Newton Law of Motion Notes

NEWTON LAW OF MOTION Page # 1 NEWTON LAW OF MOTION 1. FORCE A pull or push which changes or tends to change the state of rest or of uniform motion or direction of motion of any object is called force. Force is the interaction between the object and the source (providing the pull or push). It is a vector quantity. Effect of resultant force : • may change only speed • may change only direction of motion. • may change both the speed and direction of motion. • may change size and shape of a body kg.m unit of force : newton and (MKS System) s2 g.cm dyne and (CGS System) s2 5 1 newton = 10 dyne Kilogram force (kgf) The force with which earth attracts a 1 kg body towards its centre is called kilogram force, thus Force in newton g Dimensional Formula of force : [MLT–2] • For full information of force we require  Magnitude of force  direction of force  point of application of the force kgf = Force Electromagnetic force Gravitational force Contact force Nuclear force Normal reaction Tension friction Contact force : Forces which are transmitted between bodies by short range atomic molecular interactions are called contact forces. When two objects come in contact they exert contact forces on each other. Normal force (N) : It is the component of contact force perpendicular to the surface. It measures how strongly the surfaces in contact are pressed against each other. It is the electromagnetic force. A table is placed on Earth as shown in figure. 2 3 1 4 2. NEWTON’S FIRST LAW OF MOTION : According to this law “A system will remain in its state of rest or of uniform motion unless a net external force act on it. NEWTON LAW OF MOTION Page # 2 1st law can also be stated as “If the net external force acting on a body is zero, only then the body remains at rest.” The word external means external to the system (object under observation), interactions within the system has not to be considered. The word net means the resultant of all the forces acting on the system. Newton’s first law is nothing but Galileo’s law of inertia. Inertia means inability of a body to change its state of motion or rest by itself. The property of a body that determines its resistance to a change in its motion is its mass (inertia). Greater the mass, greater the inertia. An external force is needed to set the system into motion, but no external force is needed to keep a body moving with constant velocity in its uniform motion. Newton’s laws of motion are valid only in a set of frame of references, these frames of reference are known as inertial frames of reference. Generally, we take earth as an inertial frame of reference, but strictly speaking it is not an inertial frame. All frames moving uniformly with respect to an inertial frame are themselves inertial. We take all frames at rest or moving uniformly with respect to earth, as inertial frames. 3. NEWTON’S SECOND LAW OF MOTION : Newton’s second law states, “The rate of change of a momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts”    dp dp i.e., F  or F  k dt dt where k is a constant of proportionality.     p  mv, So F  k ( dmv) dt For a body having constant mass,    dv  F  km  k ma dt From experiments, the value of k is found to be 1.   So, Fnet  ma Force can’t change the momentum along a direction normal to it, i.e., the component of velocity normal to the force doesn’t change. Newton’s 2nd law is strictly applicable to a single point particle. In case of rigid bodies or system of particles   or system of rigid bodies, F refers to total external force acting on system and a refers to acceleration of  centre of mass of the system. The internal forces, if any, in the system are not to be included in F . Acceleration of a particle at any instant and at a particular location is determined by the force (net) acting on the particle at the same instant and at same location and is not in any way depending on the history of the motion of the particle. 4. APPLICATIONS : 4.1 Motion of a Block on a Horizontal Smooth Surface. Case (i) : When subjected to a horizontal pull : The distribution of forces on the body are shown. As there is no motion along vertical direction, hence, R = mg For horizontal motion F = ma or a = F/m R m mg a F NEWTON LAW OF MOTION Page # 3 Case (ii) : When subjected to a pull acting at an angle () to the horizontal : Now F has to be resolved into two components, F cos along the horizontal and F sin  along the vertical direction. Fsin R F  m Fcos mg For no motion along the vertical direction. we have R + F sin  = mg or R = mg – F sin  : Hence R  mg. R < mg For horizontal motion F cos  m Case (iii) : When the block is subjected to a push acting at an angle  to the horizontal : (down ward) The force equation in this case R = mg + F sin  F cos  = ma, a = R F :  R  mg, R > mg F cos = ma, a = F cos   For horizontal motion F cos  m mg F F sin  4.2 Motion of bodies in contact. Case (i) : Two body system : Let a force F be applied on mass m1 B F A f m1 f m2 4.3 Motion of a body on a smooth inclined plane : Natural acceleration down the plane = g sin  Driving force for acceleration a up the plane, F=m(a+ gsin) and for an acceleration a down the plane, F=m(a – gsin) N mg sin mgcos mg 5. NEWTONS’ 3RD LAW OF MOTION : Statement : “To every action there is equal and opposite reaction”. But what is the meaning of action and reaction and which force is action and which force is reaction? Every force that acts on body is due to the other bodies in environment. Suppose that a body A experiences   a force FAB due to other body B. Also body B will experience a force FBA due to A. According to Newton third law two forces are equal in magnitude and opposite in direction Mathematically we write it as   FAB  –FBA   Here we can take either FAB or FBA as action force and other will be the reaction force. : (i) Action-Reaction pair acts on two different bodies. NEWTON LAW OF MOTION Page # 4 (ii) Magnitude of force is same. (iii) Direction of forces are in opposite direction. (iv) For action-reaction pair there is no need of contact Rope 5.1 Climbing on the Rope : Now three condition arises. if T > mg  man accelerates in upward direction T < mg  man accelerates in downward direction T = mg  man’s acceleration is zero * F.B.D of man T a a mg Either climbing or decending on the rope man exerts force downward 6. SPRING FORCE : Every spring resists any attempt to change its length; when it is compressed or extended, it exerts force at its ends. The force exerted by a spring is given by F = –kx, where x is the change in length and k is the stiffness constant or spring constant (unit Nm–1) When spring is in its natural length, spring force is zero. +x 0 0 F=0 F Fext F = –kx F x Graph between spring force v/s x 7. WEIGHING MACHING : A weighing machine does not measure the weight but measures the force. exerted by object on its upper surface or we can say weighing machine measure normal force on the man. 7.1 Motion in a lift : (A) If the lift is unaccelerated (v = 0 or constant) In this case no pseudo force act on the man In this case the F.B.D. of the man N = mg In this case machine read the actual weight weighing machine N (B)If the lift is accelerated upward. (where a = constant) a N weighing machine F.B.D of man with respect to lift So weighing machine read N = m(g + a) Apparent weight N > Actual weight (mg) mg ma(pseudo) mg NEWTON LAW OF MOTION Page # 5 (c)If the lift is accelerated down ward. F.B.D of man with respect to lift So weighing machine read N = m(g - a) Apparent weight N < Actual weight (mg) N ma (pseudo) a weighing machine Note : mg (i) If a = g  N = 0 Thus in a freely falling lift, the man will experience a state of weightlessness (ii) If the lift is accelerated downwards such that a > g : So the man will be accelerated upward and will stay at the ceiling of the lift. (iii) Apparent weight is greater than or less than actual weight only depends on the direction and magnitude of acceleration. Magnitude and direction of velocity doesn’t play any roll in apparent weight. 8. FRICTION : Friction is a contact force that opposes the relative motion or tendency of relative motion of two bodies. Mg F F f N Consider a block on a horizontal table as shown in the figure. If we apply a force, acting to the right, the block remains stationary if F is not too large. The force that counteracts F and keeps the block in rest from moving is called frictional force. If we keep on increasing the force, the block will remain at rest and for a particular value of applied force, the body comes to state of about to move. Now if we slightly increase the force from this value, block starts its motion with a jerk and we observe that to keep the block moving we need less effort than to start its motion. So from this observation, we see that we have three states of block, first, block does not move, second, block is about to move and third, block starts moving. The friction force acting in three states are called static frictional force, limiting frictional force and kinetic frictional force respectively. If we draw the graph between applied force and frictional force for this observation its nature is as shown in figure. 8.1 Static frictional force f b flim fkin a Static region c Kinetic region d F When there is no relative motion between the contact surfaces, frictional force is called static frictional force. It is a self-adjusting force, it adjusts its value according to requirement (of no relative motion). In the taken example static frictional force is equal to applied force. Hence one can say that the portion of graph ab will have a slope of 45º. The Direction of Static Friction The direction of static friction on a body is such that the total force acting on it keeps it at rest with respect to the body in contact. The direction of static friction is as follows. For a moment consider the surfaces to be frictionless. In absence of friction the bodies will start slipping against each other. One should then find the direction of friction as opposite to the velocity with respect to the body applying the friction. NEWTON LAW OF MOTION Page # 6 8.2 Limiting Frictional Force This frictional force acts when body is about to move. This is the maximum frictional force that can exist at the contact surface. (i) The magnitude of limiting frictional force is proportional to the normal force at the contact surface. flim  N  flim = sN Here s is a constant the value of which depends on nature of surfaces in contact and is called as ‘coefficient of static firction’. 8.3 Kinetic Frictional Force Once relative motion starts between the surface in contact, the frictional force is called as kinetic frictional force. The magnitude of kinetic frictional force is also proportional to normal force. fk = kN From the previous observation we can say that k < s Although the coefficient of kinetic friction varies with speed, we shall neglect any variation i.e., when relative motion starts a constant frictional force starts opposing its motion. Direction of Kinetic Friction The kinetic friction on a body A slipping against another body B is opposite to the velocity of A with respect to B. It should be carefully noted that the velocity coming into picture is with respect to the body applying the force of friction. f f v Suppose we have a long truck moving on a horizontal road. A small block is placed on the truck which slips on the truck to fall from the rear end. As seen from the road, both the truck and the block are moving towards right, of course the velocity of the block is smaller than that of the truck. What is the direction of the kinetic friction acting on the block due to the truck ? The velocity of the block as seen from the truck is towards left. Thus, the friction on the block is towards right. The friction acting on the truck due to the block is towards left. 9. MINIMUM FORCE REQUIRED TO MOVE THE PARTICLE : A body of mass m rests on a horizontal floor with which it has a coefficient of static friction . It is desired to make the body slide by applying the minimum possible force F. F  m Fig. A Let the applied force F be at angle  with the horizontal F R  R m R = Normal force mg Fig. B For vertical equilibrium, R + F sin  = mg or, R = (mg – F sin)...(i) For horizontal equilibrium i.e. when the block is just about to slide, F cos  = R ...(ii) Substituting for R, F cos =  (mg – F sin) or F = mg / (cos  +  sin) for minimum F (cos + sin) is maximum,  Let x = cos  +  sin dx   sin    cos  d NEWTON LAW OF MOTION Page # 7 dx 0 d tan  =  and at this value of  mg Fmin  1  2 for maximum of x, 10.FRICTION AS THE COMPONENT OF CONTACT FORCE : When two bodies are kept in contact, electromagnetic forces act between the charged particles at the surfaces of the bodies. As a result, each body exerts a contact force on other The magnitudes of the contact forces acting on the two bodies are equal but their directions are opposite and hence the contact forces obey Newton’s third law. Fc=contact force N=normal force f=friction The direction of the contact force acting on a particular body is not necessarily perpendicular to the contact surface. We can resolve this contact force into two components, one perpendicular to the contact surface and the other parallel to it. The perpendicular component is called the normal contact force or normal force and parallel component is called friction. Contact force = Fc min f 2  N2 = N {when fmin = 0} Fc max =  2N2  N2 {when fmax = N} N  Fc  (  2  1) N 0  tan–1 11.MOTION ON A ROUGH INCLINED PLANE Suppose a motion up the plane takes place under the action of pull P acting parallel to the plane N = mg cos  P N Frictional force acting down the plane F = N =  mg cos  Appling Newton’s second law for motion up the plane mg sin  mg cos  P – (mg sin  + f) = ma f  mg P – mg sin  –  mg cos  = ma If P = 0 the block may slide downwards with an acceleration a. The frictional force would then act up the plane mg sin  – F = ma or, mg sin  –  mg cos  = ma 12.ANGLE OF REPOSE : Consider a rough inclined plane whose angle of inclination  with ground can be changed. A block of mass m is resting on the plane. Coefficient of (static) friction between the block and plane is . For a given angle , the FBD (Free body diagram) of the block is N f mg cos mg sin NEWTON LAW OF MOTION Page # 8 Where f is force of static friction on the block. For normal direction to the plane, we have N=mg cos As  increases, the force of gravity down the plane, mg sin , increases. Friction force resists the slide till it attains its maximum value. fmax = N =  mg cos  Which decreases with  (because cos  decreases as  increases) Hence, beyond a critical value  = c, the blocks starts to slide down the plane. The critical angle is the one when mg sin  is just equal of fmax, i.e., when mg sin c =  mg cosc or tan c =  where C is called angle of repose If  > c, block will slide down.For  < c the block stays at rest on the incline. 13.CIRCULAR MOTION When a particle moves in a plane such that its distance from a fixed (or moving) point remains constant then its motion is called as the circular motion with respect to that fixed (or moving) point. That fixed point is called centre and the distance between fixed point and particle is called radius. v A v v 1 2 B The car is moving in a straight line with respect to the man A. But the man B continuously rotate his face to see the car. So with respect to man A d 0 dt d 0 dt Therefore we conclude that with respect to A the motion of car is straight line but for man B it has some angular velocity But with respect to man B 14.KINEMATICS OF CIRCULAR MOTION : 14.1 Variables of Motion : (a) Angular Position : The angle made by the position vector with given line (reference line) is called angular position Circular motion is a two dimensional motion or motion in a plane. Suppose a particle P is moving in a circle of radius r and centre O. The position of the particle P at a given instant may be described by the angle  between OP and OX. This angle  is called the angular position of the particle. As the particle moves on the circle its angular position  change. Suppose the point rotates an angle  in time t. (b) Angular Displacement : Definition : Angle rotated by a position vector of the moving particle in a given time interval with some reference line is called its angular displacement. ______________________________________________________________________________________________ Important point : • It is dimensionless and has proper unit SI unit radian while other units are degree or revolution 2 rad = 360° = 1 rev • Infinitely small angular displacement is a vector quantity but finite angular displacement is not because the addition of the small angular displacement is cummutative while for large is not.     d1 + d 2 = d 2 + d1 but 1 + 2  2 + 1 • Direction of small angular displacement is decided by right hand thumb rule. When the fingers are directed along the motion of the point then thumb will represents the direction of angular displacement. • Angular displacement can be different for different observers NEWTON LAW OF MOTION Page # 9 ______________________________________________________________________________________________ (c) Angular Velocity  (i) Average Angular Velocity Total Angle of Rotation ; Total time taken av =  2 – 1  av = t – t = t 2 1 where 1 and 2 are angular position of the particle at time t1 and t2 respectively. (ii) Instantaneous Angular Velocity The rate at which the position vector of a particle with respect to the centre rotates, is called as instantaneous angular velocity with respect to the centre. d lim  =  = t0 dt t _______________________________________________________________________________________________ Important points : • It is an axial vector with dimensions [T–1] and SI unit rad/s. • For a rigid body as all points will rotate through same angle in same time, angular velocity is a characteristic of the body as a whole, e.g., angular velocity of all points of earth about its own axis is (2/ 24) rad/hr. • If a body makes ‘n’ rotations in ‘t’ seconds then angular velocity in radian per second will be 2n av = t If T is the period and ‘f’ the frequency of uniform circular motion 2  1 = 2f T d • If  = a – bt + ct2 then  = = – b + 2ct dt Relation between speed and angular velocity : d   = lim = dt t t The rate of change of angular velocity is called the angular acceleration (). Thus, d d2 Y = = dt dt 2 av = P' The linear distance PP’ travelled by the particle in time t is s = r S lim or t0 t  = r lim t0  t s d or =r t dt P  O or v = r r X Here, v is the linear speed of the particle It is only valid for circular motion  v v = r is a scalar quantity (   ) r (d) Relative Angular Velocity Angular velocity is defined with respect to the point from which the position vector of the moving particle is drawn Here angular velocity of the particle w.r.t. ‘O’ and ‘A’ will be different P' A O P Ref lin NEWTON LAW OF MOTION  PO  Page # 10 d d ;  PA  dt dt Definition : Relative angular velocity of a particle ‘A’ with respect to the other moving particle ‘B’ is the angular velocity of the position vector of ‘A’ with respect to ‘B’. That means it is the rate at which position vector of ‘A’ with respect to ‘B’ rotates at that instant VA A VB r B  AB   ( VAB )  rAB here VAB   Relative velocity  to position vector AB Re lative velocity of A w.r.t. B perpendicu lar to line AB Seperation between A and B ( VAB )   VA sin 1  VB sin  2 rAB  r  AB  VA sin 1  VB sin  2 r  15.CIRCULAR MOTION IN HORIZONTAL PLANE A ball of mass m attached to a light and inextensible string rotates in a horizontal circle of radius r with an angular speed  about the vertical. If we draw the force diagram of the ball. We can easily see that the component of tension force along the centre gives the centripetal force and component of tension along vertical balances the gravitation force. Such a system is called a conical pendulum. 16.MOTION OF A MOTORCYCLIST ON A CURVED PATH.  T T cos  m 2r T sin mg FBD of ball w.r.t ground A cylist having mass m move with constant speed v on a curved path as shown in figure. B A D C E We divide the motion of cyclist in four parts : (1) from A to B (2) from B to C (3) from C to D (1 and 3 are same type of motion) (A) Motion of cyclist from A to B N  mv 2 = mg cos  R mv 2 N = mg cos – R f = mg sin  (4) from D to E N+ mv 2 R f B ...(1) ..(2) (1) As cyclist move upward   decreases & cos  increases  N increases and mgsin A  s co g m   mg NEWTON LAW OF MOTION Page # 11   decreases sin  decreases  friction force required to balance mg sin  (As cyclist is moving with constant speed) also decreases (B) Motion of cyclist from B to C N B 2 f mv  mg cos  R  N = mg cos – mv 2 R mv 2 R mgcos  ...(1) mgsin f = mg sin ...(2) Therefore from B to C Normal force decrease but friction force increase becuse  increases. (C) Motion of cyclist from D to E N N mg C D f mv 2  mg cos  f = mg sin  R mv 2 R from D to E  decreases therefore mg cos  increase So N increase but f decreases E mg 17.CIRCULAR TURNING ON ROADS : When vehicles go through turnings, they travel along a nearly circular arc. There must be some force which will produce the required centripetal acceleration. If the vehicles travel in a horizontal circular path, this resultant force is also horizontal. The necessary centripetal force is being provided to the vehicles by following three ways. 1. By Friction only 2. By Banking of Roads only 3. By Friction and Banking of Roads both. In real life the necessary centripetal force is provided by friction and banking of roads both. Now let us write equations of motion in each of the three cases separately and see what are the constant in each case. 17.1 By Friction Only Suppose a car of mass m is moving at a speed v in a horizontal circular arc of radius r. In this case, the necessary centripetal force to the car will be provided by force of friction f acting towards center mv 2 r Further, limiting value of f is N or fL = N = mg (N = mg) Thus, f  Therefore, for a safe turn without sliding mv 2  fL r v2 mv 2 or  or v  rg  mg rg r Here, two situations may arise. If  and r are known to us, the speed of the vehicle should not exceed or rg and if v and r are known to us, the coefficient of friction should be greater than v2 . rg 17.2 By Banking of Roads Only Friction is not always reliable at circular turns if high speeds and sharp turns are involved to avoid dependence on friction, the roads are banked at the turn so that the outer part of the road is some what N lifted compared to the inner part. Applying Newton’s second law along the radius and the first law in the vertical direction. Nsin = mv 2 r or N cos   mg  W NEWTON LAW OF MOTION Page # 12 from these two equations, we get tan = v2 rg or v= rgtan 17.3 By Friction and Banking of Road Both If a vehicle is moving on a circular road which is rough and banked also, then three forces may act on the vehicle, of these force, the first force, i.e., weight (mg) is fixed both in magnitude and direction. N   f y mg  x Figure (ii) The direction of second force i.e., normal reaction N is also fixed (perpendicular or road) while the direction of the third i.e., friction f can be either inwards or outwards while its magnitude can be varied upto a maximum limit (fL = N). So the magnitude of normal reaction N and directions plus magnitude of friction f are so adjusted mv 2 towards the center. Of these m and r are r also constant. Therefore, magnitude of N and directions plus magnitude of friction mainly depends on the speed of the vehicle v. Thus, situation varies from problem to problem. Even though we can see that : (i) Friction f will be outwards if the vehicle is at rest v = 0. Because in that case the component weight mg sin is balanced by f. (ii) Friction f will be inwards if that the resultant of the three forces mentioned above is v> rgtan (iii) Friction f will be outwards if v< rgtan and (iv) Friction f will be zero if v= rgtan (v) For maximum safe speed (figure (ii) mv 2 r N cos – f sin = mg As maximum value of friction f = N N sin + f cos =  ...(i) ...(ii) sin    cos  v2 = cos  –  sin  rg Similarly ; vmin = : •  vmax = rg(  tan ) (1 –  tan ) rg(  – tan ) (1  tan ) The expression tan  = v2 also gives the angle of banking for an aircraft, i.e., the angle through rg which it should tilt while negotiating a curve, to avoid deviation from the circular path. • The expression tan  = v2 also gives the angle at which a cyclist should lean inward, when rounding rg a corner. In this case,  is the angle which the cyclist must make with the vertical to negotiate a safe turn. NEWTON LAW OF MOTION Page # 13 18. DEATH WELL : A motor cyclist is driving in a horizontal circle on the inner surface of vertical cyclinder of radius R. Friction coefficient between tyres of motorcyclist and surface of cylinder is . Find out the minimum velocity for which the motorcyclist can do this. v is the speed of motor cyclist and m is his mass. N= mv 2 R f = mg fmax = f mv 2 R N mg mv 2 R Cyclist does not drop down when fmax  mg  mv 2  mg R v gR  19. MOTION OF A CYCLIST ON A CIRCULAR PATH : Suppose a cyclist is going at a speed v on a circular horizontal road of radius r which is not banked. Consider the cycle and the rider together as the system. The centre of mass C (figure shown) of the system is going in a circle with the centre at O and radius r.  Mv 2 r B C  N Mg  A f D (b) Let us choose O as the origin, OC as the X-axis and vertically upward as the Z-axis. This frame is rotating v at an angular speed  = about the Z-axis. In this frame the system is at rest. Since we are working from r a rotating frame of reference, we will have to apply a centrifugal force on each particle. The net centrifugal force on the system will be M2r = Mv2/r, where M is the total mass of the system. This force will act through the centre of mass. Since the system is at rest in this frame, no other pseudo force is needed. Figure in shows the forces. The cycle is bent at an angle  with the vertical. The forces are (i) weight Mg, (ii) normal force N (iii) friction f and (iv) centrifugal force Mv 2 r In the frame considered, the system is at rest. Thus, the total external force and the total external torque must be zero. Let us consider the torques of all the forces about the point A. The torques of N and f about A are zero because these forces pass through A. The torque of Mg about A is Mg(AD) in the clockwise direction and that of Mg(AD) = or, AD v 2  CD rg Mv 2 Mv 2 (CD) in the anticlockwise direction. For rotational equilibrium, is r r Mv 2 (CD) r or, tan = v2 rg  v2  Thus, the cyclist bends at an angle tan –1   with the vertical  rg  ...(10.9)