit around the Earth.

Ans. : 7581 m s-1, 5802 sNo. 32, pg. 104,Physics for scientists and engineers with modern physics, Serway & Jewett,6th edition.

tvtvx 0x0 ==

22

c sm162rva ==

10 sm615640

10v == ..

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6.1.4 Uniform circular motion in terms of position vector,{ For example, an object is in the circular motion as shown in figure

below.

{ By using the equation of instantaneous velocity, the tangential velocity vector is

{ The position vector at point P in the circular path is

hence

where r : the radius of circular path

rrrrjrirr

y

xyx

sin cos where

==+=r

dtrdvrr =

rr

rr

O

vv

v

P

xryr tjrirr =+= where sin cos r

jtritrr sin cos +=r ( )jtitrr sin cos +=r

( )jtritrdtdv sin cos +=r

( )jtitrv cos sin +=r

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{ By using the equation of instantaneous acceleration, the centripetal (radial) acceleration vector is

Negative sign means the direction of centripetal acceleration always opposite to the direction of position vector,

( )jtritrdtdac cos sin +=rdtvdacrr =

rr( )jtitra 2c sin cos +=r

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6.2 Non-uniform Circular Motion { In non-uniform circular motion, the tangential (linear) velocity changes

both in direction and in magnitude.{ Therefore the resultant acceleration can be written as the vector sum

of the component vectors :

{ The tangential acceleration component z causes the change in the speed of the particle (object).z its direction always parallel to the tangential (linear) velocity and

given by

{ If a particle (object) is speeding up the direction of at is in the same direction of the tangential velocity, v

{ If a particle (object) is slowing up the direction of at is opposite direction of the tangential velocity, v

onaccelerati l tangentia the: onaccelerati lcentripeta the: where

t

ctc

aaaaa rrr +=

dtvd

atr

=

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{ The centripetal (radial) acceleration component z arises from the change in direction of the tangential (linear) velocity.z its direction always toward the centre of the circular path and given

by

{ The example of non-uniform circular motion is a particle moving in a vertical loop, like a roller coaster car with a varying speed as shown in figure below.

rva

2

c =

car

car

caarr =

caarr =

tar

tar

ar

ar

vrvr

vr vr

A

B

C

D

{ Because ac and at are always perpendicular to each other, the magnitude of resultant acceleration, a at any time is

{ At point D, the resultant acceleration is maximum.

{ At point B, the resultant acceleration is minimum.

2t

2c aaa +=r

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a. By using centripetal acceleration equation :

Then the speed is

{ Example 4: A particle revolves in a vertical circle with a radius of 2.50 m. At a particular instant, its total acceleration is 1.20 m s-2 in the direction that makes an angle of 30.0 with the direction of the motion. Finda. its speed at that moment.b. its speed 2.00 seconds later,Assuming constant tangential acceleration. (Csw.CD4.2.6.1.pg7)Solution: a = 1.20 m s-2 and r =2.50 m

o30ar

va2

c sin==

vr o30rav sin=1sm221v = .

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b. By using equation of motion:

where

hence we obtain

{ Example 5: (exercise) A particle starts moving in a circle of radius of 0.20 m with constant tangential acceleration, at = 5 cm s-2. After what time will the value of centripetal acceleration reach twice the value of the tangentialacceleration? (Csw.CD4.2.6.1.pg8t3)Ans. : 2.83 s

{ Example 6: (exercise) A racing car starts from rest in the pit area and accelerates at a uniform rate to a speed of 35 m s-1 in 11 s, moving on a circular track of radius 500 m. Assuming constant tangential acceleration, finda. the tangential acceleration, andb. the radial acceleration, at the instant when the speed is v=15 m s-1and again when v=30 m s-1. (Gc.121.ex.5-14)

Ans. : 3.2 m s-2, 0.45 m s-2, 1.8 m s-2

tavv t0 +=1

t sm04130aa== .cos o

( ) 1sm33002041221v =+= ....

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6.3 Centripetal Force, { Any object moving in uniform circular motion has a centripetal

acceleration directed towards the centre of the circular path.

{ From Newtons 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

rc FFrr

or

==== ccnet FFaaamFF rrrrrrr and where vr

rvaamF 2

2

ccc ==== rrr

where

mvmrr

mvF 22

c ===

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cFr

car vr

cFr

cFr

car

car

vr

vr vr vr0Fc =

r0Fc =

r0ac =r 0ac =r

car

cFr

cFr

cFrc

ar

car

vr

vr

vr

{ Its direction is in the same direction of centripetal acceleration (directed toward the centre of the circle) as shown in figure below.

{ If the centripetal force suddenly ceases to act on a body in circular motion, the body flies off in a straight line with the constant tangential (linear) velocity as show in figure below.

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{ Definition is defined as the force acting on a body causing it to move in circular path and its always directed towards the centre of the circular path.

{ Notes :z In uniform circular motion, the net force on the system is

centripetal force.z The work done by the centripetal force is zero but the kinetic

energy of the body is not zero and given by

z The figure below shows an overhead view of a ball moving in a circular path in a horizontal plane. When the string breaks, theball moves in the direction tangent to the circle.

222 mr21mv

21K ==

r

vr

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6.4 Examples of the Circular motion 6.4.1 Conical Pendulum{ Example 7:

{ The figure shows an object with mass 40 kg is attached to one end of a string 1.50 m long. The object revolves in a circle of radius, r and makes an angle, with the vertical line. The string will break if its tension exceeds 600 N. Finda. the maximum angle, b. the maximum speed,c. the minimum periodof the object can attain without breaking the string.

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{ The weight of a conical pendulum is supported by the vertical component of the tension in figure above.Hence

{ The centripetal force is contributed by the horizontal component of the tension.Hence

{ Note:z For a conical pendulum, cannot be equal to 90.

Solution : l= 1.5 m ; m =40 kg ; T 600 N

Tr

gmr gmr

cosT

sinT

mgT =cos (1)

rmvFT

2

c ==sin (2)

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a. By using equation (1),

b.

c.

o249.max =max

maxcos Tmg=

then

mTrv sinmax =l

lr=sin

By using equation (2),

1sm63v = .maxrv

T2 == and

maxmin v

r2T =s002T .min =

mTlv

2 sinmax =

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Centre of circle

6.4.2 Motion rounds a curve on a flat (unbanked) track (for car, motorcycle, bicycle, etc)

{ Example 8:A flat (unbanked) curve on a highway has a radius of 220 m. A car rounds the curve at a speed of 25.0 m s-1. What is the minimum coefficient of friction that will prevent sliding? ( Use g = 9.81 m s-2)No. 5.44, pg. 197, University Physics with Modern Physics,11th edition, Young & Freedman.Solution :

From the free body diagram above,x-component : The centripetal force is provided by the frictional force

between the wheel (4 tyres) and the road. Hence

gmr

Nr

fr

rmvf

2

=

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y-component : The normal force is balance by the weight of the car, hence

Therefore

6.4.3 Motion rounds a curve on a banked track (for car, motorcycle, bicycle, etc)

{ Example 9:A car rounding a curve on a road banked at an angle, to the horizontal with design speed 50 km h-1. If the radius of the curve is 50 m, find the angle, at which the car can travel without skidding. Neglect the friction between the car and the road. (use g = 9.81 m s-2)(Gc.120.ex.5-13)

mgN =

rmvN

2

=

rmvmg

2

=2900.=

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Solution : v = 50 km h-1 = 14 m s-1, r = 50 m

From the free body diagram above,x-component : The centripetal force is contributed by the horizontal

component of the normal force. Hence

y-component : No vertical motion, hence

Equation (1) divided by equation (2), hence

rmvN

2

=sinmgN =cos

(1)

(2)

rgv2=tan

o22=

Centre of circle

gmr

Nr

cosN

sinNcar

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{ Notes:Motion rounds a curve on a banked track with friction (for a car, motorcycle, bicycle, etc)

From the free body diagram above,x-component : The centripetal force is contributed by both the horizontal

component of the normal force and frictional force. Hence

y-component : No vertical motion, hencer

mvfN2

cossin =+

sincos fmgN +=

Centre of circle

gmr

Nr

cosN

sinN car

frcosf

sinf

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r

6.4.4 Motion in a horizontal circle{ A figure below shows a ball of mass, m whirled at the end of a thread

in a horizontal circle.

{ The centripetal force which enables the ball to move in a circle is provided by the tension in the string. Hence

{ It is an example of uniform circular motion in which the magnitude of velocity always constant.

{ Example 10:A stone with mass 0.80 kg is attached to one end of a string 0.90 m long. The string will break if its tension exceeds 600 N. The stone is whirled in a horizontal circle on a frictionless tabletop; the other end of the string remains fixed. Find the maximum speed the stone attain without breaking the string.No. 5.43, pg. 197, University Physics with Modern Physics,11th edition, Young & Freedman.

rmvT

2

=

gmrTr

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r

Solution : m = 0.80 kg, l=0.90 m=r, T 600 N

From the figure above,The centripetal force is contributed by the tension of the string. Hence

mlTv maxmax =

rlr

mvT == where2

1max 26

= smv

gmrTr

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{ The figure shows the mass and the taut string inclined at an angle, to the vertical line.

{ The forces acting on the mass, m are the tension of the string, T and its weight mg.

{ In any circular motion,

{ By resolving forces along the string, we get

6.4.5 Motion in a vertical circle{ An object of mass, m tied to an inelastic string is moving in a vertical

circle of radius, r.

== cnet FFF rrrr

OTr

gmrsinmg

cosmgr

mvmgT2

cos =

cos2

mgr

mvT +=The tension of the string changes with the position of the object because of

z the angle, z the tangential velocity, v

changes

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Avr

Bvr

Cvr

Dvr

A

B

C

D O

gmrATr

gmr

gmr

gmr

BTr

CTr

DTr

Equation of Net forcePoint

rmvmgT

2A

A =

rmvT

2B

B =

rmvmgT

2C

C =+

rmvT

2D

D =

A

B

C

D

{ For example,

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Avr

gmrATr

O

gmr

gmr

gmrBvr

Cvr

Dvr

BTr

CTr

DTr

A

B

C

D

Total EnergyPoint

2AA mv2

1K =

mgrmv21UK 2BBB +=+

( )r2mgmv21UK 2CCC +=+

mgrmv21UK 2DDD +=+

A

B

C

D

{ Notes:z The tension of the string is greatest at A and smallest at C.z The tangential (linear) speed is minimum at C and maximum at A.z The vertical circular motion is an example of the non-uniform circular motion.z If the mass is fixed to one end of a light rod, it would be possible to rotate the

mass in a vertical circle at constant speed for example Ferris wheel (uniform circular motion) where

DCBA vvvv ===

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{ Example 11:An object of mass 1.5 kg is tied to a string of length 0.50 m. The object is made to move in a vertical circle, as shown in figure below.

Solution : m = 1.5 kg, l = r = 0.50 m, vA = 4.0 m s-1a. By using the principle of conservation of energy, we get

b.

A

B

When it reaches the highest point A, the linear tangential velocity is 4.0 m s-1. Determine a. the angular velocity of the object at B,b. the tension in the string at B.(Use g = 9.81 m s-2)

m001h .=

A

Bm500 .

= BA EEBBAA UKUK +=+

mgrmr21mghmv

21 2

B22

A +=+ 1 2.10 = sradB

2B

2B

B mrrmvT ==

NTB 0.78=

BTr

gmr

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What is the magnitude of the normal force exerted on the car by the walls of the cylinder ata. point A (at the bottom of the vertical circle)?b. point B (at the top of the vertical circle)?(Use g = 9.81 m s-2)No. 5.118, pg. 205, University Physics with Modern Physics,11th edition, Young & Freedman.

{ Example 12:A small remote control car with mass 1.60 kg moves at a constant speed of v = 12.0 m s-1 in a vertical circle inside a hollow metal cylinder that has a radius of 5.00 m as shown in figure below.

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A

B

O

Solution : m = 1.60 kg, r = 5.00 m, v = 12.0 m s-1

BNr

rmvFF

2

c ==r

mvmgN2

A =N861N A .=

rmvmgN

2

B =+N430NB .=

a. At point A,

b. At point B,AN

r

gmr

gmr

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{ Example 13:A 1000 kg sports car moving at 20 m s-1 crosses the rounded top of a hill (radius = 100 m). Determinea. the normal force on the car,b. the normal force on the 70 kg driver,c. the car speed at which the normal force is zero.(Use g = 9.81 m s-2) (Gc.128.47)Solution : m = 1000 kg, r = 100 m, v = 20 m s-1a.

b.

vr

gmr

CFFrr = Nr

rvmNgm

2car

car =N10x815N 3 .=

rvmNgm

2d

d =N10x074N 2 .=

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c.

{ Example 14: A rope is attached to a bucket of water and the bucket is then rotated in a vertical circle of 0.70 m radius. Calculate the minimum speed of the bucket of water such that the water will not spill out.(Use g = 9.81 m s-1) (Csw.CD4.2.6.3.pg11t1)Solution : r = 0.70 m

0Nr

vmNgm2

carcar == where

rgv = . 1sm331v =

vrNr

gmr{ The water will spill out when the bucket at the

top of the circle where the equation of the net force is given by

At this moment, if the water is not falling out from the bucket means the speed of the bucket is minimum and normal force, N=0 therefore

2

rmvNmg =+

2min

rmvmg =

s m.rgv -1min 622==

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Example 15: (exercise)A cyclist goes around a curve of 50 m radius at a speed of 15 m s-1. The road is banked at an angle ; the cyclist travels at the right angle with the surface of the road. The mass of the bicycle and the cyclist together equals 95 kg. Finda. the centripetal acceleration of the cyclistb. the normal force which the road exerts on the bicycle and the

cyclistc. the angle, Ans. : 4.5 m s-2, 1.02 kN, 24.6(Csw.CD4.2.6.3.pg11t4)

{ Example 16: (exercise)A 4.00 kg object is attached to a vertical rod by two strings, as shown in figure below. The object rotates in a horizontal circle at constant speed 6.00 m s-1. Find the tension ina. the upper stringb. the lower string.Ans. : 108 N, 56.2 NNo. 11, pg. 172,Physics for scientists and engineers with modern physics, Serway & Jewett,6th edition.

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v

v

{ Example 17: (exercise)A small mass, m is set on the surface of a sphere as shown in figure below. If the coefficient of static friction is s = 0.60, at what angle, would the mass start sliding?(Gc.121.83)

Ans. : 31

{ Example 18: (exercise)A rider on a Ferris wheel moves in a vertical circle of radius, r = 8 m at constant speed, v as shown in figure above. If the time taken to makes one rotation is 10 s and the mass of the rider is 60 kg, find the normal force exerted on the ridera. at the top of the circle,b. at the bottom of the circle.Ans. : 399 N, 778 N