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CHAPTER 1 : CIRCULAR MOTION ( 6 MARKS)

PART 1: UNIFORM CIRCULAR MOTION

Definition: The motion of a body moving in circular path with a constant speed iscalled Uniform Circular Motion.

Characteristics (Features) of Uniform Circular Motion: The speed of the body is constant The magnitude of acceleration of the body is constant. The angular velocity of the body always remains constant. The kinetic energy of the body remains constant. The direction of the acceleration is towards the center of the circle. The linear velocity of body is along the tangent to the path.

TERMS AND DEFINITIONS RELATED TO UCM

Time taken by a body in uniform circular motion to complete one revolution iscalled the Period of Revolution.

Number of revolutions completed in unit time by a body in uniform circularmotion is called the Frequency of Revolution. The SI unit of frequency is Hz

Angle traced by radius vector in a given time is called Angular Displacement.For angular displacement the SI unit is radian and dimension formula is[M0L0T0] i.e. it is dimensionless

Angular Velocity: The rate of change of angular displacement with time iscalled angular velocity.The SI unit of angular velocity is rad/sec and dimension formula is [M0L0T-1]

Angular Acceleration: Rate of change of angular velocity with time is calledangular acceleration. The SI unit of angular acceleration is rad/sec2 andthe dimension formula is [M0L0T-2]

The directions of angular displacement, angular velocity and angular accelerationare given by the right hand rule:

angularvelocity

increasingangularvelocity

decreasingangularvelocity

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A

rB

V

s

Right Hand Rule: Consider the axis of rotation in the right hand with fingers curledround the axis and thumb stretched along the axis , then curled fingers show senseof rotation, and the thumb shows direction of angular displacement, angular velocityand angular acceleration

KINEMATICAL EQUATIONS FOR CIRCULAR MOTION For Linear Motion: Let u = initial linear velocity, v = final linear velocity, a =

linear acceleration, s = linear displacement. The kinematic equations for linearmotion are :

v = u + a t ; v2 = u2 + 2 a s ; s = u t + ½ a t2

For circular motion: Let = initial angular velocity, = final angularvelocity,= angular acceleration, = angular displacement. By analogy, the kinematicequations for angular motion are :

= 0 + t ; 2 = 2 + 2 = t + ½ t2

Questions:

1. Define Uniform Circular Motion and give its characteristics2. In Uniform Circular Motion define the following : (a) Frequency of revolution

(b) Period of revolution3. Define: Angular displacement, Angular velocity and Angular acceleration. Also,

give their directions and S.I. units.4. State kinematical equations for circular motion in analogy with linear motion.5. Give the vector representation of angular velocity.

DERIVATION: Relation between linear speed and angular speed in UCM.

The distance covered in unit time by a particle in uniform circular motion iscalled the Linear Speed. The angle traced in unit time by a particle in uniformcircular motion is called Angular Speed.

Consider a particle performing uniform circular motion asshown. Let : r = radius of the orbit ; = angular velocity ;v = linear velocity. Suppose that the particle moves distanceAB = s in time t.Using : arc AB = radius angle, we have

s = r. t

.r=ts

t

.rlim=tslim

0t0t

tlim.r=

tslim

0t0t

Also by definition we have :

ts

limdtds

v0t

andt

limdtd

0t

Hence we get : v = r i.e. linear speed = radius angular speed. This equationgives magnitude of linear velocity.

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Direction of Linear Velocity: If t then points A and B are very close to each

other. In this case displacement vector AB

is tangent to circle i.e. v

is tangent to

the circle. Thus v

and r

are in plane of circle and perpendicular to each other.

Also is perpendicular to plane of circle. Thus v

, and r

are related to

each other by right hand rule. In vector form we can write: v

=

r

DERIVATION : Linear Acceleration of particle in UCM (Vector Method)

Consider a particle performing uniform circularmotion as shown. Suppose in time t the particlemoves from point A to point B.

Let : = angular velocity of the particler = radius of the circular path

v1

= velocity vector of particle at A

v2

= velocity vector of particle at B

The speed is constant, hence |v1

| = |v2

| = v . However v1

and v2

havedifferent directions. To find the change in velocity v we use the method of

vectors. Suppose PR

represent the velocity vector at A and PQ

represent thevelocity vector at B. Then by triangle law of vectors we have:

PR

+ RQ

= PQ

RQ

= PQ

– PR

RQ

= Velocity at B – Velocity at A

Thus RQ

represents the change in velocity v

RQ = v (in magnitude) . . . (1)

Also when the angle is small i.e. we have

RQ = v. . . . (2)

Hence, from (1) and (2) we get : v = v

t

.v=tv

t.vlim=

tvlim

0t0t

tlim.v=

tvlim

0t0t

But by definition we have :tv

limdtdv

a0t

andt

limdtd

0t

linear acceleration = a = v

Ar

r

v1

v1

v2

v2

vB

P

RQ

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This gives magnitude of linear acceleration. Also using v = r , we can write :

Linear acceleration: a = v= r2

Direction of Linear acceleration: If t then points B A are very close toeach other. This means v is along radius and towards center of circle. Alsoacceleration a and v have same direction.

Hence in uniform circular motion, the acceleration of particle is along radius andtowards center of circle. This acceleration is called centripetal acceleration. Also

in vector form we write:

ra 2 where negative sign shows thata and r

have opposite directions.

DERIVATION: Linear Acceleration of particle in UCM (Calculus Method)

Consider a particle performing uniform circular motionas shown. Suppose that the particle moves from pointA to point P in time t.The angular displacement of particle is = t , wherer = radius and = angular velocity. From the figure:

OM = r.cos= r.costON = r.sin= r.sint

Let i

and j

be the unit vectors along the X-axis and

Y-axis respectively. Then position vector OP

is givenby :

r

= OM

+ ON

r

= i

r.cost + i

r.sint

r

= r( i

cost + i

sint )

To find instantaneous velocity, differentiate the above equation w.r.t. time

tcosjtsini.rtsinjtcosidtd.r

dtrd

tcosjtsini.rv

To find instantaneous acceleration, we differentiate this equation w.r.t. time

rtsinjtcosir

tsinjtcosirtcosjtsinidtd

.rdt

vd

22

ra 2

O

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This equation gives both the magnitude and direction of linear acceleration. The

magnitude of linear acceleration is a = r 2 and negative sigh shows thata and

r

have opposite directions. This acceleration is called centripetal acceleration

Questions:

1. For a body in uniform circular motion, obtain the relation between linearspeed (velocity) and angular speed (velocity) in uniform circular motion.

2. Derive the expression for (linear) acceleration of a particle performing UniformCircular Motion.

3. Using calculus method, derive the expression for linear acceleration of aparticle performing uniform circular motion.

CENTRIPETAL AND CENTRIFUGAL FORCES

A body performs circular motion with help of centripetal force. This force isdirected towards the centre of the circular path. Definition: The radial force towards the centre of circle when a particle

performs uniform circular motion is called centripetal force. Example: When a satellite moves in a circular orbit round earth the

centripetal force is provided by gravitational force of attraction between earthand satellite.

Formula: Centripetal force = mv2/r = m2r , where m = mass of body, v =linear velocity, r = radius of orbit , and = angular velocity

A body in circular motion, experiences a force in outward direction. This force isaway from the center of circle and is called centrifugal force.

Definition: The radial force away from the centre of the circle when a particleperforms uniform circular motion is called centrifugal force.

Example: Passengers inside a bus, which is taking a turn, will experience acentrifugal force. This force tries to push them in outward direction.

Formula: Centrifugal force = mv2/r = m2r , where m = mass of body, v =linear velocity, r = radius of orbit , and = angular velocity

Centrifugal force and centripetal force are equal but they have oppositedirection. The centrifugal force is not due to any interaction between bodies. Itis due to acceleration of frame of reference. Centrifugal force is an example ofpseudo force i.e. it is not a real force.

Questions:

1. What do you understand by centrifugal force?2. What do you understand by centripetal force?3. Distinguish between centripetal force and centrifugal force

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Centripetal force Centrifugal force

It is directed towards the center ofcircle along the radius

It is directed away from the center ofcircle along the radius

It is a real force It is a pseudo force

It exists in an inertial frame ofreference

It exists in an accelerated frame ofreference

BANKING OF ROADS

Banking of roads: This is an arrangement in which the outer side of a curvedroad is raised to higher level than the inner side. The angle between thehorizontal and the inclined road surface is called the angle of banking.

Need for banking: Consider a vehicle to move along a circular path Centripetal force is required to move along a curved path. This is provided by

the friction between the road and the tyres. For a fast moving vehicle, this friction force is very small and there is danger

of being thrown off the road. In addition, friction causes wear and tear oftyres.

These problems can be solved by banking of roads. The vehicle is not thrownoff the road and it can round the curve safely.

When a road is banked at a proper angle, the required centripetal force isprovided by the horizontal component of the normal reaction.

DERIVATION: Expression for Maximum Speed on banked road.Expression for Angle of Banking for a banked road.

Consider a car rounding a curve. Let : m = mass of the vehicle ; v = maximumvelocity of the car ; = angle of banking ; r = radius of the curve

As shown in figure, the two forces acting on the car are:Normal reaction (N) and the Weight (mg) of the carThe normal reaction is perpendicular to the road surface.It can be resolved in two components as:

Vertical component is N.cos: This componentbalances weight mg of car

N cos = mg . . . . . (1) Horizontal component is N.sin: This component

provides centripetal forceN sin= mv2/ r . . . . . (2)

From (1) and (2) we get:mg

rmvcosNsinN 2/

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tan = v2 / rg tanrgv .

This expression gives the maximum speed. We also have : = tan-1

rgv 2

This expression gives the Angle of Banking. We find that the angle of bankingdepends on: The maximum speed (v) the vehicle The radius (r) of curve The acceleration due to gravity (g).Also the expression for angle of banking does not involve mass (m) of the vehicle.It means that angle of banking does not depend on mass of vehicle.

Questions:1. What is banking of roads? Why is it necessary?2. Derive an expression for the maximum speed on a banked road.3. Derive an expression for the angle of banking for a banked road.4. Write the expression for angle of banking. State the factors on which the

angle of banking depends.5. Show that the angle of banking does not depend on mass of a vehicle.

CONICAL PENDULUM; MOTION IN HORIZONTAL CIRCLE

Conical Pendulum: A conical pendulum consists of a body (bob) attached to oneend of a light inextensible string. The bob moves in horizontal circle with uniformspeed and the string describes a cone of semivertical angle .

DERIVATION: Expression for Speed of bob in case of conical pendulum

Consider a conical pendulum as shown. Let : m =mass of bob ; l = length of the string ; v = speed ofcircular motion of bob ; r = radius of path.

At point P, the forces acting on bob are : TensionForce (T) in string and the Weight (mg) of bob. Thetension force can be resolved in two components as :

Vertical component – T.cos: This componentbalances weight mg of bob

T cos = mg . . . (1) Horizontal component – T.sin: This component

provides centripetal force required for circularmotion.

T sin=mv2

r . . . (2)

Divide equation (2) by equation (1) and simplify to get :

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tan =

rgv2

This expression gives the inclination of the string.We also have: v = rg tanThis expression gives the speed of bob of a conical pendulum.

Related Notes:

Expression for Period: The period of conical pendulum is defined as – thetime required by the bob to complete one revolution.

tan.gr2

tan.rgr2

vr2

SpeednceCircumfereT

Also from the diagram we have: r = l sin

Period: T = 2lcos

g Expression for tension: The tension in string is calculated as follows:

Squaring and adding equations (1) and (2) we have

T2 sin2+ T2 cos2 =m2v4

r2 + m2g2

T2 ( sin2 + cos2 ) = m2 (v4

r2 + g2 )

T2 ( 1 ) = m2 g2 (v4

r2g2 + 1 )

1gr

vmgT 22

4

This gives the string tension. We find that string tension is directlyproportional to the weight (mg) of bob

Questions:

1. In the case of a conical pendulum, show that the inclination of the string withthe vertical is given by tan = v2/rg

2. What is a conical pendulum. Derive the expression for periodic time of aconical pendulum with string length = l and bob moving in horizontal circle ofradius r

3. For a conical pendulum obtain the expression for : (a) speed of bob (b) period4. For a conical pendulum derive the expression for string tension

5. For a conical pendulum show that : T4 = 164

L2 – r2

g2

MOTION IN VERTICAL CIRCLE

DERIVATION: Expression For Minimum Velocity at highest and lowest points foruniform circular motion in a vertical circle.

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Consider a body tied to the end of a string and whirled round in a vertical path sothat it just completes the vertical circle. As shown A is the lowest point and B isthe highest point.

Let m = mass of body ( weight of body = mg )r = radius of the circular pathVA = velocity of body at point AVB = velocity of body at point BTA = tension in string at point ATB = tension in string at point B

At point A, a part of tension balances the weight ofbody. The remaining part provides the centripetalforce.

mgTr

mvA

2A . . . (1)

At highest point B, centripetal force is provided by thetension and weight.

mgTr

mvB

2B . . . (2)

Minimum Velocity at Top: For minimum velocity at top, the tension at highestpoint will be, zero. In this case, equation (2) becomes

mvB2

r = 0 + mg vB = gr

This is the minimum velocity at point B for which the body will just completethe circle. If velocity at B is less than gr , the body will not complete the circleand the string will slack.

Minimum Velocity at Bottom: To find the minimum velocity at the bottom, weuse the principle of conservation of energy.We have: P.E. at A = 0

P.E. at B = mgh (where h = 2r )K.E. at A = ½ mvA2

K.E. at B = ½ mvB2 (where vB = gr )

The total energy at A is: EA = P.E. at A + K.E. at A EA = 0 + ½ mvA2

EA = ½ mvA2

The total energy at B is: EA = P.E. at B + K.E. at B EB = mgh + ½ mvB2

EB = mg(2r) + ½ m(gr)

EB =52 mgr

The total energy is conserved, hence: EB = EB

A

O

VA

TA

TB

mg

mg

VB B highestpoint

lowestpoint

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12 mvA2 =

52 mgr mgr5mv 2

A

gr5v 2A gr5v A

Related Notes: Total Energy of Body at various points on Circle

As shown in figure, consider the position of a body atpoints A, B and C on the circle. We have to find totalenergy at these points.

At Point A : The body is at lowest point i.e. h = 0 .Hence body has only K.E.

Total energy at A = ½ mVA2 + mgh= ½ m( gr5 )2 + mg(0)

=25

mgr + 0 =25

mgr

At Point B: The body is raised through height = 2r.Hence the body has both K.E. and P.E.

Total Energy at B = ½ mVB2 + mgh = ½ m( gr )2 + mg(2r)

= ½ mgr + 2 mgr =25 mgr

At Point C : The body is raised through height = r. Hence the body has bothK.E. and P.E.

Total Energy at C = ½ mVC2 + mgh = ½ m( gr3 )2 + mg(r)

=23

mgr + mgr =25

mgr

We find that: Kinetic energy and potential energy of the body in vertical circularmotion changes but total energy at every point is same. Thus, energy is conservedin vertical circular motion.

The energy i.e.25

mgr is also the minimum energy that a body must have to be

able to perform vertical circular motion.

SHOW THAT: In vertical motion for a body tied atthe end of a string, the difference in string tension athighest and lowest points is 6 times the weight ofbody

Consider a body tied to end of a string and whirledround in a vertical path so that it is just completesthe vertical circle. As shown A is lowest point and Bis the highest point.

Let : m = mass of body ( weight of body = mg )r = radius of the circular path

A

O

VA = 5gr

VB = gr

VC = 3grB

C

A

O

VA

TA

TB

mg

mg

VB B

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VA = velocity of body at point AVB = velocity of body at point BTA = tension in string at point ATB = tension in string at point B

At point A, a part of tension balances the weight of the bogy. The remaining partprovides the centripetal force. Hence

mgTr

mvA

2A . . . (1)

At highest point B, centripetal force is provided by the tension and weight. Hence

mgTr

mvB

2B . . . (2)

Subtracting equation (2) from (1) we get:

mg2TTr

mvr

mvBA

2B

2A

mg2TT)vv(rm

BA2

B2

A

rgvand5grv:Substitute BA

mg2TT)grgr5(rm

BA mg2TT)gr4(rm

BA

mg2TTmg4 BA BA TTmg6 mg6TT BA

Questions

1. A stone is tied to end of a string and whirled round in a vertical path so that itis just able to complete the vertical circle. Find minimum velocity at thehighest and lowest points.

2. Show that, in the case of vertical motion of s body tied at the end of a string,the difference in string tension at the highest and lowest points is 6 times theweight of the body

3. In vertical motion of a body, obtain the energy at the minimum and maximumheights. Show that the energy is conserved in vertical circular motion.

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NUMERICAL PROBLEMS

SET 1 – BANKING OF ROADS: = tan-1

rgv2

; tanrgv .

1. Find the angle of banking for a curved road of radius 100 m if the limitingsafety speed of a vehicle is 108 km per hour. Neglect friction ( g = 9.8 m/s2 )

2. Calculate the maximum speed with which a car can be driven safely along asmooth curved road of radius 80m banked at 21º34’ with horizontal.

3. Calculate the maximum speed with which a car can be driven safely along acurved road of radius 100m if the coefficient of friction between the tyres andthe road is 0.2 ( take g = 9.8 m/s2 )

4. Calculate the maximum speed with which a car can be driven safely along acurved road of radius 30m if the coefficient of friction between the tyres andthe road is 0.3 ( take g = 9.8 m/s2 )

5. A road is 6 m wide and elevation of the outer edge of road is 2 m. Calculatethe limiting safety speed for a car on the road if radius of curvature path is25 m ( take g = 9.8 m/s2 )

6. A motor car is travelling at a speed of 80 km/hr along a curve of radius 40 m.If the road is banked at an angle of 30º will the car be safe? (Use g=9.8 m/s2)

Answers:(1) 42º34(2) 17.71 m/s (3) 14 m/s (4) 9.39 m/s (5) 9.035 m/s (6) The max.safe speed is 15.04 m/s. The car travels with a speed more than max. safe speed.So the car is not safe.

SET 2 – BANKING OF RAILWAY TRACKS

1. A train rounds a curve of radius 150 m at a speed of 20 m/s. Calculate (a)the angle of banking so that there is no side thrust on the rails. (b) theelevation of the outer rail over the inner rail if the distance between the rails is1 metre.( g = 9.8 m/s2 )

2. A train rounds a curve of radius 800 m for which the maximum safe speed is54 kmph. Calculate the angle of banking. Also find the elevation of the outerrail over the inner rail, if the distance between the rails is 1.8 m ( g = 9.8 m/s2)

3. A train of mass 105 kg rounds a curve of radius 150 m at a speed of 20 m/s.Find the horizontal thrust on the outer rail if the track is not banked. At whatangle must the track be banked so that there is no side thrust on the rails.( g = 9.8 m/s2 )

Answers:(1) Angle = 15º13; elevation = 0.2625 m (2) Angle = 1º39; elevation = 0.05166m (3) 2.667 105 N ; 15º 13

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SET 3: MOTION IN HORIZONTAL CIRCLE

1. A certain string breaks under a tension of 45 kg wt. A mass. A mass of 100grams is attached to this string of length 500 cm and whirled in a horizontalcircle. Find the maximum number of revolutions per second without breakingthe string

2. A body of mass 1 kg is tied to the end of a string and revolved in horizontalcircle of radius 1 metre. Calculate the maximum number of revolutions perminute performed by body so that the string should not break. The breakingtension of the string is 9.87 N

3. A certain string breaks under a tension of 10 kg wt. A mass of 400 grams istied to this string of length 1 m and whirled in a horizontal circle. Find :(a) tension in string if mass is whirled with speed of 2 m/s (b) the maximumnumber of revolutions per second without breaking the string

4. A body of mass 3 kg is tied to the free end of a string of length 2 m andrevolved in a horizontal circle making 420 rpm. Calculate linear velocity,centripetal acceleration and the centripetal force on the body

5. An object of mass 400 g is whirled in a horizontal circle of radius 2 m. If itperforms 60 rpm. Calculate the centripetal force acting on it.

6. An object of mass 0.5 kg is whirled in a horizontal circle of radius 20 cm. If itperforms 0.6 radians/second. Calculate the centripetal force acting on it.

Answers:(1) n = 4.726 Hz (2) n = 30 rev/min (3) T = 1.6 N ; n = 2.492 Hz (4) v = 87.98m/s ; a = 3870 m/s2 ; F = 11610 N (5) F = 31.55 N (6) F = 0.036 N

SET 4: MOTION IN VERTICAL CIRCLE

1. A motor cyclist rides in vertical circles in a hollow sphere of radius 4 m. Findthe minimum speed required so that he does not lose contact with the sphereat the highest point ( take g = 9.8 m/s2 )

2. A motor cyclist rides in vertical circles in a hollow sphere of radius 3 m. Findthe minimum speed required so that he does not lose contact with the sphereat the highest point ( take g = 9.8 m/s2 )

3. A small spherical body is tied to a string of length 0.5 m and revolved in avertical circle such that the tension in string is zero at the highest point.Calculate linear speed at the highest and lowest positions.

4. The vertical section of a road over a bridge is in form of an arc of radius 4.4 m.Find maximum speed to cross the bridge without losing contact with road thehighest point. The center of gravity of the vehicle is 0.5 m above the ground.

5. The vertical section of a road over a bridge is in form of an arc of radius 100m. Find maximum speed to cross the bridge without losing contact with theroad at highest point.. The center of gravity of the vehicle is 1 m above theground.

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6. A bucket containing water is tied to the end of a rope 6 m long and rotated ina vertical circle. Find the maximum number of rotations required so that waterin the bucket may not spill.

7. A bucket containing water is tied to the end of a rope 8 m long and rotated ina vertical circle. Find the maximum number of rotations required so that waterin the bucket may not spill.

8. A stone of mass 0.5 kg tied to the end of a string of length 40 cm. The stone isrevolved in a vertical circle so that the speed at the lowest point is 3 m/s. Findthe tension in the string at this point.

Answers:(1) v = 6.26 m/s (2) v = 5.422 m/s (3) v = 4.95 m/s (low) ; v = 2.214 m/s (high)(4) 6.93 m/s2 (5) 31.46 m/s2 (6) n = 12 rpm (7) n = 10.58 rpm (8) T = 16.15 N

SET 5: LEANING ANGLE WITH THE VERTICAL

1. Find the angle which the bicycle and its rider will make with the vertical whengoing round a curve at 54 km/hour on a horizontal curved road of radius 50m( g = 9.8 m/s2 )

2. A bicycle rider going at 72 km/hour on curved road. The angle, which bicycle,and its rider makes with vertical is 45º. Calculate radius of track (g = 9.8m/s2)

3. Find the angle, which a bicycle and its rider will make with vertical when goingon curved road of radius 10 m with a speed of at 18 km/hour. ( g = 9.8 m/s2 )

Answers:(1) 24º 40(2) 40.82 m (3) 14º 19

SET 6: ANGULAR VELOCITY, ANGULAR ACCELERATION

1. The minute hand of a clock is 10 cm long. Calculate the linear speed of the tipof the minute hand.

2. A body having a mass 0.2 kg moves in circle of radius 2 m and performs 120rev/min. Calculate the following: (a) period (b) angular speed (c) linear speed(d) centripetal force (e) kinetic energy of body.

3. The frequency of a particle performing uniform circular motion changes from60 rpm. to 180 rpm. in 20 seconds. The radius of circular path is 20 cm.Calculate: (i) angular acceleration (ii) tangential acceleration

4. The frequency of a particle performing uniform circular motion changes from 2Hz to 4 Hz in 2 seconds. Calculate angular acceleration of the particle

5. The frequency of a spinning particle is 10 Hz and it is brought to rest in 6.28s. Calculate the angular acceleration of the particle.

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Answers:(1) linear speed = 1.746 10–4 m/s (2) period = 0.5 sec ; angular speed = 12.56rad/s ; linear speed = 25.12 m/s ; centripetal force = 63.168 N ; kinetic energy =63.1 joule (3) angular acceleration = 0.6284 rad/s2 ; tangential acceleration =0.12568 m/s2 (4) 6.28 rad/s2 (5) – 10 rad/s2

SET 7: MISCELLANEOUS PROBLEMS

1. To stimulate acceleration in space rockets, astronauts are spun at end oflong rotating beam of radius 9.8 m. What angular velocity is required togenerate a centripetal acceleration 8 times acceleration due to gravity ? ( g =9.8 m/s2 )

2. A coin kept on a horizontal rotating disc has its center at a distance of 0.1m, from the center of the axis of rotation of the disc. If the coefficient of frictionbetween the coin and the disc is 0.25, find the angular speed of the disc atwhich the coin would be about to slip off ( take g = 9.8 m/s2 )

3. At what angular speed should the earth rotate about its own axis so that theapparent weight of the body on the equator is zero. Also calculate the periodin this case ( Radius of earth = 6400 km ; g = 9.8 m/s2 )

4. A conical pendulum has length 1 m and the angle made by the string with thevertical is 10º . The mass of the bob is 0.2 kg. Find the velocity, centripetalforce, tension in string and period of circular motion of bob. ( use g = 9.8 m/s2

; sin10º = 0.17365 ; tan 10º = 0.17633 ; cos 10º = 0.98480 )

Answers:(1) = 2.828 rad/s (2) 4.950 m/s (3) = 1.237 10–3 rad /sec ; Period = 1 hr25 min (4) Hints : r = l sin = 0.17365 m ; v = rg tan= 1.732 m/s ;

centripetal force = mv2/r = 3.456 N ; Tension = mgv4

r2g2+1 = 3.972 N ; period

= 2lcos

g = 1.99 s