Mechanics Topic B (Momentum) - 1 David Apsley

Transcription

1 TOPIC B: MOMENTUM SPRING Newton s laws of motion 2. Equivalent forms of the equation of motion 2.1 orce, impulse and energy 2.2 Derivation of the equations of motion for particles 2.3 Examples 3. Conservation of momentum 3.1 Conservation of momentum in collisions 3.2 Restitution 3.3 Oblique collisions 4. riction 4.1 Types of friction 4.2 Coefficient of friction 4.3 Object on a plane 5. Systems of particles and finite objects 5.1 Dynamics of a system 5.2 The centre of mass 5.3 Important properties of the centre of mass 5.4 inding the centre of mass 5.5 Some useful centres of mass Mechanics Topic B (Momentum) - 1 David Apsley

2 1. NEWTON S LAWS O MOTION or a particle, classical dynamics is governed by Newton s Laws of Motion: Law 1. Every particle continues in a state of rest or uniform motion in a straight line unless acted on by a resultant external force. Law 2. When acted upon by an unbalanced force a particle of mass m accelerates at a rate a such that = ma. 1 Law 3. The forces of action and reaction between interacting particles are equal in magnitude, opposite in direction and collinear. Notes (i) The laws strictly apply to particles. They may be extended by summation to rigid bodies if there is no rotation (so that each component moves with the same velocity). (ii) Law 1 requires the concept of inertial reference frames (those that aren t accelerating relative to each other). or example, an object perceived as travelling in a straight line by a stationary observer will appear to be travelling in a curved path by somebody in a rotating reference frame. u In inertial frame In rotating frame (iii) Law 3 is sometimes referred to as equal and opposite reactions. The form above is appropriate for a purely mechanical system, where the objects concerned are in direct contact, but without care it can at least appear to be violated in electrodynamics. (or example, the magnetic forces between two non-parallel current-carrying wires, or between moving charged particles, are not opposite in direction.) However, its corollary for particles in contact which in fact holds in general is that total momentum is conserved. (In the case cited this is because electromagnetic fields can be regarded as having momentum, and the wires or moving charged particles are not in direct contact but interact via the intermediate electromagnetic field.) I I I I (=BIL) 1 Newton s version was strictly = d(mv)/dt or force = rate of change of momentum, since what he described as the quantity of motion was mass velocity, what we would today call momentum. Mechanics Topic B (Momentum) - 2 David Apsley

3 2. EQUIVALENT ORMS O THE EQUATION O MOTION 2.1 orce, Impulse and Energy The Second Law is usually what is referred to as the equation of motion. Initially formulated for particles, by summation it can be applied to whole systems: for example, in continuum mechanics (e.g. fluid mechanics) and in cases of non-constant mass (e.g. rockets). The Second Law can be written in a number of forms, notably: orce-momentum: force = rate of change of momentum Impulse (force time): impulse = change of momentum Energy (force distance): work done = change of kinetic energy Other forms include the variational principles of Lagrange and Hamilton. These can be generalised to include other physics, but are too mathematically complex to consider here. 2.2 Derivation of the Equations of Motion or Particles Newton s Second Law, as we usually write it today, for individual particles is: force = mass acceleration (1) This gives the instantaneous relationship between the net force acting on a body and the resulting acceleration; if integrated it gives displacement or velocity as a function of time. orce-momentum If m is constant and dv/dt = a: force = rate of change of momentum (2) This is closer to what Newton wrote in words in Principia Mathematica. Impulse-Momentum Integrating (2) w.r.t. time: Mechanics Topic B (Momentum) - 3 David Apsley

4 impulse = change of momentum (3) This gives the total change in velocity over a period of time. Work-Energy Integrate (2) w.r.t. displacement: Since this can be written work done = change of kinetic energy (4) This gives the total change in kinetic energy (and hence velocity) in a displacement. Advantages of the energy form (to be considered in detail in Topic C) are: it is a scalar not vector equation; it relates start and end states without having to consider intermediate values; it gives velocities directly, without having to integrate acceleration; many actual forces (e.g. elasticity) depend on position, not time; the work done by important conservative forces (e.g. gravity, elasticity) can be conveniently written in terms of changes in potential energy. All forms of the equation of motion are equivalent and can be derived from each other. They can also, by summation, be extended to whole systems, rather than individual particles (see Section 5 below). Mechanics Topic B (Momentum) - 4 David Apsley

5 2.3 Examples The following examples apply = ma to multi-body systems. To solve these it is important to establish (i) how much each part moves; (ii) the forces between them. Note that this often involves exploding the diagrams: for an individual part the effect of other parts is mediated by the forces between them. Example 1. Two boxes with masses 25 kg and 35 kg are placed on a smooth frictionless surface. The boxes are in contact and a force of 30 N pushes horizontally on the smaller box. What is the force that the smaller box exerts on the larger box? 30 N 25 kg 35 kg Example 2. Masses are connected together by light, inelastic cables passing round light, smooth pulleys in different configurations, as shown below. ind the direction and magnitude of the acceleration of each mass, and the tension in each cable, for each configuration. (Give your answers in kg-m-s units, as multiples of g.) 30 kg 20 kg 30 kg 20 kg (a) (b) Mechanics Topic B (Momentum) - 5 David Apsley

6 3. CONSERVATION O MOMENTUM 3.1 Conservation of Momentum in Collisions 2 When two bodies A and B collide: the force exerted by A on B is equal and opposite to the force exerted by B on A; the change in momentum of B is equal and opposite to the change in momentum of A; the total momentum is conserved. B u B v B A u A v A Before After or the system shown, with velocities u A and u B before, and velocities v A and v B after collision, conservation of momentum gives: (5) Example 3. A vehicle A of mass 500 kg travelling at 30 m s 1 collides with a vehicle B of mass 900 kg travelling at 21 m s 1 along the same line. After the collision the vehicles remain stuck together for a short distance. What is the speed and direction of the combined assembly: (a) if the vehicles are originally travelling in the same direction? (b) if the vehicles are originally travelling in opposite directions? Momentum is a vector quantity. If there are different directions involved then conservation of momentum must be applied in each direction. If only the velocities before collision are known, this is insufficient to determine the postcollision state. urther conditions are needed: restitution and, in more than one dimension, unchanged individual momentum components in directions perpendicular to the line of approach. 2 Strictly, interactions rather than collisions there is no need for the objects actually to touch. An example is the interaction between charged particles. Mechanics Topic B (Momentum) - 6 David Apsley

7 3.2 Restitution When objects collide they are slightly compressed. Kinetic energy is given up and stored as elastic potential energy, rather like a (very stiff) spring. A B In the decompression phase this elastic energy is released as kinetic energy as the objects move apart. Because of internal friction some energy may be dissipated in both compression and decompression and the amount of kinetic energy after the collision is less than that before. This is usually modelled using a coefficient of restitution e such that: (6) In all circumstances where no external forces act:. or a perfectly elastic collision: e = 1 or a perfectly inelastic collision (i.e. objects stick together): e = 0 Before After u A A v A u B B v B or the interacting particles shown above, which are confined to travel along a straight line, momentum conservation gives whilst restitution gives (total momentum before = total momentum after) (velocity of separation = e velocity of approach) If two of the velocities are known, then these give two equations for the remaining two unknowns. Note that the signs of v A, v B etc. are important and must be consistent. Example 4. A ball of mass 0.6 kg and speed 12 m s 1 collides with a second ball of mass 0.2 kg moving in the opposite direction with speed 18 m s 1. Assuming that all motion takes place along a straight line, what is the outcome of the collision if the coefficient of restitution e is: (a) 0.5 (b) 0.8 Mechanics Topic B (Momentum) - 7 David Apsley

8 3.3 Oblique Collisions or oblique collisions (which involve all coordinate directions): restitution may be applied perpendicular to the contacting surfaces (e.g. along the line of centres for colliding spherical bodies); defining this direction as x then: total momentum (but not individual momenta) conserved in all directions, including perpendicular to the contacting surfaces: individual momentum components (and hence velocity components) unchanged in the directions parallel to the contacting surfaces (because no forces act on either object in these directions provided the surfaces are smooth). (7) (8) (9) A B A B Before x After The figure above shows an oblique collision (as seen from the initial rest frame of object B). Interaction forces act only in the x direction, so only the velocity components in this direction are changed. Example 5. A ball is dropped vertically a distance of 5 m onto a plane surface sloping at 35 to the horizontal. How far down the slope is its second bounce if the coefficient of restitution e is: (a) 1.0; (b) m 35 o Mechanics Topic B (Momentum) - 8 David Apsley

9 4. RICTION riction forces are tangential forces, generated between surfaces in contact, that oppose relative motion. riction may be sufficient to prevent relative motion, but where sliding does occur friction forces do negative work and energy is lost from the system in the form of heat. riction forces are present between all real surfaces, but are often small enough to be neglected. In the latter case, the model system or process is termed ideal. Many mechanical devices would not work without friction; examples include conveyor belts, belt drives, brake and clutch plates, screws. 4.1 Types of riction There are three types of friction forces. (1) Dry friction encountered when the unlubricated surfaces of two solids are in contact under a condition of sliding or tendency to slide. (2) luid friction developed when adjacent layers of a fluid (liquid or gas) are moving at different velocities. (3) Internal friction found in all solid materials subject to deformation. luid friction, arising from intermolecular forces of attraction, gives rise to forces between fluid layers where there are velocity gradients (particularly near solid boundaries). Adjacent layers exert equal and opposite stresses (force per unit area) τ, given by: (stress = viscosity velocity gradient) y U(y) μ is called the viscosity. luid friction was discussed in your Hydraulics course. This Mechanics unit will focus on friction between dry, solid surfaces. 4.2 Coefficient of riction riction between solid surfaces in contact arises from surface irregularities. It is: tangential to the surfaces in contact; opposed to the direction of relative motion or tendency to motion; dependent on the nature of the surface and on the normal reaction force R. When forces are applied with a tendency to cause relative motion (i.e. sliding), friction forces arise, of magnitude precisely that needed to prevent motion... up to a maximum: (10) beyond which sliding occurs. R is the normal (i.e. perpendicular to the plane of contact) reaction force. It is sometimes also called the R Mechanics Topic B (Momentum) - 9 David Apsley

10 contact force. μ s is the coefficient of static friction. Up to the point of sliding, Once relative motion occurs: (11) (12) R P max R s R k mg P Sometimes we make a distinction between: coefficient of static friction, μ s, which applies up to the point of sliding, and coefficient of kinetic friction, μ k, which applies once there is relative motion. riction tends to be slightly smaller once motion has started. Both μ s and μ k depend on the surfaces. Typical values (from Meriam and Kraige) are given below. Contacting surfaces μ s μ k Steel on steel Teflon on steel Brass on steel Brake lining on cast iron Rubber tyres on smooth pavement Metal on ice In most instances, however, we shall assume that μ k = μ s, and if you see μ without a subscript given as the coefficient of friction you may assume that this is the case. Important! The direction of the friction force is easily established by asking yourself in what direction an object would move under the other forces (the motive force ) if there were no friction. The friction force always tries to oppose relative motion. The magnitude of the friction force is given by μr only if the motive force is sufficient to make it move. If the motive force is less than max, then friction is simply what is necessary for equilibrium; i.e. equal to the motive force (see the diagram above). Example 6. (a) A block of mass 5 kg can be pushed up a slope of 1 in 3 at constant speed by a force of 45 N applied horizontally. Deduce the coefficient of friction, μ. (b) If, instead, the 45 N force is applied parallel to the slope, what is the block s acceleration up the slope? Mechanics Topic B (Momentum) - 10 David Apsley

11 4.3 Object on a Plane If an object isn t treated as a particle then there are actually three things that it might do: topple over; stay still; slide. The first must be ruled out first by determining whether the line of action of the weight (through the centre of gravity) lies outside the base of the object, so giving a turning moment about the leading edge. Provided that the object doesn t topple then determine whether motion occurs by comparing the maximum friction force with the motive force. Consider the forces on a object on an inclined plane. (Note the line of action of the normal reaction R). R( ) (normal to plane): R Maximum friction force: mg There is no sliding if the downslope component of weight can be balanced by friction; i.e. if: (13) (If you don t like inequalities, look at the limiting case where friction is just sufficient to prevent motion; at this point tan θ = μ s.) If the angle of inclination is gradually increased, the block just starts to slide at angle θ such that tan θ = μ s, dependent on the nature of the contacting surfaces, but independent of the mass of the block. This provides a means of measuring the coefficient of static friction. Similarly, if the angle of inclination is slowly decreased and, at each angle, the block is given a nudge to start it moving, then the block just stops sliding when tan θ = μ k. This provides a means of measuring the coefficient of kinetic friction. Mechanics Topic B (Momentum) - 11 David Apsley

12 Example 7. A crate of mass 30 kg is propelled along a horizontal surface at steady speed by a horizontal force of magnitude P = 120 N. (a) What is the coefficient of friction between crate and ground? or the same coefficient of friction, what minimum force must be applied to move the crate if it is directed: (b) at 25 below the horizontal; (c) at 25 above the horizontal. P 30 kg P P 25 o 25 o 30 kg 30 kg (a) (b) (c) Example 8. Exam 2014) Two cars, A of mass 800 kg and B of mass 1200 kg, collide at a 90 crossroads as shown. The collision is completely inelastic and after the collision the cars continue to move as a single body, sliding a distance 18 m at angle 30º to the original direction of A before stopping. (a) (b) If car A was travelling at 30 m s 1 before the collision, find the speed of car B before the collision and the speed of the combined vehicles after the collision. Assuming that all car wheels lock at the point of impact, find the total coefficient of kinetic friction between the tyres and the ground. A 800 kg 30 m/s 18 m 30 o B 1200 kg Mechanics Topic B (Momentum) - 12 David Apsley

13 5. SYSTEMS O PARTICLES AND INITE OBJECTS Newton s laws are formally applicable to individual point particles, which have infinitesimal size. By summation they can be applied to systems of particles: for example, finite objects. or a system of particles the internal reaction forces between particles are equal in size, opposite in direction and collinear, so make no contribution to the overall dynamics (i.e. exert no net force or moment), although they do hold it together! As far as the total momentum of the system is concerned, we need consider only external forces. 5.1 Dynamics of a System Consider a system of particles of masses m 1, m 2, m 3, etc. or any individual particle: m 1 m 4 m 3 m 2 m 5 where f i is the resultant of all (external and internal) forces on the ith particle. Summing over all particles, and noting that each m i is constant: The LHS is just the sum of all external forces (since internal forces cancel in pairs), which we may write. Hence, (14) where we have defined the centre of mass as that point where a single particle of the same total mass will have the same moment of mass: (15) Since M is constant, (16) The centre of mass moves like a single particle of the same total mass M under the resultant of the external forces. We write or V for (loosely, the velocity of the centre of mass ). Mechanics Topic B (Momentum) - 13 David Apsley

14 5.2 The Centre of Mass The centre of mass is the average position of mass. In a uniform gravitational field it coincides with the centre of gravity (the point through which the weight of a system appears to act). or a uniform-density material it coincides with the centroid or geometric centre of a body: the centre of volume (3-d), area (2-d) or length (1-d). Moments A moment of any quantity (not just a force) about a point is given by moment = quantity distance (17) The average position,, of some quantity (mass, volume, area, ) is that position which, if you placed the total quantity there, would yield the same overall moment as the original system; e.g. for mass: where M ( ) is the total mass. Hence, the position of the centre of mass is (18) Notes (1) The individual coordinates of the centre of mass are: (2) The net moment about the centre of mass is zero, since (Conversely, this is the only point about which this is true.) Mechanics Topic B (Momentum) - 14 David Apsley

15 5.3 Important Properties of the Centre of Mass Momentum (as proved in Section 5.1 above) The centre of mass moves like a single particle of mass M under the resultant of the external forces. Energy (Topic C) In uniform gravity, the total gravitational potential energy is the same as a single particle at the centre of mass. The total kinetic energy of a system is the sum of: the kinetic energy of the centre of mass ( ); plus the kinetic energy relative to the centre of mass ( ) Rotation (Topic D) or a rigid body, motion relative to the centre of mass is pure rotation (Topic D): KE 1 MV 2 2 KE MV I 2 2 V V ω 2 The total angular momentum of a system is the sum of: the angular momentum of the centre of mass; plus the angular momentum relative to the centre of mass. The equation of rotational motion ( torque = rate of change of angular momentum ) holds for the resultant of all external torques about a point which is either: fixed; or moving with the centre of mass. Mechanics Topic B (Momentum) - 15 David Apsley

16 5.4 inding the Centre of Mass (1) irst Principles Use or individual particles this can be done by direct summation. or continuous bodies integration may be necessary. (2) Centre of Area / Centre of Volume ( Centroid ) If the density of an object is constant then mass is proportional to area (2-d lamina) or to volume (3-d body). Thus, (2-d lamina) (3-d body) (3) Use of Symmetry If each contribution from +x has an equal contribution from x then the net moment is zero. Consequently the centre of mass must lie on any plane of symmetry. G (4) Addition or Subtraction of Simpler Elements Often a complex body can be broken down into simple elements (e.g. rectangles or cuboids), each of whose individual centres of mass can be found easily (e.g. by symmetry). Replace each element by an equivalent particle of the same mass at its own centre of mass. Then proceed to add or subtract moments and masses and apply first principles as in Method 1. m 2 m 1 m 2 m 1 add subtract In order for any subtracted mass, area or volume to be correctly included in a formula involving a summation (Σ), it is common for that quantity to be given a negative sign. Mechanics Topic B (Momentum) - 16 David Apsley

17 Example 9. (Method 1: first principles; direct summation) Three particles of mass 3, 3 and 4 units are at rest on a smooth horizontal plane at points with position vectors i+4j, 5i+10j and 3i+2j respectively. ind the position of their centre of mass. Example 10. (Method 4: addition of elements) ind the position of the centre of mass of the L-shaped lamina shown Example 11. (Method 4: subtraction of elements) A uniform lamina is formed by cutting quarter-circles of radius 0.5 m and 0.2 m from the topright and bottom-left corners respectively of a rectangle whose original dimensions were 0.6 m by 0.8 m (see figure). (a) ind the area of the lamina. (b) ind the position of the centre of mass relative to the top-left corner, A. (c) If the lamina is allowed to pivot freely in a vertical plane about corner A, find the angle made by side AB with the vertical when hanging in equilibrium. (The centre of mass of a quarter-circular lamina of radius R is from either straight side.) A 0.5 m 0.8 m B 0.2 m 0.6 m Mechanics Topic B (Momentum) - 17 David Apsley

18 Example 12. (Method 1: first principles; integration) ind the position of the centre of mass of a uniform solid hemisphere of radius R. 5.5 Some Useful Centres of Mass Triangle (2-d): Pyramid (any base; 3-d): y h Semi-circular arc (1-d): Semi-circular area (2-d): Hemispherical shell: Hemispherical solid (3-d): y R Mechanics Topic B (Momentum) - 18 David Apsley

19 Numerical Answers to Examples in the Text ull worked answers are given in a separate document online. Example N Example 2. (a) 30 kg mass: acceleration upward; tension 20 kg mass: acceleration downward; tension (b) 30 kg mass: acceleration 20 kg mass: acceleration upward downward; tension Example 3. (a) 24.2 m s 1 in the direction that both vehicles were travelling; (b) 2.79 m s 1 in the original direction of vehicle B. Example 4. With the positive direction that of the initial motion of the 0.6 kg ball: (a) v 1 = 0.75 m s 1 ; v 2 = m s 1 (b) v 1 = 1.5 m s 1 ; v 2 = 22.5 m s 1 Example 5. (a) 22.9 m; (b) 8.61 m Example 6. (a) 0.447; (b) 1.74 m s 2 Example 7. (a) 0.408; (b) 163 N; (c) 111 N Example 8. (a) 11.5 m s 1 and 13.9 m s 1 ; (b) Example 9. 3i+5j Example 10. measured from the bottom left corner Example 11. (a) m 2 ; (b) (across, down) = (0.258, 0.507) m; (c) 27.0 Example 12. On the axis of symmetry, a distance from the centre of the base Mechanics Topic B (Momentum) - 19 David Apsley