Centre of Mass & Collisions System of Practicles

Study of kinematics enables us to explore nature of translation motion without any consideration to forces and energy responsible for the motion. Study of kinetics enables us to explore effects of forces and energy on motion. It includes Newton’s laws of motion, methods of work and energy and methods of impulse and momentum. The r r methods of work and energy and methods of impulse and momentum are developed using equation F  m a together with the methods of kinematics. The advantage of these methods lie in the fact that they make determination of acceleration unnecessary. Methods of work and energy directly relate force, mass, velocity and displacement and enable us to explore motion between two points of space i.e. in a space interval whereas methods of impulse and momentum enable us to explore motion in a time interval. Moreover methods of impulse and momentum provides only way to analyze impulsive motion. The work energy theorem and impulse momentum principle are developed from Newton’s second law, and we have seen how to apply them to analyze motion of single particle i.e. translation motion of rigid body. Now we will further inquire into possibilities of applying these principles to a system of large number of particles or rigid bodies in translation motion. SYSTEM OF PARTICLES By the term system of particles, we mean a well defined collection of several or large number of particles, which may or may not interact or be connected to each other. As a schematic representation, consider a system of n particles of masses m1, m2,...mi...mj.... and mn respectively. They may be actual particles of rigid bodies in translation motion. Some of them may interact with each other and some of them may not. The particles, which interact with each other, apply forces on each other. The forces r r of interaction fij and f ji between a pair of ith and jth particles are f f shown in the figure. Similar to these other particles may also interact with each other. These forces of mutual interaction between the particles of the system are internal forces of the system. These internal forces always exist in pairs of forces of equal magnitudes and opposite directions. It is not necessary that all the particles interact with each other; some of them, which do not interact with each other, do not apply mutual forces on each other. Other than internal forces, external forces may also act on all or some of the particles. Here by the term external force we mean a force that is applied on any one of the particle included in the system by some other body out-side the system. In practice we usually deal with extended bodies, which may be deformable or rigid. An extended body is also a system of infinitely large number of particles having infinitely small separations between them. When a body undergoes deformation, separations between its particles and their relative locations change. A rigid body is an extended body in which separations and relative locations of all of its particles remain unchanged under all circumstances. System of Particles and Mass Center Until now we have deal with translation motion of rigid bodies, where a rigid body can be treated as a particle. When a rigid body undergoes rotation, all of its particles do not move in identical fashion, still we must treat it a system of particles in which all the particles are rigidly connected to each other. On the other hand we may have CENTRE OF MASS & COLLISIONS particles or bodies not connected rigidly to each other but may be interacting with each other through internal forces. Despite the complex motion of which a system of particles is capable, there is a single point, known as center of mass or mass center (CM), whose translation motion is characteristic of the system. The existence of this special point can be demonstrated in the following examples dealing with a rigid body. Consider two disks A and B of unequal masses connected by a very light rigid rod. Place it on a very smooth table. Now pull it horizontally applying a force at different points. You will find a point nearer to the heavier disk, on which if the force is applied the whole assembly undergoes translation motion. Furthermore you cannot find any other point having this property. This point is the mass center of this system. We can assume that all the mass were concentrated at this point. In every rigid body we can find such a point. If you apply the force on any other point, the system moves forward and rotates but the mass center always translates in the direction of the force. A A F C C B B Force applied on the mass center F Force not applied on the mass center A F C B Force not applied on the mass center In another experiment, if two forces of equal magnitudes are applied on the disks in opposite directions, the system will rotate, but the mass center C remains stationary as shown in the following figure. F A C B F Body rotates but the mass center remains stationary under action of equal and opposite forces. If the above experiment is repeated with both disks A and B of identical masses, the mass center will be the mid point. And if the experiment is repeated with a uniform rod, the mass center again is the mid point. F A C 2F B Body rotates and the mass center translates under action of unbalanced forces applied at different points. As another example let us throw a uniform rod in air holding it from one of its ends so that it rotates also. Snapshots taken after regular intervals of time are shown in the figure. The rod rotates through 360°. As the rod moves all of its particles move in a complex manner except the mass center C, which follows a parabolic trajectory as if it were a particle of mass equal to that of the rod and force of gravity were acting on it. A A A B B B A A B B B B A A B A A B A C B A B A B A B Thus mass center of a rigid body or system of particles is a point, whose translation motion under action of unbalanced forces is same as that of a particle of mass equal to that of the body or system under action the same unbalanced forces. And if different forces having a net resultant are applied at different particles, the system rotates but the mass center translates as if it were a particle of the mass same as that of the system and the net resultant were applied on it. Concept of mass center provides us a way to look into motion of the system as a whole as superposition of translation of the mass center and motion of all the particles relative to the mass center. In case of rigid bodies all of its particles relative to the mass center can move only on circular paths because they cannot changes their separations. The concept of mass center is used to represent gross translation of the system. Therefore total linear momentum of the whole system must be equal to the linear momentum of the system due to translation of its mass center. Center of Mass of System of Discrete Particles A system of several particles or several bodies having finite separations between them is known as system of discrete particles. Let at an instant particles of such a system m1, m2, ….mi, ……and mn r r r r r are moving with velocities v1 , v 1 , v2 , …… v i ,……and v n at r r r r locations r1 , r2 , …… ri ,……and rn respectively. For the sake of simplicity only ith particle and the mass center C are shown in the r r figure. The mass center C located at rc is moving with velocity v c at y vi mi vc rc z C O x this instant. As the mass center represents gross translation motion of the whole system, the total linear i.e. sum of linear momenta of all the particles must be equal to linear momentum of the whole mass due to translation of the mass center. r r r r r m 1 v 1  m 2 v 2  ...............  m i v i  ....................  m n v n  Mv c We can write the following equation in terms of masses and position vectors as an analogue to the above equation. This equation on differentiating with respect to time yields the above equation therefore can be thought as solution of the above equation. r r r r r m 1 r1  m 2 r2  ...............  m i ri  ....................  m n rn  Mrc If M  m i denotes total mass of the system, the above two equations can be written in short as r r m iv i  Mv c r r m i ri  Mrc ........(1) ........(2) CENTRE OF MASS & COLLISIONS The above equation suggests location of mass center of a system of discreet particles. r r r r r m r  m 2 r2  ...............  m i ri  ....................  m n rn m i ri r rc  1 1  M M ........(3) r Cartesian coordinate (xc, yc, zc) of the mass center are components of the position vector rc of the mass center.. xc  m i x i m i y i m i z i ; yc  ; zc  M M M ........(4) 1. Center of Mass of Two Particle System Ex. (a) Find expression of position vector of mass center of a system of two particles of masses m1 and m2 located at r r position vectors r1 and r2 . (b) Express Cartesian coordinates of mass center, if particle m1 at point (x1, y1) and particle m2 at point (x2, y2). (c) If you assume origin of your coordinate system at the mass center, what you conclude regarding location of the mass center relative to particles. (d) Now find location of mass center of a system of two particles masses m1 and m2 separated by distance r. Sol. r r (a) Consider two particles of masses m1 and m2 located at position vectors r1 and r2 . Let their mass center C at r position vector rc . From eq. , we have r m i ri r rc   M r r m 1 r1  m 2 r2 r rc  m1  m2 (b) From result obtained in part (a), we have xc  m1 x1  m 2 x 2 m1 y1  m 2 y 2 and y c  m  m m1  m 2 1 2 r (c) If we assume origin at the mass center vector rc vanishes and we have r r r m 1 r1  m 2 r2  0 Since either of the masses m1 and m2 cannot be negative, to satisfy the r r above equation, vectors r1 and r2 must have opposite signs. It is geometrically possible only when mass center C lies between the two particles on the line joining them as shown in the figure. y m1 r1 C x r2 m2 r r If we substitute magnitudes r1 and r2 of vectors r1 and r2 in the above equation, we have m1r1 = m2r2, which suggest r1 m 2  r2 m 1 Now we conclude that mass center of two particle system lies between the two particles on the line joining them and divide the distance between them in inverse ratio of masses of the particles. (d) Consider two particles masses m1 and m2 at distance r from each other. There mass center C must lie in between them on the line joining them. Let distances of these particles from the mass center are r1 and r2. Since mass center of two particle system lies between the two particles r r1 r2 C m1 m2 on the line joining them and divide the distance between them in inverse ratio of masses of the particles, we can write r1  m2r m1r r  m 1  m 2 and 2 m 1  m 2 2. Mass centre of several particles Ex. Find position vectors of mass center of a system of three particle of masses 1 kg, 2 kg and 3 kg located at r r r position vectors r1  4 ˆi  2 ˆj  3 kˆ m, r2  iˆ  4 ˆj  2 kˆ m and r3  2 ˆi  2 ˆj  kˆ m respectively..  Sol.      From eq. , we have r m i ri r r 1 4 ˆi  2 ˆj  3 kˆ  2 ˆi  4 ˆj  2 kˆ  3 2 ˆi  2 ˆj  kˆ 2 rc    rc   2 iˆ  2 ˆj  kˆ M 12 3 3       Center of Mass of an Extended Body or Continuous Distribution of Mass An extended body is collection of infinitely large number of particles so closely located that we neglect separation between them and assume the body as a continuous distribution of mass. A rigid body is an extended body in which relative locations of all the particles remain unchanged under all circumstances. Therefore a rigid body does not get deformed under any circumstances. Let an extended body is shown as a continuous distribution of mass y by the shaded object in the figure. Consider an infinitely small portion of mass dm of this body. It is called a mass element and is shown at position given by position vector rr . Total mass M of the body is dm r M   dm . The mass center C is assumed at position given by r position vector rc . Position vector of centre of mass of such a body is given by the following equation. r rdm r  rc  M rc z O C x ........(5) r Cartesian coordinate (xC, yC, zC) of the mass center are components of the position vector rc of the mass center.. xc   xdm M ; yc   ydm M ; zc   zdm M ........(6) 3. Mass centre of uniform symmetrical bodies. EX. Show that mass center of uniform and symmetric mass distributions lies on axis of symmetry. Sol. For simplicity first consider a system of two identical particles and then extend the idea obtained to a straight uniform rod, uniform symmetric plates and uniform symmetric solid objects. CENTRE OF MASS & COLLISIONS 4. r Mass Center of a system of two identical particles r/2 Mass center of a system of two identical particles lies at the C m midpoint between them on the lie joining them. 5. r/2 Mass Center of a system of a straight uniform rod A Consider two identical particles A and B at equal distances dm m B C dm from the center C of the rod. Mass center of system these two particles is at C. The whole rod can be assumed to be made of large number of such systems each having its mass center at the mid point C of the rod. Therefore mass center of the whole rod must be at its mid point. 6. Mass Center of a system of a uniform symmetric curved rod Consider two identical particles A and B located at equal distances from the line of symmetry. Mass center of system these two particles is at C. The whole rod can be assumed to be made of large number of such systems each having its mass center at the mid point C of the joining them. Therefore mass center of the whole rod must be on the axis of symmetry. 7. B A C Line of symmetry Line of symmetry Mass Center of a uniform plate (lamina) Consider a symmetric uniform plate. It can be assumed composed of A several thin uniform parallel rods like rod AB shown in the figure. All of these rods have mass center on the line of symmetry, therefore the whole lamina has its mass center on the line of symmetry. 8. Mass Center of a uniform symmetric solid object Line of symmetry A uniform symmetric solid object occupies a volume that is made by rotating a symmetric area about its line of symmetry though 180º. Consider a uniform symmetric solid object shown in the figure. It can be assumed composed of several thin uniform parallel disks shown in the figure. All of these disks have mass center on the line of symmetry, therefore the whole solid object has its mass center on the line of symmetry. 9. B C C Mass Center of uniform bodies Following the similar reasoning, it can be shown that mass center of uniform bodies lies on their geometric centers. 10. Mass Center of a system of a segment of a uniform circular rod (arc) Ex. Find location of mass center of a thin uniform rod bent into shape of an arc. Sol. Consider a thin rod of uniform line mass density  (mass per unit length) y r r and radius r subtending angle 2 on its center O. The angle bisector OP is the line of symmetry, and mass center lies on it. Therefore if we assume the angle bisector as one of the coordinate axes say x-axis, y-coordinate of mass center becomes zero. 2 O A d  P rcos  x B Let two very small segments A and B located symmetric to the line of symmetry (x-axis). Mass center of these two segments is on P at a distance x  r cos  from center O. Total mass of these two elements is dm = 2rdd. Now using eq. , we have xc  xdm  M   xc  r cos  2 rd       r  r sin   Mass center of a thin uniform arc shaped rod of radius r subtending angle 2 at the center lies on its angle bisector at distance OC from the center. OC  r sin   Ex. Find coordinates of mass center of a quarter ring of radius r placed in the first quadrant of a Cartesian coordinate system, with centre at origin. Sol. Making use of the result obtained in the previous example, distance OC of the mass center form the center is OC  y r sin   / 4  2 2 r  / 4  yc C /4  2r 2r  Coordinates of the mass center (xc, yc) are  ,    O xc x Ex. Find coordinates of mass center of a semicircular ring of radius r placed symmetric to the y-axis of a Cartesian coordinate system. Sol. The y-axis is the line of symmetry, therefore mass center of the ring y C lies on it making x-coordinate zero. Distance OC of mass center from center is given by the result obtained in example 4. Making use of this result, we have OC  r sin    yc  yc /2 O x r sin   / 2  2 r  / 2  11. Mass Center of a sector of a uniform circular plate Ex. Find location of mass center of a sector of a thin uniform plate. Sol. Consider a sector of a thin uniform plate of surface mass density mass density (mass per unit area) and radius r subtending angle 2 on its center. y dr A R r 2 x O B CENTRE OF MASS & COLLISIONS Let a thin arc of radius r and width dr be an infinitely small part of the sector. Mass dm of the arc AB equals to product of mass per unit area and area of the arc. dm    2 r dr   2  r dr Due to symmetry mass center of this arc must be on the angle bisector i.e. on x-axis at distance x  r sin   . Using above two information in eq. , we obtain the mass center of the sector. xc   xdm M   r sin       r dr   xc     Area of thre sector   R 0  R 0  r sin      2  r dr     r 2  2 r sin  3 Ex. Find coordinates of mass center of a quarter sector of a uniform disk of radius r placed in the first quadrant of a Cartesian coordinate system with centre at origin. Sol. Making use of the result obtained in the previous example, distance OC of the mass center form the center is y 2 r sin   / 4  4 2 r OC   3 / 4 3 yc  4r 4r  , Coordinates of the mass center (xc, yc) are   3 3  C /4 O xc x Ex. Find coordinates of mass center of a uniform semicircular plate of radius r placed symmetric to the y-axis of a Cartesian coordinate system, with centre at origin. Sol. The y-axis is the line of symmetry, therefore mass center of the plate y C lies on it making x-coordinate zero. yc Distance OC of mass center from center is given by the result obtained in above example. Making use of this result, we have OC  2 r sin   3 yc  /2 O x 2 r sin   / 2  4 r  3 / 2 3 Ex. Find coordinates of mass center of a nonuniform rod of length L whose linear mass density varies as =a+bx, y where x is the distance from the lighter end. Sol. Assume the rod lies along the x-axis with its lighter end on the dm = dx  O xc origin to make mass distribution equation consistent with coordinate system. Making use of eq. , we have x c  xdm  M  x c L L 0 0  x dx   x ax  b  dx   dx  ax  b  dx L 0 L 0  x dx 2 aL  3 b  L 3  aL  2b  x= L 12. Mass Center of composite bodies Ex. A composite body is made of joining two or more bodies. Find mass center of the following composite body made by joining a uniform disk of radius r and a uniform square plate of the same mass per unit area. Sol. To find mass center the component bodies are assumed particle of masses equal y to corresponding bodies located on their respective mass centers. Then we use equation to find coordinates of the mass center of the composite body. O x To find mass center of the composite body, we first have to calculate masses of the bodies, because their mass distribution is given. If we denote surface mass density (mass per unit area) by , masses of the bodies are Mass of the disk m d  Mass per unit area  Area     r 2    r 2 Mass of the square plate m p  Mass per unit area  Area    r 2    r 2 Location of mass center of the disk x d  Center of the disk  r and y d  0 Location of mass center of the square plate x p  Center of the surface p late  3 r and y d  0 Using eq. , we obtain coordinates (xc, yc) of the composite body. xc  m d x d  m s x s r   3     1  md  ms yc  and md xd  ms xs 0 md  ms  r   3   , 0 Coordinates of the mass center are      1 13. Mass Center of truncated bodies Ex. A truncated body is made by removing a portion of a body. Find mass center of y the following truncated disk made by removing disk of radius equal to half of the original disk as shown in the figure. Radius of the original uniform disk is r. Sol. O x To find mass center of truncated bodies we can make use of superposition principle that is, if we add the removed portion in the same place we obtain the original body. The idea is illustrated in the following figure. y O y y x O x O x The removed portion is added to the truncated body keeping their location unchanged relative to the coordinate frame. Denoting masses of the truncated body, removed portion and original body by mtb, mrp and mob and location of their mass centers by xtb, xrp and xob, we can write m tb x tb  m rb x rp  m ob x ob From the above equation we obtain position co-ordinate xtb of the mass center of the truncated body. x tb  m ob x ob  m rb x rp m tb .........(1) CENTRE OF MASS & COLLISIONS Denoting mass per unit area by  , we can express the masses mtb, mrp and mob.   2 r 2   3  r 2 Mass of truncated body m tb     r     4  4  m rp  Mass of the removed portion  r 2 4 Mass of the original body m ob   r 2 Mass center of the truncated body x tb x rp  Mass center of the removed portion Mass center of the original body r 2 x ob  0 Substituting the above values in equation (1), we obtain the mass the center of the truncated body. x tb  m ob x ob  m rb x rp m tb   r 2   0    r r  4   2  r  2 6 3  r 4 2  r  Mass center of the truncated body is at point   , 0  6 Body Shape of Body Position of centre of mass Uniform Ring Centre of ring cm Uniform Disc Centre of disc cm Uniform Rod Centre of Rod cm Solid sphere/ hollow sphere Centre of sphere cm Triangular plane lamina Point of intersection of the medians of the triangle i.e. centroid cm Plane lamina in the form of a square or rectangle or parallelogram Point of intersection of diagonals cm Hollow/solid cylinder Middle point of the axis of cylinder cm y Half ring ycm cm R ycm= 2R ? ycm= Rsin  x y Segment of a ring ycm cm R x CENTRE OF MASS & COLLISIONS Body Shape of Body Position of centre of mass y Half disc ycm= cm ycm 4R 3 x y ycm = cm Sector of a disc (plate) ycm ? x y Hollow hemisphere cm ycm 2Rsin 3 ycm = R 2 R x y Solid hemisphere ycm cm ycm = 3R 8 ycm = h 3 ycm = h 4 R x y Hollow cone h ycm cm x y Solid cone h ycm cm x Center of Mass Frame of Reference or Centroidal Frame Center of mass frame of reference or centroidal frame is reference frame assume attached with the mass center of the system at its origin. It moves together with the mass center. It is a special frame and presents simple interpretations and solutions to several phenomena. Let us first discuss some of its fundamental properties. In centroidal frame center of mass is assumed at the origin, therefore position vector, velocity and acceleration of the mass center in centroidal frame all become zero. • Sum of mass moments in centroidal frame vanishes. Mass moment of a particle is product of mass of the particle and its position vector. r r r r r r r m i ri / c  0 or m 1 r1 / c  m 2 r2 / c  ...............  m i ri / c  ....................  m n rn / c  0 .........(7) • Total linear momentum of the system in centroidal frame vanishes.. r r r r r r r m iv i / c  0 or m 1 v 1 / c  m 2 v 2 / c  ...............  m i v i / c  ....................  m n v n / c  0 .........(8) Ex. Motion of Mass Center in One Dimension A jeep of mass 2400 kg is moving along a straight stretch of road at 80 km/h. It is followed by a car of mass 1600 kg moving at 60 km/h. (a) How fast is the center of mass of the two cars moving? Sol. (b) Find velocities of both the vehicles in centroidal frame. r r m jeep v jeep  m car v car r vc  (a) Velocity of the mass center m jeep  m car Assuming direction of motion in the positive x-direction, we have r r m jeep v jeep  m car v car r 2 40 0  8 0  1 60 0  60 r vc  vc   72 km/h  m jeep  m car 2 40 0  1 60 0 (b) Velocity of the jeep in centroidal frame v jeep / c  8 0  72  8 km/h in positive x-direction. Velocity of the car in centroidal frame v car / c  60  72  1 2 km/h 12 km/h negative x-direction direction. Ex. Motion of Mass Center in Vector Form r A 2.0 kg particle has a velocity of v 1   2.0 iˆ  3 .0 ˆj  m/s, and a 3.0 kg particle has a velocity r v 2  1.0 iˆ  6 .0 ˆj  m/s. (a) How fast is the center of mass of the particle system moving? (b) Find velocities of both the particles in centroidal frame. Sol. (a) Velocity of the mass center r r m v  m 2v 2 r vc  1 1 m1  m2  (b) r r m v  m 2v 2 r vc  1 1 m1  m2 2  2 .0 iˆ  3 .0 ˆj   3 1.0 ˆi  6 .0 ˆj  r vc   1.4 ˆi  2 .4 ˆj  m/s 2 3 Velocity of the first particle in centroidal frame r r r r v 1 / c  v 1  v c  v 1 / c   2.0 iˆ  3.0 ˆj   1 .4 iˆ  2.4 ˆj   0 .6  iˆ  ˆj  m/s Velocity of the second particle in centroidal frame r r r r v 2 / c  v 1  v c  v 2 / c  1.0 iˆ  6.0 ˆj   1.4 iˆ  2 .4 ˆj     0 .4 iˆ  3.6 ˆj  m/s APPLICATION OF NEWTON’S LAWS OF MOTION TO A SYSTEM OF PARTICLES In order to write equation of motion for a system of particles, we begin by applying Newton’s second law to an individual particle. Consider ith particle of mass mi. Internal force applied on it by jth r particle is shown by fij . Other particles of the system may also apply y y Fi mi mi mi ai f ij internal forces on it. One of them is shown in the figure by an unlabeled O x O x vector. In addition to these internal forces, external forces may also z z be applied on it by bodies out side the system. Resultant of all these r r external forces is shown by vector Fi . If under the action of these forces this particles has acceleration ai relative to an inertial frame Oxyz, its free body diagram and kinetic diagram can be represented by the following figure and CENTRE OF MASS & COLLISIONS Newton’s second law can be written by the following equation. r r r Fi   fij  m i ai In similar fashion, we can write Newton’s second law for all the particles of the system. These equations are r r r For 1st particle F1   f1 j  m 1 a1 r r r For 2nd particle F2   f2 j  m 2 a2 ..................... For ith particle ..................... For nth particle ..................... r r r Fi   fij  m i ai ..................... r r r Fn   fnj  m n an r r Every internal force fij on particle mi due to particle mj and f ji on the particle mj due to particle mi constituting Newton’s third law pair must be equal in magnitude and opposite in direction, therefore the sum all these internal forces for all the particles must be zero. Keeping this fact in mind and denoting the mass of the whole system by r M and acceleration of the mass center C by aC relative to the inertial frame, Newton’s second law representing translation motion of the system of particles particle can be represented by the following equation. r r r .........(9)  Fi    m i ai   MaC r r dp r .........(10)  Fi  MaC  dtc Ex. Newton’ Laws of Motion and System of Particles A ladder of mass 20 kg is hanging from ceiling as shown in figure. Three men A, B and C of masses 40 kg, 60 kg, and 50 kg are climbing the ladder. Man A is climbing with upward retardation 2 m/s2, B is climbing up with a constant speed of 0.5 m/s and C is climbing with upward acceleration of 1 m/s2. Find the A B C tension in the string supporting the ladder. Sol. External forces acting on the system are weights of the men, weight of the ladder and tension supporting the ladder. Denoting masses of men A, B, C and ladder by mA, mB, mC and mL, acceleration due to gravity by g , tension in the string by T and accelerations of the men A, B, C and ladder by aA, aB, aC and aL respectively, we can write the following equation according to equation . r r Fi    m i ai   T  m A g  m B g  m C g  m L g  m A a A  m B aA  m C aA  m L aA Substituting given values of masses m A  40 kg, m B  60 kg, m C  50 kg, m L  20 kg, given values of accelerations g  1 0 m/s2, a A  2 m/s2, aB  0 m/s2, aC  1 m/s2, and aL  0 m/s2, we obtain T  400  600  500  200  80  0  50  0 T = 1670 N Ex. Simple Atwood Machine as System of Particles The system shown in the figure is known as simple Atwood machine. Initially the masses are held at rest and then let free. Assuming mass m2 more than the mass m1, find acceleration of mass center and tension in the string supporting the pulley. Sol. We know that accelerations a1 and a2 are given by the following equations. m1 m2 m2  m1 m  m1 g  and a1  2 g  m2  m1 m2  m a2  Making use of eq. , we can find acceleration aC of the mass center. We denote upward direction positive and downward direction negative signs respectively. r r MaC    m i ai    m 1  m 2  ac  m 1 a1  m 2 a2 Substituting values of accelerations a1 and a2, we obtain aC  2 m 1 m 2   m 12  m 22   m 1  m 2 2 g To find tension T in the string supporting the pulley, we again use eq. (9) r r T  m 1 g  m 2 g   m 1  m 2  ac Fi  MaC  Substituting expression obtained for aC, we have T  Ex. 4 m1m2 g m1  m 2 Two blocks each of mass m, connected by an un-stretched spring are kept at rest on a frictionless horizontal surface. A constant force F is applied on one of the blocks pulling it away from the other as shown in figure. A B F (a) Find acceleration of the mass center. (b) Find the displacement of the centre of mass as function of time t. (c) If the extension of the spring is xo at an instant t, find the displacements of the two blocks relative to the ground at this instant. Sol. (a) Forces in vertical direction on the system are weights of the blocks and normal reaction from the ground. They balance themselves and have no net resultant. The only external force on the system is the applied force F in the horizontal direction towards the right. r r F   m  m  ac Fi  MaC  ac  F towards right 2m (b) The mass center moves with constant acceleration, therefore it displacement in time t is given by equation of constant acceleration motion. x  ut  12 at 2  x c  Ft 2 4m CENTRE OF MASS & COLLISIONS (c) Positions xA and xB of particles A and B forming a system and position xC mass center are obtained by following eq. r r Mrc  m i ri xA  xB 2 Substituting values wee obtain 2 m x c  m x A  m x B xc  Now using result obtained in part (b), we have xA  xB  Ft 2 2m Extension in the spring at this instant is x o  x B  x A From the above two equations, we have xA    1  Ft 2 1  Ft 2  x x   x0  0 B and    2 2m 2 2m   Application of Methods of Impulse and Momentum to a System of Particles In a phenomenon, when a system changes its configuration, some or all of its particles change their respective locations and momenta. Sum of linear momenta of all the particles equals to the linear momentum due to translation of mass center. Principle of impulse and moment suggests net impulse of all the external forces equals to change in momentum of mass center. r r r .........(11)   Fi dt  p cf  p ci Conservation of Linear momentum The above event suggests that total linear momentum of a system of particle remains conserved in a time interval in which impulse of external forces is zero. Total momentum of a system of particles cannot change under the action of internal forces and if net impulse of the external forces in a time interval is zero, the total momentum of the system in that time interval will remain conserved. r p initial r   p final or r r p ci  p cf .........(12) The above statement is known as the principle of conservation of momentum. Since force, impulse and momentum are vectors, component of momentum of a system in a particular direction is conserved, if net impulse of all external forces in that direction vanishes. During an event the net impulse of external forces in a direction is zero in the following cases. • When no external force acts in a particular direction on any of the particles or bodies. • When resultant of all the external forces acting in a particular direction on all the particles or bodies is zero. • In impulsive motion, where time interval is negligibly small, the direction in which no impulsive forces act. No external force: Stationary mass relative to an inertial frame remains at rest Ex. A man of mass m is standing at on end of a plank of mass M. The length of the plank is L and it rests on a frictionless horizontal ground. The man walks to the other end of the plank. Find displacement of the plank and man relative to the ground. Sol. Denoting x-coordinates of the man, mass center of plank and mass center of the man-plank system by xm, xp and xc, we can write the following equation. r r  m  M  xr c  m xr m  Mxr p  m i  rc  m i ri  Net force on the system relative to the ground is zero. Therefore mass center of the system which is at rest before r the man starts walking, remains at rest  x c  0  after while the man walks on the plank. r  x c  0  r r m x m  M  x p  0 .......(1) r The man walks displacement x m / p  Liˆ relative to the plank. Denoting displacements of the man and the r r plank relative to the ground by x m and x p , we can write  r r r x m / p  x m  x p   r r x m  x p  Liˆ .......(2) From the above equations (1) and (2), we have MLiˆ r x m   m M The man moves a distance ML towards left relative to the ground. m M m Liˆ r x p  m M The plank moves a distance Ex. mL towards right relative to the ground. m M No external force: Mass center moving relative to an inertial frame moves with constant velocity Two particles of masses 2 kg and 3 kg are moving under their mutual interaction in free space. At an instant they     were observed at points (2 m, 1 m, 4 m) and (2 m, 3 m, 6 m) with velocities 3 ˆi  2 ˆj  kˆ m/s and ˆi  ˆj  2kˆ m/ s respectively. If after 10 sec, the first particle passes the point (6 m, 8 m, 6 m), find coordinate of the point where the second particle passes at this instant? Sol. System of these two particles is in free, therefore no external forces act on them. There total linear momentum remains conserved and their mass center moves with constant velocity relative to an inertial frame. Velocity of the mass center r m i v i 2 3 ˆi  2 ˆj  kˆ  3 ˆi  ˆj  2 kˆ 3 ˆi  ˆj  4 kˆ vc    m/s m i 23 5     r Location rco of the mass center at the instant t = 0 s r m i ri r rc   m i     2 2 ˆi  ˆj  4 kˆ  3 2 ˆi  3 ˆj  6 kˆ r 2 ˆi  7 ˆj  2 6 kˆ rco   2 3 5 r New location rc of the mass center at the instant t = 10 s r 2 ˆi  7 ˆj  26 kˆ 3 ˆi  ˆj  4 kˆ 32 ˆi  17 ˆj  14 kˆ r r r  10  rc  rco  v c t  rc  5 5 5 New location (x, y, z) of the second particle. r r m iri 3 2 ˆi  17 ˆj  1 4 kˆ 2 6 ˆi  8 ˆj  6 kˆ  3 xiˆ  yjˆ  zkˆ rc    m i 5 2 3     Solving the above equation, we obtain the coordinates of the second particle (20/3, –11, –2/3) CENTRE OF MASS & COLLISIONS APPLICATION OF METHODS OF WORK AND ENERGY TO A SYSTEM OF PARTICLES In a system of particles, all the particles occupy different locations at every instant of time and may change their locations with time. At an instant of time set of locations of all the particles of a system is known as configuration of the system. We say something has happened with the system only when some or all of its particles change their locations. It means that in every event or phenomena the system changes its configuration. Methods of work and energy equips us to analyze what happens when a particle moves form one point of space to other. Now we will apply these methods to analyze a phenomenon in which a system of particle changes its configuration. Kinetic Energy of a System of Particle y Kinetic energy of a system of particles is defined as sum of kinetic energies of all mi the particles of the system. vi If at an instant particles of masses m1, m2, ….mi….mj……….. and mn are r r r r observed moving with velocities v 1 , v 2 , …. v i , …….. v n respectively relative z to a reference frame, the kinetic energy of the whole system relative to the reference frame is given by the following equation. K  1  m iv i2 2 x O .......(13) y vi If the system consists of continuous distribution of mass, instead of dm discrete particles, expression of kinetic energy becomes K  1 2 v dm 2 .......(14) z O x Kinetic Energy of a System of Particle using Centroidal Frame Centroidal frame of reference or center of mass frame is reference frame y v i/c v i attached with the mass center of the system. m r Let velocity of ith particle of mass mi is moving with velocity v i v O relative to frame Oxyz. Mass center C and hence the centroidal frame r Cxyz is moving with velocity v c . Therefore velocity of ith particle r relative to the centroidal frame is v i / c . z v i c c C x Kinetic energy of the whole system is given by the following equation. K  1 1 1 m i v i2   m iv c2   m i v i2/ c  2 2 2 .......(15) Here the first term on the right hand side is kinetic energy due to translation of the mass center and the second term is kinetic energy of the system relative to the centroidal frame. Kinetic Energy of a Two Particle System using Centroidal Frame A two particle system consists of only two particles. Let a two particle system consists of particles of masses m1 and m2 moving with r r velocities v 1 and v 2 relative to a frame Oxyz. Their mass center C lies on the line joining them and divides separation between them in reciprocal ratio of masses m1 and m2. The mass center and hence the r centroidal frame is moving with velocity v c . Kinetic energy of this two particle system relative to a frame Oxyz is given by the following equation. y v1 m1 O z vc m2 v2 C x 1 1  m  m 2  v c2  2 v rel2 .......(16) 2 1 The first term on the right hand side is kinetic energy due to translation of the mass center and the second term is kinetic energy of the system relative to the centroidal frame. K  Here symbol  is known as reduced mass of the two particle system and symbol vrel is magnitude of velocity of either of the particles relative to the other.  m1 m 2 r r r r v  v1  v 2  v 2  v1 m 1  m 2 and rel Work Energy Theorem for a System of Particles The work energy theorem can be applied to each particle of the system. For ith particle of the system, we can write K i, i  W i,i f  K i,f r r Here Wi,i’!f is total work done by all the internal forces fij and resultant external force Fi on the ith particle, when the system goes from one configuration to other. Adding kinetic energies of all particles, we can write kinetic energies Ki and Kf of the whole system in the initial as well as the final configuration. Adding work done W i, i  f by internal as well as external forces on every particle we find total work done W i f by all the internal as well as external forces on the system. Now we can write work energy theorem. K i  W i f  K f .......(17) While applying the above equation to a system, care must be taken in calculating W i f . In spite of the fact that the r r internal forces fij and f ji being equal in magnitude and opposite in direction, the work done by them on the ith and the jth particles will not, in general, cancel out, since ith and the jth particles may undergo different amount of displacements. The above description at first presents calculating of W i f as a cumbersome task. However for systems, which we usually encounter are not as complex as a general system of large number of particles may be. Systems which we usually face to analyze have limited number of particles or bodies interacting. For these systems we can simplify the task by calculating work of conservative internal forces as decrease in potential energy of the system. Total work of internal forces other than internal conservative forces vanishes, if these forces are due connecting inextensible links or links of constant length. These forces include string tension and normal reaction at direct contacts between the bodies included I the system. Work of internal forces of the kind other than these and work of external forces, can be calculated by definition of work. CENTRE OF MASS & COLLISIONS Conservation of Mechanical Energy If total work of internal forces other than conservative is zero and no external forces act on a system, total mechanical energy remains conserved. .......(18) Ki U i  Kf  Uf Since external forces are capable of changing mechanical energy of the system, under their presence total mechanical energy changes by amount equal to work Wext, i’!f done by all the external forces. W ext , i f  E f  E i   K f  U f    K i  U i  .......(19) Total work of pseudo forces in centroidal frame. Ex. Show that total work done in centroidal frame on all the particles of a system by pseudo forces due to acceleration of mass center is zero. Sol. r Let acceleration of mass centre relative to an inertial frame is a c . Pseudo force on ith particle of mass mi in r r centroidal frame is   m i a c  . Let displacement of ith particle in a time interval is  ri relative to the centroidal frame. Total work of pseudo forces on all the particles in centroidal frame can now be expressed by the expression r r r r r r   m i a c    ri   a c    m i  ri   a c  0  0 Ex. Two blocks of masses m and M connected by a spring are placed on frictionless horizontal ground. When the spring is relaxed, a constant force F is applied as shown. Find maximum extension of the spring during subsequent motion. M Sol. m F If we use ground as inertial frame as we usually do, solution of the problem becomes quite involved. Therefore, we prefer to use the centroidal frame, in which mass center remains at rest. CM M x2  mx m M m x1  M F Mx m M m F In the adjacent figure is shown horizontal position of mass center (CM) by dashed line. It remains unchanged in centroidal frame. Mass center of two particle system divides separation between them in reciprocal ratio of the masses; therefore displacements x1 and x2 of the blocks must also be in reciprocal ratio of their masses. The extension x is sum of displacements x1 and x2 of the blocks as shown in the figure. When extension of the spring achieves its maximum value, both the block must stop receding away from the mass center, therefore, velocities of both the blocks in centroidal frame must be zero. During the process when spring is being extended, total work done by pseudo forces in centroidal frame become zero, negative work done by spring forces becomes equal to increase in potential energy and work done by the applied force evidently becomes Fx1. Using above fact in applying work energy theorem on the system relative to the centroidal frame, we obtain K i  W i f  K f    0  12 kx 2   0  W i  f , springforce  W i  f , F  0 2 FM FMx  0  x  k m  m  m m