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CIRCULAR MOTION (KINEMATICS)LECTURE

In XIII 3 Lectures in Bull's Eye 4+1 Lectures & ACME 4 Lectures including discussion.1. INTRODUCTION: Lets say you are watching F-1 rate Michael Schumacher is going on a curved path

and you want to sheet him by your handycam. If you are given two positions to stand O1 & O2 whichone will you choose?

Ans. O2 why? 2 < 1 hence easy manouvering at camera you have to cover less angle in same time. Althoughyou may not have much idea about circular motion but your decision was based on analysis of angularvariables.You decided by thinking that you have to cover smaller angle in same time.(2 and 1 are anglesubstended by M.S. while going from P1 to P2 on O2 and O1)

2. ANGULAR VARIABLES:

Angular Displacement: Angle substended by a moving particle on a fixed point is called angulardisplacement about the fixed point. Thus in above discussion angular displacement about O1 is 1 & aboutO2 is 2.

Few Facts:

Dimensionless quantity

Units radian / Never Degree

Angular displacement depends on reference frame same as linear displacement depends on refrence frame.Angular displacement will be different if we observer the car from another car.

Imp. (but angular displacement is different for different observers in the same frame). (The linear displacement issame for two observers at different positions in same frame) e.g. O1 & O2 will observe same lineardisplacement but different angular displacement although both points are in the same ground frame.

(V.Imp) It may be a bit shocking for you but it is a vector quantity. Direction of angular displacement vector isdecided by right hand rule.i.e. move your right hand fingers in sense of motion and direction of your thumbwill be the direction of angular displacement.

It sounds quite confusing that direction of vector is nowhere near the actual motion. But if you pay attentionto a few facts you will understand that perhaps this is the best way to represent angular displacement. If avector represents angular displacement it holds three informations which can completely describe the angulardisplacement.(Take a pen show it to students tell them this is the angular displacement vector and thendeduce these three facts giving them full idea of 3D view)

This is the teaching module for circular motion. This module has to be followed in the class. VK Bansal

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The unique plane perpendicular to the line represents the plane of motion of particle.Now place your right hand thumb along the vector and direction of your fingers will give you the sense ofrotation.Length of the vector will represent the magnitude angular displacement. i.e. length of vector representingfour complete rotations will be 8 and representation of one complete revolution will be by length of 2.

Note: Above discussion was for increasing comfort level of students with angular variables tell them following aswell.

In JEE syllabus plane of circular motion is fixed so the direction of angular variables remains same (althoughsense may become +ve or ve). Thus even though they are vectors but we need not give much thought toit as there directions will remain unchanged. Much like one-dimensional motion where variables are vectorsbut we do not use vector notation, we simply use signs +ve & ve to represent sense. In short we will studyangular kinematics of one-D. i.e. will never use vector notation but stick to sign +ve or ve by defining our+ve sense.

Angular velocity: Rate of angular displacement is called angular velocity.

= dt

d

Unit rad s1 / Dimension T1

Relation between angular velocity & linear velocity.

V sin

V cos

V

OP'

P

aQ

A particle P is moving with speed V along a curve & observer is located at O - is angle between linejoining OP and velocity.Note that V sin (comp. of velocity perpendicular to OP) is the cause of angular displacement. i.e. if onlyV cos existed we need not turn our head to always look at particle. Hence

PQ = (V sin ) tPQ = OP ()

Thus )OP(sinV

t

= joininglineoflength

joininglinetolarperpendicuvelocityofComp.

Asking QuestionIn circular motion find angular velocity of particle moving with speed V wrt center

= R

Valso, rv

Ex. A projectile (u,) is launched from horizontal plane, find angular velocity as observed from the point ofprojection at the time of landing.

= R

sinu

or = cosu2

g

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Objective : explain = Rv =

joininglineofLengthjoininglinetolarperpendicu

velocityofComponent

Ex. A spotlight S rotates in a horizontal plane with a constant angular velocity of 0.1rad/sec. The spot of light P moves along the wall at a distance of 3m. The velocity of the spot P when = 45 is _____ m/sec. 3m

S

Pv

Ans: V = 0.6m/s

Sol. 2d

= 3

d = m23Let velocity of spot at = 45 in V, then

= joininglineofLength

joininglinetolarperpendicuvelocityofComponent

45

45

3m

S

d

VP

= dV

0.1 = 33

2V

V = 0.3 2 2V = 0.6 m/s

Angular Acceleration:Rate of change angular velocity

= dtd

Unit rad s2 / Dimensions T2

Direction is along angular velocity if it is increasing otherwise opposite.

For uniform angular acceleration

= 0 + t , = 0t + 21

t2 , 2 = 20 + 2

Ex. 0 = 10 rad s1 and = 5 rad s2 (uniform).Find angular displacement and number of turns at t = 6 sec.

Sol. = 60 21

5p 36 = 30

Number of turns 25.

Objective : Explain why not 15 (students will give this as answer).Explain why distance is not equal to displacement.

[Home Work : Read HCV Chapter-7)

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3. KINEMATICS OF CIRCULAR MOTION:

NOTE : for XI class only (Derivation from HCV can also be used)Derivation of centripetal acceleration for a particle moving in a circle with constant speed.

2V

= 1V

= V

acceleration vector a , a = t

VV 12

V = )V(V 12

V = 2 V sin

2

or V = Vfor small ()

a = a = tV

= V = RV2

=2R

We can also prove that it is directed towards centre as angle with tangent is

2

90 in lim t 0.

lim 0 and angle 90 i.e. towards the centre Va which is towards the centre.

What we observe here is that when magnitude of velocity is constant but only direction is changing accelerationis directed perpendicular to velocity. In other words component of acceleration perpendicular to velocitycauses change in direction not magnitude.Now lets take a case of particle moving in straight line, with changing speed. Its acceleration will be alongthe line of velocity. This acceleration will change only magnitude.

Imp. Conclusion: Comp. of acceleration along velocity called tangential acceleration changes magnitude andcomp. of acceleration perpendicular to velocity called radial acceleration changes direction of velocity.

Hence taraa

.

Derivation of tangential acceleration for circular motion.

a = dt

dv

dt

Rd= R

dt

)R(d = R

dt

dv = R

understand that v is speed or magnitude of velocity, v = v & rate of change of magnitude of velocity is

called tangential acceleration. at = dtdv

= dt

Vd

= R. or )Ra( t

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Also note that dtVd

= a = 2r2t aa

i.e. dtVd

= )R()R( 222 ( while dt

Vd

= R.)

Note : One is a magnitude of rate of change of velocity & other is rate of change of magnitude of velocity (at).

dtvd|a|

and dt|v|d|a| t

Note : for XIII class only

rdtd

dtdv

= r

rdtd

dtrd

dtvda

= tc aa

Where , vac

and ra t

4. CENTRIPETAL CONDITION: When component of acceleration perpendicular to velocity is R

V2 andalways directed towards a fixed point then particle will undergo circular motion about that fixed point.

ac =ar = RV2

Ex. If a particle is undergoing circular motion with speed V and radius R angle between acceleration & V is find magnitude of tangential acceleration in terms of V, R & .

ac = RV2 = a sin

hence a = sinR

V2 at = a cos

at=

sinRcosV2 =

RV2 cot

Objective : at and ac are components of a and v&a may be at any angle

between 0 to 180

Note: dtVd

= a cos (Component of acceleration along velocity)

Valid for any type of motion (circle or not circle)

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Asking QuestionIs this diagram possible?

Ans. This acceleration is not possible as perpendicular comp. of acceleration is away from curvature.

Asking QuestionIs this diagram possible?

Ans This is possible but speed will be decreasing.

5. RADIUS OF CURVATURE: For general curvilinear motion. When the particle crosses this point A, it is

satisfying condition of moving on this imaginary circle at this instant, if a = c

2

RV / where Rc is radius of

curvature at this instant.

Rc= a

V2

Rc= velocitytolarperpendicuonacceleraticomp.of(speed)2

Ex. A projectile is launched horizontally with 20 ms1 from some height. Find Rc at t = 2 sec.Sol. 20 m/s

after t = 2 sec20

g

45

20m/s

gcos 45

220

RC = a

V2=

1022400

RC = 80 2 m

Objective : Explain Rc= a

V2 = velocitytolarperpendicuonacceleraticomp.of

(speed)2

Note: U,VVelocity / v, u speed

t)a(uvtaUV

t

but aUV

t is wrong.

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CIRCULAR DYNAMICSINTRODUCTION: Component of Net force perpendicular to velocity should be inwards and its magnitude

should be equal to

RmV 2

. This is condition for circular motion.

STEPS FOR SOLVING PROBLEMS.

Identify the plane of circular motion.

Locate the centre and calculate the radius.

Make F.B.D.

Resolve forces only and always along these three directions:(a) In the plane along radial direction.(b) In the plane along tangential direction.(c) Perpendicular to the plane of circular motion.

(a) Add the forces assuming radially inward direction as positive

rF = RmV2

(b)

(c) If plane is not accelerating then F = 0. If plane is accelerating then actually it will not be circularmotion from ground frame. But still F= ma , where a is acceleration of particle perpendicular to plane.

Ex. Find tension in OA before and after AB is cut.

Sol. Before cutting a

= 0 in all directionsRevolving vertically & horizontallyT1 cos = mgT1 sin = T1

T1 = cosmg

Objective : This discussion is being done to explain what happens when we do notfollow step (4) and resolve forces along other directions.

After cutting

T2 cos mg = 0

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The above equation is wrong because acceleration of a particle in vertical direction is not zero.while T2 mg cos = 0 is true as

T2 mg cos = m 2 l (as = 0)What if we want to write equation in vertical direction.at = g sin / ac = 0 mg T2 cos = m at sin T2 cos = mg (1 sin2 )T2 = mg cos

Ex Two different masses are connected to two light andinextensible strings as shown in the figure. Both massesrotate about a central fixed point with constant angularspeed of 10 rad s1 on a smooth horizontal plane. Findthe ratio

of tensions 2

1TT

in the strings.

[Ans. 89

]

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Sol. Drawing the FBDs for masses M1 and M2

T1 T2 = M1R12 T2 = M2R22

2

21T

TT =

2

1

2

1RR

MM

= 21

41

2

1TT

= 1 + 81

= 89

]

Objective : Centripetal force is the net force directed towards centre.

Ex. A car is moving in a circular path of radius 50 m, on a flat rough horizontal ground. The mass of the car is1000 kg. At a certain moment, when the constant speed of the car is 5 m/s, find the force of friction acting

on it? [Note : grv for uniform speed]

[Ans. N500 ]

Objective: Tell that only ext. force which can drive a car is friction. In case of car the direction of friction is not decided by velocity but according to need.

Ex. A car is moving in a circular path of radius 50 m, on a flat rough horizontal ground. The mass of the car is1000 kg. At a certain moment, when the speed of the car is 5 m/s, the driver is increasing speed at the rateof 1 m/s2. Find the value of static friction on tyres (total) at this moment, in Newtons.

[Sol. Fnet = 222

dtdv

rvm

= m 2

2

)1(21

= 52m

= 500 5 N Ans.

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Objective : Friction is providing tangential as well as centripetal acceleration both

Ex. A particle suspended from the ceiling by inextensible light string is moving along a horizontal circle of radius0.05 m as shown. The string traces a cone of height 0.1 m. Find the speed.

Conical pendulum

0.05m

0.1m

[Sol. T sin = r

mv2; & T cos = mg

v = tanrg = 0.5 m ]

Objective : Explain the STEPS FOR SOLVING PROBLEMSalso expalin why T cos = mg & not T = mg cos

Ex. In a rotor, a hollow vertical cyclindrical structure rotates about its axis and a person rests against the innerwall. At a particular speed of the rotor, the floor below the person is removed and the person hangs restingagainst the wall without any floor. If athe radius of the rotor is 2m and the coefficient of static frictionbetween the wall and the person is 0.2, find the minimum speed at which the floor may be removed. Takeg = 10 m/s2.

Nmg

fs

Ans. v = s

rg = 2.0

s/m10m2 2 = 10 m/s.

[Sol. Refer to HCV part-1 : Q.10 page-110 ]

Objective : N is not always equal to mg but any force directed towards centreis called centripetal force.

[Home work : HCV (part-1) Ex. 1 17 ]

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Note : Give the following example for reading from HCV part-1 : Q.12 page-110Ex. A metal ring of mass m and radius R is placed on a smooth horizontal table and is set rotating about its own

axis in such a way that each part of the ring moves with a speed v. Find the tension in the ring.[Sol. Refer to HCV part-1 : Q.12 page-110 ]

Sol. Consider a small part ACB of the ring that subtends an angle at the centre as shown in figure. Let thetension in the ring be T.

/2/2 C

B

T

T

A

O

The forces on this small part ACB are(a) tension T by the part of the ring left to A,(b) tension T by the part of the ring right to B,(c) weight (m)g and(d) normal force N by the table.The tension at A acts along the tangent at A and the tension at B acts along the tangent at B. As the small partACB moves in a circle of radius R at a constant speed v, its acceleration is towards the centre (along CO)and has a magnitude (m)v2 / R.Resolving the forces along the radius CO,

T cos

2

90 + T cos

2

90 = (m)

Rv2

or, 2T sin 2

= (m)

Rv2

... (i)

The length of the part ACB is R. As the total mass of the ring is m, the mass of the part ACB will be

m = R2m

R =

2m

.

Putting m in (i),

2T sin 2

= 2

m

Rv2

or, T = R2mv2

)2/(sin2/

1 and T = R2

mv2

]

Objective : Net force on the ring is zero but still it has tension.

Q. A block of mass m is sitting on a rotating frictionless wedge. The wedge rotates with constant angularvelocity around the axis shown in figure.

Calculate the value of such that the block stays at constant height h. (express your answer in terms of g,h,)

Sol. The velocity of P point on the wedge. Velocity of wedge at point P = r = tan

h

Drawing a free body diagram for the block.

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Resolving forces as indicated in step 2 .4

rh

= tantan

h = r

Ncos mg = 0 (ydirection)Nsin = m2r

tan = gr2

=

tang

h2 2 =

htang 2

= tanhg

Ans.]

Obejctive : Explain the concept of banking.

Explain banking then take following as example [Refer HCV Ch.-7 (7.5) ]What is the speed required to negotiate the turn shown in figure. Frictionless and radius of curvature 'R' andbanking ''.

mgf

RC

O N

inside

N cos

vN sin

N cos mg = 0N cos = mg

N sin = R

mv2

mg tan = R

mv2

v = tanRg

Q. A block of mass 25 kg rests on a horizontal floor ( = 0.2). It is attached by a 5m long horizontal rope to apeg fixed on floor. The block is pushed along the ground with an initial velocity of 10 m/s so that it moves ina circle around the peg. Find

(a) Tangential acceleration of the block (b) Speed of the block at time t.(c) Time when tension in rope becomes zero.

[Ans. (a) 2 m/s2, (b) 10 2t, (c) 5 sec ][Sol.(a) Tangential acceleration is the retardation produced by the friction

a = f/m = mg/mat = 0.2 10 = 2 m/s2

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(b) dtdv

= at = 2

v

10

dv = t

0

dt2

v 10 = 2tv = 10 2t

(c) Tension in the rope will become zero when centripetal acceleration becomes zeroi.e. when speed becomes zerov = 0 10 2t = 0 t = 5 sec

Objective : Force directed towards centre is centripetal & force in the velocitydirection is tangential force.Tangential acceleration is responsible for change in speed.If friction is kinetic it is opposite to velocity.If there is slipping on surface then net friction is kinetic.

[Refer HCV Ch.-7 (7.6) ]CENTRIFUGAL FORCE: If reference frame is particle (undergoing circular motion) itself then it willexperience pseudo force which will be radially outward and equal to m(w2R).Do not use it to explain any of the e.g. and let students solve all problems w/o it.For man outside N = mw2R

For man inside N mw2R = 0.

This is called centrifugal force.Solved e.g. 13 H.C.V. Chap.7 take illustration in class without mentioning coriolis force{Home work : HCV 18 30 }Leave Effect of rotation of earth (to be discussed with gravitation.)

Note : For any suggestion or correction please contact Amit Gupta or give it in computer room.