Lecture L9 - Linear Impulse and Momentum. Collisions

Transcription

1 J. Peraire, S. Widnall Dynaics Fall 009 Version.0 Lecture L9 - Linear Ipulse and Moentu. Collisions In this lecture, we will consider the equations that result fro integrating Newton s second law, F = a, in tie. This will lead to the principle of linear ipulse and oentu. This principle is very useful when solving probles in which we are interested in deterining the global effect of a force acting on a particle over a tie interval. Linear Moentu We consider the curvilinear otion of a particle of ass,, under the influence of a force F. Assuing that the ass does not change, we have fro Newton s second law, dv d F = a = = (v). dt dt The case where the ass of the particle changes with tie (e.g. a rocket) will be considered later on in this course. The linear oentu vector, L, is defined as L = v. Thus, an alternative for of Newton s second law is F = L, (1) which states that the total force acting on a particle is equal to the tie rate of change of its linear oentu. Principle of Linear Ipulse and Moentu Iagine now that the force considered acts on the particle between tie and tie t. Equation (1) can then be integrated in tie to obtain t F dt = t Here, L 1 = L( ) and L = L(t ). The ter t I = F dt = ΔL = (v) (v) 1, L dt = L L 1 = ΔL. () is called the linear ipulse. Thus, the linear ipulse on a particle is equal to the linear oentu change ΔL. In any applications, the focus is on an ipulse odeled as a large force acting over a sall tie. But 1

2 t in fact, this restriction is unnecessary. All that is required is to be able to perfor the integral F dt. If the force is a constant F, then ΔL = t F dt = F (t ). If the force is given as a function of tie, then t ΔL = t F (t) dt 1 Note Units of Ipulse and Moentu It is obvious that linear ipulse and oentu have the sae units. In the SI syste they are N s or kg /s, whereas in the English syste they are lb s, or slug ft/s. Exaple (MK) Average Drag Force The pilot of a 90, 000-lb airplane which is originally flying horizontally at a speed of 400 ph cuts off all engine power and enters a glide path as shown where β = 5 o. After 10 s, the airspeed of the plane is 360 ph. We want to calculate the agnitude of the tie-averaged drag force. Aligning the x-axis with the flight path, we can write the x coponent of equation () as follows 10 (W sin β D) dt = L x (10) L x (0). The tie-averaged value of the drag force, D, is D = D dt Therefore, (W sin β D )10 = (v x (10) v x (0)). Substituting and applying the appropriate unit conversion factors we obtain, 90, (90, 000 sin 5 o D )10 = 3. ( ) 3600 D = 9, 10 lb.

3 Ipulsive Forces We typically think of ipulsive forces as being forces of very large agnitude that act over a very sall interval of tie, but cause a significant change in the oentu. Exaples of ipulsive forces are those generated when a ball is hit by a tennis racquet or a baseball bat, or when a steel ball bounces on a steel plate. The table below shows typical tie intervals over which soe of these ipulses occur. Tie interval Δt [s] Racquet hitting a tennis ball Bat hitting a baseball Golf club hitting a golf ball Pile Driver Shotgun Steel ball bouncing on steel plate Exaple (MK) Baseball Bat A baseball is traveling with a horizontal velocity of 85 ph just before ipact with the bat. Just after the 1 ipact, the velocity of the 5 8 oz. ball is 130 ph, at an angle of 35 o above the horizontal. We want to deterine the horizontal and vertical coponents of the average force exerted by the bat on the baseball during the 0.0 s ipact. If we consider the tie interval between the instant the baseball hits the bat,, and the instant after it leaves the bat, t, the forces responsible for changing the baseball s oentu are gravity and the contact forces exerted by the bat. We can use the x and y coponents of equation () to deterine the average force of contact. First, consider the x coponent, t v x1 + F x dt = v x. Inserting nubers and using appropriate conversion factors, we obtain 5.15/ (85 ) / F (0.0) = (130 cos 35 o x ) F x = lb

4 For the y direction, we have which gives, t v y1 + F y dt = v y, 0 + F y(0.0) 5.15 (0.0) = 5.15/16 ( sin 35 o ) F y = 54.7 lb We note that gδt = lb-s, whereas F y(δt) = lb - s. Thus, gδt is 0.59% of the total ipulse. We could have safely neglected it. In this case we would have obtained F y = 54.4 lb. Conservation of Linear Moentu We see fro equation (1) that if the resultant force on a particle is zero during an interval of tie, then its linear oentu L ust reain constant. Since equation (1) is a vector quantity, we can have situations in which only soe coponents of the resultant force are zero. For instance, in Cartesian coordinates, if the resultant force has a non-zero coponent in the y direction only, then the x and z coponents of the linear oentu will be conserved since the force coponents in x and z are zero. Consider now two particles, 1 and, which interact during an interval of tie. Assue that interaction forces between the are the only unbalanced forces on the particles. Let F be the interaction force that particle exerts on particle 1. Then, according to Newton s third law, the interaction force that particle 1 exerts on particle will be F. Using expression (), we will have that ΔL 1 = ΔL, or ΔL = ΔL 1 + ΔL = 0. That is, the changes of oentu of particles 1 and are equal in agnitude and opposite in sign, and the total oentu change equals zero. Recall that this is true if the only unbalanced forces on the particles are the interaction forces. The ore general situation in which external forces can be present will be considered in future lectures. We note that the above arguent is also valid in a coponentwise sense. That is, when two particles interact and there are no external unbalanced forces along a given direction, then the total oentu change along that direction ust be zero. Exaple Ballistic Pendulu The ballistic pendulu is used to easure the velocity of a projectile by observing the axiu angle θ ax to which the box of sand with the ebedded projectile swings. Find an expression that relates the initial velocity of a projectile v 0 of ass to the axiu angle θ ax reached by the pendulu. The ass of the sand box is M and the length of the pendulu is L. 4

5 We consider the equation of conservation of linear oentu along the horizontal direction. The initial oentu of the projectile is v 0. Since the sand box is initially at rest its oentu is zero. Just after the projectile penetrates into the box, the velocity of the sand box and the projectile are the sae. Therefore, if v 1 is the velocity of the sand box (with the ebedded projectile) just after ipact, we have fro conservation of oentu, v 0 = (M + )v 1. After ipact, the proble reduces to that of a siple pendulu. The only force doing any work is gravity and therefore we can apply the principle of conservation of work and energy. At the point when θ is axiu, the velocity will be zero. Fro energy conservation we finally obtain, 1 (M + )v 1 = (M + )gh ax, and since h = L(1 cos θ), we have θ ax = cos 1 v 1 0. M + Lg Note that energy is not conserved in the collision. The initial kinetic energy of the syste is the kinetic energy of the projectile T 0 = 1 v 0 (taking the reference height as zero). After the collision the kinetic energy of the syste is Since v 1 = v 0 (+M) we have which is less than T 0 for M > 0. 1 T 1 = ( + M)v 1. (3) 1 T 1 = (M + ) v 0 (4) Collisions We consider now, the situation of two isolated particles colliding. The only forces on the particles are due to their utual interaction. When the velocity vector of the two particles is parallel to the line joining the two particles, we say that we have one-diensional collisions, otherwise we say that the collision is oblique. 5

6 1D Collisions Here, we are dealing with rectilinear otion and therefore the velocity vector becoes a scalar quantity. In order to understand the collision process we consider five different stages. I) Before Ipact Let particle 1, of ass 1, occupy position x 1 and travel with velocity v 1 along the direction parallel to the line joining the two particles. And let particle, of ass, occupy position x and travel with speed v also in the sae direction. We assue that v 1 > v so that collision will occur. We can introduce the position of the center of ass x G = ( 1 x 1 + x )/, where = 1 +. The velocity of the center of ass is then given by v G = ( 1 v 1 + v )/. Note that since conservation of oentu requires ( 1 v 1 + v ) = ( 1 v 1 + v ), the velocity of the center of ass reains unchanged by the collision process between the particles. If we define the relative velocity g = v 1 v we can express v 1 and v as a function of v G and g as, II) Deforation v 1 = v G + g v = v G 1 g. The particles establish contact and the force between the F d increases until the instant of axiu deforation. III) Maxiu Deforation The contact force is at its axiu and the two particles travel at the sae velocity v G. Thus the deforation force has slowed 1 down to a velocity of v G and sped up to a velocity of v G. The ipulse applied to particle 1 and fro the deforation force equals the change in oentu in the deforation process. 6

7 For particle 1 F d dt = 1 v G 1 v 1 (deforation phase) and for particle F d dt = v G v (deforation phase) IV) Restoration The contact force F r decreases and the particles ove apart. The ipulse applied to particle 1 and fro the restoring force equals the change in oentu in the restoration process. After the restoration process the velocity of 1 is v 1 and the velocity of is v. For particle 1 F r dt = 1 v 1 1 v G (restoration phase) and for particle F r dt = v v G (restoration phase) V) After Ipact The particles travel with a constant velocity v 1 and v. 7

8 Fro oentu conservation, the total oentu before and after ipact should reain the sae, i.e. 1 v 1 + v = 1 v 1 + v, and therefore the velocity of the center of ass will reain unchanged v = v G, thus v G = 1 v + v. If g = v G 1 1 v is the relative velocity of the two particles after ipact, then v 1 v = v G + g (5) = v G 1 g. (6) Coefficient of Restitution We define the coefficient of restitution as the ratio between the restoration and deforation ipulses. F r dt e = F d dt For physically acceptable collisions 0 < e < 1. The value of e = 1 corresponds to an elastic collision, whereas the value of e = 0 corresponds to a totally inelastic collision in which the restoration ipulse is equal to zero. We can consider each particle separately and set the ipulse on the particle equal to the change of linear oentu Since for particle 1 F d dt = 1 v G 1 v 1 (deforation phase) F r dt = 1 v 1 1 v G (restoration phase) Thus, e = (v 1 v G )/(v G v 1 ), or (e + 1)v G = v + ev 1. 1 and for particle F d dt = v G v (deforation phase) F r dt = v v G (restoration phase) Thus, e = (v v G )/(v G v ), or (e + 1)v G = v + ev. Cobining the above expressions, we have F r dt v 1 v g e = = =, v1 v F d dt g which expresses the coefficient of restitution as the ratio of the relative velocities after and before the collision. Finally, if we know e we can use equations (5) and (6) to obtain 8

9 v 1 v = v G eg = v G + 1 eg. Exaple Siple Collisions As an exaple, we consider a siple case of a ass at rest and a ass of 1 traveling at velocity v 1 headed for a collision with. After the collision, the velocities of the two asses will in general be non-aero. We consider the cases e = 1,.5, and0, the is a range of coefficient of restitution between 1, perfectly elastic, to zero, perfectly inelastic. The result for the final velocities of 1 and, noralized by the initial velocity v 1, are given by the equations above and shown in the figure. For e = 0, these is only one curve since the particles stick together: v 1 /v 1 = v /v 1. For the elastic collision, note the result for equal asses: v 1 = 0 and v /v 1 = 1 9

10 Energy We can now look at the kinetic energy before and after the ipact. The kinetic energy before ipact is, 1 1 T = 1 v 1 + v = 1 1 (v + gv G + g ) + 1 (v 1 gv G + 1 G g ) G = v G + (v 1 v ) = v G + µg. Here we introduce the notation µ = ( 1 )/; this is often referred to as the reduced ass of the pair. The reduced ass approaches 1/ of each ass when the two asses are siilar, but it approaches the lighter ass if the asses are dissiilar. The kinetic energy after ipact will be T = 1 v G (v 1 v ) = v G + g = 1 1 v G + µe g. 1 ) since g = eg and = µ. Therefore, the kinetic energy lost during the collision is ΔE = T T = 1 µ(1 e )g, which is zero for e = 1 (elastic collision), and axiu when e = 0. Moentu exchange We can also look at the oentu which is exchanged between particle 1 and particle, M 1 = 1 v 1 1 v 1 = 1 (v G + g v G + eg) = (1 + e)µg. This, is axiu when e = 1 (elastic collision) and iniu when e = 0. It is particularly illuinating to exaine a special case of collisions, that for which the two asses are equal. We consider two cases shown in the figure. In both cases one particle is at rest, the other approaching with velocity v 1. In one case, we assue an elastic collision, e = 1; in the other a copletely inelastic collision, 10

11 e = 0. In the first case, kinetic energy is conserved so it is obvious although you can work out the algebra that the first particle copletely transfers its oentu to the second particle, where upon it continues to the right with the original velocity v = v 1. In the second case of a copletely inelastic collision, there is no rebounding force and the particles stick and ove together. Conservation of oentu requires the v = v 1 /. Kinetic energy is not conserved. Exaple Bouncing Ball If a ball, initially at rest, is released on a flat surface fro a height h and it rebounds to a height h, the coefficient of restitution is given by v h e = =, v h where v and v are the velocities before and after ipact. Oblique Collisions In the case of oblique collisions, we consider the instant of ipact and define the noral direction, n, along the line that connects the two ass centers, and the tangential direction, t, along the line tangent to the surfaces at the point of contact. We write the equations of conservation of oentu in the tangent and noral directions. Since the force of contact is assued to act along the noral direction, the conservation of linear oentu along the tangential direction iplies 1 (v 1 ) t = 1 (v 1 ) t (v ) t = (v ) t (v 1 ) t = (v 1 ) t (v ) t = (v ) t In the noral direction, we solve a 1D collision proble, that is, which deterine (v 1 ) n and (v ) n. 1 (v 1 ) n + (v ) n = 1 (v 1 ) n + (v ) n (v 1 ) n (v ) n e = (v1 ) n (v ) n 11

12 ADDITIONAL READING J.L. Meria and L.G. Kraige, Engineering Mechanics, DYNAMICS, 5th Edition 3/8, 3/9 1

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