System of Particles and Rotational Motion

System of Particles and Rotational Motion 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. Let P be the point where mass ‘m’ is located. Take position vector of point P with respect to origin O. The moment of point mass m about origin O is defined as  M  =mr The physical significance of moment of mass is that when differentiated with respect to time it gives momentum of the particle. It is worth noting that moment of point mass depends on choice of origin. Centre of mass of system of N point masses is that point about which moment of mass of the system is MOMENT OF SYSTEM OF N POINT zero. It means that if about a particular origin the MASSES ABOUT AN ORIGIN ‘O’ Consider a system of N point masses m 1, m 2, moment of mass of system of N point masses is zero m 3 ........... mn whose position vectors from origin then that particular origin is the centre of mass of the     O are given by r1 , r2 , r3 ,............... rn respectively.. system. MOMENT OF POINT MASS ‘M’ ABOUT AN ORIGIN ‘O’ Mass Moment : It is defined as the product of mass of the particle and distance of the particle from the point about which mass moment is taken. It is a vector quantity and its direction is directed from the point about which it is taken to the particle The moment of system of point masses about origin O is the sum of individual moment of each point mass about origin O.  M     = m1 r1 + m2 r2 + m3 r3 + ............ + mn rn System of Particles and Rotational Motion or m1r1 = m2r2 CENTRE OF MASS OF A SYSTEM OF 'N' DISCRETE PARTICLES Consider a system of N point masses m 1, m 2, 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.  rcm =  m2   r and   m 2  m1   m1 m  1  m2 or r1 =  r2 =  Here, r1 = distance or COM from m1 and r2 = distance or COM from m2  r   From the above discussion, we see that    m1r1  m 2 r2  ........  mn rn m1  m 2  ........  m n r1 = r2 = 1 2 if m1 = m2, i.e., COM lies midway between the two particles of equal masses. n  m r i i  ; rcm =  rcm i 1 n m = 1 M n  Similarly, r1 > r2 if m1 < m2 and r1 < r2 if m2 < m1, i.e., COM is nearer to the particle having larger mass.  mi ri i 1 i i 1 Ex.1 Two particles of mass 1 kg and 2 kg are located at x = 0 and x = 3 m. Find the position of their centre  n    m of mass. where M   i  is the total mass of the system.   i 1 Sol. Since, both the particles lies on x-axis, the COM will also lie on x-axis. Let the COM is located at x = x, then r1 = distance of COM from the particle of mass 1 kg = x and r2 = distance of COM from the particle POSITION OF COM OF of mass 2 kg = (3 – x) Using TWO PARTICLES Centre of mass of two particles of mass m1 and m2 separated by a distance r lies in between the two particles. The distance of centre of mass from any of the particle (r) is inversely proportional to the mass of the particle (m) i.e. r or r1 r2 1 m m2 = m 1 or x 3x = 2 1 r1 r2 m2 = m 1 or x = 2 m Thus, the COM of the two particles is located at x = 2 m. Ans. Ex.2 The position vector of three particles of mass m 1 = 1 kg, m 2 = 2 kg and m 3 = 3 kg are   r1  ( î  4 ĵ  k̂ ) m , r2  ( î  ĵ  k̂ ) m  and r3  ( 2 î  ĵ  2k̂ ) m respectively. Find the position vector of their centre of mass. Sol. The position vector of COM of the three particles will be given by     m r  m 2 r2  m 3 r3 rCOM  1 1 m1  m 2  m 3 System of Particles and Rotational Motion Ex.4 Three particles of masses 0.5 kg, 1.0 kg and 1.5 kg are placed at the three corners of a right angled triangle of sides 3.0 cm, 4.0 cm and 5.0 cm as shown in figure. Locate the centre of mass of the system. Substituting the values, we get  (1)( î  4 ĵ  k̂ )  (2)( î  ĵ  k̂ )  (3)(2î – ĵ – 2k̂ ) rCOM  1 2  3 = 9ˆi  3ˆj  3kˆ 6  1 rCOM  (3ˆi  ˆj  kˆ ) m 2 Ans. Ans. The centre of mass is 1.3 cm to the right and 1.5 cm above the 0.5 kg particle. Ex.5 Consider a two-particle system with the particles having masses m1 and m2. If the first particle is pushed towards the centre of mass through a distance d, by what distance should the second particle be moved so as to keep the centre of mass at the same position? Sol. Assuming D as the origin, DC as x -axis and DA as Sol. Consider figure. Suppose the distance of m1 from the centre of mass C is x1 and that of m2 from C is y-axis, we have x2. Suppose the mass m2 is moved through a distance m1 = 1 kg, (x1, y1) = (0, 1m) d towards C so as to keep the centre of mass at C. m2 = 2 kg, (x2, y2) = (1m, 1m) m3 = 3 kg, (x3, y3) = (1m, 0) and m4 = 4 kg, (x4, y4) = (0, 0) Then, m1x2 = m2x2 .........(i) Co-ordinates of their COM are and m1(x1 – d) = m2 (x2 – d) .........(ii) Ex.3 Four particles of mass 1 kg, 2 kg, 3 kg and 4 kg are placed at the four vertices A, B, C and D of a square of side 1 m. Find the position of centre of mass of the particles. xCOM = m1x1  m 2 x 2  m3m3  m 4 x 4 m1  m 2  m 3  m 4 = (1)(0)  2(1)  3(1)  4(0) 1 2  3  4 = 5 10 = m1 Subtracting (ii) from (i) m1d = m2 d or, d = m d, 2 1 m = 0.5 m 2 Similarly, yCOM = m1y1  m 2 y 2  m3 y 3  m 4 y 4 m1  m 2  m 3  m 4 = (1)(1)  2(1)  3(0 )  4(0) 1 2  3  4 = 3 m = 0.3 m 10 CENTRE OF MASS OF A CONTINUOUS MASS DISTRIBUTION 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 xcm =  dm  xCOM, yCOM) = (0.5 m, 0.3 m)  rcm Ans. Thus, position of COM of the four particles is as shown in figure.  x dm , y  dm = cm  y dm , z  dm = cm  z dm  dm = M (mass of the body) = 1 M   r dm . Note: If an object has symmetric uniform mass distribution about x axis than y coordinate of COM is zero and vice-versa System of Particles and Rotational Motion CENTRE OF MASS OF A UNIFORM ROD L 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 x = L. Mass per unit length of the rod =  x dm =  dm 0 L xCOM 0 M L L = Hence, dm, (the mass of the element dx situated at x  (x)(Rx) dx  (Rx) dx 0 L 0 = x is) = M dx L L L  2 R x dx The coordinates of the element PQ are (x, 0, 0). Therefore, x-coordinate of COM of the rod will be = 0 L =  R x dx xCOM =  L 0 x dm 0 = = L 1 L  L x dx  0  y dm  dm  y dm  dm = 0 Similarly, x   2  L     0 2L 3 (as y = 0) zCOM = 0 Hence, the centre of mass of the rod lies at L 2  2L   3 ,0, 0    The y-coordinate of COM is yCOM = 2 The y-coordinate of COM of the rod is y COM  dm M  ( x )  dx  L  = 0 M   x3     3   0 Ans. Ex. 7 The density of a straight rod of length L varies as  = A + Bx where x is the distance from the left end. Locate the centre of mass. =0 Similarly, zCOM = 0 L  i.e., the coordinates of COM of the rod are  2 , 0, 0  .   Or it lies at the centre of the rod. Sol. 2AL  2BL2 3(2A  BL ) Ex.6 A rod of length L is placed along the x-axis between x = 0 and x = L. The linear density (mass/length)  of the rod varies with the distance x from the origin CENTRE OF MASS OF A as  = Rx. Here, R is a positive constant. Find the SEMICIRCULAR RING position of centre of mass of this rod. Figure shows the object (semi circular ring). By observation we can say that the x-coordinate of the Sol. Mass of element dx situated at x = x is dm =  centre of mass of the ring is zero as the half ring is dx = Rx dx symmetrical on both sides of the origin. Only we are The COM of the element has coordinates (x, 0, 0). required to find the y-coordinate of the centre of Therefore, x-coordinate of COM of the rod will be mass. , System of Particles and Rotational Motion To find ycm we use ycm = 1 M  dm y ...(i) Here for dm we consider an elemental arc of the ring at an angle  from the x-direction of angular width d. If radius of the ring is R then its y coordinate will be R sin, here dm is given as dm = M R Now the y-coordinate of the element is taken as 2x  , as in previous section, we have derived that the centre of mass of a semi circular ring is × R d concentrated at 2R  So from equation ---(i), we have R ycm = 1 M   0 M Rd (R sin) R Here ycm is given asycm = = ycm =  sin  d = 0 2R  ...(ii) 1 M  dm 2x  4M x 2 dx 0 R  R  1 M  R 2 0 ycm = 4R 3 CENTRE OF MASS OF SEMICIRCULAR DISC Figure shows the half disc of mass M and radius R. Ex.8 Find the centre of mass of an annular half disc shown Here, we are only required to find the y-coordinate in figure. 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 x 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 = 2M R 2 ( x)dx System of Particles and Rotational Motion Sol. Let  be the mass per unit area of the object. To find its centre of mass we consider an element as a half ring of mass dm as shown in figure of radius r and width dr and there we have dm = r dr Now, Centre of mass of this half ring will be at height R2 2r ycm = 1 M ycm = 2  2  (R 2  R12 ) 2 R2 r R1 2 dr 2r   (. r dr ).  R1 4(R 32  R13 ) = 3(R 2  R 2 ) 2 1 The mass dm of this disc can be given as Alternative solution : We can also find the centre of mass of this object by considering it to be complete half disc of radius R2 and a smaller half disc of radius R1 cut from it. If ycm be the centre of mass of this disc we have from the mass moments.  R12   4R1    2   .   . (R  R12 )     3   2 2 2     R 22   4R 2     (ycm ) =    2  ×   3    4(R 32  R13 ) ycm = 3(R 2  R 2 ) 2 1 CENTRE OF MASS OF A SOLID HEMISPHERE dm = = 3M ×  r2 dy 2R3 3M 2R 3 (R2 – y2) dy ycm of the hemisphere is given as R ycm = 1 M  dm y 0 R = 1 M 3M  2R 3 3 (R2 – y2) dy y 0 R 2 2 = 2R 3  (R  y ) y dy 0 ycm = 3R 8 CENTRE OF MASS OF A The hemisphere is of mass M and radius R. To find HOLLOW HEMISPHERE its centre of mass (only y-coordinate), we consider A hollow hemisphere of mass M and radius R. Now an element disc of width dy, mass dm at a distance y we consider an elemental circular strip of angular from the centre of the hemisphere. The radius of this width d at an angular distance  from the base of the hemisphere. This strip will have an area. elemental disc will be given as r = R 2  y 2 dS = 2R cos  Rd System of Particles and Rotational Motion 3M dm = × r2 dy R 2H here ycm can be given as H ycm = = 1 M R 1 M  0  y dm 0  3M  Ry  2     dy  y  R 2H  H     H = Its mass dm is given as dm = M 2R 2 2R cos  Rd 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. y cm = = 1 M  2 0  M   2R 0 2  2 R 2 cos  d   y H3 3 dy 0 3H 4 Ex.9 Find out the centre of mass of an isosceles triangle of base length a and altitude b. Assume that the mass of the triangle is uniformly distributed over its area. Sol. To locate the centre of mass of the triangle, we take a strip of width dx at a distance x from the vertex of the triangle. Length of this strip can be evaluated by similar triangles as  = x. (a/b)  dm R sin   2 1 M = 3 R sin  2 = R  sin  cos  d Mass of the strip isdm = 2M dx ab Distance of centre of mass from the vertex of the triangle is 0 ycm = R 2 xCM = 1 M b CENTRE OF MASS OF A SOLID CONE =  x dm 2x 2 b 2 dx 0 A solid cone has mass M, height H and base radius R. Obviously the centre of mass of this cone will lie somewhere on its axis, at a height less than H/2. To locate the centre of mass we consider an elemental disc of width dy and radius r, at a distance y from the apex of the cone. Let the mass of this disc be dm, which can be given as = 2 3 b Proceeding in the similar manner, we can find the COM of certain rigid bodies. Centre of mass of some well known rigid bodies are given below : 1. System of Particles and Rotational Motion Centre of mass of a uniform rectangular, square or Ex.10 Find the position of centre of mass of the uniform circular plate lies at its centre. Axis of symmetry plane lamina shown in figure. of symmetry. Sol. Here, A1 = area of complete circle = a2 2 a a 2 A2 = area of small circle =    = 2 2. For a laminar type (2-dimensional) body with uniform negligible thickness the formulae for finding the position of centre of mass are as follows :      m1r1  m 2 r2  .... A 1t r1  A 2 t r2  .... rCOM   m1  m 2  .... A 1t  A 2 t  .... ( m = At) or    A r  A 2 r2  .... rCOM  1 1 A 1  A 2  .... Here, A stands for the area, 3. (i)    m1r1  m 2 r2 rCOM  m1  m 2 or    A 1r1  A 2 r2 rCOM  A1  A 2 x COM  m1x1  m 2 x 2 m1  m 2 or x COM  A 1x 1  A 2 x 2 A1  A 2 y COM  m1y1  m 2 y 2 m1  m 2 or y COM  A 1y 1  A 2 y 2 A1  A 2 and z COM  m1z1  m 2 z 2 m1  m 2 or z COM  A 1z1  A 2 z 2 A1  A 2 (ii) (x1, y1) = coordinates of centre of mass of large circle = (0, 0) and (x2, y2) = coordinates of centre Using    xCOM = A 1x 1  A 2 x 2 A1  A 2 xCOM = a 2  a    4 2 a 2 a 2  4  =  1   8 3   4 a=– a 6 and yCOM = 0 as y1 and y2 both are zero. Therefore, coordinates of COM of the lamina shown  a  6  in figure are   , 0   Ans. Ex.11 Half of the rectangular plate shown in figure is made of a material of density 1 and the other half of density 2The length of the plate is L. Locate the centre of mass of the plate.  Here, m1, A1, r1 , x1, y1 and z1 are the values for the  a of mass of small circle =  2 , 0  we get If some mass of area is removed from a rigid body, then the position of centre of mass of the remaining portion is obtained from the following formulae: 4  whole mass while m2, A2, r2 , x 2 , y2 and z2 are the values for the mass which has been removed. Let us Ans. see two examples in support of the above theory. (1  3 2 ) X = 4 (   ) L 1 2 System of Particles and Rotational Motion Ex.12 The centre of mass of rigid body always lie inside the body. Is this statement true or false? Ans.  A semi-circular ring yc = 2R  xc = O False Ex.13 The centre of mass always lie on the axis of symmetry if it exists. Is this statement true of false? Ans. True Ex.14 If all the particles of a system lie in y-z plane, the x-coordinate of the centre of mass will be zero. Is this statement true or not? 4R  A semi-circular disc yc = 3 xc = O Ans. True CENTRE OF MASS OF SOME COMMON SYSTEMS  A system of two point masses m1 r1 = m2 r2  A hemispherical shell yc = R 2  A solid hemisphere 3R 8 xc = O The centre of mass lies closer to the heavier mass.  Rectangular plate (By symmetry) xc = b 2 yc = L 2  A circular cone (solid)  A triangular plate (By qualitative argument) at the centroid : yc = h 3 yc = yc = xc = O h 4 System of Particles and Rotational Motion  A circular cone (hollow) yc =  where Fext is the sum of the 'external' forces acting h 3 on the system. The internal forces which the particles exert on one another play absolutely no role in the motion of the centre of mass. If no external force is acting on a system of particles, the acceleration of centre of mass of the system will be zero. If ac = 0, it implies that vc must be a constant and if vcm is a constant, it implies that the total momentum of the system must remain constant. It leads to the principal of conservation of momentum in absence of external forces. MOTION OF CENTRE OF MASS AND  CONSERVATION OF MOMENTUM Velocity of centre of mass of system  v cm = = dr dr dr dr m1 1  m 2 2  m3 3 .......... ....  mn n dt dt dt dt M     m1 v 1  m 2 v 2  m 3 v 3 ..........  m n v n M Here numerator of the right hand side term is the total momentum of the system i.e., summation of momentum of the individual component (particle) of the system Hence velocity of centre of mass of the system is the ratio of momentum of the system per unit mass of the system. Acceleration of centre of mass of system dv dv dv dv m1 1  m 2 2  m3 3 .......... ....  m n n  dt dt dt dt a cm = = “If no external force is acting on the system, net momentum of the system must remain constant”. Motion of COM in a moving system of particles: (1) COM at rest : If Fext = 0 and Vcm = 0, then COM remains at rest. Individual components of the system may move and have non-zero momentum due to mutual forces (internal), but the net momentum of the system remains zero. (i) All the particles of the system are at rest. (ii) Particles are moving such that their net momentum is zero. example: M     m1a1  m 2 a 2  m 3 a 3 ..........  mn an M = Net force on system M = Net External Force  Net internal Force M = Net External Force M (action and reaction both of an internal force must be within the system. Vector summation will cancel all internal forces and hence net internal force on system is zero)   If Fext  0 then v cm = constant  Fext  = M a cm (iii) A bomb at rest suddenly explodes into various smaller fragments, all moving in different directions then, since the explosive forces are internal & there is no external force on the system for explosion therefore, the COM of the bomb will remain at the original position and the fragment fly such that their net momentum remains zero. (iv) Two men standing on a frictionless platform, push each other, then also their net momentum remains zero because the push forces are internal for the two men system. System of Particles and Rotational Motion (iii) Man jumping from cart or buggy also exert (v) A boat floating in a lake, also has net momentum internal forces therefore net momentum of the system zero if the people on it changes their position, because and hence, Motion of COM remains conserved. the friction force required to move the people is (iv) Two moving blocks connected by a light spring internal of the boat system. of spring constant on a smooth horizontal surface. If (vi) Objects initially at rest, if moving under mutual the acting forces is only due to spring then COM forces (electrostatic or gravitation)also have net will remain in its motion and momentum will remain momentum zero. conserved. (vii) A light spring of spring constant k kept copressed (v) Particles colliding in absence of external impulsive between two blocks of masses m1 and m2 on a forces also have their momentum conserved. smooth horizontal surface. When released, the blocks acquire velocities in opposite directions, such (3) COM moving with acceleration : If an external force is present then COM continues that the net momentum is zero. its original motion as if the external force is acting on (viii) In a fan, all particles are moving but com it, irrespective of internal forces. is at rest Example: Projectile motion : An axe is thrown in air at an angle  with the horizontal will perform a complicated motion of rotation as well as parabolic motion under the effect of gravitation (2) COM moving with uniform velocity : If Fext = 0, then Vcm remains constant therefore, net momentum of the system also remains conserved. Individual components of the system may have variable velocity and momentum due to mutual forces (internal), but the net momentum of the system remains constant and COM continues to move with the initial velocity. (i) All the particles of the system are moving with same velocity. Example: A car moving with uniform speed on a straight road, has its COM moving with a constant velocity. (ii) Internal explosions / breaking does not change the motion of COM and net momentum remains conserved. A bomb moving in a straight line suddenly explodes into various smaller fragments, all moving in different directions then, since the explosive forces are internal & there is no external force on the system for explosion therefore, the COM of the bomb will continue the original motion and the fragment fly such that their net momentum remains conserved. Hcom = u2 sin2  2g Rcom = 2u sin  u2 sin2  T= g g example: Circular Motion : A rod hinged at an end, rotates, than its COM performs circular motion. The centripetal force (Fc) required in the circular motion is assumed to be acting on the com. FC  m 2 R com Ex.15 A projectile is fired at a speed of 100 m/s at an angle of 37º above the horizontal. At the highest point, the projectile breaks into two parts of mass ratio 1 : 3, the smaller coming to rest. Find the distance from the launching point to the point where the heavier piece lands. System of Particles and Rotational Motion Ex.17 A block A (mass = 4M) is placed on the top of a wedge B of base length  (mass = 20 M) as shown in figure. When the system is released from rest. Find the distance moved by the wedge B till the block A reaches ground. Assume all surfaces are frictionless. Sol. Internal force do not effect the motion of the centre of mass, the centre of mass hits the ground at the position where the original projectile would have Ans. /6 landed. The range of the original projectile is, Ex.18 An isolated particle of mass m is moving in a 3 4 horizontal xy plane, along x-axis, at a certain height 2  10 4   2u2 sin  cos  5 5 above ground. It suddenly explodes into two xCOM = = m g 10 fragments of masses m/4 and 3m/4. An instant later, the smaller fragment is at y = + 15 cm. Find the position of heaver fragment at this instant. Ans. y = – 5 cm Momentum Conservation : = 960 m The centre of mass will hit the ground at this position. As the smaller block comes to rest after breaking, it falls down vertically and hits the ground at half of the range, i.e., at x = 480 m. If the heavier block hits the ground at x2, then xCOM = 960 =  (m)( 480)  (3m)( x 2 ) (m  3m) Ans.  and the velocity of its centre of mass. P = M v cm  Fext If m1x 1  m 2 x 2 m1  m 2 x2 = 1120 m The total linear momentum of a system of particles is equal to the product of the total mass of the system  Fext  = dP dt =0  dP  dt  = 0 ; P = constant When the vector sum of the external forces acting on a system is zero, the total linear momentum of the system remains constant.     Ex.16 In a boat of mass 4 M and length  on a frictionless P1 + P1 + P + ...............+ Pn = constant. water surface. Two men A (mass = M) and B (mass 2M) are standing on the two opposite ends. Now A Ex.19 A shell is fired from a cannon with a speed of 100 m/s at an angle 60º with the horizontal (positive xtravels a distance /4 relative to boat towards its direction). At the highest point of its trajectory, the centre and B moves a distance 3/4 relative to boat shell explodes into two equal fragments. One of the and meet A. Find the distance travelled by the boat fragments moves along the negative x-direction with on water till A and B meet. a speed of 50 m/s. What is the speed of the other Ans. 5/28 fragment at the time of explosion. System of Particles and Rotational Motion Ex.23 A block at rest explodes into three equal parts. Sol. As we know in absence of external force the motion Two parts starts moving along X and Y axes of centre of mass of a body remains unaffected. respectively with equal speeds of 10 m/s. Find the Thus, here the centre of mass of the two fragment initial velocity of the third part. will continue to follow the original projectile path. The velocity of the shell at the highest point of Ans. 10 2 m/s 135° below the X-axis. trajectory is Ex.24 A boy of mass 25 kg stands on a board of mass 10 kg which in turn is kept on a frictionless horizontal vM = ucos = 100 ×cos60º = 50 m/s. ice surface. The boy makes a jump with a velocity Let v1 be the speed of the fragment which moves component 5 m/s in horizontal direction with respect along the negative x-direction and the other fragment to the ice. With what velocity does the board recoil? has speed v2,. which must be along + ve x-direction. With what rate are the boy and the board separating Now from momentum conservation, we have from each other ? mv = m 2 v1 + m 2 v2 or 2v = v2 – v1 or v2 = 2v + v1 = (2 × 50) + 50 = 150 m/s Ex.20 A shell is fired from a cannon with a speed of 100 m/s at an angle 30º with the vertical (y-direction). At the highest point of its trajectory, the shell explodes into two fragments of masses in the ratio 1 : 2. The lighter fragments moves vertically upwards with an initial speed of 200 m/s. What is the speed of the heavier fragment at the time of explosion. Ans. 17.5 m/s. Ex.25 A man of mass m is standing on a platform of mass M kept on smooth ice. If the man starts moving on the platform with a speed v relative to the platform, with what velocity relative to the ice does the platform recoil ? Sol. Consider the situation shown in figure. Suppose the man moves at a speed w towards right and the platform recoils at a speed V towards left, both relative to the ice. Hence, the speed of the man relative to the platform is V + w. By the question, V + w = v, or w = v – V .............(i) 125 m/sec Ex.21 A shell at rest at origin explodes into three fragments of masses 1 kg, 2 kg and mkg. The fragments of masses 1 kg and 2 kg fly off with speeds 12 m/s along x-axis and 8 m/s along y-axis respectively. If mkg files off with speed 40 m/s then find the total mass of the shell. Ans. Sol. v = 12.5 m/s; 3.5 kg Taking the platform and the man to be the system, there is no external horizontal force on the system. The linear momentum of the system remains constant. Initially both the man and the platform were at rest. Thus, 0 = MV - mw or, MV = m (v – V) [Using (i)] Ex.22 A block moving horizontally on a smooth surface with a speed of 20 m/s bursts into two equal parts mv continuing in the same direction. If one of the parts or, V = . Mm moves at 30 m/s, with what speed does the second part move and what is the fractional change in the Ex.26 A flat car of mass M is at rest on a frictionless floor with a child of mass m standing at its edge. If kinetic energy? child jumps off from the car towards right with an 1 initial velocity u, with respect to the car, find the Ans. v = 10 m/s, . 4 velocity of the car after its jump. System of Particles and Rotational Motion Sol. Let car attains a velocity v, and the net velocity of the child with respect to earth will be u – v, as u is its velocity with respect to car. According to momentum conservation (M + m)v1 = Mv2 + m (u + v2) Velocity of car after jump is v2 = (M  m)v 1  mu Mm Velocity of child after jump is Initially, the system was at rest, thus according to momentum conservation, momentum after jump must be zero, as m (u – v) = M v v= mu mM Ex.27 A flat car of mass M with a child of mass m is moving with a velocity v1. The child jumps in the direction of motion of car with a velocity u with respect to car. Find the final velocities of the child and that of the car after jump. u + v2 = (M  m)v 1  (M)u Mm Ex.28 Two persons A and B, each of mass m are standing at the two ends of rail-road car of mass M. The person A jumps to the left with a horizontal speed u with respect to the car. Thereafter, the person B jumps to the right, again with the same horizontal speed u with respect to the car. Find the velocity of the car after both the persons have jumped off. Sol. This case is similar to the previous example, except now the car is moving before jump. Here also no external force is acting on the system in horizontal m 2u Ans. (M  2m)(M  m) direction, hence momentum remains conserved in this direction. After jump car attains a velocity v2 in Ex.29 Two identical buggies move one after the other the same direction, which is less than v1, due to due to inertia (without friction) with the same velocity backward push of the child for jumping. After jump v0. A man of mass m jumps into the front buggy from child attains a velocity u + v2 in the direction of motion the rear buggy with a velocity u relative to his buggy. of car, with respect to ground. Knowing that the mass of each huggy is equal to M, find the velocities with which the buggies will move after that. Ans. Mmu mu vF = v0 + (M  m)2 ; vA = v 0  (M  m) System of Particles and Rotational Motion Sol. (a) Velocity of centre of mass is Ex.30 Each of the blocks shown in figure has mass 1 kg. The rear block moves with a speed of 5  14  2  0 vcm = = 10 m/s 52 2 m/s towards the front block kept at rest. The spring (b) Due to kick on 5 kg block, it starts moving with attached to the front block is light and has a spring a velocity 14 m/s immediately, but due to inertia 2 constant 50 N/m. Find the maximum compression kg block remains at rest, at that moment. Thus of the spring. Velocity of 5 kg block with respect to the centre of mass is v1 = 14 – 10 = 4 m/s ///////////////////////////////////////////////////////////////// and the velocity of 2 kg block w.r.t. to centre of mass is v2 = 0 – 10 = –10 m/s Sol. Maximum compression will take place when the Ex.32 A light spring of spring constant k is kept blocks move with equal velocity. As no net external compressed between two blocks of masses m and horizontal force acts on the system of the two blocks, M on a smooth horizontal surface. When released, the total linear momentum will remain constant. If V the blocks acquire velocities in opposite directions. is the common speed at maximum compression, we The spring loses contact with the blocks when it have, acquires natural length. If the spring was initially (1 kg) (2 m/s) = (1 kg)V + (1 kg)V compressed through a distance x, find the final speeds of the two blocks. or, V = 1 m/s. Sol. Consider the two blocks plus the spring to be the 1 2 Initial kinetic energy = (1 kg) (2 m/s) = 2 J. system. No external force acts on this system in 2 horizontal direction. Hence, the linear momentum will Final kinetic energy remain constant. Suppose, the block of mass M moves with a speed V and the other block with a 1 1 = (1 kg) (1m/s)2 + (1 kg) (1 m/s)2 = 1 J speed v after losing contact with the spring. From 2 2 conservation of linear momentum in horizontal The kinetic energy lost is stored as the elastic energy direction we have in the spring. m MV – mv = 0 or V = v, ....(i) M 1 Hence, (50 N/m) x2 = 2J – 1J = 1 J 1 2 Initially, the energy of the system = kx2 2 or, x = 0.2 m. Finally, the energy of the system Ex.31 Figure shows two blocks of masses 5 kg and 2 kg placed on a frictionless surface and connected with a spring. An external kick gives a velocity 14 m/s to the heavier block in the direction of lighter one. Deduce (a) velocity gained by the centre of mass and (b) the separate velocities of the two blocks with respect to centre of mass just after the kick. = 1 2 mv 2 + 1 2 MV 2 As there is no friction, mechanical energy will remain conserved. Therefore, 1 1 1 mv2 + MV2 = kx2 2 2 2 ...(ii) Solving Eqs. (i) and (ii), we get  kM     m(M  m )  1/ 2 or, V= and  km  V =  M (M  m)    x 1/ 2 x Ans. Ex.33 Blocks A and B have masses 40 kg and 60 kg respectively. They are placed on a smooth surface and the spring connected between them is stretched by 2m. If they are released from rest, determine the speeds of both blocks at the instant the spring becomes unstretched. System of Particles and Rotational Motion Thus on equating we get the maximum extension in the spring, as after this instant the spring starts contracting. 1 2 mF mM k(x1 + x2)2 = . (x1 + x2) 2mF xmax = x1 + x2 = k(m  M) Ans. 3.2 m/s, 2.19 m/s Ex.34 A block of mass m is connected to another block of mass M by a massless spring of spring constant k. The blocks are kept on a smooth horizontal plane and are at rest. The spring is unstretched when a constant force F starts acting on the block of mass M to pull it. Find the maximum extension of the spring. Ex.35 Two blocks of equal mass m are connected by an unstretched spring and the system is 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) Find the displacement of the centre of mass at time t (b) if the extension of the spring is x0 at time t, find the displacement of the two blocks at this instant. Sol. (a) The acceleration of the centre of mass is F aCOM = Sol. We solve the situation in the reference frame of centre 2m of mass. As only F is the external force acting on the The displacement of the centre of mass at time t will system, due to this force, the acceleration of the 1 Ft 2 be x = aCOM t2 = Ans. centre of mass is F/(M + m). Thus with respect to 2 4m centre of mass there is a Pseudo force on the two (b) Suppose the displacement of the first block is x1 and that of the second is x2. Then, masses in opposite direction, the free body diagram of m and M with respect to centre of mass (taking mx 1  mx 2 x= centre of mass at rest) is shown in figure. 2m Taking centre of mass at rest, if m moves maximum by a distance x 1 and M moves maximum by a distance x2, then the work done by external forces (including Pseudo force) will be W= = MF   mF . x1 +  F  m  M    mM mF . (x1 + x2) mM U= k(x1 + x2)2 or, x1 + x2 = 2 m Ft 2 ...(i) Further, the extension of the spring is x 1 – x2. Therefore, x1 – x2 = x0 ...(ii) . x2 This work is stored in the form of potential energy of the spring as 1 2 or, Ft 2 x1  x 2  4m 2 From Eqs. (i) and (ii), and x1 = 1 2  Ft 2    x0   2m    x2 = 1 2  Ft 2    x 0  Ans.  2m    System of Particles and Rotational Motion IMPULSE Impulse of a force F action on a body is defined as : J =  tf Fdt ti  J =  Fdt =  J = m(v2 – v1)  dv m dt dt =  m dv It is also defined as change in momentum   J  ΔP (impulse - momentum theorem) Instantaneous Impulse : There are many occasions 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 momenta. Thus, we can write.      J  F dt  P  Pf  Pi  Ex.36 The hero of a stunt film fires 50 g bullets from a machine gun, each at a speed of 1.0 km/s. If he fires 20 bullets in 4 seconds, what average force does he exert against the machine gun during this period? Sol. The momentum of each bullet = (0.050 kg) (1000 m/s) = 50 kg-m/s. The gun is imparted this much of momentum by each bullet fired. Thus, the rate of change of momentum of the gun = (50 kg  m / s)  20 4s = 250 N. In order to hold the gun, the hero must exert a force of 250 N against the gun. Impulsive force : A force, of relatively higher magnitude and acting for relatively shorter time, is called impulsive force. An impulsive force can change the momentum of a body in a finite magnitude in a very short time interval. Impulsive force is a relative term. There is no clear boundary between an impulsive and Non-Impulsive force. Note: Usually colliding forces are impulsive in nature. Since, the application time is very small, hence, very little motion of the particle takes place. Important points : 1. Gravitational force and spring force are always nonRegarding the impulse it is important to note that Impulsive. impulse applied to an object in a given time interval can also be calculated from the area under force 2. Normal, tension and friction are case dependent. 3. An impulsive force can only be balanced by another time (F-t) graph in the same time interval. impulsive force. Important Points : 1. Impulsive Normal : In case of collision, normal (1) It is a vector quantity. forces at the surface of collision are always impulsive (2) Dimensions = [MLT–1] (3) S unit = kg m/s (4) Direction is along change in momentum. eq. (5) Magnitude is equal to area under the F-t. graph. (6) J =  Fdt = Fav  dt = Fav t (7) It is not a property of any particle, but it is a measure of the degree, to which an external force changes the momentum of the particle. Ni = Impulsive; Ng = Non-impulsive 3. Both normals are Impulsive N1, N3 = Impulsive; N2 = non-impulsive 2. Impulsive Friction : If the normal between the two objects is impulsive, then the friction between the two will also be impulsive. System of Particles and Rotational Motion Impulsive Tensions : When a string jerks, equal and opposite tension act suddenly at each e n d . Consequently equal and opposite impulses act on the bodies attached with the string in the direction of the string. There are two cases to be considered. (a) One end of the string is fixed : The impulse which acts at the fixed end of the string cannot change the momentum of the fixed object there. The object attached to the free end however will undergo a change in momentum in the direction of the string. The momentum remains unchanged in a direction perpendicular to the string where no impulsive forces act. (b) Both ends of the string attached to movable objects : In this case equal and opposite impulses act on the two objects, producing equal and opposite changes in momentum. The total momentum of the system therefore remains constant, although the momentum of each individual object is changed in the direction of the string. Perpendicular to the string however, no impulse acts and the momentum of each particle in this direction is unchanged. Both normals are Impulsive Note : In case of rod Tension is always impulsive In case of spring Tension is always non-impulsive. Friction due to N2 is non-impulsive and due to N3 is impulsive System of Particles and Rotational Motion Ex.37 A block of mass m and a pan of equal mass are COLLISION OR IMPACT connected by a string going over a smooth light Collision is an isolated event in which a strong force pulley. Initially the system is at rest when a particle acts between two or more bodies for a short time, of mass m falls on the pan and sticks to it. If the which results in change of their velocities. particle strikes the pan with a speed v, find the speed with which the system moves just after the collision. Note : (a) In collision particles may or may not come in physical contact. (b) The duration of collision, t is negligible as compared to the usual time intervals of observation of motion. (c) In a collision the effect of external non impulsive forces such as gravity are not taken into a account Sol. Let the required speed is V. as due to small duration of collision (t) average Further, impulsive force responsible for collision is much let J1 = impulse between particle and pan larger than external forces acting on the system. and J2 = impulse imparted to the block and the pan by the string The collision is in fact a redistribution of total Using impulse = change in momentum momentum of the particles. Thus, law of For particle J1 = mv – mV ....(i) conservation of linear momentum is indispensable in For pan J1 – J2 = mV .....(ii) dealing with the phenomenon of collision between For block J2 = mV .....(iii) particles. v Line of Impact: Solving, these three equation, we get V = 3 Ex.38 Two identical block A and B, connected by al massless string are placed on a frictionless horizontal plane. A bullet having same mass, moving with speed u strikes block B from behind as shown. If the bullet gets embedded into the block B then find: 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. (a) The velocity of A,B,C after collision. (b) Impulse on A due to tension in the string (c) Impulse on C due to normal force of collision. (d) Impulse on B due to normal force of collision. Sol. (a) By Conservation of linear momentum v = (b)  T dt  mu 3 (c) u (d)  (N  T) dt =  Ndt   T dt    N dt  2mu 3   N dt  m 3  u  = mu 3 u 3 2mu 3 If one particle is stationary before the collision then the line of impact will be along its motion after collision. Classification of collisions (a) On the basis of line of impact (i) Head-on collision : If the velocities of the particles are along the same line before and after the collision. (ii) Oblique collision : If the velocities of the particles are along different lines before and after the collision. System of Particles and Rotational Motion (b) On the basis of energy : (ii) Two balls A and B are approaching each other such that their centre are moving along dotted lines as (i) Elastic collision : In an elastic collision, the shown in figure. particle regain their shape and size completely after collision. i.e., no fraction of mechanical energy remains stored as deformation potential energy in the bodies. Thus, kinetic energy of system after collision is equal to kinetic energy of system before collision. Thus in addition to the linear momentum, kinetic energy also remains conserved before and after collision. Oblique Collision (ii) Inelastic collision : In an inelastic collision, the particle do not regain their shape and size (iii) Ball is falling on a stationary wedge. completely after collision. 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. However, in the absence of external forces, law of conservation of linear momentum still holds good. (iii) Perfectly inelastic : If velocity of separation just after collision becomes zero then the collision is perfectly inelastic. Collision is said to be perfectly inelastic if both the particles stick together after collision and move with same velocity, Note : Actually collision between all real objects are neither perfectly elastic nor perfectly inelastic, its inelastic in nature. Oblique Collision COEFFICIENT OF RESTITUTION (e) The coefficient of restitution is defined as the ratio of the impulses of recovery and deformation of either body. e= Im pulse of reformation Im pulse of deformatio n  F dt  F dt r = d Velocity of seperation along line of impact Illustrations of line of impact and collisions based on line of impact (i) Two balls A and B are approaching each other such that their centres are moving along line CD. Head on Collision = Velocity of approach along line of impact The most general expression for coefficient of restitution is e= velocityof separationof pointsof contactalongline of impact velocityof approachof point of contactalongline of impact Illustration 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. System of Particles and Rotational Motion Just Before collision Note : e is independent of shape and mass of object but depends on the material. The coefficient of restitution is constant for two Line of impact particular objects. C D A (a) e = 1  Impulse of Reformation Just After collision v1 v2 = Impulse of Deformation  Velocity of separation = Velocity of approach Line of impact  Kinetic Energy may be conserved C D A B  Elastic collision. (b) e = 0  Impulse of Reformation = 0  Velocity of separation = 0  Kinetic Energy is not conserved  Perfectly Inelastic collision. (c) 0 < e < 1  Fext = 0 momentum is conserved for the system.  Impulse of Reformation  m1u1 + m2 u2 = (m1 + m2)v = m1v1 + m2v2 < Impulse of Deformation m1u1  m 2u 2 m1v 1  m 2 v 2  Velocity of separation < Velocity of approach  v = m m = m m ......(1) 1 2 1 2  Kinetic Energy is not conserved Impulse of Deformation :  Inelastic collision. JD = change in momentum of any one body during Note : In case of contact collisions e is always less than deformation. unity. = m2 (v – u2) for m2  0e1 = m1 (–v + u1) for m1 u1 u2 Impulse of Reformation : Important Point : JR = change in momentum of any one body during Reformation. = m2 (v2 – v) for m2 = m1 (v – v1) for m1 e=  Impulse of Reformation ( JR )  Impulse of Deformation ( JD ) v 2  v1 = u u 1 2 In case of elastic collision, if rough surface is present then kf < ki (because friction is impulsive) v2  v = vv 1 ( substituting v from (1)) Velocity of separation along line of impact = Velocity of approach along line of impact A particle ‘B’ moving along the dotted line collides with a rod also in state of motion as shown in the figure. The particle B comes in contact with point C on the rod. System of Particles and Rotational Motion To write down the expression for coefficient of Special Case : restitution e, we first draw the line of impact. Then (1) e=0 v1 = v2 we resolve the components of velocities of points of  for perfectly inelastic collision, both the bodies, contact of both the bodies along line of impact just move with same vel. after collision. before and just after collision. (2) e=1 and m1 = m2 = m, we get v1 = u2 and v2 = u1 i.e., when two particles of equal mass collide elastically and the collision is head on, they exchange their velocities., e.g. Then v 2 x  v 1x e = u u 1x 2x Collision in one dimension (Head on) u1 > u 2 (3) m1 >>>> m2 v2 > v1 v 2  v1 e = u u 1 2  m m1 + m2  m1 and 2  0 m1  v1 = u1 No change and v2 = u1 + e(u1 – u2) (u1 – u2)e = (v2 – v1) By momentum conservation, m1u1 + m2u2 = m1v1 + m2v2 v2 = v1 + e(u1 – u2) and v1 = m1u1  m 2u 2  m 2 e(u1  u 2 ) m1  m 2 v2 = m1u1  m 2u 2  m1e(u1  u 2 ) m1  m 2 Ex.39 Two identical balls are approaching towards each other on a straight line with velocity 2 m/s and 4 m/ s respectively. Find the final velocities, after elastic collision between them. Sol. The two velocities will be exchanged and the final motion is reverse of initial motion for both. Ex.40 Three balls A, B and C of same mass ‘m’ are placed on a frictionless horizontal plane in a straight line as shown. Ball A is moved with velocity u towards the middle ball B. If all the collisions are elastic then, find the final velocities of all the balls. System of Particles and Rotational Motion B and C collides elastically and exchange their velocities to move in opposite directions Now, B and C collides elastically with A and D respectively and come to rest but A and D starts moving with velocity u away from each other as shown  Final velocities VA = u (  ); VB = 0; VC = 0 and VD = u (  ) Ans. Sol. A collides elastically with B and comes to rest but B starts moving with velocity u Ex.42 If A is moved with velocity u and D is moved with 2u as shown. What will be the final velocities now be? After a while B collides elastically with C and comes Ans. to rest but C starts moving with velocity u Ex.43 Two particles of mass m and 2m moving in opposite directions collide elastically with velocity v and 2v respectively. Find their velocities after  Final velocities VA = 0; VB = 0 and VC = u Ans. collision. \Ex.41 Four identical balls A, B, C and D are placed in a line on a frictionless horizontal surface. A and D are moved with same speed ‘u’ towards the middle as shown. Assuming elastic collisions, find the final Sol. Let the final velocities of m and 2m be v1 and v2 velocities. respectively as shown in the figure: Sol. A and D collides elastically with B and C respectively and come to rest but B and C starts moving with velocity u towards each other as shown By conservation of momentum: m(2v) + 2m(–v) = m(v1) + 2m (v2) or 0 = mv1 + 2mv2 or v1 + 2v2 = 0 .........(1) and since the collision is elastic: v2 – v1 = 2v –(–v) or v2 – v1 = 3v .........(2) Solving the above two equations, we get, v2 = v and v1 = –2v Ans. System of Particles and Rotational Motion Ex.47 A ball of mass m moving at a speed v makes a head on collision with an identical ball at rest. The kinetic energy of the balls after the collision is 3/4th of the original. Find the coefficient of restitution. Sol. As we have seen in the above discussion, that under the given conditions : i.e., the mass 2m returns with velocity v while the mass m returns with velocity 2v in the direction shown in figure:  1 e  v 1'   v  2  Ex.44 Find the fraction of kinetic energy lost by the colliding particles after collision in the above situation. Given that Ans.The collision was elastic therefore, no kinetic energy is lost, or KE loss = KEi - KEf 3 Ki 4 1 1 mv1’2 + 2 2  1 e     2  =0 Ans. 2 collisions Kf  mv2’2 = 3  1 mv 2     4 2 Substituting the value, we get 1 1 1  1   m( 2v ) 2  ( 2m)(  v ) 2    m( 2v ) 2  ( 2m)v 2  2 2 2  2  Ex.45 Three balls A, B and C are placed on a smooth horizontal surface. Given that mA = mC = 4mB. Ball B collides with ball C with an initial velocity v as shown in figure. Find the total number of collisions between the balls. All collisions are elastic.  1 e  v2'  v  2  and 2 +  1 e     2  2 = or (1 + e)2 + (1 – e)2 = 3 or 2 + 2e2 = 3 or e2 = 1 2 or 3 4 e= 1 2 Ans. Ex.48 A block of mass m moving at speed v collides with another block of mass 2 m at rest. The lighter block comes to rest after the collision. Find the coefficient of restitution. Ex.46 Two balls shown in figure are identical. Ball A is 1 moving towards right with a speed v and the second Ans. 2 ball is at rest. Assume all collisions to be elastic. Show Ex.49 A block of mass 2 kg is pushed towards a very that the speeds of the balls remains unchanged after heavy object moving with 2 m/s closer to the block all the collisions have taken place. (as shown). Assuming elastic collision and frictionless surfaces, find the final velocities of the blocks. System of Particles and Rotational Motion Sol. Let v1 and v2 be the final velocities of 2kg block and Ex.51 Two balls of masses 2 kg and 4 kg are moved heavy object respectively then, towards each other with velocities 4 m/s and v1 = u1 + 1 (u1 – u2) = 2u1 – u2 2 m/s respectively on a frictionless surface. After colliding the 2 kg balls returns back with velocity = –14 m/s 2m/s. Then find: v = –2m/s 2 (a) velocity of 4 kg ball after collision (b) coefficient of restitution e. (c) Impulse of deformation JD. Ex.50 A ball is moving with velocity 2 m/s towards a heavy wall moving towards the ball with speed 1m/ s as shown in fig. Assuming collision to be elastic, find the velocity of the ball immediately after the collision. (d) Maximum potential energy of deformation. (e) Impulse of reformation JR. Sol. (a) By momentum conservation, 2(4) – 4(2) = 2(–2) + 4(v2)  Sol. The speed of wall will not change after the collision. So, let v be the velocity of the ball after collision in the direction shown in figure. Since collision is elastic (e = 1), v2 = 1 m/s velocity of separation 1  ( 2) 3 (b) e = velocity of approach = 4  ( 2)  6 = 0.5 (c) At maximum deformed state, by conservation of momentum, common velocity is v = 0. JD = m1(v – u1) = m2(v – u2) = 2(0 – 4) = –8 N -s = 4(0 – 2) = – 8 N - s or = 4(0 – 2) = – 8 N - s (d) Potential energy at maximum deformed state U = loss in kinetic energy during deformation. 1 1  1 2 2 or U =  2 m1u1  2 m 2u 2  – (m1 + m2)v2   2 1 separation speed = approach sped or v–1=2+1 or v = 4 m/s 1  1 2 2 =  2 2( 4)  2 4(2)  – (2 + 4) (0)2   2 or U = 24 Joule Ans. (e) JR = m1(v1 – v) = m2 (v – v2) = 2 (–2 – 0) = –4 N-s = 4(0 – 1) = –4 N-s or or 2. JR e = J D 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  JR = eJD = (0.5) (–8) = –4 N-s Ex.52 A block of mass m moving at a speed v collides with another block of mass 2m at rest. The lighter block comes to rest after the collision. Find the coefficient of restitution. Ans. 1 . 2 System of Particles and Rotational Motion 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. Relative speed of separation = e (relative speed of approach) Ex.55 A ball of mass m hits a floor with a speed v0 making Ex.53 A block of mass 1.2 kg moving at a speed of 20 an angle of incidence  with the normal. The cm/s collides head-on with a similar block kept at coefficient of restitution is e. Find the speed of the rest. The coefficient of restitution is 3/5. Find the reflected ball and the angle of reflection of the ball. loss of the kinetic energy during the collision. Sol. The component of velocity v0 along common tangent Ans. 7.7 × 10–3 J. direction v0 sin  will remain unchanged. Let v be Ex.54 The sphere of mass m1 travels with an initial velocity the component along common normal direction after u1 directed as shown and strikes the stationary collision. Applying sphere of mass m2 head on. For a given coefficient m1 of restitution e, what condition on the mass ratio m 2 ensures that the final velocity of m2 is greater than u1? Ans. m1 m2 > 1 e Collision in two dimension (oblique) 1. A pair of equal and opposite impulses act along common normal direction. Hence, linear momentum of individual particles do 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. Relative speed of separation = e (relative speed of approach) along common normal direction, we get v = ev0 cos  Thus, after collision components of velocity v’ are v0 sin  and ev0 cos   v '  ( v 0 sin  )2  (ev 0 cos  )2 and tan  = ev cos  0 Ans. v 0 sin  System of Particles and Rotational Motion becomes zero after collision, while that of 2 becomes v cos . While the components along common tangent direction of both the particles remain unchanged. Thus, the components along common tangent and common normal direction of both the balls in tabular form are given below. Ball 1 2 or tan  = tan  e Ans. Note : For elastic collision, e = 1  v’ = v0 and  =  Component along common tangent direction Before After collision collision v sin  v sin    Compoent along common normal direction Before After collision collision v cos   v cos   From the above table and figure, we see that both the balls move at right angle after collision with velocities v sin  and v cos . Note : When two identical bodies have an oblique elastic collision, with one particle at rest before Ex.56 A ball of mass m makes an elastic collision with collision, then the two particles will go in  directions. nother identical ball at rest. Show that if the collision is oblique, the bodies go at right angles to each other after collision. Ex.57 Two spheres are moving towards each other. Both Sol. In head on elastic collision between two particles, they exchange their velocities. In this case, the component of ball 1 along common normal direction, v cos  have same radius but their masses are 2 kg and 4 kg. If the velocities are 4 m/s and 2 m/s respectively and coefficient of restitution is e = 1/3, find. (a) The common velocity along the line of impact. (b) Final velocities along line of impact. (c) Impulse of deformation. (d) impulse of reformation. (e) Maximum potential energy of deformation. (f) Loss in kinetic energy due to collision. System of Particles and Rotational Motion JD = m1(v – u1) (c) = 2(0 – 4 cos 30º) = – 4 3 N-s (d) Sol. JR = eJD = 1 (–4 3 3 4 )=– N-s 3 (e) Maximum potential energy of deformation is equal to loss in kinetic energy during deformation upto maximum deformed state, In ABC sin  = BC AB = R 2R = 1 2 or  = 30º U= 1 2 m 1(u 1 cos ) 2 + 1 2 m 2(u 2 cos ) 2 (a) By conservation of momentum along line of impact. – = 1 2 2(4 cos 30º) 2 + 1 2 1 2 (m1 + m2)v2 4(–2cos 30º) 2 – 1 2 (2 + 4) (0) 2 or U = 18 Joule. (f) Loss in kinetic energy, 2(4 cos 30º) – 4(2cos 30º) = (2 + 4)v KE = or v = 0 (common velocity along LO) 1 2 m 1 (u 1 cos ) 2 + 1 2 m 2 (u 2 cos ) 2 1 1  2 2 –  2 m1v 1  2 m 2 v 2    = (b) Let v1 and v2 be the final velocity of A and B respectively then, by conservation of momentum along line of impact, 2(4cos 30º) – 4(2cos 30º) = 2(v1) + 4(v2) or 0 = v1 + 2v2 ......... (1) By coefficient of restitution, e or 1 3 velocity of separation along LO  veloctiy of approach along LO  = 1 2 2(4 cos 30º)2 + 1 2 4(–2 cos 30º)2  1  2 2 1  1 2    –  2 2 3   2 4 3         KE = 16 Joule Ex.58 Two point particles A and B are placed in line on a friction less horizontal plane. If particle A (mass 1 kg) is moved with velocity 10 m/s towards stationary particle B (mass 2 kg) and after collision the two move at an angle of 45º with the initial direction of motion, then find : v 2  v1 4 cos 30 º 2 cos 30 º or v2 – v1 = 3 ........ (2) (a) Find velocities of A and B just after collision. from the above two equations, v1 = 2 3 m/s and v2 = 1 3 m/s . (b) Coefficient of restitution. System of Particles and Rotational Motion Sol. The very first step to solve such problems is to find Ex.59 A smooth sphere of mass m is moving on a the line of impact which is along the direction of force horizontal plane with a velocity 3 î  ĵ when it collides applied by A on B, resulting the stationary B to move. with a vertical wall which is parallel to the vector ĵ . Thus, by watching the direction of motion of B, line If the coefficient of restitution between the sphere of impact can be determined. In this case line of impact is along the direction of motion of B. i.e. 45º 1 and the wall is , find with the initial direction of motion of A. 2 (a) the velocity of the sphere after impact, (b) the loss in kinetic energy caused by the impact.  (c) the impulse J that acts on the sphere.  Sol. Let v be the velocity of the sphere after impact. (a) By conservation of momentum, along x direction: mA uA = mAvA cos 45º + mBvB cos 45º or 1(10) = 1(vA cos 45º) + 2(vB (cos 45º) or vA + 2vB = 10 2 ............ (1) along y direction 0 = mAvA sin 45º + mBvB sin 45º or 0 = 1(vA sin 45º) – 2(vB sin 45º) or vA = 2vB  To find v we must separate the velocity components parallel and perpendicular to the wall. Using the law of restitution the component of velocity parallel to the wall remains unchanged while component perpendicular to the wall becomes e times in opposite direction.  3 v   î  ĵ 2 Thus, ............ (2) solving the two equations, vA = 10 2 m/s and vB = 5 2 m/s. velocity of separation along LO  (b) e  veloctiy of approach along LO  (a) Therefore, the velocity of the sphere after impact 3 2 is =  î  ĵ Ans. (b) The loss in K.E. = 2 2 (3 + 1 ) – or e= 5 =  v B  v A cos 90 º u A cos 45 º   3 2 = 1 2 2   1   3   12   27 m m   2   8 2     Ans.  (c) J = P  Pf  Pi  m(v )  m(u)  9 = m   2 î  ĵ  – m 3 î  ĵ =  m î   2 0 2 10  1 1 1 mv2 – mu2 = m 2 2 2 Ans. Ans. Ex.60 A sphere of mass m is moving with a velocity 4 î  ĵ when it hits a wall and rebounds with velocity î  3 ĵ . Find the impulse it receives. Find also the coefficient of restitution between the sphere and the wall. Sol.  J = m(– 3 î + 4 ĵ ) and e = 9 16 System of Particles and Rotational Motion Solving these equations, we have u = 2 and v = 3 The velocities of A and B after impact are therefore, 2 î  2 ĵ respectively and 3 î  ĵ Ans. Before impact the kinetic energy of A is Ans. Ex.61 Two smooth spheres, A and B, having equal radii, lie on a horizontal table. A is of mass m and B is of mass 3m. The spheres are projected towards each other with velocity vector 5 î  2 ĵ and 2 î  ĵ respectively and when they collide the line joining 1 29 m(52 + 22) = 2 2 1 15 (3m)(22 + 12) = 2 2 and of B is 1 , find the velocities after impact and the loss in 3 1 m(22 + 22) = 4m 2 î , the velocity components of A and B perpendicular to î are unchanged by the impact. 1 (3m) (32 + 12) = 15 m 2 and of B is Therefore, the loss in K.E. at impact is kinetic energy caused by the collision. Find also the magnitude of the impulses that act at the instant of impact. Sol. The line of centres at impact, is parallel to the vector m After impact the kinetic energy of A is their centres is parallel to the vector î . If the coefficient of restitution between A and B is m 29 2 m+ 15 m – 4m – 15m = 3m 2 Ans. To find value of J, we consider the change in momentum along î for one sphere only.. For sphere B or J = 3m ( 3 – 2) J = 3m Ans. Applying conservation of linear momentum and the law of restitution, we have in x direction mv and 5m + (3m) (2) = mu + 3 .... (i) 1 (5 – 2) = v– u 3 .....(ii) Ex.62 A small steel ball A is suspended by an inextensible thread of length  = 1.5 from O. Another identical ball is thrown vertically downwards such that its surface remains just in contact with thread during downward motion and collides elastically with the suspended ball. If the suspended ball just completes vertical circle after collision, calculate the velocity of the falling ball just before collision. (g = 10 ms–2) System of Particles and Rotational Motion Sol. Velocity of ball A just after collision is 5gl Let radius of each ball be r and the joining centres of the two balls makes an angle  with the vertical at the instant of collision, then r 1 sin  = 2 r = or  = 30º 2 Let velocity of ball B (just before collision) be v0. This velocity can be resolved into two components, (i) v0 cos 30º, along the line joining the centre of the two balls and (ii) v0 sin 30º normal to this line. Head -on collision takes place due to v0 cos 30º and the component v0 sin 30º of velocity of ball B remains unchanged. Since, the balls collide elastically, therefore, coefficient of restitution is e = 1. Hence, e= v 2 sin 30 º ( v 1 ) 1 v 0 cos 30 º 0 ...(iii) Solving Eqs. (i), (ii), and (iii), J = 1.6 mv0 cos 30º Since, ball A is suspended by an inextensible string,  v1 = 0.6 v0 cos 30º and v2 = 0.8 v0 cos 30º therefore, just after collision, it can move along Since, ball A just completes vertical circle, therefore horizontal direction only. Hence, a vertically upward impulse is exerted by thread on the ball A. This means v2 = 5g that during collision two impulses act on ball A  0.8v0 cos 30º = 5g or v0 = 12.5 ms–1 simultaneously. One is impulsive interaction J between the balls and the other is impulsive reaction J’ of the thread. VARIABLE MASS SYSTEM Velocity v1 of ball B along line of collision is given by If a mass is added or ejected from a system, at rate  J – mv0 cos 30º = mv1 or v1 = J m – v0 cos 30º ...(i) J v2 = 2 m  has magnitude  v rel . Horizontal velocity v2 of ball A is given by J sin 30º = mv2 or  kg/s and relative velocity v rel (w.r.t. the system), then the force exerted by this mass on the system ...(ii)  Thrust Force ( Ft )    dm  Ft  v rel    dt  Suppose at some moment t = t mass of a body is m 3.  and its velocity is v . After some time at t = t + dt its mass becomes (m – dm) and velocity becomes   v  dv . The mass dm is ejected with relative velocity   v r . Absolute velocity of mass ‘dm’ is therefore ( v +  v r ). If no external forces are acting on the system, 4. System of Particles and Rotational Motion Find net force on the mass and apply   dv Fnet  m dt (m = mass at the particular instant) Integrate it with proper limits to find velocity at any time t. the linear momentum of the system will remain conserved, or or Note : Problems of one-dimensional motion (which are mostly asked in JEE) can be solved in easier manner   just by assigning positive and negative signs to all Pi  Pf vector quantities. Here are few example in support     m = (m – dm) ( v + d v ) + dm ( v + v r ) of the above theory. or m v = m v + md v – (dm) v – (dm)        (d v ) + (dm) v + v r dm  The term (dm) (d v ) is too small and can be neglected.  or Here, 1. 2.   md v = – v r dm   dv  m   dt    dv  m   dt  dm dt  dm   = v r   dt    Ex.63 A flat car of mass m0 starts moving to the right due to a constant horizontal force F. Sand spills on the flat car from a stationary hopper. The rate of loading is constant and equal to  kg/s. Find the time dependence of the velocity and the acceleration of the flat car in the process of loading. The friction is negligibly small.  = thrust force Ft and – = rate at which mass is ejecting or    dm  Ft  v r    dt  Problems related to variable mass can be solved Sol. Initial velocity of the flat car is zero. Let v be its in following four steps velocity at time t and m its mass at that instant. Then Make a list of all the forces acting on the main mass and apply them on it.  Apply an additional thrust force Ft on the mass, the  dm   magnitude of which is v r   dt  and direction is    given by the direction of v r in case the mass is  increasing and otherwise the direction of – v r if it is decreasing. At t = 0, v = 0 and m = m0 at t = t, v = v and m = m0 + t Here, vr = v dm dt  (backwards) = Ft = vr dm dt = v (backwards) Net force on the flat car at time t is Fnet = F – Ft dv dt = F – v or m or (m0 + t) v dv dt dv = F  v ....(i) Vr = 0  t Thus, dt m 0  t   –  [n (F – v)]0v =  [n (m0 + t)]0t   F  n  F  v     or 0 1 1  m 0  t     m  0   = n Ft m0  t From Eq. (i), dv dt or F  v m = a= Ft = 0 dm    as Ft  v r  dt   F m  t = 0 F  v m0 v= Sol. In this problem the sand through a hole in the bottom of the cart. Hence, the relative velocity of the sand vr will be zero because it will acquire the same velocity as that of the cart at the moment. =F – v or 0 System of Particles and Rotational Motion Ex.64 A cart loaded with sand moves along a horizontal floor due to a constant force F coinciding in direction with the cart’s velocity vector. In the process sand spills through a hole in the bottom with a constant rate kg/s. Find the acceleration and velocity of the cart at the moment t, if at the initial moment t = 0 the cart with loaded sand had the mass m0 and its velocity was equal to zero. Friction is to be neglected. and the net force will be F only.  Fnet = F or m  dt  = F But here m = m0 – t  (m0 – t) Ans.        Fm 0 a = (m  t )2 0 dv dt =F t F dt 0 m 0  t v    v =   [ln (m0 – t)]0t or v =  ln  m  t   0  dv 0 = F F  m0  Ans. From Eq. (i), acceleration of the cart a= or ...(i)  or = acceleration of flat car at time t Ft  F  m 0  t   m  t 0    dv  Ans. or dv dt = F F m a = m  t 0 Ans. System of Particles and Rotational Motion Rocket propulsion : vr 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. or dv = m  dm – g dt or  v u dv  v r  m m0  dm –g m  t dt 0 Thus, m  0 v = u – gt + vr n  m     dm    dt  Note : 1. F t = v r   downwards and   dm    dt  Further, let  be the mass of the gas ejected per unit time and vr the exhaust velocity of the gases   dm    dt  with respect to rocket. Usually  1. 2. is negative. 2. If gravity is ignored and initial velocity of the rocket  m0  .  m  u = 0, Eq. (i) reduces to v = vr ln  and vr are m  0 v = u – gt + vr ln  m    (downwards) Net force on the rocket Fnet = Ft – W   dm    dt  or Fnet = vr  4. is upwards, as v r is Ex.65 A rocket, with an initial mass of 1000 kg, is kept constant throughout the journey of the rocket. launched vertically upwards from rest under gravity. Now, let us write few equations which can be used The rocket burns fuel at the rate of 10 kg per second. in the problems of rocket propulsion. At time t = t, The burnt matter is ejected vertically downwards with a speed of 2000 ms–1 relative to the rocket. If Thrust force on the rocket burning after one minute. Find the maximum velocity   dm  of the rocket. (Take g as at 10 ms–2) Ft = vr  dt  (upwards)   Sol. Using the velocity equation Weight of the rocket W = mg 3. dm dt ...(i) (upwards) –mg Net acceleration of the rocket a= or F m dv dt v   dm  r = m  dt  –g   Here u = 0, t = 60s, g = 10 m/s2, vr = 2000 m/s, m0 = 1000 kg and m = 1000 – 10 × 60 = 400 kg We get v = 0 – 600 + 2000 ln  400    or v = 2000 ln 2.5 – 600  1000  The maximum velocity of the rocket is 200(10 ln 2.5 – 3) = 1232.6 ms–1 Ans. System of Particles and Rotational Motion Ex.66 A uniform chain of mass m and length  hangs on Ex.67 If the chain is lowered at a constant a thread and touches the surface of a table by its speed v = 1.2 m/s, determine the normal lower end. Find the force exerted by the table on reaction exerted on the floor as a function the chain when half of its length has fallen on the of time. The chain has a mass of 80 kg and table. The fallen part does not form heap. a total length of 6 m. Sol. 1. Weight of the portion BC of the chain lying on the table, W = mg (downwards) 2 Using v = 2gh Ans. (19.2 + 16 t) N LINEAR MOMENTUM CONSERVATION IN PRESENCE OF EXTERNAL FORCE. Fext = 2. Thrust force Ft = vr  dm     dt  dp dt  Fext dt = dP  dP = Fext) impulsive dt  If Fext)impulsive = 0  dP = 0 or P is constant vr = v dm dt Note: Momentum is conserved if the external force present is non-Impulsive. eg. Gravitation or Spring force = v Ft = v2 (where,   m , is mass per unit length of chain)  2 v = ( g 2 = g m  Ft =    (g) = mg   (downwards)  Net force exerted by the chain on the table is F = W + Ft = mg 3  mg  mg 2 2 So, from Newton’s third law the force exerted by the table on the chain will be upwards). 3 2 mg (vertically Ex.68 Two balls are moving towards each other on a vertical line collides with each other as shown. Find their velocities just after collision. System of Particles and Rotational Motion Sol. Let the final velocity of 4 kg ball just after Sol. Let the speed of the bullet be v. Let the common collision be v. Since, external force is gravitational velocity of the bullet and the bob, after the bullet is which is non - impulsive, hence, linear momentum embedded into the bob, is V. By the principle of will be conserved. conservation of the linear momentum, (0.05 kg) v v V = 0.45 kg  0.05 kg = 10 The string becomes loose and the bob will go up with a deceleration of g = 10 m/s2. As it comes to rest at a height of 1.8 m, using the equation v2 = u2 + 2ax, 1.8 m = Applying linear momentum conservation: ( v / 10 )2 2  10 m / s 2 or, v = 60 m/s. 2(–3) + 4(4) = 2(4) + 4(v) 1 m/s 2 Ex.71 A small ball of mass m collides with a rough wall having coefficient of friction  at an angle with the normal to the wall. If after collision the ball moves Ex.69 A ball is approaching ground with speed u. If the with angle  with the normal to the wall and the coefficient of restitution is e then find out: coefficient of restitution is e then find the reflected velocity v of the ball just after collision. or v = (a) the velocity just after collision. (b) the impulse exerted by the normal due to ground on the ball. Ans (a) v = eu; (b) J = mu(1 + e) Ex.70 A bullet of mass 50g is fired from below into the bob of mass 450g of a long simple pendulum as shown in figure. The bullet remains inside the bob and the bob rises through a height of 1.8 m. Find the speed of the bullet. Take g = 10 m/ s2 . Sol. mv cos  – (m (–u cos )) =  N dt mv sin  – mu sin  = –  N dt and e = v cos  u cos   v cos  = eu cos  or mv sin  – mu sin  = –(mv cos  + mu cos ) or v = u sin  [sin  –  cos  (e + 1)] Ans. System of Particles and Rotational Motion ROTATION OF MOTION Ex : A box is being pushed on a horizontal surface. 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 ground are VA and VB respectively.   Vcm  V of any particle   a cm  a of any particle   S cm  S of any particle For pure translation motion :–    Fext = + m2a 2 + m3a3 + ...................... A Where m1,m2,m3,....... are the masses of different particles of the body having accelerations B    a1, a 2 , a 3 ,..........respectively But acceleration of all the particles are same so If the above body is rigid     a1  a 2  a 3  ........  a VA cos 1 = VB cos 2 = M a Where M = Total mass of the body NOTE : With respect to any particle of rigid body the motion of any other particle of that rigid body is circular.  a = acceleration of any particle or of centre of mass    or of body = m1v1 + m2v 2 + m3v 3 + ...................... VBA = relative velocity of B with respect to A. Where m1,m2,m3,....... are the masses of different particles of the body    having velocities v 1, v 2 , v 3 ..........respectively But velocities of all the particles are same so     v 1  v 2  v 3  ........  v  = Mv  I. 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 howsoever small or large. In this motion all the particles have same    s ,v & , a at an instant. Where v = velocity of any particle or of centre of mass. Total Kinetic Energy of body = m2v22 + .................... = 1 2 m1v12 + 1 2 Mv2 1 2 System of Particles and Rotational Motion II. Pure Rotational Motion : A body is said to be in pure rotational motion if the Ex.72 A bucket is being lowered down into a well through perpendicular distance of each particle remains a rope passing over a fixed pulley of radius 10 cm. constant from a fixed line or point and do not move Assume that the rope does not slip on the pulley. Find parallel to the line, and that line is known as axis of the angular velocity and angular acceleration of the pulley rotation. In this motion all the particles have same at an instant when the bucket is going down at a speed     ,  & ,  at an instant. Eg :– a rotating ceiling fan, of 20 cm/s and has an acceleration of 4.0 m/s2. arms of a clock. Sol. Since the rope does not slip on the pulley, the linear For pure rotation motion : – speed v of the rim of the pulley is same as the speed  = s r Where  = angle rotated by the particle of the bucket. The angular velocity of the pulley is then s = length of arc traced by the particle. r = distance of particle from the axis of rotation. = d dt Where  = angular speed of the body.. = d dt Where  = angular acceleration of the body..  = v/r = 20 cm / s 10 cm = 2 rad/s and the angular acceleration of the pulley is  = a/r = 4.0 m / s 2 10 cm = 40 rad/s2. Ex.73 A disc of radius 10 cm is rotating about its axis at an angular speed of 20 rad/s. Find the linear speed of (a) a point on the rim, (b)the middle point of a radius All the parameters ,  and  are same for all the particles. Axis of rotation is perpendicular tothe plane of rotationof particles. Special case : If  = constant ,  = 0 + tWhere 0 = initial angular speed  = 0t + 1 t2 2 t = time interval  = 02 + 2 Total Kinetic Energy = = = 1 1 m1v12 + m2v22 + .......................... 2 2 1 [m1r12 + m2r22 + ..........................] 2 2 1 2 Where  = Moment of Inertia 2 = m1r12 + m2r22 + .............  = angular speed of body. Sol. 2 m/s, 1 m/s Ex.74 A wheel rotates with a constant acceleration of 2.0 rad/s2. If the wheel starts from rest, how many revolutions will it make in the first 10 seconds? Sol. The angular displacement in the first 10 seconds is given by  = 0t + 1  t2 2 = 1 2 (2.0 rad/s2) (10 s)2 = 100 rad. As the wheel turns by 2 radian in each revolution, the number of revolutions in 10 s in n= 100 2 = 16. Ex.75 The wheel of a motor, accelerated uniformly from rest, rotates through 2.5 radian during the first second. Find the angle rotated during the next second. Sol. As the angular acceleration is constant, we have Thus,  = 0t + 1  2 t2 =  t2. 2.5 rad = 1  2 (1s)2  = 5 rad/s2 1 2 =100 rev/minute = (10/3) rad/s. Final angular velocity = 0. Time invertial = 15 s. or  = 5 rad/s2 The angle rotated during the first two seconds is = System of Particles and Rotational Motion Sol. The initial angular velocity 1 × (5 rad/s2) (2s)2 = 10 rad. 2 Let the angular acceleration be . Using the equation  = 0 + at, we obtain  = ( – 2p/9) rad/s2 The angle rotated by the motor during this motion is Thus, the angle rotated during the 2nd second is 10 rad – 2.5 rad = 7.5 rad. Ex.76 A wheel is making revolutions about its axis with a uniform angular acceleration. Starting from rest, it reaches 100 rev/sec in 4 seconds. Find the angular acceleration. Find the angle rotated during these four seconds. Ans. 25 rev/s2, 400  rad  =0t + 1 2 t 2  10  rad   s  = 3  (15s) – 1 2  2 rad     9 s2  (15s)2 = 25 rad = 12.5 revolutions. Hence the motor rotates through 12.5 revolutions before coming to rest. Ex.77 Starting from rest, a fan takes five seconds to attain the maximum speed of 400 rpm (revolution per minute). III. Combined Translational and Rotational Motion Assuming constant acceleration, find the time taken by A body is said to be in translation and rotational the fan in attaining half the maximum speed. motion if all the particles rotates about an axis of Sol. Let the angular acceleration be . According to the rotation and the axis of rotation moves with respect question, to the ground. 400 rev/min = 0 +  5 s ...........(i) Let t be the time taken in attaining the speed of 200 rev/min which is half the maximum. Then, 200 rev/min = 0 + t ...........(ii) Dividing (i) by (ii), we get, 2 = 5 s/t or t = 2.5 s. Ex.78 The motor of an engine is rotating about its axis with an angular velocity of 100 rev/minute. It comes to rest in 15 s, after being switched off. Assuming constant angular deceleration, calculate the number of revolutions made by it before coming to rest. MOMENT OF INERTIA (I) Definition : Moment of nertia is defined as the capability of system to oppose the change produced in the rotational motion of a body. Moment of Inertia depends on : (i) density of the material of body (ii) shape & size of body (iii)axis of rotation Combinedly we can say that it depends upon distribution of mass relative to axis of rotation. System of Particles and Rotational Motion Note : Moment of inertia does not change if the mass : (i) is shited parallel to the axis of the rotation. (ii) is rotated with constant radius about axis of rotation. Moment of nertia is a scalar positive quantity.  = mr12 + m2 r22 +......................... Sol. (i) Moment of inertia of particle on left is 1 = m1r12. =  +  +  +......................... Moment of Inertia of particle on right is 2 = m2r22. 2 S units of Moment of nertia is Kgm . Moment of Inertia of the system about AB is Moment of Inertia of :  =1+ 2 = m1r22 + m2r22 (I) A single particle :  = mr2 (ii) Moment of inertia of particle on left is 1 = 0 where m = mass of the particle Moment of Inertia of particle on right is r = perpendicular distance of the particle from the 2 = m2 (r1 + r2)2. axis about which moment of Inertia is to be calculated Moment of Inertia of the system about AB is  = (II) For many particles (system of particles) : 1+ 2 = 0 + m2(r1 + r2)2 n 2  =  m iri (iii) Moment of inertia of particle on left is  = 0 1 i 1 Moment of Inertia of particle on right is 2 = 0 Moment of Inertia of the system about AB is Ex.79 Four particles each of mass  = 1+ 2 = 0 + 0 m are kept at the four corners Q.81 Four point masses are connected by a massless of a square of edge a. Find the rod as shown in figure. Find out the moment of inertia moment of inertia of the system of the system about a line perpendicular to about axis CD ? the plane of the square and passing through the centre of the square. Sol. The perpendicular distance of every particle from the given line is a/ 2 . The moment of inertia of one particle 1 is, therefore, m(a/ 2 )2 = ma2. The moment of inertia Ans. 10 ma2 2 1 2 of the system is, therefore, 4 × ma2 = 2ma2. Ex.80 Two heavy particles having masses m1 & m2 are situated in a plane perpendicular to line AB at a distance of r1 and r2 respectively. (i) What is the moment of inertia of the system about axis AB ? (ii) What is the moment of inertia of the system about an axis passing through m1 and perpendicular to the line joining m1 and m2 ? (iii) What is the moment of inertia of the system about an axis passing through m1 and m2 ? Ex.82 Three particles, each of mass m, are situated at the vertices of an equilateral triangle ABC of side L (figure). Find the moment of inertia of the system about the line AX perpendicular to AB in the plane of ABC. System of Particles and Rotational Motion Sol. Perpendicular distance of A from AX = 0 Sol. = Perpendicular distance of B from AX = L Perpendicular distance of C from AX = L/2 5 mL2 4 2 Because each element is equally distanced from the axis so r = R Thus, the moment of inertia of the particle at A = 0, of the particle at B - mL2, and of the particle at C = m(L/2)2. The moment of inertia of the threeparticle system about AX is 0 + mL2 + m(L/2)2 =  (dm)r = R 2  dm  MR 2  = MR2 (Note : Answer will remain same even if the mass is dm  M always.) Note that the particles on the axis do not contribute nonuniformly distributed because  to the moment of inertia. Ex.83 Three point masses are located at the corners of an equilibrium triangle of side 1 cm. Masses are Ex.85 Calculate the moment of inertia of a uniform rod of1,2,&3kg respectively and kept as shown in figure. of mass M and length  about an axis passing through Calculate the moment of Inertia of system about an an end and perpendicular to the rod. axis passing through 1 kg mass and perpendicular  M  2 to the plane of triangle ? M 2 2 Sol.  =  ( dm) r =    dx  x = 0 –4 Sol. 5 × 10 kgm 2 (III) For a continuous object :  =  dmr 2 where dm = mass of a small element 3 Ex.86 Using the above formula or other wise find the moment of inertia about an axis which is perpendicular to the rod and passing through centre of rod. Also calculate the moment of inertia of rod about an axis passing through rod and parallel to the rod. M 2 r = perpendicular distance of the particle from the Ans. 12 , 0 axis  m 2  Ex.87 Using the above result    3  or otherwise Ex.84 Calculate the moment of inertia of a ring having   determined the moment of inertia of a uniform mass M, radius R and having uniform mass rectangular plate of side 'b' and ''' about an axis distribution about an axis passing through the centre passing through the edge 'b' and in the plane of of ring and perpendicular to the plane of ring ? plate. Ans. M 2 3 System of Particles and Rotational Motion Ex.88 Find the moment of inertia of uniform rod of length ''about the axis parallel to the rod and 'd' distance apart Sol. =  d ring d = dmr2 element - ring Ans. Md2 Ex.89 Find out the moment of Inertia of figures shown each having mass M, radius R and having uniform mass distribution about an axis passing through the centre and perpendicular to the plane ? dm = R 2 2rdr (here we have used the uniform mass distribution) R =  (a) M (b) M  R 2 . (2rdr).r2 0 =  (c) Ans. MR2 (infact M.I. of any part of mass M of a ring of radius R about axis passing through geometrical centre and perpendicular to the plane of the ring is = MR2 ) MR 2 2 Ex.92 Find the moment of inertia of the uniform square plate of side 'a' and mass M about the axis AB. Ans. ma 2 3 Ex.90 Find out the moment of Inertia of a hollow cyllinder of mass M, radius R and having uniform mass Ex.93 Calculate the moment of inertia of a uniform hollow distribution about its axis. Will the answer change if cylinder of mass M, radius R and length  about its mass is nonuniformly distributed? axis. Ans. MR2, No Sol. Moment of inertia of a uniform hollow cylinder is (IV) For a larger object :  =  d element where d = moment of inertia of a small element Element chosen should be :   (dm) R 2  = mR2 (i) as larger as possible among all types of elements. (ii) as much symmetric as possible (iii) moment of inertia of element should be known earlier. Ex.91 Determine the moment of Inertia of a uniform disc having mass M, radius R about an axis passing through centre & perpendicular to the plane of disc ? Ex.94 Calculate the moment of inertia of a uniform solid cylinder of mass M, radius R and length  about its axis. System of Particles and Rotational Motion Sol. Let AB and CD be two mutually perpendicular diameters of the ring. Take them as X and Y-axes and the line perpendicular to the plane ofthe ring through the centre as the Z-axis. The moment of inertia of the ring about the Z-axis is  = MR2. As the ring is uniform, all of its diameters are equivalent and so x = y, From perpendicular axes theorem, z = x + y. Ans.  = I. MR 2 2 Hence x = z 2 = MR 2 2 . Similarly, the moment of inertia of a uniform disc about a diameter is MR2/4. Two Important Theorems on Moment of Inertia : Ex.96 Two uniform identical rods each of mass M and Perpendicular Axis Theorem [Only applicable to length  are joined to form a cross as shown in figure. plane lamina (that means for 2-D objects only)]. Find the moment of inertia of the cross about a bisector as shown dotted in the figure. Sol. Consider the line perpendicular to the plane of the figure through the centre of the cross. The moment z = x + y (when object is in x-y plane). Where z = moment of inertia of the body about z axis. x = moment of inertia of the body about x axis. y = moment of inertia of the body about y axis. y = x + z (when object is in x-z plane) x = y + z (when object is in y-z plane) M 2 and 12 M 2 hence the moment of inertia of the cross is . 6 of inertia of each rod about this line is The moment of inertia of the cross about the two bisector are equal by symmetry and according to the theorem of perpendicular axes, the moment of inertia of the cross about the bisector is M 2 12 . Note : Defined for any 3 perpendicular concurrent axis out of which two lie in the plane of object. Ex.97 In the figure shown find moment of inertia of a plate having mass M, length  and width b about axis 1,2,3 and 4. Assume that see is centre and mass Ex.95 Find the moment of inertia of a uniform ring of is uniformly distributed mass M and radius R about a diameter. System of Particles and Rotational Motion Sol. Moment of inertia of the plate about axis 1 (by taking II. Parallel Axis Theorem (Applicable to any type rods perpendicular to axis 1) I1 = Mb2 / 3 of object): Moment of inertia of the plate about axis 2 (by taking rods perpendicular to axis 2) I2 = M2 / 12 Moment of inertia of the plate about axis 3 (by taking rods perpendicular to axis 3) I3 = Mb2 / 12 Moment of inertia of the plate about axis 4 (by taking rods perpendicular to axis 4) I4 = M2 / 3  = cm + Md2 Where Ex. 98 Find the moment of Inertia of a uniform ring of cm = Moment of Inertia of the object about an axis mass M and radius R about a diameter. passing through centre of mass and parallel to axis  Sol. Consider x & y two mutually perpendicular diameters of the ring. x + y = z x = y (due to symmetry) z = MR2 x = y = MR 2 2  = Moment of Inertia of the object about axis AB M = Total mass of object d = perpendicular distance between axis about which moment of nertia is to be calculated & the one passing through the centre of mass Ex.102 Find the moment of inertia of a solid cylinder of mass M and radius R about a line parallel to the axis Ex.99 Find the moment of inertia of a uniform disc having of the cylinder and on the surface of the cylinder. mass M, radius R about a diameter. MR 2 Ans. 4 Ex.100 Find the moment of inertia of a uniform rectangular plate of mass M, edges of length '' and 'b' about its axis passing through centre and perpendicular to it. Sol. The moment of inertia of the cylinder about its axis = MR 2 2 . Using parallel axes theorem,  = 0 + MR2 = MR 2 2 + MR2 = 3 2 MR2. Similarly, the moment of inertia of a solid sphere about a tangent is 2 5 MR2 + MR2 = 7 5 MR2. Ex.103 Find out the moment of inertia of a ring having uniform mass distribution of mass M & radius R about an axis which is tangent to the ring and (i) in the plane Ex.101 Find the moment of inertia of a uniform square of the ring (ii) perpendicular to the plane of the ring. plate of mass M, edge of length '' about its axis passing through P and perpendicular to it. Ans. M( 2  b 2 ) 12 (i) Ans. 2m  3 2 Ans. (i) (ii) 3MR 2 2 (ii) 2MR2 Ex.104 Calculate the moment of inertia of a rectangular frame formed by uniform rods having mass m each as shown in figure about an axis passing through its centre and perpendicular to the plane of frame ? Also find moment of inertia about an axis passing through PQ ? System of Particles and Rotational Motion Ring. MR2 (Uniform or Non Uniform) Disc Ans. (i) 2m 2 (  b 2 ) 3 (ii) 5mb 2 3 MR 2 2 Ex.105 Find out the moment of inertia of a semi circular disc about an axis passing through its centre of mass and perpendicular to the plane? Ans. (Uniform) Hollow cylinder 2  MR 2  4R     M    2  3   Ex.106 Find the moment of inertia of the two uniform joint rods about point P as shown in figure. Using parallel axis theorem. Ans. 10 m  2 3 MR2 (Uniform or Non Uniform) List of some useful formula : Solid cylinder Object Moment of Inertia Solid Sphere MR 2 2 2 MR 2 5 (Uniform) (Uniform) Hollow Sphere ML2 3 2 MR 2 3 (Uniform) (Uniform) System of Particles and Rotational Motion Radius of Gyration 2 ML 12 (Uniform) 2m  2 3 As a measure of the way in which the mass of a rotating rigid body is distributed with respect to the axis of rotation we can define a new parameter, the radius of gyration. It is related to the moment of intertia and total mass of the body.  = MK2 (Uniform) where Rectangular Plate  = Moment of Inertia of a body M = Mass of a body K = Radius of gyration 2  = K= 2 M(a  b ) 12 (Uniform)  M Length K is the geometrical property of the body and axis of rotation. Square Plate S.. Unit of K is meter. Ma2 AB = CD = DF = 12 Ex.107 Find the radius of gyration of a solid uniform sphere of radius R about its tangent. Sol.  = (Uniform) 2 mR 2  mR 2 5 = 7 mR 2 5 = mK2  K = 7 R 5 Ex.108 Find the radius of gyration of a hollow uniform sphere of radius R about its tangent. Square Plate Ans. Ma 2 6 5 R 3 (Uniform) Moment of inertia of Bodies with cut Cuboid M(a 2  b 2 ) 12 (Uniform) Ex.109 A uniform disc of radius R has a round disc of radius R/3 cut as shown in Fig. .The mass of the remaining (shaded) portion of the disc equals M. Find the moment of inertia of such a disc relative to the axis passing through geometrical centre of original disc and perpendicular to the planeof the disc. System of Particles and Rotational Motion 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. Sol. Let the mass per unit area of the material of disc be . Now the empty space can be considered as having density – and . Now 0 =  + –  = ( R2)R2/2 = M.I. of  about o – =  (R / 3)2 (R / 3)2 + [–(R/3)2] 2   = r F sin = rF = rF  = angle between the direction Where 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 S unit of torque is Nm Torque is a vector quantity and its direction is determined using right hand thumb rule. (2R/3)2 = M.I. of – about o 0 = 4 R4 9 Ans. Ex.110 Find the moment of inertia of a uniform disc of radius R1 having an empty symmetric annular region of radius R2 in between, about an axis passing through geometrical centre and perpendicular to the disc. 2 Ans. M R1  R 2 2 2 Ex.111 A particle of mass M is released in vertical plane from a point P at x = x0 on the x-axis it falls vertically along the y-axis. Find the torque  acting on the particle at a time t about origin ? Sol. Torque is produced by the force of gravity.   or TORQUE = r F sin  k̂  = rF  x 0mg x = r mg 0 = mgx0 k̂ Torque represents the capability of a force to r produce change in the rotational motion of thebody. Ex.112 A particle having mass m is projected with a velocity v0 from a point P on a horizontal ground making an angle  with horizontal. Find out the torque about the point of projection acting on the particle when (a) it is at its maximum height ? (b) It is just about to hit the ground back ? Torque about point :  Torque of force F about a point  Where F = force applied      rF P = point of application of force 2 Ans. (a) mv 0 sin 2 2 (b) mv 0 2 sin 2 Ex.113 In the previous question, during the motion of particle from P to Q. Torque of gravitational force about P is : (A) increasing (B) decreasing System of Particles and Rotational Motion Force Couple : A pair of forces each of same magnitude and acting in opposite direction is called a force couple. Torque due to couple = Magnitude of one force × distance between their lines of action. (C) remains constant (D) first increasing then decreasing Ans. Increasing Torque about axis :      r F where  = torque acting on the body about the axis of rotation.  r = position vector of the point of application of force about the axis of rotation.  F = force applied on the body.. Note : The direction of torque is calculated using right hand thumb rule and it is always perpendicular to the plane of rotation of the body. Magnitude of torque == F (2d) A couple does not exert a net force on an object even though it exerts a torque. Point of Application of Force : Point of Application of force is the point at which if nett force is assumed to be acting then it will produce same effect of both translation & rotational nature, as was produced earlier. OR If nett force is applied at the point of application in the opposite direction, then the body will be in equilibrium. (translational and rotational both) Ex.114 Determine the point of application of force, when forces of 20 N & 30 N are acting on the rod as shown in figure. If F1 or F2 is applied body applied body revolves in anti-clockwise direction F3 makes body revolve in clockwise direction. If all three are applied. resultant = F1 r1 + F2r2 + F3r3 (in anti-clockwise direction) Note :– Torque produced by a force can be zero in cases. If force vector :– (i) is parallel to the axis of rotation. Sol. Nett force acting on the rod Frel = 10N Nett torque acting on the rod about point C c = (20 × 0) + ( 30 × 20) = 600 clockwise Let the point of application be at a distance x from point C 600 = 10 x  x = 60 cm (ii) passes through the axis of rotation.  70 cm from A is point of Application System of Particles and Rotational Motion Ex.115 Determine the point of application of force, when Ex.116 A wheel of radius r and moment of inertia  about forces are acting on the rod as shown in figure. ts axis is fixed at the top of an inclined plane of inclination  as shown in figure. A string is wrapped round the wheel and its free end supports a block of mass M which can slide on the plane. Initially, the wheel is rotating at a speed  in a direction such that Ans. 4.375 cm left on the rod from the point where 10 the block slides up the plane. How far will the block N force is acting. move before stopping ? Note : (i) Point of application of gravitational force is known as the centre of gravity. (ii) Centre of gravity coincides with the centre of  mass if value of g is assumed to be constant. (iii) Concept of point of application of force is imaginary, as in some cases it can lie outside the body. Relation between '' & '' (for hinged object or pure rotation) Where  ext Hinge  ext Hinge = Hinge  Sol. Suppose the deceleration of the block is a. The linear deceleration of the rim of the wheel is also a. The angular deceleration of the wheel is  = a/r. If the tension in the string is T, the equations of motion are as follows: Mg sin  – T = Ma = nett external torque acting = moment of Inertia of body about Hinge point F1t = M1a1t = M1r1 F2t = M2a2t = M2r2 Mg sin –  resultant ) external x= =  1 2   P  M v CM   Fexternal  M a CM Rotational Kinetic Energy = .  . 2 Net external force acting on the body has two parts tangential and centripetal. v2  FC = maC = m r = m2 rCM CM  Ft = mat = m rCM a r2 = Ma giving, a = Mg r 2 sin    Mr 2 The initial velocity of the block up the incline is v =  r. Thus, the distance moved by the block before stopping is t resultant = F1t r1 + F2t r2 + .... = M1  r12 + M2  r22 + ..... Tr =  = a/r. Eliminating T from these equations, on the body about Hinge point Hinge and v2 2a = 2r 2 (   Mr 2 ) 2M r 2 sin  (  Mr 2 )2 = 2M g sin  Ex.117 The pulley shown in figure has a moment of inertia  about its axis and its radius is R. Find the magnitude of the acceleration of the two blocks. Assume that the string is light and does not slip on the pulley. System of Particles and Rotational Motion Sol : Suppose the tension in the left string is T1 and that Ex.120 A uniform rod of mass m and length  can rotate in the right string in T2. Suppose the block of mass in vertical plane about a smooth horizontal axis hinged M goes down with an acceleration  and the other at point H. Find force exerted by the hinge just after block moves up with the same acceleration. This is the rod is released from rest, from an initial horizontal also the tangential acceleration of the rim of the wheel position ? as the string does not slip over the rim. The angular acceleration of the wheel is, therefore,  = a/R. The Sol. Suppose hinge exerts normal reaction in component form as shown In vertical direction equations of motion for the mass M, the mass m and the pulley are as follows : Fext = maCM Mg – T1 = Ma .........(i) 3g  mg – N1 = m. 4 T2 – mg = ma .........(ii) (we get the value of aCM from previous example) T1R – T2R = I = I /R .........(iii) mg Putting T1 and T2 from (i) and (ii) into (iii),  N1 = 4 a [(Mg – a) – m(g + a)] R =  R which gives a = (M  m)gR 3   (M  m)R 2 . Ex.118 A uniform rod of mass m and length  can rotate in vertical plane about a smoot h horizontal axis hinged at point H. In horizontal direction Fext = maCM  N2 = 0 ( aCM in horizontal = 0 as  = 0 just after release). Ex.121 A uniform rod of mass m and length  can rotate in vertical plane about a smooth horizontal axis hinged at point H. Find force exerted by the hinge just after the rod is released from rest, from an initial position making an angle of 370 with horizontal ? (i) Find angular acceleration  of the rod just after it is released from initial horizontal position from rest ? (ii) Calculate the acceleration (tangential and radial) Ans. of point A at this moment. Sol. (i) H = IH   2 m 2  3 ROTATIONAL EQUILIBRIUM If nett external torque acting on the body is zero, then the body is said to be in rotational equilibrium. 3g 3g (ii) atA =  = . = The centre of mass of a body remains in equilibrium if 2 2 the total external force acting on the body is zero. aCA = 2 r = 0. = 0 (  = 0 just after release) Similarly, a body remains in rotational equilibrium if Ex.119 A uniform rod of mass m and length  can rotate the total external torque acting on the body is zero. in vertical plane about a smooth horizontal axis hinged at point H. Find angular acceleration  of the rod just after it is released from initial position making an angle of 370 with horizontal from rest ? mg. Ans. = 6g / 5 = 3g 2 For translational equilibrium. and Fx  0 ............. (i) Fy  0 ............. (ii) The condition of rotational equilibrium is z  0 The equilibrium of a body is called stable if the body tries to regain its equilibrium position after being slightly displaced and released. It is called unstable if it gets further displaced after being slightly displaced and released. If it can stay in equilibrium even after being slightly displaced and released, it is said to be in neutral equilibrium. System of Particles and Rotational Motion Sol. The forces acting on the ladder are shown in figure. They are (a) Its weight W, (b)normal force N1 by the vertical wall, (c) normal force N2 by the floor and (d)frictional force f by the floor. Taking horizontal and vertical components, and N1 = f ..........(i) N2 = W ..........(ii) Taking torque about B, N1(AO) = W(CB) Ex.122 Two small kids weighing 10 kg and 15 kg are trying to balance a seesaw of total length 5.0 m, with the fulcrum at the centre. If one of the kids is sitting at an end, where should the other sit ? or, N1(AB) cos 53° =W or N1 3 5 or, N1 = = 2 3 AB 2 sin 53° W 4 2 5 W ..........(iii) The normal force by the floor is Sol. It is clear that the 10 kg kid should sit at the end and N2 = W = (10 kg) (9.8 m/s2) = 98 N. the 15 kg kid should sit closer to the centre. Suppose The frictional force is his distance from the centre is x. As the kids are in 2 equilibrium, the normal force between a kid and the f = N1 = W = 65 N. 3 seesaw equals the weight of that kid. Considering the rotational equilibrium of the seesaw, the torque Ex.124 The ladder shown in figure has negligible mass and rests on a frictionless floor. The crossbar of the forces acting on it should add to zero. The connected the two legs of the ladder at the middle. forces are The angle between the two legs is 60°. The fat person (a) (15 kg) g downward by the 15 kg kid, sitting on the ladder has a mass of 80 kg. Find the (b) (10 kg) g downward by the 10 kg kid, constant forces exerted by the floor on each leg and (c) weight of the seesaw and the tension in the crossbar. (d) the normal force by the fulcrum. Taking torques about the fulcrum, (15 kg)g x = (10 kg)g (2.5 m) orx = 1.7 m. Ex.123 A uniform ladder of mass 10 kg leans against a smooth vertical wall making an angle of 53° with it. The other ends rests on a rough horizontal floor. Find the normal force and the friction force that the floor exerts on the ladder. System of Particles and Rotational Motion Sol . The forces acting on different parts are shown in Q.126 A uniform rod of length  , mass m is hung from figure. Consider the vertical equilibrium of “the ladder two strings of equal length from a ceiling as shown in plus the person” system. The forces acting on this figure . Determine the tensions in the strings ? system are its weight (80 kg)g and the contact force N + N = 2 N due to the floor. Thus 2 N = (80 kg) g or N = (40 kg) (9.8 m/s2) = 392 N. Next consider the equilibrium of the left leg of the Ans. T = mg/3, T = 2mg/3 A B ladder. Taking torques of the forces acting on it about Ex.127 A uniform rod of mass m and length is kept  the upper end, vertical with the lower end clamped. It is slightly N (2m) tan 30° = T(1 m) pushed to let it fall down under gravity. Find its angular speed when the rod is passing through its 2 2 or T = N = (392 N) × = 450 N. lowest position. Neglect any friction at the clamp. 3 3 What will be the linear speed of the free end at this Ex.125 A stationary uniform rod of mass 'm' , length '' instant? leans against a smooth vertical wall making an angle  Sol. As the rod reaches its lowest position, the centre of with rough horizontal floor. Find the normal force & mass is lowered by a distance  . Its gravitational frictional force that is exerted by the floor on the rod ? potential energy is decreased by mg  . As no energy is lost against friction, this should be equal to the increase in the kinetic energy. As the rotation occurs about the horizontal axis through the clamped end, the moment of inertia is  = m  2/3. Thus, Sol . As the rod is stationary so the linear acceleration and angular acceleration of rod is zero. i.e. acm = 0 ;  = 0. N2  f    a cm  0 N1  mg 2  1 2 1  m  2  = mg   = mg  2 2  3  or  = 6g  . Torque about any point of the rod should also be zero.   = 0 A = 0  mg cos N1 cos= sinf + f =  + f  sin = N1cos.  2 mg cos  2 mg cos  mgCot = 2 sin  2 The linear speed of the free end is v =   = 6g ANGULAR MOMENTUM Angular momentum of a particle about a point.    L = r P  L = rpsin  L = r × P  L = P × r Where System of Particles and Rotational Motion     L= r × p = mr × v  P = momentum of particle   r = position of vector of particle with respect to point    = ( i x + j y) × ( i vx + j vy)  about which angular momentum is to be calculated .  = m ( k xuu – k yux)   = angle between vectors r & P  v 2  v v2 v    0  0 0 0 = m k  2 g  2 (1  2 )  2g ( 2  1) 2     r= perpendicular distance of line of motion of particle from point O. P= perpendicular component of momentum.  mv 30 = –k SI unit of angular momentum is kgm2/sec. 2 2g . mv 30 Thus, the angular momentum of the particle is 2 2g Ex.128 A particle is projected at time t = 0 from a point in the negative Z-direction i.e., perpendicular to the P with a speed v0 at an angle of 45° to the horizontal. plane of motion, going into the plane. Find the magnitude and the direction of the angular Q.129 A particle of mass m starts moving from origin momentum of the particle about the point P at time t with a constant velocity u î find out its angular = v0/g. momentum about origin at this moment. What will Sol. Let us take the origin at P, X-axis along the horizontal be the answer later time ? and Y-axis along the vertically upward direction as shown in figure. For horizontal motion during the time 0 to t, Ans. Ex.130 A particle of mass 'm' starts moving from point (o,d) with a constant velocity u î . Find out its angular momentum about origin at this moment what will be the answer at the later time? vx = v0 cos 45° = v0/ 2 and x = bxt = v0 2 . v0 g v 20 = 2g . For vertical motion, vy = v0 sin 45° = v0 2 and y = (v0 sin 45°) t – = v 02 v 02 – v0 = 1 2 (1  2 ) 2 v0 gt2 v 02 – 2 g = 2 g ( 2 – 1). 2g The angular momentum of the particle at time t about the origin is Sol. L = mud direction is always clockwise same. Q.131 A particle of mass 'm' is projected on horizontal ground with an initial velocity of u making an angle  with horizontal . Find out the angular momentum of particle about the point of projection when . (i) it just starts its motion (ii) it is at highest point of path. (iii) it just strikes the ground. 2 Ans. (i) O ; (ii) mu cos 2 u sin  2g u2 sin 2 (iii) mu sin  g ; For system of particles : Considering a system of particles with both external and internal forces acting we can add the angular momentum of the indivizual particles to obtain the angular momentum L. System of Particles and Rotational Motion Sol. Consider the situation shown in figure. The velocity of the ball A with respect to the centre O is v = d . 2 The angular momentum of the ball with respect to the axis is  d  d  L  L   = r1  p1 +   r2  p 2 +   r3  p 3 + ........................    = L 1 + L 2 + L 3 + ................................... about the same point. Ex.132 A particle of mass'm' is projected on horizontal ground with an initial velocity of u making an angle  with horizontal . Find out the angular momentum at any time t of particle p about : 1 L1 = mvr = m  2   2  = md2. The same     4 the angular momentum L2 of the second ball. The angular momentum of the system is equal to sum of these two angular momenta i.e., L = 1/2 md2. Ex.134 Two particles of mass m each are attached to a light rod of length d, one at its centre and the other at a free end. The rod is fixed at the other end and is rotated in a plane at an angular speed . Calculate the angular momentum of the particle at the end with respect to the particle at the centre. Sol. The situation is shown in figure. The velocity of the particle A with respect to the fixed end O is vA =  (i) y axis (ii) z-axis Sol. (i) O d/2 and that of B with respect to O is vB = d. Hence the velocity of B with respect to A is vB – vA (ii) – 1/2 u cos . gt2 =  d / 2 . The angular momentum of B with respect Angular momentum of a rigid body rotating about fixed axis : to A is, therefore,  LH L = mvr = m  2   =  H LH = angular momentum of object about axis H. IH = Moment of Inertia of rigid , object about axis H.  = angular velocity of the object. d   d 2 = 1 4 md2 along the direction perpendicular to the plane of rotation. Ex.133 Two small balls A and B, each of mass m, are attached rigidly to the ends of a light rod of length d. Ex.135 A uniform circular disc of mass 200 g and radius The structure rotates about the perpendicular bisector 4.0 cm is rotated about one of its diameter at an of the rod at an angular speed . Calculate the angular angular speed of 10 rad/s. Find the kinetic energy of momentum of the individual balls and of the system the disc and its angular momentum about the axis of about the axis of rotation. rotation. System of Particles and Rotational Motion Sol. The moment of inertia of the circular disc about its (a) The linear momentum before the collision is mv and that after the collision is MV. diameter is Thus, mv = MV, = 1 1 Mr2 = 4 4 (0.200 kg) (0.04 m)2 = 8.0 × 10–5 kg-m2. The kinetic energy is K= 1 2 1  = 2 2 (8.0 × 10–5 kg - m2) (100 rad2/s2) = 4.0 × 10–3 J and the angular momentum about the axis of rotation is L =  = (8.0 × 10–5 kg-m2) (10 rad/s) or m M v.. (b) Let A be the centre of the rod when it is at rest. Let AB be the line perpendicular to theplane of the figure. Consider the angular momentum of “the rod plus the particle” system about AB. Initially the rod is at rest. The angular momentum of the particle about AB is L = mv (a/4) After the collision, the particle comes to rest. The angular momentum of the rod about A is  L = 8.0 × 10–4 kg-m2/s = 8.0 × 10–4 J-s. Conservation of Angular Momentum As Angular momentum of a particle or a system remains constant if  ext = 0about that point or axis of rotation. Thus, Ex.136 A uniform rod of mass M and length a lies on a smooth horizontal plane. A particle of mass m moving at a speed v perpendicular to the length of the rod strikes it at a distance a/4 from the centre and stops after the collision. Find (a) the velocity of the centre of the rod and (b) the angular velocity of the rod about its centre just after the collision. V=    L cm + M r0 × V =     r0 || V , r0 × V   L = L cm =0 Hence the angular momentum of the rod about AB is L = = M 2 . 12 Thus, mva 4 Ma 2  12 or,  = = 3mv Ma Sol. The situation is shown in figure. Consider the rod Ex.137 A uniform rod of mass m and length can rotate and the particle together as the system. As there is freely on a smooth horizontal plane about a vertical no external resultant force, the linear momentum of axis hinged at point H. A point mass having same the system will remains constant. Also there is no mass m coming with an initial speed u perpendicular to the rod, strikes the rod in-elastically at its free resultant external torque on the system and so the end. Find out the angular velocity of the rod just angular momentum of the system about the any line after collision ? will remain constant. Suppose the velocity of the centre of the rod is V and the angular velocity about the centre is . Sol . Angular momentum is conserved about H because no external force is present in horizontal plane which is producing torque about H.  m 2  2 mul =  3  m      = 3a 4 System of Particles and Rotational Motion Q.138 A uniform rod of mass m and length  can rotate freely on a smooth horizontal plane about a vertical axis hinged at point H. A point mass having same mass m coming with an initial speed u perpendicular For velocities to the rod, strikes the rod in-elastically at a distance of 3/4 from hinge point. Find out the angular velocity of the rod just after collision ? VP  VQ 2  r 2  2 VQ r cos  For acceleration : Relation between Torque and Angular Momentum    dL = dt Torque is change in angular momentum , ,  are same about every point of the body (or Impulse of Torque : any other point outside which is rigidly attached to the body).  dt  J Dynamics : J  d Charge in angular momentum. Combined Translational and Rotational motion of a rigid body : If the axis of rotation is moving w.r. to ground then the motion is combined translational and rotational motion.   Psystem  Mv cm ,      cm   cm  , Fext  Ma cm 1 1 Mv cm2 +  cm  2 2 2   Angular momentum axis AB = L about C.M. + L     of C.M. about AB L AB   cm   rcm  Mv cm Total K.E. = Kinematics : The most general motion of a rigid body can be thought of as a sum of two independent motions. A translation of some point of the body Ex.139 A uniform sphere of mass 200 g rolls without plus a rotation about this point . This is called slipping on a plane surface so that its centre moves Chasle's Theorem. A convenient choice of the point at a speed of 2.00 cm/s. Find its kinetic energy. is the centre of mass of the body. One good example Sol. As the sphere rolls without slipping on the plane of the type of motion is rolling of a wheel. surface, its angular speed about the centre is The general motion of the body can be thought of as v cm = . The kinetic energy is the result of a translation of the point Q and the r motion of the body about Q. Let us choose another 1 1  K = cm2 + Mvcm2 point P in the body with position vector rP . Let 2  rP / Q denote the position vector of P with respect to    Q , then rP = rQ + rP / Q .   1 2 2 5 1 2 = . Mr22 + Mvcm2  By differentiating we get,  P   Q   P / Q . For a rigid body, the distance between the particles remain unchanged during its motion i.e. rP/Q = constant 2 = = 1 5 Mvcm2 + 1 Mvcm2 2 = 7 Mvcm2 10 7 (0.200 kg) (0.02 m/s)2 = 5.6 × 10–5 J. 10 System of Particles and Rotational Motion Ex.140 A wheel of perimeter 220 cm rolls on a level Sol. Let the mass of the cylinder be m and its radius r. Suppose the linear speed of the cylinder when it road at a speed of 9 km/h. How many revolutions reaches the bottom is v. As the cylinder rolls without does the wheel make per second ? slipping, its angular speed about its axis is  = v/r. Sol. As the wheel rolls on the road, its angular speed  The kinetic energy at the bottom will be about the centre and the linear speed v of the centre are related as v = r. 1 1 1  1 mr 2  2 1    + mv2 K = 2 + mv2 = 2  = = 9 km / h 220 cm / 2 900 22  36 Ex.141 = 2  9  10 5 220  3600 rev/s = 25 22 2 rad/s. = 1 4 2 2  mv2 + 1 2 mv2 = 3 4 2  mv2. This should be equal to the loss of potential energy mg  sin. Thus, rev/s. Find  VA  VB  VC  VD Sol . wrt. centre wrt. ground Result PURE ROLLING 3 mv2 = mg  4 v= 4 g sin  3 . Ex.143 Figure shows two cylinders of radii r1 and r2 having moments of inertia 1 and 2 about their respective axes. Initially, the cylinders rotate about their axes with angular speed 1 and 2 as shown in the figure. The cylinders are moved closed to touch each other keeping the axes parallel. The cylinders first slip over each other at the contact but the slipping finally ceases due to the friction between them. Find the angular speeds of the cylinders after the slipping ceases. Sol. When slipping ceases, the linear speeds of the points of contact of the two cylinders will be equal. If 1 and 2 be the respective angular speeds, we have 1 r1 and 2 r2 ................(i) The change in the angular speed is brought about by the frictional force which acts as long as the slipping exists. If this force f acts for a time t, the torque on the first cylinder is fr1 and that on the second is fr2. Assuming 1 > 2, the corresponding angular impulses are – fr1t and fr2t, We, there fore, have – f r1 t = 1 (1 –1) and fr2 t = 2 (2 –2) or, – Ex.142 A cylinder is released from rest from the top of an incline of inclination  and length  . If the cylinder rolls without slipping, what will be its speed when it reaches the bottom ? sin or  1 (1 –1) = 2 (2 –2) r2 r1 Solving (i) and (ii) 1 = and 2 = ................(ii) 1 1 r2   2 2 r1  2 r12  1 r22 1 1 r2   2 2 r1  2 r12  1 r22 r2 r1. System of Particles and Rotational Motion Ex.144 A cylinder of mass m is suspended through two Sol. The situation is shown in figure. As the force F rotates the sphere, the point of contact has a tendency to strings wrapped around it as shown in figure. Find slip towards left so that the static friction on the (a) the tension T in the string and (b) the speed of sphere will act towards right. Let r be the radius of the cylinder as it falls through a distance h. the sphere and a be the linear acceleration of the centre of the sphere. The angular acceleration about the centre of the sphere is  = a/r, as there is no slipping. For the linear motion of the centre, F + f = ma Sol. The portion of the strings between the ceiling and the cylinder is at rest. Hence the points of the cylinder where the strings leave it are at rest. The cylinder is thus rolling without slipping on the strings. Suppose the centre of the cylinder falls with an acceleration a. The angular acceleration of the cylinder about its axis is  = a/R, as the cylinder does not slip over the strings. The equation of motion for the centre of mass of the cylinder is mg – 2T - ma and for the rotational motion about the centre, 2  a 2 Fr – f r =  =  5 mr   r      or, F – f = 2 5 ma, ...........(iii) From (i) and (ii), 2F = or a= .............(i) and for the motion about the centre of mass, it is 1 ............(i)  1 2 2 Tr  2 mr   = mra   2 From (i) and (ii), 1 = ma. 2 2 mg a = 3 g and T = 6 . or 2T = 7 5 ma 10 F . 7m Ex.146 A sphere of mass M and radius r shown in figure slips on a rough horizontal plane. At some instant it has translational velocity v0 and rotational velocity v0 about the centre 2 r . Find the translational velocity after the sphere starts pure rolling. As the centre of the cylinder starts moving from rest, the velocity after it has fallen through a distance h is given by 2  v2 = 2  3 g  h   or v= 4 gh 3 . Ex.145 A force F acts tangentially at the highest point of a sphere of mass m kept on a rough horizontal plane. If the sphere rolls without slipping, find the acceleration of the centre of the sphere. Sol. Velocity of the centre = v0 and the angular velocity v0 about the centre = 2 r . Thus v0 > 0r. The sphere slips forward and thus the friction by the plane on the sphere will act backward. As the friction is kinetic, its value is µN = µMg and the sphere will be decelerated by acm = f/M. Hence, v(t) = v0 – f t. M .............(i) System of Particles and Rotational Motion This friction will also have a torque  = fr about the centre. This torque is clockwise and in the direction of 0. Hence the angular acceleration about the centre will be r  = f 5f = 2 Mr ( 2 / 5 )Mr and the clockwise angular velocity at time t will be 2 5f (t) = 0 + 2 Mr v0 5f t = 2 r + 2 Mr t. Pure rolling starts when v(t) = r (t) v0 5f i.e., v(t) = 2 + 2 M t. ............(ii) Eliminating t from (i) and (ii), 5 2 v(t) + v(t) = or, v(t) = 2 7 v0 5 v0 + 2 2 × 3v0 = 6 v. 7 0 Thus, the sphere rolls with translational velocity 6v0/ 7 in the forward direction.