ch#5 circular motion

8/12/2019 Ch#5 Circular Motion

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Unit # 5

Circular motionAbrar Ahmad

(ZAAS Academy 0333-5307019)

Q1 Define 1) Angular M!in ") #ircular M!inAns. 1. ANGULAR MOTION

The motion of a body along a curvilinear path is called angular motion.EXAMPLES1)Motion of a car taking turn.2)Motion of snake.2. CIRCULAR MOTIONIt is special case of angular motion in which a body moves in a circular path.EXAMPLES:1)Motion of motorcyclist in the well of death.

2)Motion of the moon around the earth (in an approximate circular motion because in fact it ellipticalangular motion. !ame is true for the motion of electron around the nucleus.

Q2 a) Define and explain angular displacement?b) Establish a relation between linear and angular displacement? ! "ro#e that $%rc) "ro#e that& 1 radian %'(.

Ans. *a) ANGULAR DISPLACEMENTIt is defined as the angle traced at the center of the circle by the line "oining the body with the center ofthat circle.

NATURE:#or small values the angular displacement can be regarded as a vector $uantity otherwise it is ascalar $uantity.SYMBOL:The angular displacement is represented by or $

SI UNIT: The !I unit of angular displacement is radian denoted by rad.

DIMENSIONS:%ngular displacement is a dimensionless $uantity.EXPLANATION

&onsider a particle attached to one end of a mass less rigid rod of length 'r with its other end pivoted atpoint ') taken as origin. !uppose the particle rotates anticlockwise in xy*plane+ about ,*axis serving as

axis of rotation. %s the particle is moving along the circular path+ the rod %&rotates in the plane of thecircle. &onsider a system of as shown in fig(a.

#or position -1+ let 1is angle subtended by the rod with reference line %&+ at time '!1. If " is anglesubtended by the rod for position&"at time '!" with reference line %&then angle traced by the rod inmoving from position&1to position&"is given by

&1%&"' " - 1---+++++++++++++*1)

Time interval taken by particle in going from position&1to position&"is ! ' !" - !1******************(/

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The angle given by e$ (0 is defined as the angular displacement of the particle in time interval !. #orsmall values of !+ the angular displacement can be regarded as vector $uantity otherwise it is a scalar$uantity.

DIRECTION OF ANGULAR DISPLACEMENTThe direction of angular displacement is taken along the axis of rotation and is determined by a rule knownas right hand rule. It is described as under

RIGHT HAND RULE

%ccording to right hand rule1rasp the axis of rotation in the right hand side with fingers curling in the direction of rotation+ the thumb

points in the direction of angular displacement.

UNITS OF ANGULAR DISPLACEMENTThe angular displacement is measured in three units including two conventional units and one !I unit.(a CONVENTIONAL UNITS:Degree

The angle subtended by 02345th

part of circumference of circle at its center is called one degree.REVOLUTIONThe complete round trip of the body along the circumference of the circle is called one revolution

0 rev6 345(b SI UNIT:Radia)ne radian angle is defined+ as the angle made at the center of the circle by an arc whose length is e$ualto the radius of the circle.It can be proved that

/rad 6 0rev 6 345

(bRELATION BET!EEN LINEAR DISPLACEMENT AND ANGULAR DISPLACEMENTOR PROOF OF S"r&onsider a body moving on circle of radius 'r and centered at point ). !uppose arc %7 makes an angle at the center of the circle. If ! is length of arc %7 then

%rcA ' S *********************(0

A% ' ***********************2)

&onsider another arc %& of the circle whose length is e$ual to the radius 'r of the circle. Then this arc %&makes an angle of 1 radat the center of the circle so that

%rcA# ' r **********************(3

A%# ' 1 rad******************(8%s clear from the figure that the length of an arc is directly proportional to angle made by it at center ofcircle i.e.

A A%

/


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A ' A%********************(9:here 'k is proportionality constant.

!imilarly A# A%# A# ' A%# ********************(4

;ivide e$ (9 by e$ (4 to get

)(...*....................A%#)

A%()

A#

A(

we get

1

r

S

S ' r :here angle is measured in radian.

(c PROOF OF 1 radia "#$.%?elation between arc length ! and radius of circle r is

S ' r #or one complete revolution ! 6 /r+ the circumference of circle

' 3*0 !o that"r ' r

"r rad ' r 3*0 "rad ' 3*0

rad ' 1+0 1 rad '1+0, 1 rad ' 57$3

Q *a) Define and explain the angular #elocit,?*b) Establish a relation between linear #elocit, and angular #elocit,? ! "ro#e that&

= r@%ns. ANGULAR VELOCITY

&The angular velocity is defined as the time rate of change of angular displacement.NATURE:#or small values of angular displacement or time interval+ the angular velocity can be regardedas @A&T)? B


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#or position -1+ let 1is angle subtended by the rod with reference line %&+ at time '!1. If " is anglesubtended by the rod for position&"at time '!" with reference line %&then angle traced by the rod inmoving from position&1to position&"is given by

&1%&"' " - 1

Time interval taken by particle in going from position&1to position&"is

! ' !" - !1Then ratio is defined as magnitude of angular velocity

!

DIRECTION OF ANGULAR VELOCITYThe direction of angular velocity is taken along the axis of rotation and is determined by a rule known asright hand rule.

RIG'T 'AND RULE%ccording to right hand rule1)!tretch right hand with thumb to palm.2)1rasp the axis of rotation in stretched right hand and curl the fingers along the sense of rotation thenthumb indicates the direction of angular velocity.#igure (b illustrates for the use of right hand rule for the direction of angular velocity of a bodyundergoing the anticlockwise angular motion along a circle in xy*plane.

UNITS OF ANGULAR VELOCITY

CONVENTIONAL UNITS: There are two conventional units of angular velocity as under(a ?evolution per second (rev2sec(b ;egree per second (deg2secSI UINT:The !I unit is rad2sec.

TYPES OF ANGULAR VELOCITY%ngular velocity has following four types.

1( AVERAGE ANGULAR VELOCITY:It is defined as the ratio of total angular displacement and total time interval.

SYMBOL:It is donated by av.FORMULA: If (total is total angular displacement of body in total time interval t(total thenmagnitude of average angular velocity is given by

)*

)*

!!al

!!al

!

2( INSTANTANEOUS ANGULAR VELOCITYIt is defined as the angular velocity of a boy at a particular instant.SYMBOL:It is denoted by ins.

#)?M


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Def& If a body covers e$ual angular displacements in une$ual time intervals or vice versa+ how short these timeintervals may be+ the body has variable angular velocity.ote& If a body has variable angular velocity then

av F ins

RELATION BETWEEN LINEAR VELOCITY AND ANGULAR VELOCITY

If ! is length of the arc that subtends an angle at the center of the circle of the radius r then!6r

%pply the differential change on both sides to get!6(r!6r

;ivide both sides by t to get

%pply limit to gettt

S

=

rlimlim

t5Eimit ! 6 magnitude of instantaneous linear velocity.t5 t

6 @ (sayEimit 6 magnitude of instantaneous angular velocity.t5 t

6 (say@6r

In @ector form it can be written as

= rV

Q.o9. a) Define and explain angular acceleration?

b) Establish a relation between linear acceleration and angular acceleration

!

"ro#e that&

= ra

A$:E!& A-3A! A55E3E!A6/

Def&The angular acceleration is defined as the time rate of change of angular velocity.

A6-!E&

#or small values of change in angular velocity or time interval+ the angular acceleration can beregarded as a vector $uantity otherwise it is a scalar $uantity.

$783& The angular acceleration is denoted by (alpha.The angular acceleration is denoted by (alpha.0!-3A& If is the change in angular velocity during the time interval t+ then magnitudeof angular acceleration is given by

t

a

=

$/ -/6& The !I unit of angular acceleration is rad 2 secG.

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EXPLANATION

&onsider a particle attached to one end of a mass*less rigid rod of length r with its other endpivoted at point ) taken as origin. !uppose the particle is set in anti*clockwise circular motion inxy* plane+ about ,*axis+ the axis of rotation.

Eet - is the initial position of the particle so that line )- serves as reference line coincident withHx*axis. #or position -+ let is the angular velocity of the particle at time t . If /is the angularvelocity of the particle for position -/at time t/+ then the change in angular velocity in going fromposition -to position -/is

6 * /

The time interval for this change in angular velocity is

t 6 t/* tThen ratio 2 t is defined as the angular acceleration. Thus+

6 2 t#or small values of or t the angular acceleration can be regarded as vector $uantity

otherwise it is a scalar $uantity.

DIRECTION OF ANGULAR ACCELERATION

The direction of angular acceleration is taken along that of angular velocity+ which in turn is takenalong axis of rotation and is determined by right hand rule. If the angular speed of the body1) Increases+ then angular acceleration will be like parallel to the angular velocity.2);ecreases+ then angular acceleration will be anti*parallel to the angular velocity.

UNITS OF ANGULAR ACCELERATION

a 54E6/A3 -/6$& There are two conventional units as0 Two conventional units as

a ?evolution per second s$uare. (?ev 2 secG.b ;egree per second s$uare (deg 2 secG.

/ !I unit as radian per second s$uare J ?ad 2 secG.

TYPES OF ANGULAR ACCELERATION

The angular acceleration can be put into following six types.1) A4E!AE A-3A! A55E3E!A6/

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Def&The average angular acceleration is defined as the ratio of total change in angular velocityand total time interval.

SYMBOL: The average angular acceleration is denoted by av.

0!-3A& If (total is change in angular velocity in total time interval t (total+ then

angular acceleration is given by

total

total

5av

tlimK

=

t

2) /$6A6AE-$ A-3A! A55E3E!A6/

Def& It is defined as the angular acceleration of the body at particular instant.

$783& It is denoted by ins.

0!-3A&If is change in angular velocity during time interval t then limiting value ofthe ratio 2 t then magnitude of angular acceleration. Thus+

tlimK

5av

=

t

) -niform Angular Acceleration

;ef If e$ual changes take place in angular velocity of a body in e$ual time intervals+ how shortor how long these time intervals may be+ the angular acceleration is said to be uniform.6E&If a body has uniform angular acceleration then

av6 ins9) 4A!/A83E A-3A! A55E3E!A6/

Def If e$ual changes take place in angular velocity of a body in une$ual time intervals or viceversa+ how short or how long these time intervals may be+ the angular acceleration is said to bevariable or non*uniform.') "$/6/4E A-3A! A55E3E!A6/Def&It is defined as the time rate of increase of magnitude of angular velocity.

;) EA6/4E A-3A! A55E3E!A6/

DefIt is defined as the time rate of decrease of magnitude of angular velocity.b) !E3A6/ 8E6:EE 3/EA! A55E3E!A6/ AD A-3A!

A55E3E!A6/

OR

"!0 0& ra =

The relation between linear speed v of a body moving on circle of radius r with angular speed isgiven by

v 6 r

Take differential (small change on both sides to get v 6 (r v6 r

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;ivide both sides by t to get

tlim

t

@lim

55

=

tt

7ut at

=

t

@lim

5magnitude of instantaneous linear acceleration

6 a (say

t

lim5

t6 magnitude of instantaneous angular acceleration

6 K (say!o that a 6 r KIn vector form+ it can be written as

ra =

Q.o' i.e cm/arin f 2ua!in f linear m!in and angular m!in$A$:E!& EQ-A6/$ 0 A-3A! 6/There is a strong similarity between e$uations of uniformly accelerated linear motion and those of

uniformly accelerated angular motion. #ollowing table gives the comparison of A$uations of twokind of motion.

E


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In the absence of net external force+ a body at rest and a moving body continue its rectilinearmotion with constant speed.

#ollowing two important conclusions can be drawn from Cewtons first law of motion1) :hen a net external force acts on a moving body parallel to the direction of motion then

body speeds up and slows down when this force acts anti*parallel to the direction of motionbut direction remains the same.

2) :hen net external force acts on the body at an angle different from 5Oor 0N5O+ then bothspeed and direction of motion change.

) :hen net external force acts on the body at right angle to the direction of its motion+ thendirection of motion changes but speed remains constant.

It means that to bend the rectilinear path of body moving at constant speed+ into curvilinear path+ anet external force must act on it at right angle to its direction of motion. !ince the direction ofmotion of a body undergoing circular motion continuously changes at each and every point of the

circular path+ therefore+ it is acted upon by a net external force parallel to its direction of motion.This force is known as centripetal force.

It is very force+ which keeps the body moving on circular path+ because if it vanishes at any pointof the circular path+ the body would fly off along the tangent to the path. Thus+ centripetal force isthe force+ which bends the rectilinear path of motion into curvilinear path. %s shown above in #igthat centripetal force is always directed towards the canter of the circle.If m is the mass of the body moving on circle of radius r+ with constant speed v then magnitudeof centripetal force is given by

rvmFc

/

=

%s v 6 r+ therefore+ # 6 mrG%ccording to Cewtons second law of motion+ the centripetal force produces an acceleration+ in itsown direction+ known as centripetal acceleration. Thus centripetal acceleration is directed towardsthe center of the circle. Its magnitude is given as

ac6#c2m 6 (mvG2r2m

6 @G2r 6 rGb E>A"3E$ 0 5E6!/"E6A3 0!5E

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!ource of centripetal force may differ from situation to situational as evident from followingexamplesE>A"3E 1 *$6E A6 6E 0 $6!/)&onsider a stone of mass m and being whirled at the end of a string of length r. Then tension instring provides centripetal force

#c6 m@G2r6T

E>A"3E 2 *"3AE6A!7 6/)

:hen earth revolves around the sun in elliptical orbit+ the gravitational pull provides centripetalforce thus #c 6 #g 6 me ms2rG:here meis mass of earth msis mass of the sun is gravitational &onstant

r is center*to*center distance between the earth and the sunE>A"3E *6/ 0 E3E56! A!-D -53E-$)

:hen an electron revolves around the nucleus then electrical force of attraction providescentripetal force

#c6 #e 6 Q eGR2rG:here @ is 5oulombs constant

eis charge on an electron or protonBis atomic number of nucleusris the distance between the nucleus and electron

E>A"3E 9& *6/ 0 5A!ED "A!6/53E / "A!6/53E A55E3E!A6!)

-article accelerator is a machine that can accelerate the charged particle to impart them Q.A. e.g.cyclotron+ betatron etc. %ll particle accelerators except the linear accelerator employ magneticfields to set the charged particles in circular motion. The magnetic force then provides centripetal

force to the charged particles #c6 #b #c6 $ v 7:here $ is charge on the charged particle v is speed of the charged particle b magnetic field

Q$7$ a) Define cen!ri/e!al accelera!in and deri.e a ma!hema!ical rela!in fr i!4

b) Define cen!ri/e!al frce and deduce i! frmula frm !he e/rein f cen!ri/e!alaccelera!in4

A$:E!& 5E6!/"E6A3 A55E3E!A6/Def&The acceleration produced by the centripetal force towards the center of the circle is calledcentripetal acceleration

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E>"!E$$/ If @ is speed of body moving along a circle of radius r then centripetalacceleration is given by ac6 vG2r

;A?I@%TI)C )# AS-?A!!I)C&onsider a particle of mass Cmand moving with constant linear speed C#along a circle of radiusCrwith center at point C. %s the particle moves on the circle+ constantly undergoes accelerationdue to change of direction of velocity vector at each and every point of the circle.If 4is change in the velocity of the particle in going from point % to point 7 during timeinterval t+ then magnitude of acceleration is given by

a 6 @2t************0

DE6E!/A6/ 0 t

If %7 6 ! (say is length of arc from point % to point 7 then time taken by particle to go frompoint % to point 7 is given by formula ! 6 @t t 6 !2@****************/DE6E!/A6/ 0 4

Eet @0is velocity of the particle at point % and @/be that at point 7. !ince the speed of theparticle is constant e$ual to @ therefore|@0|6 @ 6 |@/|

It implies that @06 @ 6@/The vector U%)7 is an isosceles such that

)%6)7 6 r and %)76

&onstruct a vector UB-? such that is

PQ e$ual and parallel to0

V and

PR is e$ual and parallel

/

V to i.e.

PQ6 0

V -B6@0

PR 6 /

V -? 6 @/

-B6-?6@@06@/In this vector UB-?+ the head to tail rule of vector addition gives that

+= QRPQPR

+= QRVV0/

V

VVQR

=

=

0/

Its magnitude is

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B?6@Thus+ the vector UB-? is an isosceles triangle is with

-B6-? 6 @ and B-?6B?6@

7oth U%)7 and UB-? are such that )% @0and )7 @/+ so that+ therefore+

B-?6%)76

%s %)7 and B-? are two isosceles triangles with e$ual angles between their e$ual armstherefore they are similar triangles. Thus+ the ratio of their corresponding sides is e$ual.

Q!"!%A8A

:hen t5 then chord %76%7 so that@2@6%72r

@2@6!2r %76! @6!@2r ************3


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#c6 mrGQ.o.a) Define Mmen! f iner!ia4 Deri.e an e/rein fr !he mmen! f iner!ia f a

/ar!icle and generali6e i! fr a rigid bdy4

b) i! !he mmen! f iner!ia f unifrm lid rd8 h/8 lid circular di8 lid cylinder8 and

unifrm lid /here4

%C:!A?M)MACT )# ICA?TI%Def&It is defined as the product of mass and s$uare of its perpendicular distance from the axis ofrotation

A6-!E

The moment of inertia is a scalar $uantity$783

The moment of inertia is denoted by 'I.0!-3A

If 'm is mass distributed at perpendicular distance r from the axis of rotation then moment ofinertia is given by I 6 mrG

$/ -/6

The !I unit of moment of inertia is kg*mGD /E$/$&

The dimensions of moment of inertia are WMEGX.

AS-E%C%TI)C

&onsider a body of mass 'm is attached to a mass less rigid rod of length 'r whose other end ispivoted at point ') taken as center of circular path. % tangential force '#t acts on the particle toset it into circular path. If at is tangential acceleration produced by tangential force '# then beCewtons !econd Eaw of Motion

#t6 mat **************(0If is angular acceleration then

at6 r **************(/


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Moment of inertia also known as rotational inertia is direct measure of inertness or inability ofexternal to change the state of uniform angular motion. It is not an intrinsic property of a particleor a body.

E6 0 /E!6/A 0 !//D 8D7

% rigid body is regarded as system of particles. &onsider a rigid body composed of n particles ofmasses m0+ m/+ZZZZ..mnrespectively at distance r0+ r/+ZZZ.rn+ from the axis of rotation ')+as shown in #ig below

!uppose the rigid body is set into rotational motion with angular acceleration in anticlockwisesense of rotation. Then each constituent particle undergoes some angular acceleration as shownin above figure. Vence the tor$ue acting on constituent particle of mass m0 is given by e$ 8 as

[6 m r0G!imilarly [/6 m/r/G

\n6 mnrnG

Total tor$ue acting on a rigid body is e$ual to some of tor$ue acting on its all*constituent particlesi.e.

[(total 6 [0H [/H [3H [ n-utting values of these tor$ues in the above e$uation we get [ (total6 m0r0GH m/r/GH mnrnG [ (total6 (m0r0G H m/r/G H mnrnG G

[ (total6 (miriG [ (total6 Ir

:hereas+ (miriG is the moment of inertia of the rigid body.

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b E6 0 /E6/A$ 0 $E :E33 @: 8D/E$#ollowing table gives the moment of inertias of some well*known bodies

!.C)

1

2

9

Came of the body

$3/D !D

6/ "

5/!5-3A! $3/D

D/$5

$3/D $"E!E

0igure#ormula

/%12mrF

/%mrF

/%12mrF

/%2'mrF

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Q.o a) Define and explain angular momentum of a particle and extend the mathematical

expression for a rigid bod,?

b) Explain difference between spin angular momentum and orbital angular momentum?

%C!:A? %C1


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If 'r is position vector of the particle w.r.t. The axis of rotation passing through origin thenangular momentum of the particle would be e$ual to the cross product of its position vector andlinear momentum that is

E 6 r ^p % r p n:here n is unit vector normal containing vector r and -. r is the perpendicular distance betweenthe line of motion (line of action of vector - and axis of rotation. %s shown in the that if isangle between r and - then

r 6 r sin !o that E 6 r - sin ***********0Cote that the angular momentum is a vector $uantity whose magnitude and direction are given asbelow

A/6-DE

#rom e$0 the magnitude of angular momentum vector is given byE 6 r - sin 7ut p 6 m@ so that

3 % rm4 sin

If particle is moving on circle then r@ and hence r@ so that 6 P5 E 6 rm@ sinP5 E 6 rm@ *********/

If is angular speed of the particle then v 6 rso that e$ / may becomeE 6 r m r E 6 mrG!ince mrG6I+ the moment of inertia of the particle therefore

E 6 I *********3A$ 3 gives magnitude of angular momentum e$ual to product of moment of inertia (angular massand angular speed

D/!E56/

The direction of angular momentum is taken along that of angular velocity + which in turn istaken along axis of rotation. The use of right hand rule tells that angular momentum E will beperpendicular to r and p i.e. normal to plane containing the vectors r and -.

A-3A! E6- 0 !//D 8D7

% rigid body is regarded as system of particles. &onsider a rigid body is composed of n particlesof masses m0m/m3Z mnat distances r0+ r/+ r3Z rnrespectively from the axis of rotation passing

0>


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perpendicularly through center of mass. If is constant angular speed of rigid body then each ofthe constituent particles would be the same e$ual to .

Vowever the linear speed of each would be different depending on the distance from the axis ofrotation. Thus for ith particle of distance rifrom axis of rotation we have

@i6 riAxtending this e$uation from0+ /+3+8+9Znwe have

@6 r@/6 r/

@i6 ri

@n6 rnThe magnitude of angular momentum of particle 0 is

E06 m0r0v0$imilarl,&

E/6 m/r/v/E36 m3r3v3

E06 m0r0v0

-utting values of @+ @/+ @_Z@n+ we haveE06 m0r0G E/6 m/r/G

En 6 mnrnG The magnitude of angular momentum of the rigid body is e$ual to sum of magnitudes of angularmomenta of its all constituent particles

E6 E0H E/HZZZ.En

-utting @alues of E0+ E/+ZZZ.Enwe getE6 m0r0G H m/r/G ZZZ..H mnrnG

0N


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E6 (m0r0G H m/r/GZZH mnrnG E6(miriG

7ut I6(miriG the moment of inertia of the rigid body thereforeE6I

$/ -/6$

The !I units of angular momentum is Qg*m2secG. It can be expressed in another form as ]*sec.;ivide and multiply by sec to get

Qg*mG2sec6(Qg*mG2secG sec 6(Qg*m2sec(msec

6C*m sec 6]*sec

D/E$/$

The dimensions of angular momentum are 6WMEG2TX

b D/00E!E5E 8E6:EE $"/ A-3A! E6- AD !8/6A3A-3A! E6-

% spinning body possesses the spin angular momentum about the axle through it e.g. the spinningtop and spinning earth about its own axis.

%n orbiting body possesses the orbital momentum about an axis of rotation or center of rotatione.g. the electron orbiting around the nucleus and the earth orbiting around the sun have orbitalangular momenta.

Q.No.a! !tate and explain the law of conservation of angular momentum.b) i.e me a//lica!in f la f cner.a!in f angular mmen!um$

AS:;


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E>"3AA6/

The time rate of change of angular momentum is e$ual to the net external tor$ue acting on a bodyor a system. %ccording to Cewtons second law of motion+ the net external force acting on asystem is e$ual to the time rate of change of its linear momentum that is

#6p2t

Take cross product of position vector r w.r.t. axis of rotation passing through origin ) to get

r ^#6 r ^ p 2t

6r ^ p2t

7ut r x # 6[extand r^-6E the change in angular momentum.

[ext6p 2t**************0

If net external tor$ue acting on the system is ,ero then [ext65 so that e$0may give

)6E2t

E6)^t

E6)**********/

E6constant

E6 E0H E/HZZZ.En6constant

%lternatively+ if Eiis initial angular momentum and Efis the final momentum then E6 Ef`Eisothat e$ / may give

Ef`Ei65

Ef6Ei

A""3/5A6/$ 0 3A: 0 5$E!4A6/ 0 A-3A! E6-

The law of conservation of angular momentum is one of the fundamental laws of -hysics. Thetruth of its conservation has been verified from cosmological level to submicroscopic level. It hasmany applications in practical -hysics as well as in the Cature as explained below

6E $"!/ 8A!D D!/4E!

#igure below shows driver leaving the springboard.

The diver pushes spring board to ac$uire a small angular speed iabout a hori,ontal axis passingthrough his center of gravity+ so that his initial angular momentum is

Ei6 Ii i:here Iiis initial moment of inertia. Initially+ the divers legs and arms are fully extendedtherefore+ Iihas large value. :hen the diver pulls his legs and arms to adopt a &E)!A; T


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Ef6 Ei-utting values of Eiand Ef+ we get

Iff6 Iiif 6 (Ii2If i

553-$/&!ince IfIi+ therefore+ fi. Thus+ the diver lands into swimming pool withhigh angular speed which enables him to take extra revolutions.

/. 6E $"// A5!8A6#igure shows an acrobat sitting on a turntable+ holding heavy weights in his extended arms.

In this configuration+ the system (%crobatH turntable H weights has large moment of inertia. If I iis initial moment of inertia of acrobat having initial angular speed ithen initial angularmomentum of the system is

Ei6 Iii:hen acrobat pulls his arms inward+ then final moment of inertia Ifof system reduces to small

value. If fis final angular speed of the acrobat then final angular momentum of the system isgiven thatEf6 Iff

%ccording to the law of conservation of angular momentum+ we haveEf6 Ei

-utting values of Eiand Ef+ we haveIff 6Iii

f6 (Ii2If i553-$/&!ince IfIi+ therefore+ fi. Thus+ the acrobat spins faster when he pulls his

arms close his body and slows down+ when he extends them.. 6E !6A6/ 8/5753E :EE3#igure shows a student seated on a turntable that is free to rotate about vertical axis.

/0


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The student holds a bicycle wheel that has been set spinning anticlockwise+ such that the turntableis initially at rest. :hen student turns the spinning bicycle wheel upside down+ then to conservethe angular momentum+ the turntable begins to rotate anticlockwise while now the spinningbicycle wheel is continuing its spin motion+ now clockwise.8. 6E $6A8/3/67 0 $"// 8HE56$The spin motion provides orientation stability on the basis of the law of conservation of angularmomentum. In simple words+ it means that greater is the spin angular momentum+ it would bemore difficult for external tor$ue to change the orientation of the spinning ob"ect.E>A"3E$

0. Veavenly bodies e.g. the earth has orbital angular momentum due to orbital motion and spinangular momentum due to spin motion. The spin motion stabili,es the orientation of earthoverall motion.

/. The atomic electron spins in addition to orbiting around the nucleus "ust to stabili,e itsorientation.

3. )ne of the causes of the stability of the nucleus is the spin motion of its nucleons.

8. ?ider less bicycle given a slight push is able to remain upright over afar longer distance than expected+ due to spin motion of its wheels

9. %rtificial satellites are set into spin motion to stabili,e their orientationE43-6/ 0 $6A!$ AD 533A"$/ $6A!$The evolution of the stars in the result of law of conservation of angularmomentum. The stars are formed due to self gravitational pull of cloudsof dust and gas in interstellar space. These clouds gain a small angularvelocity due to rotation of galaxy. The gravitational pull shrinks the cloudsof dust and gas to a dense ob"ect. In order to conserve the angular

momentum+ the angular velocity of such an ab"ect increases a great value.The process of gravitational contraction heats up the stars and thus theybecome luminous. The internal temperature becomes as 05054 Q to starta nuclear fusion reaction in the star to provide it long lasting source of heatand light. In stars of large si,e+ the high speed motion its particles makethem unstable+ at end of revolution. %s a result supernova explosion takesplace+ which causes the outer layer of star to be blown out in space.)nly dense pore is left behind in star in which nuclei come close to eachother thus reducing its radius and increasing its angular velocity. !uch stars+

infect become one giant atomic nucleus and are known as CA


23/43

b) A circular dic and h/ f e2ual mae and radii are alled ! rle dn frm !he

ame inclined /lane$ #alcula!e !he linear /eed f dic and h/ n reaching !he b!!m f

inclined /lane and decide hich ne ill reach !he b!!m earlier4

A$:E! a)!6A6/A3 @/E6/5 EE!7Def& The ability of a body by virtue of its spin motion about the axis of rotation is called itsrotational Q.A.0!-3A if I is rotational inertia and : is angular speed of a rotating or spinning body thenits rotational Q.A is given by

/rot/

0Q.A I=

DE!/4A6/ 0 0!-3A 0 !6A6/A3 @.E

&onsider a rigid body spinning about an axis %% with constant angular speed in anticlockwisesense of rotation+ as shown in #ig.The rigid body is regarded as system of particles.

Eet mibe the mass of the ith

constituent particle of the rigid body rotating at a distance rifrom theaxis of rotation. This ithparticle has the same rotational speed as that of the rigid body. If @ iisthe linear speed of the particle then

@i6 ri ***** (0Cow translational Q.A of the ithparticle is

Q.Ai rot 6 L mi@i/**********(/:hen e$.0 is used in e$./+ then we get rotational Q.A of the ith particle of the rigid body as Q.Ai rot 6 L mi(ri/

6 L miri// ******* (3

7ut miri/

6 Ii+ the rotational inertia of the ith constituent particle of the rigid bodyQ.Arot 6 L Ii/ ********** (8If m0+ m/+ m3Z.are masses of constituent particles rotating at distances r0+ r/+ r3Z respectivelythen by the symmetry of e$.3+ their rotational Q.As are given by

Q.A0 rot 6 L m0r0//Q.A3rot 6 L m/r///Q.A/rot 6 L m3r3//

Q.Ai rot 6 L miri//

7ut $uantities m0r0/+ m/ r//+ m3 r3/+Z+ mi ri/ZZ give moments of inertia of the constituentparticles. These moments of inertial are denoted by I0+ I/+ I3+ Z IiZ so that

Q.A0rot6 L I0 /

Q.A/rot6 L I//

Q.A3rot6 L I3/

Q.Ai rot 6 L Ii /

/3


24/43

The total rotational Q.A of the body is e$ual to the sum of the rotational kinetic energies of its allconstituent particles i.e.

Q.A rot6 Q.A0rotH Q.A/rotH Q.A3rot H **** HL Ii / H***** 6L I0/H L I//H L I3/H*****L Ii / H***** Q.A rot6 L (I0HI/HI3HZ.. /

Q.A rot6 L ( Ii / 7ut Ii6I+ the moment of inertia of the rigid body.

Q.A rot6 L I /

This is the rotational Q.A of a rigid body.

3/EA! $"EED$ 0 D/$5 AD " A6 6E 866 0 /53/ED "3AE

&onsider a disc and hoop of e$ual masses and radii+ let m be mass and r be the radius of each disc

and hoop. !uppose both the disc and hoop is allowed to roll down simultaneously from sameinclined plane of height h+ as shown in the #igure.

%t the top of the inclined both the disc and hoop have gravitational -.A given by-.Agrav 6 mgh********(0

:hen both the disc and hoop roll down along the inclined plane then gravitational -.A converseinto rotational Q.A and translational Q.A+ in general given by+

Q.Arot6 L I /

****** (/:here I is the moment of inertiaQ.Atran6Lm@/ ********* (3

:here @ is the translational speed at the bottom of the inclined plane.7y the law of conservation of energy+ we have -.A 6 Q.ArotH Q.Atran********(8


25/43

/mgh @/6 ************

m H I2r/

/

/@

r

Im

m"#

+

=

***** (9

D/$5&

#or disc+ the moment of inertia isI 6 L mr/

-utting this value of I in e$.9+ we get"#09.0@

disc =

"&

#or hoop+ the moment of inertia isI 6 mr/

-utting this value in e$.9+ we get@hoop6 gh

#rom e$.4 and e$.>+ it can be concluded that@disc @hoop

553-$/&The disc will reach the bottom of inclined plane earlier than hoop.

Q$11$a) Define

1) a!ural Sa!elli!e

") Ar!ificial Sa!elli!e

b) Deri.e an e/rein fr !he rbi!al /eed f an ar!ificial a!elli!e$ Define cri!ical /eed (.elci!y) f an ar!ificial a!elli!e and calcula!e i!

c) %b!ain an e/rein fr !he /erid f re.lu!in f a a!elli!e$ #alcula!e !he /erid f

re.lu!in f an ar!ificial a!elli!e rbi!ing !he ear!h i!h cri!ical /eed$

d) i.e main !y/e f ar!ificial a!elli!e

A$:E!& a) 0.A6-!A3 $A6E33/6EDef&% body that revolves around another body without any fuel is called a natural satellite.E>A"3E$&The moon is a natural satellite of the earth+ which inurn is the same for the sun.!imilarly+ the atomic electron can be regarded as a natural satellite of the nucleus.

2. A!6/0/5/A3 $A6E33/6E;ef % man made device that revolves around the earth without any fuel is called the artificialsatellite.

b E>"!E$$/ 0! !8/6A3 $"EED 0 A A!6/0/5/A3 $A6E33/6E

&onsider an artificial satellite of mass 'm revolving around the earth in a circular orbit concentricwith the centre of the earth. The centripetal force is provided by the gravitational pull of the earthon the satellite i.e.

#c6 #gIf @ is orbital speed of the satellite and r is its orbital radius then

/

/

sm

r

$mG

r

V%

=

/9


26/43

:here 1 is gravitational constant e$ual to 4.4>^05*00. C*m/2kg/and M is mass of the earth

@ 6 1M2r****** (07y the definition+ the gravitational pull is the weight of the satellite+ therefore

#g6 :1mM2r/6 mg

1M2r/6 gMultiplying both sides by r+ to get

1M2r 6 gr****** (/

A$> gives the period of revolution of the artificial satellite around the earth. If h is height of theartificial satellite from the surface of the earth then r 6 hH?+ so that e$.> may give

/4


27/43

T 6 / (hH?2@ ******* (N:here ? is radius of the earth. If h? then e$.N may reduce to T 6 /?2 @-utting ?6 4.8 054 m and @6 >P55 m2s+ we get T 6 / ^3.08 ^4.8^054 >P55 6 9545 sec 6 9545245 min 6 N8 min.d 67"E$ 0 A!6/0/5/A3 $A6E33/6E&%rtificial satellites are launched in space for many purposes e.g. research and information. In thisway they fall into following categories.

1) 5-/5A6/ $A6E33/6E$These satellites mean telephone calls facilitate internet services and T@ transmission. They areplaced in geo*stationary orbit which is 34555 km above the surface of earth. Threecommunication satellites in the geostationary orbits positioned properly+ can cover whole

populated regions of the earth.2) :EA6E!$A6E33/6E$&

These satellites remain changing their positions in their orbit. They give useful informationabout the weather conditions.

) !E$EA!5 $A6E33/6E$ These satellites are specially designed to keep an eye at the changes

taking place in the upper atmosphere of the earth.9) $"7 $A6E33/6E$

These satellites are used to spy the military activities of enemy.

Q.o12 a Define geostationar, orbit and calculate its height from the surface of the earth. b) 5alculate the orbital speed of geostationar, satellite.A$:E! a E$6A6/A!7 !8/6&Def& /) The orbit in which the period of revolution of the artificial satellite becomes e$ual to theperiod of daily rotation of the earth about its own axis is called geostationary orbit.;ef II %n orbit in which the orbital motion of an artificial satellite is synchroni,ed with thedaily rotation of the earth about its own axis is called geostationary orbit.

6E&%nother name for geostationary orbit is synchroni,ed orbit.E>"3AA6/&onsider a geostationary satellite of mass m and revolving around the earth in a geostationaryorbit of radius r. The gravitational pull of the earth provides centripetal force to the geostationarysatellite for its orbital motion in the geostationary orbit. i.e.

#c6 #gms@/ 6 1msM

r r/

:here @ is orbital speed of the geostationary satellite and M is mass of the earth. 1 isgravitational constant.

@/6 1M2r@ 6 1M2r ********** (0

/>


28/43

;uring one revolution+ the geostationary satellite covers distance s e$ual to the circumference ofthe geostationary orbit. i.e. ! 6 circumference of geostationary orbit 6 /rThen by formula ! 6 @T+ we get /r 6 @t @ 6 /r2t ************* (/:here t is period of revolution of the geostationary satellite is /8 hours+ the period of dailyrotation of the earth around its own axis.&omparing e$.0 and e$./+ we get /r2t 6 1M2rTaking s$uare on both sides to get 8/r/2t/6 1M2r r36 1Mt/28/

Taking cube root on both sides we get

r 6 (1Mt/28/023****** (3

A$.3 gives the radius of geostationary orbit.

5A35-3A6/ 0 E$6A6/A!7 !8/6

#or e$.3 values of constants are 1 6 4.4>^05*00C*m/2kg/

M 6 4^05/8kg t 6 /8 V

6 /8^3455 sec 6 N4855 sec 6 3.08

:ith these values e$.3 becomes r 6 4^05 /8 ^4^05 /8 (N4855 j 023 8^ (3.08/ 6 8./3^05>m 6 8./3^058km

r 6 8/3+ 555+ 55 m

5A35-3A6/ 0 E/6 0 E$6A6/A!7 !8/6If ? is the radius of the earth then height of then height of the geostationary orbit from the surfaceof the earth is given by h 6 r J ?Vere r 6 8./3^05>m ? 6 4.8^054m!o that h 6 8./3^05>J 4.8054

6 39P+ 555+ 55 m 6 39P55 km h 6 34555 km

/N


29/43

5A35-3A6/ 0 !8/6A3 $"EED E$6A6/A!7 $A6E33/6E&

#rom e$.0+ the orbital speed of a geostationary satellite can be calculated as @ 6 4.4>^05*00 ^4^05 /8

8.38^05>

6 35>9 m2s 6 3.5>9053m2s

6 3.0 km2s-$E$ 0 E$6A6/A!7 $A6E33/6E

1eostationary satellites are used for following purposes

1) :orldwide communication2) :eather observations) Cavigation9) )ther military uses

Q.o1 a Define a//aren! eigh!$ Dicu !he .aria!in f a//aren! eigh! under fllingcae$

1) :hen lif! i a! re! r i! m.e u/ r dn i!h cn!an! /eed (r 6er accelera!in)

") :hen lif! acend i!h accelera!in$

3) :hen lif! decend i!h accelera!in$

b) ha! ha//en hen cable carrying !he lif! brea u/4

%;

Define a//aren! eigh! and dicu m!in f lif! r ele.a!r.A$:E!&A""A!E6 :E/6

;ef* The reading of spring weighing machine is called apparent weight of the body.E>"3AA6/

The gravitational pull of the earth on the body placed on it surface is called !EA3 :E/6.The weight of body is measured with the help of spring weighing machine (or springbalance.:hen body is at rest or moves up or down with constant speed then reading weighingmachine speed is e$ual to the ?A%E :AI1VT of the body. Vowever+ when body moves up ordown with some acceleration+ then spring balance would read more or less then the real weight.

The ?A%;IC1 of spring balance+ for a body having accelerated upward or downward motionwith acceleration is called %--%?ACT :AI1VT. The apparent weight denoted by : ande$uals the tension T in the suspension string that is : 6 T.?emember that apparent weight of the body may vary while the real weight cant. The apparentweight :of a body may be such that0 : 6 :(real weight/ : :3 : :8 : 65

The variation of apparent weight can be discussed by considering the motion of lift or elevator.6/ 0 3/06 *E3E4A6!)&onsider a block of mass in suspended by means of a spring balance whose upper end attached tothe ceiling of a lift (elevator.The real weight of the block is :6mg+ while the reading of the

/P


30/43

spring balance gives apparent weight : of the block which in turn e$uals the tension T in thesuspension string that is. :6T5ase 1 :E 3/06 /$ A6 !E$6 ! 4E$ -" ! D: :/6 5$6A6 $"EED

#ollowing two forces act on the block1) #up6 T (tension in string

2) #down6 : (real weight

In this case the net force on the block is#net 6 #up * #down

6T*:If a is acceleration of the block then by Cewtons /ndlaw of motion #net6ma+ so that

ma6#*T%s the block is at rest or it moves up or down with constant speed+ therefore+ a65+ so that

mo6T*:o 6 T*:

T6:Vowever+ T6:+ the apparent+ weight of the block so that

:6 :553-$/ The apparent weight is e$ual to the real weight+ when lift is at rest or it movesup or down with constant speed.

5ase 2& :E 3/06 A$5ED$ :/6 -/0! A55E3E!A6/

#ollowing two forces act on the block

0 #up6 T(tension in string/ #down6: (real weight!ince the block along with elevator is ascending+ therefore+ #up #down+ so that the net force onthe block is

#net6 #up* #down

6 T J :

If a is acceleration of the block then by Cewtons seconds law of motion the net force is # net 6ma+ so that

ma 6 T*: T 6:Hma

Vowever+ T6:+ the apparent+ weight of the block so that

35


31/43

:6 :Hma 6mgHma

553-$/&:hen lift ascends with uniform acceleration the apparent weight of the blockincreases by an amount ma.

5ase :E 3/06 DE$ED$ :/6 -/0! A55E3E!A6/.

#ollowing two forces act on the block

0 #up6 T(tension in string/ #down6: (real weight

!ince the block along with elevator is ascending+ therefore+ #up #down+ so that the net force on

the block is#net6 #down * #up

6 : *T

If a is acceleration of the block then by Cewtons seconds law of motion the net force is # net 6ma+ so that

ma 6 :*T

T 6:*ma

Vowever+ T6:+ the apparent+ weight of the block so that:6 :*ma 6mg*ma

553-$/ :hen lift descends with uniform acceleration the apparent weight of the blockdecreases by an amount ma.

$"E5/A3 5A$E& :E 5A83E 5A!!7/ 6E 3/06 8!E@$ -"

:hen cable carrying the lift breaks up+ then block along with the lift falls freely towards theearth with an acceleration g that is in this case

a6g

:hen lift moves down the apparent weight of the block is

:6mg*mg

:65

T 65

Thus+ the pointer of spring balance will read ,ero apparent weight. In simple words the blockbecomes apparently weightless+ when lift falls freely. Thus+ the lift becomes gravity freesystem when it falls freely. This situation is known as :E/63E$$E$$J. To overcomethe problem of weightlessness in gravity free system+ the artificial gravity is created+ by sitting

30


32/43

the system into the revolution about its own axis of symmetry. ?emember that theweightlessness can be observed only within gravity free system.

Q.o12 :ha! i mean! by !he eigh!lene in ar!ificial a!elli!e4 Dicu in de!ail=:eigh!lene in a!elli!e and gra.i!y free y!em>$

A$:E!&:E/63E$$E$$ / A!6/0/5/A3 $A6E33/6E

Def& The absence of restraining force in a free falling system is called weightlessness.

:E/63E$$E$$ / A!6/0/5/A3 $A6E33/6E AD !A4/67 0!EE $7$6E

:hen a satellite revolves around the earth in its specified orbit+ it falls freely towards thecenter of the earth all times. Vence+ the motion of this artificial satellite is similar to thedownward motion of a (elevator which falls freely towards the earth with an acceleration of

a6g+ due to breaking up of its cable. %s the block inside the free falling lift becomes weightlessin the same way every ob"ect including astronaut becomes weightless.

:hen a pro"ectile is thrown parallel to the hori,ontal surface of the earth in the absence of theair+ its falls to the surface of earth moving along a curved path. The curvature of the path ofpro"ectile depends inversely upon its speed of pro"ection. Thus+ if a pro"ectile is pro"ected athigh speed+ the curvature of its path would decrease. It means that there exists a speed forwhich the curvature of path of pro"ectile becomes e$ual to the curvature of the earth. ;uringhence it begins to orbital motion+ the pro"ectile falls freely all times towards the earth. Thistype of pro"ectile that orbits the earth in free fall manner is called the artificial satellite. In fact

the space ship that orbits the earth falls freely towards it+ but the curvature of the earth preventsit from hitting the surface of the earth. It should be remembered that the :AI1VTEA!!CA!!could be observed only in the frame of reference of the satellite or spaceship. Theweightlessness is created due to the fact that all ob"ects with in the space ship fall at the samerate towards the earth.

"!0 0 :E/63E$$E$$ / A!6/0/5/A3 $A6E33/6E

&onsider a satellite of mass ms and revolving around the earth in an orbit of radius r.!uppose a block of mass m is suspended from a spring balance whose upper end is attached tothe ceiling of the artificial satellite.

Then following two forces act on the block.

0 #up6T (tension in string/ #down6: (real weight of block

!ince the block along with elevator is ascending+ therefore+ #up #down+ so that the net force on theblock is

#net6 #down * #up

6 : *T

3/


33/43

Vowever+ the tension in the suspension string is e$ual to the apparent weight : of the block thatis T6: + so that

#net6:*:

This net force provicles necessary centripetal force to the block is+

#c6 #net

6:*:

6mg*:

If @ is orbital speed of the artificial satellite and hence that of the block then # c6 m@/2r+ sothat

m@/6mg*:

;ivided by m to get

@/2r 6g*:2m******0

The gravitational pull e$ual to weight of the artificial satellite provides it the necessarycentripetal force that is for the satellite+ we have+

#c6weight of the satellite

m!@/2r6m!g

@/2r6gZZZZ /


34/43

R

"

/

0f=

A$:E!&a A!6/0/5/A3 !A4/67

Def&The reaction force which gives the impression of real gravity in the artificial satellite fallingfreely is called artificial gravity.

E>"3AA6/

:hen an artificial satellite revolves around the earth then weightlessness is produced inside it+due to its free fall towards the earth. Vence+ every ob"ect and astronauts too+ inside theartificial satellite becomes weightless. In simple words the real gravity appears to be

disappears. The absence of real gravity i.e the weightlessness gives rise to many problems inperformance of experimental work to be carried out by astronauts. Therefore+ to over come theproblem produced by the weightlessness inside the satellite+ the artificial gravity must becreated e$ual to the real gravity of the earth.

: 6 5!EA6E 6E A!6/0/5/A3 !A4/67 /$/DE 6E A!6/0/5/A3

$A6E33/6E

%ccording to space engineering+ the method to create the artificial gravity inside the artificialsatellite is to set it into rotation (spin motion about its own axis of symmetry with

predetermined fre$uency of revolution. :hen this is done then due to reaction of centripetalforce the astronauts press the walls and flow of artificial satellite outward. This outwardlydirected reaction force that give the impression of real gravity is known as artificial gravity.

b) "!0 0&R

"

/

0f =

&onsider a ring*shaped artificial satellite of radius ?+ set into rotation (spin rotation about itsaxis of symmetry+ with constant angular fre$uency :. If T is time period of rotational motionthen fre$uency of revolution necessary to create the artificial gravity e$ual to real gravity ofthe earth is given by

f602TZZZZ (0

7y relation T6/2 + we have T6/ 2+ so that e$ 0 may become

38


35/43

/

0f = ***** (/

If @ is linear speed og the spinning artificial satellite b then by relation @6r+ we have

@6?

6@2?

-utting this value of in e$ / we get

f6@2/?ZZZZ. (3

If acis centripetal acceleration then

a&6@/2?

@/6a&?

@6a&?ZZZZ(8


36/43

A$:E!& E:6$ AD E/$6E/$ 4/E:$ 0 !A4/6A6/

%ccording to Cewtons point of view is the natural (intrinsic property of the matter that everyparticle of matter attracts every other particle with a force that acts along the line "oining themsuch that the magnitude of the force is 0 directly proportional to the product of their masses/ Inversely proportional to the s$uare of distance between them.

%ccording to Ainsteins point of view space time is curved near massive bodies. In order toobserve it+ we might thick of space as thin rubber sheet if a heavy ball is hung from it+ it curves asshown in the #ig.

The weight corresponds to the huge mass that causes the space itself to curve. The heavier themass+ the greater is the curve or dent. Ainsteins theory does not speak about the force of gravityacting on the bodies but it says that bodies and light travel along straight line in curved spacetime. Thus+ a body moving or at rest near a heavy body will follow straight line path (geodesictowards the heavy mass. Ainsteins theory physically explains how does gravity operates. Cewtondiscovered universal law of gravitation but does not explain the reason for obeying this law.Ainsteins theory not only explains that gravity obeys inverse s$uare law over short distances butalso explains the reason that why is it so. Vence+ Ainsteins theory is better than Cewtons theory.Ainsteins concluded that if gravity and acceleration are e$uivalent then gravity must bend lightby an amount that can be calculated. Cewtons theory based on the idea of light as stream of tinyparticles also explains that the light ray is bent near heavy body due to gravity. Vowever+deviation of light from straight line path near heavy body is double in Ainsteins theory than thatin Cewtons theory. :hen in 0P0P+ the bending of light ray due to gravity of the sun wasmeasured during sun eclipse measurements matched that Ainsteins prediction rather thanCewtons. Thus+ Ainsteins theory was considered to be scientific triumph.

A$:E!$ 6 $!6 Q-E$6/$

$.QK'.1&/lain !he difference be!een !he !angen!ial and angular .elci!y$ Bf ne f !hee i

gi.en fr a heel !hen h ill yu find !he !her4

A$:E!&DBCC;# : ? ABA AD AEA; F%#BG

ABA F%#BG AEA; F%#BG

1 It is defined as the rate of change lineardisplacement.

1 It is defined as the rate of change angulardisplacement.

2 It is vector $uantity. 2 It is vector $uantity only when time is asshort as Ut 5

Its symbol is

V . Its symbol is.9

Its formula ist

&V

=

.9

Its formula ist

=

.

34


37/43

' Its !I unit is m2s. ' Its !I unit is rad2s.; Its direction is given by direction angle. ; Its direction is given by right hand rule.( It is always perpendicular to the radius

vector in plane of circle.( It is taken along the axis of rotation which

may or may not be perpendicular to planeof circle.

Its direction continuously changes as thebody moves on the circle.

Its direction does not change until thesense of rotation is not changed.

L It is always along the tangent. L It is never along the tangent.

If v is the magnitude of the linear velocity and is that of the angular speed then@6r **** (0

If one of the tangential or angular velocity is known then other can be determined from the e$0.$.QK'.2&/lain ha! i mean! by !he cen!ri/e!al frce and i! mu! be furnihed ! an bHec! if

!he bHec! i ! fll !he circular /a!h$

A$:E!& 5E6!/"E6A3 0!5E

Def& It is radial inward directed force that bends straight path of the into circular.

It is given byr

vmF

c

/

=

%s v 6 r+ therefore+ # 6 mrG%ccording to Cewtons first law of motion to change the direction of motion of a body movingwith constant speed+ it must be acted upon by a force that is perpendicular to the direction ofmotion. !uch a force compels the body to move along the circular path.

:7 6E 5E6!/"E6A3 /$ E5E$$A!7 6 "!4/DEIf centripetal force vanishes at a point then body will shoot along the tangent at that point. Vencethe centripetal is necessary to keep the circular motin to continue.$.QK'.& :ha! i mean! by !he mmen! f iner!ia4 /lain i! ignificance$

A$:E!& E6 0 /E!6/A

Def& It is defined as the product of mass and s$uare of perpendicular distance of the mass fromaxis of rotation.It is denoted by I. If m is mass of the particle rotating about an axis of rotation at distance r then

I6mr/

% rigid body is regarded as system of particles+ thereforeI6mr/

The moment of inertia is measured in Qg*m/.

"7$/5A3 $//0/5A5E& In linear motion+ the property of body due to which it opposesany change in its state of rest or uniform motion is called linear inertia. Thus+ the linear inertia isthe measure of opposition offered to the net external force that tends to produce the linearacceleration in it. In simple words+ it is the mass of the body that decide the amount of the linearacceleration of produced in it by the constant external force. Vence+ in linear motion+ mass is

linear inertia+ given by m6#2a!imilarly+ rotational inertia is the measure of opposition offered to the net external tor$ue thatproduces or tends to produce the angular acceleration in it. It is given by

3>


38/43

I6Y2K

$.QK'.9& :ha! i mean! by !he angular mmen!um4 /lain !he la f cner.a!in f

angular mmen!um$

%C!:A? %C1


39/43

D/!E56/

The direction of angular momentum is taken along that of angular velocity + which in turn istaken along axis of rotation. The use of right hand rule tells that angular momentum E will beperpendicular to r and p i.e. normal to plane containing the vectors r and -.

3A: 0 5$E!4A6/ 0 A-3A! E6-

$6A6EE6& The law of conservation of angular momentum states that for an isolated systemthe total initial angular momentum is e$ual to the total final angular momentum. !

In the absence of net external force acting on a system the total momentum remains constant.

E>"3AA6/

The time rate of change of angular momentum is e$ual to the net external tor$ue acting on a body

or a system. %ccording to Cewtons second law of motion+ the net external force acting on asystem is e$ual to the time rate of change of its linear momentum that is

#6p2t

Take cross product of position vector r w.r.t. axis of rotation passing through origin ) to get

r ^#6 r ^ p 2t

6r ^ p2t

7ut r x # 6[extand r^-6E the change in angular momentum.

[ext6p 2t**************0

If net external tor$ue acting on the system is ,ero then [ext65 so that e$0may give)6E2t

E6)^t

E6)**********/

E6constant

E6 E0H E/HZZZ.En6constant%lternatively+ if Eiis initial angular momentum and Efis the final momentum then E6 Ef`Eisothat e$ / may give

Ef`Ei65

Ef6Ei

A""3/5A6/$ 0 3A: 0 5$E!4A6/ 0 A-3A! E6-

The law of conservation of angular momentum is one of the fundamental laws of -hysics. The

truth of its conservation has been verified from cosmological level to submicroscopic level. It hasmany applications in practical -hysics as well as in the Cature. #or example+ the motion springboard diver+ motion of an acrobat spinning wheel+ stability of atom and planetary system.

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Q.o'.9& !how that the orbital angular momentum is given by Eo 6mvr.A$:E!& !8/6A3 A-3A! E6-

Def= The angular momentum of a body having orbital motion is called orbital angular momentum.The angular momentum of the body is given is

E6mvr !inqIn case of orbital motion+ the orbital velocity is always perpendicular to the radius vector+therefore+ q6P55 so that

E6mvr !inP5560Q.o'.;&Decribe ha! huld be !he minimum .elci!y fr a a!elli!e ! rbi! cle ! !he

ear!h arund i!4

A$:E!& 1ravity provides centripetal force so that#g6#c

If m is the mass of the satellite and v is its orbital speed thenm@/ 2r6 mg

@6gr

:hen the satellite is very close to the earth then r?+ the radius of the earth+ so that@6g?

@6P.N^4.8^054

6>.P Qm2sQ.o'.(&S!a!e !he direc!in f !he flling .ec!r in !he im/le i!ua!in$

a) Angular mmen!um b) Angular .elci!y

A$:E!& D/!E56/ 0AEA; M%MEM

The angular momentum is given as

= 'rL

The angular momentum vector is perpendicular to the plane containing the both vectors

'a(&r .It is shown in the #ig.

D/!E56/ 0AEA; M%MEM

The angular momentum is given as

= 'rL

The angular momentum vector is perpendicular to the plane containing the both vectors

'a(&r . Itis shown in the #ig. It is determined right hand rule!/6 AD !-3E

%ccording to right hand rule1)!tretch right hand with thumb perpendicular to palm2)1rasp the axis of rotation in the stretched right hand and curl the fingers along the sense ofrotation then thumb indicates the direction of angular momentum.

D/!E56/ 0AEA; F%#BG

The angular velocity is given as

= rV

The direction of the angular velocity is along the axis of the rotation. It is determined by the righthand rule as described above.Q.o'.& /lain hy an bHec! rbi!ing !he ear!h i aid ! be falling freely$ Ee yur

e/lana!in ! /in! u! hy i a//ear eigh!le under cer!ain circum!ance$A$:E!& :hen ob"ect is pro"ected along the surface of the earth+ with as high as >.P Qm2s + thecurvature of the path of the ob"ect becomes e$ual to that of then ob"ect begins to revolve around

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the earth and falls to the freely at an acceleration e$ual to g6P.N m2s / . !uch an orbiting ob"ectdoes not hit the surface of the earth because of the tangential component of the velocity.:hen an ob"ect falls freely towards the earth+ its acceleration is always a6g6P.N m2s / + in theobservers frame of reference. The apparent weight of the ob"ect then becomes

:6mg*ma

:6mg*mg:65

T 65

Thus+ the absence of restraining force in the frame of reference of the ob"ect is calledweightlessness.Q.o'.L& :hen mud flie ff !he !yre f a m.ing bicycle8 in ha! direc!in de i! fly4

/lain$

A$:E!& The mud sticks the tyre of the bicycle due to the adhesive forces between it and the

tyre. :hen wheel of the bicycle is set into rotational motion+ then centrifugal force comes intoplay as reaction of the centripetal force. %s the rotational speed of the bicycle wheel increases+ thecentrifugal force increases accordingly+ because the centripetal force does so. %t certain stage thecentrifugal force exceeds the adhesive force then mud flies off along the tangent to the tyre at thepoint of adhesion.

Q.o'.1I&A dic and h/ !ar! m.ing dn frm !he !/ f an inclined /lane a! !he ame

!ime$ :hich ne ill m.e fa!er n reaching !he b!!m4

A$:E!& If h is height of the inclined plane then it can be proved that"#09.0@

disc =

@hoop6 gh

@disc @hoop553-$/&The disc will reach the bottom of inclined plane earlier than hoop.Q.o'.11&:hy de !he di.er change hi bdy /i!in befre di.ing in !he /l4

A$:E!& %fter leaving the spring board diver shrinks his arms and legs to reduce his momentof inertia and thereby increasing his rotational speed before diving into the pool. Vis rotationalspeed increases according to the following formula

f 6(Ii 2If i 6(ri 2rf /i

Q.o'.12& % student holds two dumb*bells without stretched arms+ while sitting on turntable. Veis given a push until he is rotating with certain angular velocity. The student then pulls the dumb*bells towards his chest. :hat will be the effect on rate of rotation

A$:E!If Iiis initial moment of inertia of student with stretched arms and iinitial angularspeed then initial angular momentum of the student is

Ei6 Iii

:hen student pulls his arms inward+ then his final moment of inertia Ifreduces to small value. Iffis final angular speed of the acrobat then final angular momentum of the system is given that

Ef6 If f%ccording to the law of conservation of angular momentum+ we have

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Ef6 Ei-utting values of Eiand Ef+ we have

Iff 6Iiif6 (Ii2If i

553-$/&!ince IfIi+ therefore+ fi. Thus+ the student spins faster when he pullsdumb*bells close his chest.Q.o'.12& Axplain how many minimum number of geostationary satellites are re$uired for globalcoverage of T@*transmission.A$:E!& The properly positioned geostationary satellites are re$uired for globalcoverage ofT@*transmission. The angular separation between two consecutive geostationary satellites mustbe 0/55 for the whole populated area of the earth.

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by

Abrar Ahmad

(ZAAS Academy 0333-5307019)

83

ch#5 circular motion

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