PHYSICS VOL.1

240(iii) About a tangentThe moment of inertia of the ring about a tangent EF parallel to ABis obtained by using the parallel axes theorem. The moment of inertia ofthe ring about any tangent is,I = I + M RTd2+ 221= 2 MRMR22 T3IMR =3 Moment of inertia of a circular disc(i) About an axis passing through itscentre and perpendicular to its planeConsider a circular disc of mass Mand radius R with its centre at asOshown in Fig. 4. Let be the mass perσunit area of the disc. The disc can beimagined to be made up of a large numberof concentric circular rings of radiivarying from to .Let us consider oneORsuch ring of radius and width rdr.The circumference of the ring = 2 r π .The area of the elementary ring = 2πr drMass of the ring= 2 r dr = 2 r dr πσπ σ...(1)Moment of inertia of this elementary ring about the axis passingthrough its centre and perpendicular to its plane isdI= mass × ( distance )2= (2 r dr) rπ σ2...(2)The moment of inertia of the whole disc about an axis passingthrough its centre and perpendicular to its plane is,I= ∫ RO2πσr 3dr = 2πσ∫ ROr 3dr = 2πσ44ROr⎡⎤ ⎢⎥ ⎣⎦(or)I=() 42221 = R42RRπσπσ = 212MR...(3)where M = Rπ2σ is the mass of the disc.drrROFig 4 Moment of inertia of acircular disc

241(ii) About a diameterSince, the disc is symmetrical aboutany diameter, the moment of inertia aboutthe diameter AB will be same as its momentof inertia about the diameter CD. Let it beId(Fig. 5). According to perpendicular axestheorem, the moment of inertia I of the disc,about an axis perpendicular to its plane andpassing through the centre will be equal tothe sum of its moment of inertia about twomutually perpendicular diameters ABand CD .Hence,I = I + I dd= 12MR2= MR142(iii) About a tangent in its planeThe moment of inertia of the disc about the tangent EF in the planeof the disc and parallel to AB can be obtained by using the theorem ofparallel axes (Fig. 3.15).IT= I + MRd 2 =212MRMR4+∴ Ι = Τ54 MR24 Moment of inertia of a sphere(i) About a diameterLet us consider a homogeneous solidsphere of mass M, density and radius Rρwith centre O (Fig. 6). AB is the diameterabout which the moment of inertia is to bedetermined. The sphere may be consideredas made up of a large number of coaxialcircular discs with their centres lying onAB and their planes perpendicular to AB.Consider a disc of radius PO = ′ y andthickness dx with centre O and at a′ distance x from O,ABCDEFRORPO /dx xyFig 6 Moment of inertia of asphere about a diameterABCODEFRFig 5 Moment of inertia of adisc about a tangent line

242Its volume = πy dx 2...(1)Mass of the disc = π y dx . 2ρ...(2)From Fig. 6, R = y + x 222 (or) y = R – x 222...(3)Using (3) in (2),Mass of the circular disc = π ( R – x ) dx 22ρ...(4)The moment of inertia of the disc about the diameter AB is,dI= (mass) × (radius)122= 221(). ()ρ2Rx dxyπ2−= () π−2221ρ 2Rxdx... (5)The moment of inertia of the entire sphere about the diameter AB isobtained by integrating eqn (5) within the limits x = –R to x = + R.∴ = I22 21ρ ()2 RRRxdxπ+−−∫ I = 2 × 22 21 (ρ)()2RORxd xπ−∫ = 4422(ρ)(2)RORxR xd xπ+− ∫ = 5552ρ53RRR π⎡⎤ +−⎢⎥ ⎣⎦ = 532842ρ=1535RRRππ ρ ⎛⎞⎛⎞ ⎛⎞⎜⎟⎜⎟ ⎜⎟⎝⎠⎝⎠ ⎝⎠ = 2222.=55MRMR⎛⎞ ⎜⎟ ⎝⎠whereM = 34ρ3Rπ = mass of the solid sphere∴I = 225MR(ii)About a tangentThe moment of inertia of a solid sphere about a tangent EF parallelto the diameter AB (Fig. 7) can be determined using the parallel axestheorem,ABODEFRFig 7 Moment of inertia of asphere about a tangent

243IT= IAB + MR2 = 2252MRMR+∴ I =T75 MR25. Moment of inertia of a solid cylinder(i) about its own axisLet us consider a solid cylinder of mass , radius and length ItMRl.may be assumed that it is made up of a large number of thin circulardiscs each of mass and radius placed one above the other.mRMoment of inertia of a disc about an axis passing through its centrebut perpendicular to its plane = 22 mR∴ Moment of inertia of the cylinder about its axis = IΣ22 mRI = 222222RRMRmM⎛⎞ ==⎜⎟ ⎝⎠∑(ii) About an axis passing through its centre and perpendicular toits lengthMass per unit length of the cylinder = Ml...(1)Let O be the centreof gravity of the cylinderand YOY be the axis′passing through thecentre of gravity andperpendicular to thelength of the cylinder(Fig. 8).Consider a smallcircular disc of width dxat a distance from thexaxis YY .′∴ Mass of the disc = mass per unit length × width = Ml⎛⎞ ⎜⎟ ⎝⎠ dx...(2)l/ 2xOdxXY /YX /Y 1Y 1/Fig.8 Moment of inertia of acylinder about its axis

244Moment of inertia of the disc about an axis parallel to YY (i.e) about′its diameter = (mass) 24radius⎛⎞ ⎜⎟ ⎝⎠ = 2244MRMRdxdxll⎛ ⎞⎞⎛=⎜ ⎟⎟⎜ ⎝⎠⎝⎠...(3)By parallel axes theorem, the moment of inertia of this disc aboutan axis parallel to its diameter and passing through the centre of thecylinder (i.e. about YY ) is′dI =24 MRl⎛⎞ ⎜⎟ ⎝⎠ dx + Mdxl⎛⎞ ⎜⎟ ⎝⎠ (x )2...(4)Hence the moment of inertia of the cylinder about YY is,′I = /222/24llMRM dxx dxll+−⎛⎞+⎜⎟ ⎝⎠ ∫ = I/2/222/2/24llllMRM dxx dxll++ −−+∫∫ I = []/223/2/2/24llllMRMxxll3++−−⎛⎞ + ⎜⎟ ⎝⎠ I =⎡⎤ ⎛⎞ ⎛ ⎞⎢⎥−−⎜⎟ ⎜ ⎟⎡⎤⎛ ⎞⎛ ⎞⎝ ⎢⎠ ⎝ ⎠⎥ −−+⎜ ⎢⎟⎜ ⎟⎥ ⎢⎥⎝ ⎣⎠⎝ ⎠⎦ ⎢⎥ ⎢⎥ ⎣⎦332224223llMRllMll = I () 24 MRll3 224Mll⎡ ⎛⎤ ⎞ + ⎢ ⎜⎥ ⎟ ⎝⎠ ⎣⎦ = 22412MRMl+I = M22412Rl⎛⎞+⎜⎟ ⎝⎠...(5)

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m u11 – m1 1v = m u21 – 2m2u + m v221(m + m )v = m u12111 – m2u + 2m u122(m + m )v = u (m12111 – m2) + 2m u22Special cases Case ( i) :If the masses of colliding bodies are equal, i.e. m = m12v = u and v = u ...(9)1221After head on elastic collision, the velocities of the colliding bodies are mutually interchanged.Case (ii) : If the particle B is initially at rest, (i.e) u = 0 then2and 2.9.2 Inelastic collisionDuring a collision between two bodies if there is a loss of kinetic energy, then the collision is said to be an inelastic collision.Since there is always some loss of kinetic energy in any collision,collisions are generally inelastic.In inelastic collision, the linear momentum is conserved but the energy is not conserved. If two bodies stick together, after colliding, the collision is perfectly inelastic but it is a special case of inelastic collision called plastic collision. (eg) a bullet striking a block of wood and being embedded in it. The loss of kinetic energy usually results in the form of heat or sound energy.Let us consider a simple situation in which the inelastic head on collision between two bodies of masses m Aand m Btakes place. Let the colliding bodies be initially move with velocitiesu Aand u B. After collision both bodies stick together and moves with common velocity . vTotal momentum of the system before collision = m u + m uAABBTotal momentum of the system after collision = mass of the composite body common velocity ×= (m + m ) vABBy law of conservation of momentum91v2=(m2 – m1 )(m + m )12u 22m u11+(m + m )12v2=(m + m )122m u11v1=(m1 – m2 )(m + m )12u 1 ........(8) ........(7)v1= um 1 – m2(m + m )12m + m1212m u22+[ [

m u + m u = (m + m ) v (or) v =. A AB BABThus,knowing the masses of the two bodies and their velocities before collision, the common velocity of the system after collision can be calculated.If the second particle is initially at rest i.e. uB = 0 thenv = kinetic energy of the system before collision E = K1 [ u = 0] Bkinetic energy of the system after collisionEK2 =Hence Substituting the value of v in the above equation, (or) < 1It is clear from the above equation that in a perfectly inelastic collision, the kinetic energy after impact is less than the kinetic energy before impact. The loss in kinetic energy may appear as heat energy.92Kinetic energy before collisionE K2E K1Kinetic energy after collision=m u + m uA AB B(m + m )AB m uA A212(m + m ) vAB212 m uA A2(m + m )vAB2=E K1E K2 m + mAB mA=E K1E K2E K1E K2m u A A(m + m )AB