Rotational . Motion ! Axis of Rotation Centre of Mars - : Centre of mars of a booly or ayatem 438 & point which mows such they all the maws 46 Concentyeted there anol all externned forcas ane applied Harr * The position of Centye of mass clipercls upon the Shape of the bocly and Astribution of mast. *« In Symmetrical boclies in which ckahibution of mass Js homogenenus, the centre of mars Coincides with the geometrical centre of the body, ¥ Centre of mars moy be insick or outsich the body. eq. Circular diac —> Within the body Cirewon Ring —> Outside the body Position Vector of Centre of mass for n particle system Lit a System consist of n Perticles of masses M,mMa, .... ™n Whose porition vectors are T1y Vay oe respectively + Then position vector of Centre of mass y= Mit, + mv, +... Minin Mm) + Mate Mn Pesition vector of Certve of mard for two particle system i m +mIMPORTANT POINTS A) If the origin 22 at the centre of mars, then the sum of the “moments of the masses of the system about the centre of mass dd zero. 2) Tp a system of particles of massed mM, > mM y--- Mn move with velocities Vio Vays Vn then the velocity of centre of mons 1s rm rum => Vem = 31 Th @ system of particles of massed m, 9 m72,..- Mn move with accelerations , 925.14 then the accelevation of centre of mars is Bron cma =m 4 Force on a rigial bady FE = e ay ma M oe arounel Cenbe of Mars Colculate RAT SS Poi 3 Origin (00,0) WL asbume BX on aPosition of Centre of mans of olifferent bodies Body | Position of Centre of Mars Uniform hollous/Salid Sphee| Centre of sphere Uniform civeway ving isc] Centre of ring /dlisc Uniform Red Centre of rod Squane, Rectangle, Parallaarem| Paint of intersection of liagonts Triangular Plane Lamina Point of Intersection of mections —$——— | Rectangular [Cubical block | Points of intersection of diagonal Hollow / Solid Cylinder Midldle point off the axis . A On the anis the cone ot Solid Cone or Pyramid pot Teeny from the Vertex y h = height of the Cone Equations of Linens MotHon + Rotational Motion Linean, Mom Retational Motion Ch) Ib a =0, ten U=tmast. |G) Th & <0, then w = Gm. Gi) vosutat (id ton = Ww tet qin s = ut +Lae WH) 8 = wt + batt dv) Vv WwW) S =U + $alQ2n-1) (Y) GO = wW, + da (2n-1) Wid v = ae @) w- ae Wit) a ay = oe Wil) a= dia = aie ‘ 2 = ur+ 205 dv) Wr = we +200Moment of Inte - Ist Ja a property of a body cue + which ut Opposes any change in the State of vest oy Uniform rotator. Moment of Inerha plays the Same role ‘im rotatonal motion ab mars plays in Linean motion. The moment of inertia of ae Q particle with respect JD an axis of rotation is equal to the product of the mass of the particle and Square of distance from rotatonal Gis. Moment of Treatia of System ob Particles I= myP+myie... axis r or I = mr om e -2.. Foy Continuous distribution of mass Va Lik} Note - (i) SZ Unit = gem Gi) Moment of and lensi ty Bimensions- Cm re] mata cepenals on mars, shape, site ve oF the body, T+ also erls Upon (ity Me Festi ofy axis oy rotation and a Aigtribution. a 1, . Moment ob imatia does not dupencls upon Angular | velocity, avioular ° Momentum el the bea? torque and angular _4Radius of Gyration (K) The radius of gyration of A body 1b the perpendialar distance from axis of Totation, the Squane of this clistance when multiplieel by the mars of body then it gives the moment of inanta of the bedy (T= MKY) about seme axis of yotadion. Il=Mk* » From the formula of oliscyete lia tribution Lo mint + mary teres mat TEM, =M_ = --- My = M then Lo= M(44 +e Ye) Hone MK™ =m (rt-artee 13) (am)k* = m(yP4rte= W) [TL = Total No. of Pasties} Hence raoliud of gyration of a body about a given axis 4 equal do the root mean Squars olistanee of the consti tutnt Pantides of the body from the given axis. at SZ unit - Metre Dimensions - Cm’? ] 2) Radius eb gyration depends on axis of rotation and distribution of masd of the body (shape and size). 3d Radius op gyration oes not depend on mars of the body .Theorem of Ponollul axed- Moment ob inertia of a body about an axis is equal to the moment of Inutia about a parallel axis passing ne toy the centre of mars plus product of maw of the and the square of distance bebwen thae two Norre: axis. Note Applicable for bodies of any type amd shape’ Theorem of Pe diculan Axes — Moment of Inertia of a plane daming about an axis perpendicular 4o its plane 4s equal to the sum of the moment of inarHa of the lamina about any to mutually perpendicular ads im its own plane pie secting each ether at the peint through which the perpendicular axis Pardes. Iz = Ixy + Ty How Ix, Ty and Tr ane moment of intra about KY and Z axis Yedpectvely, Note - Applicable only fox dum cimensional bedies and Gn not be used for three climensional bodies.rough the centre ob ring lane - ie fet 2] About the diameter of the ving - Let moment of inertia of the ring about each ameter (xx! and yy!) 4s Ta- Both the ciameter ane perpenctiaular to the axis zz! which ds passing the Centre of the ring ard perpendicwar 40 its plane - By theorem of Perpendicular axed Iz, = Ix + Ty or Iz = Ia +Ta = 20g MR* = 274 ES Ty = MR— diameter of the ving 3) About an axis tangential and poralll to the By theorem of parallel axis It = Ia + MR* It = 2 = MR™ 4 mR* a2 = 2MR* [tee | Ul About the the cmtre of nt ponaltel to the axis parting thigh and Perpendicular to its plane- By theorem of parallel axis Iy = I + me? Ir = mMR> 4+ mer Ir = 2meR———————— Moment of Inertia of some standard Bodies ——— Shape of Body Axis of Rotation Figure Moment of Inertia K Circular Dise Mass = M Radius = R | | | = About an axis passing through the centre and perpendicularto the plane of dise About a diametric axis wv OD 1 yy? gM 1 wR? qm About an axis tangential to the rim and lying in the plane of the dise 4 Sup? a About an axis tangential tothe rimand perpendicular to the plane of dise € MR? NI® Circular Ring Mass = M Radius = R About an axis passing through the centre and perpendicular to the plane of Ring <4 MR? About a diametric axis a D Amp?Axis of Rotation Aboutits | geometrical axis | Ez Solid Cylinder 5 T Mass=M ‘tangential to the le a Radius=R & parallel to its Length=L | Aboutan axis passing through the centre of and perpendicular to its length Aboutits geometrical axis Hollow Cylinder About an axis Mass = M perpendicular to Redius=R | its lengthand passes through Length=L the centre Aboutan axis perpendicular to its length & Passing through one end Bl * [| | =|Shape of Body | Axtsof Rotation Figure Moment of Inertia K About an axis al Rp+R the centre and Mp2, pe: - SIRF +R: a Anmular Disc | perpendicularto 3 it + Re} Mass = M. the plane of disc Internal Radius = R, Outer Radius = R, About a diametric axis Ry +R; Mint +R “~ 2 2 CO MR? ee 5 Ss Se gue SR 2 ur? © gMR 2Thin Rod Mass = M Length = L _Auls of Rotation About an axis passing through centre & perpendicular to its length | ‘Moment of = | ML? 2 About an axis passing through one end & perpendicular to its length Rectangular Pate Mass = M Length =/ Breadth = b Equiateral Triangular plane Side =a About an axis passing through centre and perpendicular to plane About an axis passing through centre & perpendicular to side | in its plane e About an axis passing through | centre & perpendicular to side /in its plane About an axis passing through centre & perpendicular to its plane “3 12Shape of Body | Axis of Rotation siptcal disc of semumajor wos = @ and senuminor axis = b Passing through | CM and perpendicular to | the plane | | | 4 | Solid cone of radius) Axis joining the Randheighth | vertex and centre of the base Torque (or Moment of force) / The twening effect of a force about the axis of rotation Callal torque oy moment of force out +0 the given fore. ’ Rotation Rotation Radial Component of force Np oe Fez F Cos f Trartverse Component of Fore Cy i Fi= FSin@ (a) (a) As T = TFSiNO = TF T= Position Vector xX Transverse Component of force @ Hence Toque 44 clue tO trarsverse Component + fore only, orDivecton - terque is an axial vector. T4s direction ds alw. perpendicular to the plane. Containing vector VW and Fy ae direction id given by ‘Right hand sere Rule’. Gi) For clockwise yotetion Ts -ve Wi) Foy Awhelockwise votation te +ve Note - Aj For & given force and angle, magnitude of forque depends on v. The more the value of 1, the more Will be tre forque and Uts easy to votate the body. Couple - ; ae == Tt WB olepined oF the Gmbinadtion of two equal but oppositely Airvecteal Forces Not acting along the same Line» FE Cs Tx er ——— F Note - | The bosic olipperence between tovque anc comple is rata ‘pelied” couple both the forces are : extern ied while in case UL One i force 48 Bressary applied and wl once ja Yeachonany-Angular Momentum - Tre angulan. momentum of a body about 2 given axis Ls the product ots its LUnean momentum and per- Pendiculan olistence of Line off acton of Unear mometum vector from te axils of yotatton. Angular Momentum B/S Plo. 6 1) Lo=mvxrsine ee Tm vec form , or T = mPxv) s SI Unit — Toule-sec. Dimensons — (M2174 =~ In Contesion Coordinates ity Voz xtasyf azk = P= At +A) + Pek and “Vz Vet + Wt 4VeR ~ tor Tan Lely y 2 = (VA-Z2R)E - (xP -2P If (xR YER Fey ad T tT aR t=m e y z = m[(vve “RUE (eva~zmh + (eW-VIR]— — Note ~ 4) lan momentum is an axial vector» Its obirecHon is given by !Right hand Screw Rule’. Z) Fr 0-0 oreo, Lim =0 For @ = 30° > Lima = mvr 3); In Case of Circular motion Tae = m(PeV) = mvrsine Lo=mvr zmrtw [As Y¥LV J oy | L= Iw (As re mr] Tre rode of change of angulan momentum 4s equal 4s the not torque acting on the particle, This expression is also Known a> Newton's 2) Law of rotational motion. BJ In Core ok Yotational Motion Angulan Momentum lL = Tw Rotational Kinetic Energy En = fiw > Ee = 42[4]” > £ = 64 Argilan momentum of a System of panticies is equal to vector sum oh angular momentum of each partcle. sccsikar. T st4ts+...Th www.—— | Ang lan Impulse - Ih @ large torque acts on a body for a small tme then Angular Impulse = Tat =. oo = At As Ts RE Angular Impulse = AL Hence angular Impulse 48 equal to the change in a momentum. It has same unit, dimension andl clivecton as angular, momentum. Law of Conservation of Angular Momentum - Th the resultant torque acting on a body is zew than total angulon momentum of the system yemains Constant. By Newton's 274 law of rotational Motion T= a Ty Tso then At =o D> Also Henee change in angular. momentum =0, oY Le= Constent, or Li = Ls > TW, = Lor &3. When a person havi weights im his hands and standing On @ Yotating platform, suddenly folds his arma, ther its moment of imaha clecreases and in accorclance the angulan speed increases. ra 4Note - Ip external torque of the system is zero, ten the angular momentum Js Consedved. Homer the rotatimal Kinetic energy ts not Corvservecd + T= then Tw, = LWe > 4( dv) = L(Gn) batt] ttt ov phe = Phe then Tike = Inka Hene if >IT, then Key < Ke, So ik moment of Imeatiq decreases, the retahonal Kinadtic energy Increases ane vice-versa,Rototional Kinetic Enengy - The enargy clu 40 rotaHonal moHon of a body 3b Known as rotobbonal Kinehe ensngy lat a rigid body is rotating about an axis with uniform angular Velocity WO. Ib the becky Js composed oF panticles of masses my, may... trun Kinetic energy of rotating bedy Eg = dmyp+ Amis. As tne Linear velocities off He poueticles ten ec, = E (mph tind -.- )uo* “ Me Werk in Retotonal Motion - Let a tangential Force Fr acts at the rim of pivoted clisk. W Laxis The disc *ototes trough a small GD angle dé during ~ small dime alt. The usork clone by the force Fy while Q point on the rim moves a clistence as 4s dw = Frds i Th 6 48 angular clsplacement then as = Rdo i dw = FRdob| The torque duc to the force Fo ia TPH RR dw = tdo During an angular displacement from 6, to 6. +the total work clone by the torque is 6, It the torque is constant while the angle Changes by 8; 40 82 ten weet s[ar ety Hen +he work done by a constant torque id the Product of torque and the angular diuphroment. Let T represents the net torque on the beady so thot Ts 1a Pessuming thot the bedy is rigid So that the moment of imatia I is Gmstant , then Tap = (Ia)dep = 1 dw do = Twoaw Hente total Work done on the rotading body iA Wd: w =ftwdw > [we Z (at -wFJ Ww) & When a torque cloes work on a rotating body > the KineHe energy changes by an amount equal +o the work clone,Work _Enevgy theorem in RotaHonal Moton- The change in the rotational Kinetic energy of a vigid body dd equal to the Work clone by external torque. We A&e = 410, - £1uy Rotational Power - TH Gs te Power awsotiated with he work done by a terqua acting on a rotating body. Work done by the torque T luring Amal cbisplacemint ae is dw = tae dividing beth Sick by the me interval ot during which the angular dlisplacemant oecurs, we get Where FR 4d instantaneous Power Rotational Equilbrum - A body 4s Zoid do be in votatioal equilibrium 1b resultant torque acting on it is zeyo. Eg, Im case of beam balance the System will be in yotational equilibrium if op» Tut =0 Or by = TM a Fily = FleSlippin Spinning and Rollin 4d SHPPIMA- ionen the body slides on a surface without Yoram Hun its mohon ud Catlol Slipping motion. @=9 Friction betwen the booly and surface Fro ¢ \ Bool Possess ordy tronslatocy KN ] ae Kinetic Energy Kr = mv \ 24 Spinning - = When the booky rotates in such a manner that its axis of rotatory aes not move then ‘ts mortion) ib Called spinning motion. In His tendition the axis ef yotathon oa body 44 fixed. In Spinning the body possess only votedory Kinetic energy Ka = $ Iu -2 2zy2 1 en oy Ke = $mk = Lm) Hence Rotatary KE. = x x Transitory KE. Rolling Motion -— When a body perkorms combined translatory and Yotatory motion then ut is Known aa rolling. The. velotity of Centre of maw yepresents linear motion tule angular Velocity represents yotetvry motion.For put rolling Vv = OR By Law of conservation of ennrgy Totol Energy = Transladory Kinutic Energy + Rotatory Kinsie Enurgy Toto) Energy E = Er +Er Eo: Amy? + Arw* Es amy + dm (4) E+ dmv*[1 +5] 2v Ve > Vtvi2v = [vtsyvi = favRotatory (Kr) 1 got XE zm [15] 2” a Ring and Cylinderical mv? Shell Dise and Solid 342 Cylinder Solid Sphere 7 ay? Hollow Sphere Rolling motion on an inclined Plane - Let a bedy of mars ‘mm eee and reading ‘R’ rolls down on inclined plane ° b height * h’ and angle of Pehnatten 8. As the body rolls down its Potentiou ennrgy doses and it gain Kinetic ener of translation and roteH By Comsewasion of mechanical enrgy 2 mah = mv*(1 + BP) Neloaty at the Lowest Point__ Acceleration in moton Gem eguahon v2 = ur+205 by Substituting uso, S= hang v= [23h g Sme Toe KE we gat Re Time of descent - Frm cesucton veur+at By Aubatrtucting U=0 and valu of ‘V’and’a’ we ge ee 2h 2 t= sine eo +8] Note - tj) Hee factor KY 4s a measure of moment of inertia. Lesser the moment of inertia of the rolling body lesser will be the Value of K7R2. So areates wall be its velocity and acceleration and fesser wil) be the time of olexent. 24 From Figuae 0; > 0. >6s A>a >a ti< tr < ts Vi= My = MsNote - Foy re without slipping “the minimum Value of coeffich of fricton Amin = — tan G+ &) Important Points - ft.) For pure velling there shod be frletHim on the Surface because Jt provides torque to He body - 2) On Smooth sunkace rolling does not take pte. 3) The clibplacerrant of point of Contact with respect fo the surface is equal 40 zero in pure rolling So Work done i equot 4o zero Work is done agamat frletion force in Slicing and it re negative: 5) Angular momentum Ja Corservedl about point of Gadact.Comparison behoeen formula of transitory and Rotational Motion Translatory Motion Rotatonel Moton = _ = SF = ma Forte F = $f Lynean Momentum P= mv Linean Kinetic Energy K= dmy? Work done by Constord Force we Fe Work clone, by variable. Force. Sra Powen P = as = ev we Work Energy theorem m TM. WwW = dmvp = dmy> Linear. Impulse = AB = Fat Torque v: x 21x Angular momerdum TE = 1a Retational Kinetic Energy Ep =4 10+ Work done by Constand Torquc we 8 Work dene by varioble Terque- Ww = ft.a6 Power p= dw = Tit Work Energy theorem in RM. W = drut -AIuy Angular Impulse = SL TA