Rotational Motion & Center of Mass Notes

ROTATIONAL MOTION & CENTER OF MASS Page # 1 ROTATIONAL MOTION & CENTER OF MASS 1. RIGID BODY Rigid body is defined as a system of particles in which distance between each pair of particles remains constant (with respect to time) that means the shape and size do not change, during the motion. Eg. Fan, Pen, Table, stone and so on. Our body is not a rigid body, two blocks with a spring attached between them is also not a rigid body. For every pair of particles in a rigid body, there is no velocity of seperation or approach between the particles.   In the figure shown velocities of A and B with respect to ground are VA and VB respectively A VA sin1 A A VA cos 1  1 VA B B VB  2 VBA B VB sin 2 VB cos 2 If the above body is rigid VA cos 1 = VB cos 2 Note : With respect to any particle of rigid body the motion of any other particle of that rigid body is circular. VBA = relative velocity of B with respect to A. Types of Motion of rigid body Pure Translational Motion Pure Rotational Motion Combined Translational and Rotational Motion 1.1. Pure Translational Motion : A body is said to be in pure translational motion if the displacement of each particle is same during any    time interval however small or large. In this motion all the particles have same s, v & a at an instant. Example. A box is being pushed on a horizontal surface. 10 6 6 10 16     a cm  a of any particle Vcm  V of any particle,   Scm  S of any particle For pure translational motion :v m1 v m4 v m2 v m7 v m5 v m3 v vm6 m8 m2 m1 m3 m4 m5 m7 m8 m6     Fext  m1a1  m 2 a 2  m 3 a 3 ............. Where m1, m2, m3, ......... are the masses of different particles of the body having accelerations    a1, a 2 , a 3 ,............... respectively..     But acceleration of all the particles are same So, a1  a 2  a 3  .........  a   Fext  Ma ROTATIONAL MOTION & CENTER OF MASS Page # 2 Where M = Total mass of the body  a = acceleration of any particle or of centre of mass of body     P  m1v1  m 2 v 2  m 3 v 3 .............    Where m1, m2, m3 ...... are the masses of different particles of the body having velocities v1, v 2 , v 3 ............. respectively     But velocities of all the particles are same so v1  v 2  v 3 ..........  v   P  Mv  Where v = velocity of any particle or of centre of mass of the body.. Total Kinetic Energy of body = 1 1 1 m1v 12  m 2 v 22  .......... .  Mv 2 2 2 2 1.2. Pure Rotational Motion : A body is said to be in pure rotational motion if the perpendicular distance of each particle remains constant from a fixed line or point and do not move parallel to the line, and that line is known as axis of    rotation. In this motion all the particles have same ,  and  at an instant. Eg. : - a rotating ceiling fan, arms of a clock. For pure rotation motion :m3 m1 m5 m6 m1 m4 m4 d Where  = angular speed of the body.. dt  m3 m6  m5 r = distance of particle from the axis of rotation. m2 m2 s Where  = angle rotated by the particle r s = length of arc traced by the particle.  d Where  = angular acceleration of the body.. dt All the parameters ,  and  are same for all the particles. Axis of rotation is perpendicular to the plane of rotation of particles.  Special case : If  = constant,  = 0 +  t Where 0 = initial angular speed 1 2 t 2 2 = 02 + 2     0t  t = time interval Total Kinetic Energy  1 m1v12  1 m 2 v 22 ................. 2 2 1  [m1r12  m 2r22 ................]  2 2 1  I 2 Where I = Moment of Inertia = m1r12  m 2r22 ....... 2  = angular speed of body. 1.3 Combined translation and rotational Motion A body is said to be in translation and rotational motion if all the particles rotates about an axis of rotation and the axis of rotation moves with respect to the ground. 2. MOMENT OF INERTIA Like the centre of mass, the moment of inertia is a property of an object that is related to its mass distribution. The moment of inertia (denoted by I) is an important quantity in the study of system of particles that are rotating. The role of the moment of inertia in the study of rotational motion is analogous to that of mass in the study of linear motion. Moment of inertia gives a measurement of the resistance of a body to a change in its rotaional motion. If a body is at rest, the larger the moment of inertia of a body the more difficuilt it is to put that body into rotational motion. Similarly, the larger the moment of inertia of a body, the more difficult to stop its rotational motion. The moment of inertia is calculated about some axis (usually the rotational axis). ROTATIONAL MOTION & CENTER OF MASS Page # 3 Moment of inertia depends on : (i) density of the material of body (ii) shape & size of body (iii) axis of rotation In totality we can say that it depends upon distribution of mass relative to axis of rotation. Note : Moment of inertia does not change if the mass : (i) is shifted parallel to the axis of the rotation (ii) is rotated with constant radius about axis of rotation r 2.1 Moment of Inertia of a Single Particle For a very simple case the moment of inertia of a single particle about an axis is given by, I = mr2 ...(i) Here, m is the mass of the particle and r its distance from the axis under consideration. 2.2 Moment of Inertia of a System of Particles The moment of inertia of a system of particles about an axis is given by, I= m r 2 i i ...(ii) i r1 r2 m3 r3 m1 m2 where ri is the perpendicular distance from the axis to the ith particle, which has a mass mi. 2.3 Moment of Inertia of Rigid Bodies For a continuous mass distribution such as found in a rigid body, we replace the summation of I  m r 2 i i by y i an integral. If the system is divided into infinitesimal element of mass dm and if r is the distance from a mass element to the axis of rotation, the moment of inertia is, I = r 2 dm r where the integral is taken over the system. (A) Uniform rod about a perpendicular bisector Consider a uniform rod of mass M and length l figure and suppose the moment of inertia is to be calculated about the bisector AB. Take the origin at the middle point O of the rod. Consider the element of the rod between a distance x and x + dx from the origin. As the rod is uniform, A x dx Mass per unit length of the rod = M / l so that the mass of the element = (M/l)dx. 0 B The perpendicular distance of the element from the line AB is x. The moment of inertia of this element about AB is M dI  dx x 2 . l When x = – l/2, the element is at the left end of the rod. As x is changed from – l/2 to l/2, the elements cover the whole rod. Thus, the moment of inertia of the entire rod about AB is l/2 I  M x3  M 2 x dx    l  l 3  l / 2 l/2  –l / 2 Ml 2 12 (B) Moment of inertia of a rectangular plate about a line parallel to an edge and passing through the centre The situation is shown in figure. Draw a line parallel to AB at a distance x from it and another at a distance x + dx. We can take the strip enclosed between the two lines as the small element. ROTATIONAL MOTION & CENTER OF MASS Page # 4 x A b dx B l It is “small” because the perpendiculars from different points of the strip to AB differ by not more than dx. As the plate is uniform, its mass per unit area = M bl M M b dx  dx . bl l The perpendicular distance of the strip from AB = x. Mass of the strip = The moment of inertia of the strip about AB = dI = M dx x 2 . The moment of inertia of the given plate is, l therefore, l/2 I  M 2 Ml 2 x dx  l 12 l / 2 The moment of inertia of the plate about the line parallel to the other edge and passing through the centre may be obtained from the above formula by replacing l by b and thus, Mb2 . 12 (C) Moment of inertia of a circular ring about its axis (the line perpendicular to the plane of the ring through its centre) Suppose the radius of the ring is R and its mass is M. As all the elements of the ring are at the same perpendicular distance R from the axis, the moment of inertia of the ring is I    I  r 2 dm  R2 dm  R2 dm  MR2 . (D) Moment of inertia of a uniform circular plate about its axis Let the mass of the plate be M and its radius R. The centre is at O and the axis OX is perpendicular to the plane of the plate. X 0 dx x R Draw two concentric circles of radii x and x + dx, both centred at O and consider the area of the plate in between the two circles. This part of the plate may be considered to be a circular ring of radius x. As the periphery of the ring is 2 x and its width is dx, the area of this elementary ring is 2xdx. The area of the plate is  R2. As the plate is uniform, M Its mass per unit area =  R2 M 2 M xdx 2 xdx   R2 R2 Using the result obtained above for a circular ring, the moment of inertia of the elementary ring about OX is  2 Mx dx  2 dI   x  R2  . Mass of the ring  ROTATIONAL MOTION & CENTER OF MASS Page # 5 The moment of inertia of the plate about OX is R I 2M R 2 MR 2 . 2 x 3 dx  0 (E) Moment of inertia of a hollow cylinder about its axis Suppose the radius of the cylinder is R and its mass is M. As every element of this cylinder is at the same perpendicular distance R from the axis, the moment of inertia of the hollow cylinder about its axis is   I  r 2 dm  R 2 dm  MR 2 (F) Moment of inertia of a uniform solid cylinder about its axis Let the mass of the cylinder be M and its radius R. Draw two cylindrical surface of radii x and x + dx coaxial with the given cylinder. Consider the part of the cylinder in between the two surface. This part of the cylinder may be considered to be a hollow cylinder of radius x. The area of cross-section of the wall of this hollow cylinder is 2 x dx. If the length of the cylinder is l hollow cylinder is 2 x dxl. The volume of the solid cylinder is  R2 l and it is uniform, hence its mass per unit volume is M   R2 l The mass of the hollow cylinder considered is dx M 2 x dx l  2M xdx . R l R2 As its radius is x, its moment of inertia about the given axis is 2 x  2M  dI   2 x dx x2 . R  The moment of inertia of the solid cylinder is, therefore, R I 2M R 0 2 x 3 dx  MR2 2 . Note that the formula does not depend on the length of the cylinder. (G) Moment of inertia of a uniform hollow sphere about a diameter Let M and R be the mass and the radius of the sphere, O its centre and OX the given axis (figure). The mass is spread over the surface of the sphere and the inside is hollow. Let us consider a radius OA of the sphere at an angle  with the axis OX and rotate this radius about OX. The point A traces a circle on the sphere. Now change  to  + d and get another circle of somewhat larger radius on the sphere. The part of the sphere between these two circles, shown in the figure, forms a ring of radius R sin. The width of this ring is Rd and its periphery is 2R sin. Hence, the area of the ring = (2R sin) (Rd). x Mass per unit area of the sphere  The mass of the ring  M M 4 R 2 R sin A Rd . (2R sin )(Rd)  2 R M sin  d. 2 4 R The moment of inertia of this elemental ring about OX is 0 d M  d I   sin  d. (R sin ) 2  M R2 sin3  d 2  2 As  increases from 0 to , the elemental rings cover the whole spherical surface. The moment of inertia of the hollow sphere is, therefore,   M 2 MR2   (1  cos 2 )sin  d R sin3  d  2 2   0 0   I    MR2  cos 3   2 2  cos     MR 2  3  3 0    MR 2   (1  cos 2 ) d (cos ) 2     0   ROTATIONAL MOTION & CENTER OF MASS Page # 6 (H) Moment of inertia of a uniform solid sphere about a diameter Let M and R be the mass and radius of the given solid sphere. Let O be centre and OX the given axis. Draw two spheres of radii x and x + dx concentric with the given solid sphere. The thin spherical shell trapped x between these spheres may be treated as a hollow sphere of radius x. The mass per unit volume of the solid sphere dx 0 x M 3M  3 4 R3 4 R 3 The thin hollow sphere considered above has a surface area 4x2 and thickness dx. Its volume is 4  x2 dx and hence its mass is  3M  3M  (4  x2 dx) = 3 x 2 dx =  3 R  4 R  Its moment of inertia about the diameter OX is, therefore, = 2  3M 2  x dx x2 3  R3  2M x 4 dx R3 If x = 0, the shell is formed at the centre of the solid sphere. As x increases from 0 to R, the shells cover the whole solid sphere. The moment of inertia of the solid sphere about OX is, therefore, dl = R I= 2M R 3 x 4 dx = 0 = 2 MR 2 . 5 3. THEOREMS OF MOMENT OF INERTIA There are two important theorems on moment of inertia, which, in some cases enable the moment of inertia of a body to be determined about an axis, if its moment of inertia about some other axis is known. Let us now discuss both of them. 3.1 Theorem of parallel axes two parallel axes, one of which passes through the centre of mass. Two such axes are shown in figure for a body of mass M. If r is the distance COM between the axes and ICOM and I are the respective moments of inertia about them, these moments are related by, I = ICOM + Mr2 r * Theorem of parallel axis is applicable for any type of rigid body whether it is a two dimensional or three dimensional 3.2 Theorem of perpendicular axes The theorem states that the moment of inertia of a plane lamina about an axis perpendicular to the plane of the lamina is equal to the sum of the moments of inertia of the lamina about two axes perpendicular to each other, in its own plane and intersecting each other, at the point where the perpendicular axis passes through it. Let x and y axes be chosen in the plane of the body and z-axis perpendicular, to this plane, three axes being mutually perpendicular, then the theorem states that. z y ri O xi P yi x Iz = I x + Iy Important point in perpendicular axis theorem (i) This theorem is applicable only for the plane bodies (two dimensional). (ii) In theorem of perpendicular axes, all the three axes (x, y and z) intersect each other and this point ROTATIONAL MOTION & CENTER OF MASS Page # 7 may be any point on the plane of the body (it may even lie outside the body). (iii) Intersection point may or may not be the centre of mass of the body. 4. TORQUE : Torque represents the capability of a force to produce change in the rotational motion of the body Line of action of force P F r r sin Q 4.1 Torque about point :     Torque of force F about a point   r  F  where F = force applied P = point of application of force Q = point about which we want to calculate the torque.  r = position vector of the point of application of force from the point about which we want to determine the torque.    rF sin  = rF = rF where  = angle between the direction of force and the position vector of P wrt. Q. r = perpendicular distance of line of action of force from point Q. F = force arm SI unit to torque is Nm Torque is a vector quantity and its direction is determined using right hand thumb rule. 5. BODY IS IN EQUILIBRIUM : We can say rigid body is in equillibrium when it is in (a) Translational equilibrium  i.e. Fnet x = 0 and Fnet y = 0 and Fnet  0 , (b) Rotational equillibrium   net  0 i.e., torque about any point is zero Note : (i) If net force on the body is zero then net torque of the forces may or may not be zero. example. A pair of forces each of same magnitude and acting in opposite direction on the rod. F B C A F 2  A  2F (2) If net force on the body is zero then torque of the forces about each and every point is same  about B  B  F + F  B  2F  about C  C  2F 6. RELATION BETWEEN TORQUE AND ANGULAR ACCELERATION The angular acceleration of a rigid body is directly proportional to the sum of the torque components along the axis of rotation. The proportionality constant is the inverse of the moment of inertia about that axis, or   I Thus, for a rigid body we have the rotational analog of Newton's second law ; ROTATIONAL MOTION & CENTER OF MASS Page # 8    I ...(iii) Following two points are important regarding the above equation. (i) The above equation is valid only for rigid bodies. If the body is not rigid like a rotating tank of water, the angular acceleration  is different for different particles. (ii) The sum  in the above equation includes only the torques of the external forces, because all the internal torques add to zero. 7. ANGULAR MOMENTUM 7.1 Angular momentum of a particle about a point.    L = r p sin  L  r P   |L|  r  P  |L|  P  r  Where P = momentum of partilcle  r = position of vector of particle with respect to point about which angular momentum is to be calculated.    = angle between vectors r & p Pcos    r  P P sin O r = perpendicular distance of line of motion of particle from point O. P = perpendicular component of momentum. SI unit of angular momentum is kgm2/sec. 7.2 Angular Momentum of a rigid body rotating about a fixed axis Suppose a particle P of mass m is going in a circle of radius r and at some instant the speed of the particle is v. For finding the angular momentum of the particle about the axis of rotation, the origin may be chosen  anywhere on the axis. We choose it at the centre of the circle. In this case  r and P are perpendicular     to each other and r  P is along the axis. Thus, component of r  P along the axis is mvr itself. The angular momentum of the whole rigid body about AB is the sum of components of all particles, i.e., L= m r v i i i i Here, vi = ri   L= m r i i 2  i or L =  i m r i i 2 i or L = I Here, I is the moment of inertia of the rigid body about AB.  Note : Angular momentum about axis is the component of I along the axis. In most of the cases angular momentum about axis is I. 8. CONSERVATION OF ANGULAR MOMENTUM : The time rate of change of angular momentum of a particle about some referenence point in an inertial frame of reference is equal to the net torques acting on it. or   dL net  dt ....(i)  Now, suppose that  net    0 , then d L  0 , so that L = constant. dt "When the resultant external torque acting on a system is zero, the total vector angular momentum of the system remains constant. This is the principle of the conservation of angular momentum. For a rigid body rotating about an axis (the z-axis, say) that is fixed in an inertial reference frame, we have Lz = I ROTATIONAL MOTION & CENTER OF MASS Page # 9 It is possible for the moment of inertia I of a rotating body to change by rearrangement of its parts. If no net external torque acts, then Lz must remains constant and if I does change, there must be a compensating change in . The principle of conservation of angular momentum in this case is expressed. I = constant. 9. ANGULAR IMPULSE The angular impulse of a torque in a given time interval is defined as  t2  t1  dt Here,   is the resultant torque acting on the body. Further, since   dL       dt  d L dt t2     dt = angular impulse = L 2 – L 1 or  t1 Thus, the angular impulse of the resultant torque is equal to the change in angular momentum. Let us take few examples based on the angular impulse. 9.2 Kinetic Energy of a Rolling Body If a body of mass M is rolling on a plane such that velocity of its centre of mass is V and its angular speed is , its kinetic energy is given by 1 1 Mv 2  I  2 2 2 I is moment of inertia of body about axis passing through centre of mass. In case of rolling without slipping. KE = KE = 1 1 M  2 R2 + I  2 [ v = R] 2 2   1 1 MR2  I  2 = Ic  2 2 2 Ic is moment of inertia of the body about the axis passing through point of contact.  10.CENTRE OF MASS : Every physical system has associated with it a certain point whose motion characterises the motion of the whole system. When the system moves under some external forces, then this point moves as if the entire mass of the system is concentrated at this point and also the external force is applied at this point for translational motion. This point is called the centre of mass of the system. 10.1 Centre of Mass of a System of ‘N’ Discrete Particles : Consider a system of N point masses m1, m2, m3, .................... mn whose position vectors from origin O     are given by r1, r2 , r3 .............. rn respectively. Then the position vector of the centre of mass C of the system is given by. Y i 1 n m i  1 rcm  M n  m r i i i 1  r1   . . . .. . .. . . . . . . ..m...1 .. ..C. ..m..2 .. .. . . .. . . . . .. .. . . . .. .. . . . r2.. . .. . . . . . .. . .m. .. . .. . .. .. .n . rn m      m r  m 2 r2 ...........mn rn ; rcm  rcm  1 1 m1  m 2 .........mn  mi ri rc n i 1  where, mi ri is called the moment of mass of particle with respect to origin. O  M     n  m  i is the total mass of the system. i 1   r COM  xCOM i  y COM j  zCOMk and r i  xi i  y i j  zik So, the cartesian co-ordinates of the COM will be Further, x ROTATIONAL MOTION & CENTER OF MASS Page # 10 n m x i i xCOM m1x1  m 2 x2 ......mn xn  = m1  m 2 ....... mn i 1 n m i i 1 n m x i i or xCOM = i 1 M n Similarly, yCOM =  n mi y i i 1 M m z i i z COM  and i 1 M Note : n • If the origin is taken at the centre of mass then  m r i i = 0. hence, the COM is the point about which the i1 • sum of “mass moments” of the system is zero.     If we change the origin then r1, r2 , r3 ....... changes. So rcm also changes but exact location of center of mass does not change. 10.2 Position of COM of two particles : Consider two particles of masses m1 and m2 separated by a distance l as shown in figure. m2 m1 C l Let us assume that m1 is placed at origin and m2 is placed at position (l, 0) and the distance of centre of mass from m1 & m2 is r1 & r2 respectively. So xCOM = m1x1  m 2 x2 m1  m 2 0  m2l r1 = m  m 1 2 m2l = m m 1 2 r1 ...(1) m1l m2l r2 = l – = m m ...(2) 1 2 m1  m 2 From the above discussion, we see that C r2 (0,0) m1 m2 l l if m1 = m2, i.e., COM lies midway between the two particles of equal masses. 2 Similarly, r1 > r2 if m1 < m2 and r1 < r2 if m2 < m1 i.e., COM is nearer to the particle having larger mass. From equation (1) & (2) m1r1 = m2r2 ...(3) Centre of mass of two particle system lie on the line joining the centre of mass of two particle system. For continuous mass distribution the centre of mass can be located by replacing summation sign with an integral sign. Proper limits for the integral are chosen according to the situation r1 = r2 = xcm   xdm , y  dm cm   y dm , z  dm cm   zdm  dm ...(i)  dm = M (mass of the body) here x,y,z in the numerator of the eq. (i) is the coordinate of the centre of mass of the dm mass. 1   rdm rcm = M Note : If an object has symmetric mass distribution about x axis then y coordinate of COM is zero and vice-versa  • ROTATIONAL MOTION & CENTER OF MASS Page # 11 (a) Centre of Mass of a Uniform Rod Suppose a rod of mass M and length L is lying along the x-axis with its one end at x = 0 and the other at M x = L. Mass per unit length of the rod  = L Hence, dm, (the mass of the element dx situated at x = x is) =  dx The coordinates of the element dx are (x, 0, 0). Therefore, x-coordinate of COM of the rod will be dx x=x x=0 L xCOM = x=L L  xdm  dm 0 = L 0  ( x)dx  1 =  L  dx  L 0 L 0 x dx  L 2 0 The y-coordinate of COM is yCOM =  y dm  dm = 0 Similarly, zCOM = 0 L  i.e., the coordinates of COM of the rod are  , 0, 0 , i.e, it lies at the centre of the rod. 2  \(c) Centre of mass of Semicircular Disc : Figure shows the half disc of mass M and radius R. Here, we are only required to find the y-coordinate of the centre of mass of this disc as centre of mass will be located on its half vertical diameter. Here to find ycm, we consider a small elemental ring of mass dm of radius r on the disc (disc can be considered to be made up such thin rings of increasing radii) which will be integrated from 0 to R. Here dm is given as dm   rdr where  is the mass density of the semi circular disc. M 2M  = 2 R / 2 R 2 Y Y ycm dr ycm r R X R X 2r , (as in previous section, we have derived that the  2R centre of mass of a semi circular ring is concentrated at )  R dm . y Now the y-coordinate of the element is taken as  y cm  0 R  dm 0 Here y is the position COM of dm mass. R 2r  dm  0 Here ycm is given as y cm  R  rdr 0 R 4  R  0 2 r 2 dr  y cm  4R 3 ROTATIONAL MOTION & CENTER OF MASS Page # 12 (d) Centre of mass of a Hollow Hemisphere : A hollow hemisphere of mass M and radius R. Now we consider an elemental circular strip of angular width d at an angular distance  from the base of the hemisphere. This strip will have an area. dS  2R cos  Rd Y Y Rcos Rd ycm X R X R Its mass dm is given as dm  2RcosRd Here  is the mass density of a hollow hemisphere M = 2 R 2 Here y-coordinate of this strip of mass dm can be taken as R sin. Now we can obtain the centre of mass of the system as.  2  2  dmR sin  y cm  0 /2  dm 0   2R 2   2 cos d R sin 0 /2  2R 2 cos d R   sin  cos  d  y cm  0 R 2 0 11.MOTION OF CENTRE OF MASS AND CONSERVATION OF MOMENTUM: The position of centre of mass is given by     m r  m 2 r 2  m 3 r 3  ........ r COM  1 1 m1  m 2  m 3  ....... ....(1)    Here m1, m2, m3 ..... are the mass in the system and r1, r2 , r3 ......... is the corresponding position vector of m1, m2, m3 respectively 11.1Velocity of C.O.M of system : To find the velocity of centre of mass we differentiate equation (1) with respect to time    dr1 dr2 dr3  m1  m2  m3 ....... drcom dt dt dt  dt m1  m 2  m 3 .........   Vcom  m1    dr d r1 dr  m 2 2  m3 3  ....... dt dt dt m1  m 2  m3  .........     m v  m2 v 2  m3 v 3  ...... Vcom  1 1 ...(2) m1  m2  m3  ... 11.2 Acceleration of centre of mass of the system : To find the acceleration of C.O.M we differentiate equation (2)    dv dv dv  m1 1  m 2 2  m 3 3 ...... dVcom dt dt dt   dt m1  m 2  m 3 ........     m a  m 2 a2  m 3 a 3 ...... a com  1 1 m1  m 2  m 3 ........ ...(3) ROTATIONAL MOTION & CENTER OF MASS Page # 13     Now (m1 + m2 + m3) a com = m1a1  m 2 a 2  m 3 a 3 ......     Fnet( system)  F1net  F2 net  F3 net ........ The internal forces which the particles exert on one another play absolutely no role in the motion of the centre of mass. 12. IMPULSE :  Impulse of a force F acting on a body for the time interval t = t1 to t = t2 is defined as t2     I = F dt I = F dt   t1   dv  m dt  mdv dt      I = m(v 2  v1 )   P  change in momentum due to force F    Also IRe = *  t2  t1  FRe s dt   P (impulse - momentum theorem) Note : Impulse applied to an object in a given time interval can also be calculated from the area under force time (F-t) graph in the same time interval. 12.1 Instantaneous Impulse : There are many cases when a force acts for such a short time that the effect is instantaneous, e.g., a bat striking a ball. In such cases, although the magnitude of the force and the time for which it acts may each be unknown but the value of their product (i.e., impulse) can be known by measuring the initial and final momentum. Thus, we can write.      I = F dt  P  Pf  Pi  Important Points : (1) It is a vector quantity. (3) SI unit = kg m/s (2) Dimensions = [MLT–1] (4) Direction is along change in momentum.     (5) Magnitude is equal to area under the F-t. graph. (6) I = F dt  Fav dt  Fav t   (7) It is not a property of a particle, but it is a measure of the degree to which an external force changes the momentum of the particle. 13. COEFFICIENT OF RESTITUTION (e) The coefficient of restitution is defined as the ratio of the impulses of reformation and deformation of either body. e= Im pulse of reformation = Im pulse of deformation  F dt  F dt r e d Velocity of separation of point of contact Velocity of approach of point of contact Example for calculation of e : Two smooth balls A and B approaching each other such that their centres are moving along line CD in absence of external impulsive force. The velocities of A and B just before collision be u1 and u2 respectively. The velocities of A and B just after collision be v1 and v2 respectively. Just Before collision u2 u1 Just After collision v2 v1 u1>u2 C A D B v 2 – v1 e = u –u 1 2 C A D B v2 > v1 ROTATIONAL MOTION & CENTER OF MASS Page # 14 Note : Coefficient of restitution is a factor between two colliding bodies which is depends on the material of the body but independent of shape. We can say e is a factor which relates deformation and reformation of the body. 0  e1 13.1 Line of Motion The line passing through the centre of the body along the direction of resultant velocity. 13.2 Line of Impact The line passing through the common normal to the surfaces in contact during impact is called line of impact. The force during collision acts along this line on both the bodies. Direction of Line of impact can be determined by : (a) Geometry of colliding objects like spheres, discs, wedge etc. (b) Direction of change of momentum. If one particle is stationary before the collision then the line of impact will be along its motion after collision. Examples of line of impact Line ofthat impact and (i) Two balls A and B are approaching each other such their centres are moving along line CD. line of motion C D A B (ii) Two balls A and B are approaching each other such that their centre are moving along dotted lines as shown in figure. B Line of motion of ball A Line of motion of ball B D A Line of impact (iii) Ball is falling on a stationary wedge. Line of motion of ball Line of impact Note : In previous discussed examples line of motion is same as line of impact. But in problems in which line of impact and line of motion is different then e will be e= velocity of seperation along line of impact velocity of approach along line of impact 14. COLLISION OR IMPACT • • • Collision is an event in which an impulsive force acts between two or more bodies for a short time, which results in change of their velocities. Note : In a collision, particles may or may not come in physical contact. The duration of collision, t is negligible as compared to the usual time intervals of observation of motion. In a collision the effect of external non impulsive forces such as gravity are not taken into account as due to small duration of collision (t) average impulsive force responsible for collision is much larger than external forces acting on the system. ROTATIONAL MOTION & CENTER OF MASS Page # 15 The collision is in fact a redistribution of total momentum of the particle : Thus law of conservation of linear momentum is indepensible in dealing with the phenomenon of collision between particles. Consider a situation shown in figure. Two balls of masses m 1 and m2 are moving with velocities v1 and v2 (< v1) along the same straight line in a smooth horizontal surface. Now let us see what happens during the collision between two particles. v1 ' v2 ' v1 v2 N N N m1 m2 N figure (a) figure(b) figure(c) figure (a) : Balls of mass m1 is behind m2. Since v1 > v2, the balls will collide after some time. figure (b) : During collision both the balls are a little bit deformed. Due to deformation two equal and opposite normal forces act on both the balls. These forces decreases the velocity of m1 and increase the velocity of m2 figure (c): Now velocity of ball m1 is decrease from v1 to v1 and velocity of ball m2 is increase from v2 to v2. But still v1 > v2 so both the ball are continuously deformed. figure(d) : Contact surface of both the balls are deformed till the velocity of both the balls become equal. So at maximum deformation velocities of both the blocks are equal v1 '' v 2 '' figure(d) at maximum deformation v1 ''  v 2 '' figure(e) : Normal force is still in the direction shown in figure i.e. velocity of m1 is further decreased and that of m2 increased. Now both the balls starts to regain their original shape and size. v1 '' v 2 '' N N v 2 ''  v1 '' figure(e) figure (f) : These two forces redistributes their linear momentum in such a manner that both the blocks are separated from one another, Velocity of ball m2 becomes more than the velocity of block m1 i.e., v2 > v1 v1 v2 m1 m2 v2>v1 figure(f) The collision is said to be elastic if both the blocks regain their original form, The collision is said to be inelastic. If the deformation is permanent, and the blocks move together with same velocity after the collision, the collision is said to be perfectly inelastic. 14.1 Classification of collisions (a) On the basis of line of impact (i) Head-on collision : If the velocities of the colliding particles are along the same line before and after the collision. (ii) Oblique collision : If the velocities of the colliding particles are along different lines before and after the collision. ROTATIONAL MOTION & CENTER OF MASS Page # 16 (b) On the basis of energy : (i) Elastic collision : (a) In an elastic collision, the colliding particles regain their shape and size completely after collision. i.e., no fraction of mechanical energy remains stored as deformation potential energy in the bodies. (b) Thus, kinetic energy of system after collision is equal to kinetic energy of system before collision. (c) e = 1 (d) Due to Fnet on the system is zero linear momentum remains conserved. (ii) Inelastic collision : (a) In an inelastic collision, the colliding particles do not regain their shape and size completely after collision. (b) Some fraction of mechanical energy is retained by the colliding particles in the form of deformation potential energy. Thus, the kinetic energy of the particles no longer remains conserved. (c) However, in the absence of external forces, law of conservation of linear momentum still holds good. (d) (Energy loss)Perfectly Inelastic > (Energy loss)Partial Inelastic (e) 0 < e < 1 (iii) Perfectly Inelastic collision : (i) In this the colliding bodies do not return to their original shape and size after collision i.e. both the particles stick together after collision and moving with same velocity (ii) But due to Fnet of the system is zero linear momentum remains conserved. (iii) Total energy is conserved. (iv) Initial kinetic energy > Final K.E. Energy (v) Loss in kinetic energy goes to the deformation potential energy (vi) e = 0 14.2 Value of Velocities after collision : Let us now find the velocities of two particles after collision if they collide directly and the coefficient of restitution between them is given as e. m2 m2 m1 m1 u1 u2 v1 v2 (a) Before Collision (b) After Collision u1 > u2 v2 > v1 v 2 – v1 e= u –u 1 2  (u1 – u2)e = (v2 – v1) By momentum conservation m1u1 + m2u2 = m1v1 + m2v2 v2 = v1 + e(u1 – u2) from above equation m 1u1  m 2u 2  m 2 e(u 2 – u1 ) v1 = m1  m 2 m 1u1  m 2 u 2  m 1e(u 1 – u 2 ) v2 = m1  m 2 Special cases : 1. If m1 >> m2 and u2 = 0 and u1 = u ...(i) ...(ii) ...(iii) ...(iii) ...(iv) u and e = 1 m1 = m2 from eq. (iii) & (iv) m1 m1u – m 2u u(m1 – m 2 ) v1 = m  m = m m 1 2 1 2 v1 ~– u , 2. If m1 = m2 = m and e = 1 then from eq. (iii) & (iv) v2 = m1u  m 2u 2m1u = m1  m 2 m1  m 2 ; v2 = 2u m2 ROTATIONAL MOTION & CENTER OF MASS Page # 17 m(u1  u 2 )  m(u2 – u1 ) 2m v 1 = u2 In this way v2 = u1 i.e when two particles of equal mass collide elastically and the collision is head on, they exchange their velocities. 14.3 Collision in two dimension (oblique) : 1. A pair of equal and opposite impulses act along common normal direction. Hence, linear momentum of individual particles change along common normal direction. If mass of the colliding particles remain constant during collision, then we can say that linear velocity of the individual particles change during collision in this direction. 2. No component of impulse act along common tangent direction. Hence, linear momentum or linear velocity of individual particles (if mass is constant) remain unchanged along this direction. 3. Net impulse on both the particles is zero during collision. Hence, net momentum of both the particles remain conserved before and after collision in any direction. 4. Definition of coefficient of restitution can be applied along common normal direction, i.e., along common normal direction we can apply Relative speed of separation = e (relative speed of approach) 14.4 Rocket Propulsion Let m0 be the mass of the rocket at time t = 0. m its mass at any time t and v its velocity at that moment. Initially let us suppose that the velocity of the rocket is u. v1 = u u At t=0 v=u m = m0 At t=t m=m v=v Exhaust velocity = vr   dm   be the mass of the gas ejected per unit time and vr the exhaust velocity of the gases. Further, let  dt    dm   and vr are kept constant throughout the journey of the rocket. Now, let us write few Usually  dt  equations which can be used in the problems of rocket propulsion. At time t = t 1. Thrust force on the rocket  dm  Ft  v r     dt  2. Weight of the rocket W = mg 3. Net force on the rocket Fnet = Ft – W or (upwards) (downwards) (upwards)  – dm  Fnet  v r   – mg  dt  4. Net acceleration of the rocket a  or dv v r   dm     g dt m  dt  or  dv  v  Thus,  m0   v = u – gt + vr In  m v u r m m0 F m t dm  g dt 0 m  ...(i) or   dm  dv  v r    g dt  m  or m  v – u = vr In  0   gt  m