WORKBOOK COMPUTER SIMULATIONS OF PHYSICAL PROCESSES IN MECHANICS LUBLIN 14
Author: Jarosław Borc, Wiesław Polak Desktop publishing: Paweł Droździel Technical editor: Paweł Droździel Figures: Jarosław Borc, Wiesław Polak Coer and graphic design; Paweł Droździel All rights resered. No part of this publication may be scanned, photocopied, copied or distributed in any form, electronic, mechanical, photocopying, recording or otherwise, including the placing or distributing in digital form on the Internet or in local area networks, without the prior written permission of the copyright owner. Publikacja współfinansowana ze środków Unii Europejskiej w ramach Europejskiego Funduszu w ramach projektu Inżynier z gwarancją jakości dostosowanie oferty Politechniki Lubelskiej do wymagań europejskiego rynku pracy Copyright by Jarosław Borc, Wiesław Polak, Lublin Uniersity of Technology Lublin 14 First edition
TABLE OF CONTENTS Eercise I Simulation of an projectile motion in the field of graity 7 Eercise II Simulation of body collision 13 Eercise III Eercise IV Eercise V Eercise VI Eercise VII Simulation of object motion with friction Simulation of oscillating motion Simulation of the motion in the force field Statics and dynamics of a many-body system Simulation of the motion of a rocket Bibliography
Eercise I. Simulation of an projectile motion in the field of graity Theory Projectile motion concerns an object starting from a certain initial position with an initial elocity of horizontally directed at angle α. We assume that the constant force of graity F g = mg, directed ertically downwards, where m is mass of the object, and g = 9.81m/s is graitational acceleration acts upon the object. Initial elocity has two components: horizontal o and ertical oy, while acceleration has only one ertical component with the alue of g. Figure 1.1. shows that o = o cosα, while oy = o sinα. Horizontal component of the object elocity during motion has constant alue of o cosα, while the ertical one changes as in the retarded motion: y = oy gt (1.1) Dependencies of the horizontal coordinate and the ertical y coordinate of the object on time t are defined by the equations: (uniform motion a = ): ( = o t (1.a) (uniformly accelerated motion where a y = g): y( = oy t gt / (1.b) The aboe equations can be used for isualization of motion by calculating and y coordinates with the gien time t and for marking the object s position in the XY coordinate system. The trajectory is a parabola of a following equation: g y ( ) = + tgα (1.3) cos α Fig. 1.1 The object s motion path in projectile motion.
The aboe equations proide: (a) Horizontal range (substituting y() = in equation 1.3): (b) Flight time (substituting y = in equation 1.1): sin α =, (1.4a) g sin α t ma = g, (1.4b) sin α (c) Maimum height (we use equation 1.b y(t ma /)): yma =. (1.4c) g Special case of projectile motion is ertical throw, where projection angle is α = 9. For this case equations 1.4b and 1.4c proide the alue of maimum height = g. tma yma = and flight time g We will get horizontal throw from height h by substituting oy =, o = o, α = into the equations (1.a and b) and adding initial height h to y coordinate. Dependencies of the horizontal coordinate and the ertical y coordinate on time t will be defined by the equations: : uniform motion: ( = o t y: uniformly accelerated and decelerated motion (a =, a y = g): y( = h gt / (a) Horizontal range (we substitute y( = ): (b) Flight time (we substitute y( = ): t = h =. g h g In simulations of projectile motion without friction we usually establish the starting point as = and y = and then we use numerical procedure for equations of motion for eample Euler s method: ( t + = ( t + = y ( t + = ( + y( t + = y( + y g t, t, y t. Projectile motion with resistance happens when the motion of an object takes place in gaseous or liquid medium. The substance of the medium flows around an object, depending on the size and elocity of the object, in the form of: (a) layered/laminar flow for low elocities or small size, and (b) (1.5)
turbulent flow for higher alues. Relatiely big objects from our surrounding such as human body, a bike and a car in the air or a paddle or a boat in the water, een moing relatiely slowly, cause eddies. Producing eddies in the medium inoles losing energy by the moing object and that is why the motion of an object will be slowed down. The drag force of a medium eerted on a moing object is opposite to the object elocity ector and its alue is gien by the formula: C F ( ) = o ρ S, (1.6) where - object elocity, S - cross section area of an object perpendicular to the elocity, ρ - density of a medium and C - drag coefficient dependent on an object shape. When an object of mass m moes in a straight line and the only force acting upon it is the force of resistance of a medium, the object acceleration is a = F o /m and it is directed opposite to the elocity. Its alue is: CρS a( =. (1.7) m Fig. 1.. Drag force F and force of graity mg acting upon the object of mass m thrown into the air with elocity at a certain angle to the direction of graitational acceleration g. When an object is thrown into the air at a certain angle, two forces act upon this object: the drag force of a medium and graity. At the same time the drag force has two components: horizontal and ertical. Thus the object s acceleration will hae two different components: y a = a( and a y = a g. (1.8) In this case the following equations must be used for the simulations:
( t + = ( t + = y + a + a ( t + = ( + y( t + = y( + y t, t, t, y y t. (1.9) Problem 1 Perform simulations of the motion of an object thrown ertically into the air with the elocity o, gien by superisor, in the area of graitational acceleration g, assuming that this is the motion without friction. Compare achieed results of maimum height and flight time with theoretical alues. Then modify the program in such a way that you can describe ertical projection of a ball with the radius of R = 1 cm and mass m =,55 kg in the air. Problem a) Perform simulations of the motion of an object with elocity and different alues of an angle α gien by superisor. Graitational acceleration g is constant. b) Compare achieed results of maimum height, flight time and horizontal range with theoretical alues. c) Perform isualization of the motion. d) Does the size of time step affect the accuracy of simulation results? Problem 3 a) Perform simulations of the motion of an object thrown horizontally with the initial elocity and height h gien by superisor. Graitational acceleration g is constant. b) Compare achieed results of flight time and horizontal range with theoretical alues. c) Perform isualization of the motion. d) Does the size of time step affect the accuracy of simulation results? Problem 4 a) Perform simulations of the motion of a sphere-shaped object with, gien by superisor, initial elocity and different alues of an angle with regard to the drag force of a medium gien by formula 1.6. b) Compare the shape of a trajectory, horizontal range, maimum height and flight time, results of the simulation with resistance of the motion and calculations for the motion without resistance.
c) Perform isualization of the motion. Eercise II. Simulation of body collision Theory Behaiour of a physical point during the collision with the wall on the eample of trapping a ball in a rectangular bo with rigid walls A ball with the radius R moes without friction with elocity with and y components in a rectangular bo of the gien dimensions L and L y. Let us assume that initially the ball moes with the elocity V o directed at a certain angle α to -ae and is in a position r(t = ) = [L /; L y /]. In two dimensions the ball will be treated as a disc with the radius R. We assume that the collision is perfectly elastic. The ball moes and collides with the wall and changes its own position and elocity as follows: (t+ = ( + t; y(t+ = y( + t when ( L R then replace with {horizontal rebound}; when y( L y R then replace y with y ; {ertical rebound} Inelastic central collisions of two balls The motion happens in one dimension. The balls hook together after the collision. According to the principle of conseration of momentum the alues of the elocity of the hooked balls is as follows: 3 m1 = m 1 1 ± + m m Assuming that initially the balls moe with the elocities 1 and and their positions are r 1 (t = ) and r (t = ), the positions of both balls 1, are as follows: 1 (t+ = 1 ( + 1 t; (t+ = ( ± t; When 1 ( ( R collision occurs. Position of the hooked balls 3 (t+ = 3 ( ± 3 t Elastic central collisions of two balls
The motion happens in one dimension. We assume that the collision is perfectly elastic. According to the principle of conseration of momentum and kinetic energy, the alues of the elocity and position of the balls after the collision are as follows: k1 k m1 m = m + m 1 1 m m1 = m + m m 1 + m + m 1 1 m1 + m + m 1 We assume the same initial conditions and collisions as in the inelastic collision. After the collision moing balls change their position as follows: 1 (t+ = 1 ( ± k1 t (t+ = ( ± k t Elastic non-central collisions of two balls We assume that the motion happens in two dimensions. Stages of non-central collisions are shown in the figure.1. Fig..1. Positions and parameters of balls during non-central elastic collision in systems: (a) laboratory and (b) centre of mass. We assume that initially the balls moe with elocities 1 and and are in positions r 1 (t = ) and r (t = ). According to the principle of conseration of momentum and energy, elocity and positions of the balls are defined by the following algorithm: if (r1kw<=sum R) then begin length_r:= Sqrt(r1kw);
rj1:= r1/length_r; rj1y:= r1y/length_r; projection1:= V1*rj1 + V1y*rj1y; {radial and transersal component V1} Vr1:=projection1*rj1; Vr1y:=projection1*rj1y; Vt1:=V1 - Vr1; Vt1y:=V1y - Vr1y; Projection := V*rj1 + Vy*rj1y; {radial and transersal component V} Vr:=projection*rj1; Vry:=projection*rj1y; Vt:=V - Vr; Vty:=Vy - Vry; Vr1:=Sqrt(Vr1*Vr1 + Vr1y*Vr1y); {radial elocity before collision} Vr:=Sqrt(Vr*Vr + Vry*Vry); if (projection1<) then Vr1 := -Vr1; if (projection <) then Vr := -Vr; Vr1prim:=(m1*Vr1+m*(*Vr-Vr1))/(m1+m);{radial elocity after collision } Vrprim:=(m*Vr+m1*(*Vr1-Vr))/(m1+m); Vr1:=rj1*Vr1prim; Vr1y:=rj1y*Vr1prim; Vr:=rj1*Vrprim; Vry:=rj1y*Vrprim; V1:=Vt1 + Vr1; {elocity 1 after collision} V1y:=Vt1y + Vr1y; V:=Vt + Vr; {elocity after collision} Vy:=Vty + Vry; Problem 1 a) Perform a simulation of two-dimensional motion of a ball with the radius R trapped in a rectangular bo with rigid walls. b) Perform isualization of the motion. Problem
a) Perform a simulation of one-dimensional motion of balls with the radius R and with different mass m 1 and m, for inelastic collision when their elocities are: in opposite directions, in the same directions and one of them is at rest. b) Calculate elocity 3 of the balls hooked together as a result of inelastic collision for the cases described in point (a). c) Perform isualization of the motion for the aboe cases. Problem 3 a) Perform a simulation of one-dimensional motion of balls with the radius R and with different mass m 1 and m, when their initial elocities are: in opposite directions, in the same directions and one of them is at rest. b) Calculate elocities k1 and k after elastic collision of balls for the cases described in point (a). c) Perform isualization of the motion for the aboe cases. Problem 4 a) Perform a simulation of two-dimensional motion of balls with the radius R 1 and R and with different mass m 1 and m, after non-central elastic collision for different initial elocities 1 and of the balls and when one of them is at rest. b) Calculate elocities k1 and k for the cases described in point (a). c) Perform isualization of the motion for the aboe cases.
Eercise III. Simulation of object motion with friction Theory Surface acts upon an object with the force of friction F t and holds back its motion. Friction force: F t = m N µ - coefficient of friction (static is usually 1-% bigger than kinetic) depends on the type of friction surfaces, N - force pressing by an object onto the surface (always perpendicular to the surface). Types of friction: (a) Kinetic friction motion of friction surfaces (b) Static friction rest Motion on the inclined plane We eamine an object starting slipping off the plane under the force of graity. Fig. 3.1. shows distribution of forces acting upon this object on the surface of an inclined plane. Fig. 3.1. Distribution of forces acting upon the object on the surface of an inclined plane. Object acceleration will be defined by the formula a F Ft = where F is a component of the force m of graity F g = mg along the direction of the motion, and F t is the force of friction of an object with mass m eerted on the surface of the plane. As we see in figure 3.1. F = F = F g sinα and F y = F g cosα, therefore F t = µf g cosα. Time of coering the distance L will be:
L t = g (sinα µ cosα) The object will remain at rest when F F t we will hae µ tgα. We will get isualization of the motion by changing the time abruptly according to: g(sinα µ cosα) t ( t + = ( + We will get motion on the horizontal plane by substituting α = to the equation 3.1. This is what we find: (3.1) Position of the accelerated object without additional force ( = t gµ t Stopping distance ( k = ) ma = µ g When constant horizontal force F acts upon the object, its acceleration will be: a Ft + F = m p We will get isualization of the motion by calculating position, changing the time abruptly according to : ( t + = + a t ( t + = + t Friction force and drag force hae alues: F p F =µmg t ρ ( ) = C S. where elocity, S - cross section area of surface perpendicular to elocity, ρ - density of a medium and C drag coefficient dependent on an object s shape. Under the constant horizontal force F the object will reach the final elocity k after sufficiently long time k ( F µ mg) =. C Sρ Motion of a body on semi-circle is presented in Fig. 3.. Position of the body on the circle of radius R is determined by alue of angle α(, which is connected by dependence a( = l(/r with the arc
length l( measured from the lowest point of the circle. The force Fs tangential to the arc is responsible for the body acceleration gien by the equation: F s = mg sinα F t, where the frictional force Ft is a result of the friction between the surface and the body pressed by the perpendicular force which is the sum of the normal component of body weight and centrifugal force F t =µ(mg cosα+m / R). Because the position of the body on the arc will change, according II Newton's law, with the acceleration a( = F s (/m, its new elocity and position can be calculated numerically from the equation: (t+ t )= (+a (t ) t l (t+ t )=l (+ (t ) t. Fig. 3.. Forces acting on the body on the circle of radius R. Problem 1 a) Perform a simulation of the motion of an object on an inclined plane for different alues of coefficient of friction µ (also for µ = ) and inclination angle of a plane α. Graitational acceleration g is constant. b) Compare results of an angle α by which the object starts slipping off and the time of slipping off the gien path with theoretical alues. c) Perform isualization of the motion. Problem
a) Perform a simulation of one-dimensional motion of a car on the horizontal surface with elocity o for different alues of coefficient of friction µ and for different alues of drag coefficient C, mass m and cross section area of the surface S. Graitational acceleration g will be constant. b) Find the distance ma made by the car before stopping. c) Find the time t 1/ after which the car elocity decreases twice and the time t z after which the car stops. d) Perform isualization of the motion. Problem 3 Perform a simulation and isualization of the motion of an object accelerated from rest by the constant force F for different alues of coefficient of friction µ (also for µ = ) and for different alues of drag coefficient C, mass m and cross section area of the surface S. Graitational acceleration g is constant. Find the time after which the object reaches gien elocity k. Problem 4 (a) Perform a simulation of the motion of a body located on a semi-circle for different alues of the coefficient of friction µ (also for µ = ) and the circle radius R. Graitational acceleration g is constant. (b) Compare the angle α at which the body begins to slide with the theoretical alue and obtain in the simulations the time dependence of the position angle α(. (c) Perform isualization of the motion.
Eercise IV. Simulation of oscillating motion Theory Simple harmonic motion An eample of harmonic oscillator is an object with mass m attached to an ideal spring of spring constant k, while the other end of the spring is fied in some position i.e. to the wall (Fig. 4.1). Such system is called spring pendulum. After the displacement of an object from its equilibrium to position o, the object moes in the form of harmonic oscillation. Analytical solution for the equation of motion (theory) proides position of the object, elocity and acceleration a as follows: ( = o sin (ω = Aω cos( ω a = Aω sin( ω where k ω = m π ω = T F z Fig. 4.1 Eample of the harmonic motion of spring pendulum Period of oscillation is T = π m k We will get isualization of the motion by calculating position and changing the time gradually according to: (t+ = ( + o sin (ω Mechanical energy E m = E k + E p mv = k +
where E k and E p are kinetic energy and potential energy, respectiely. Gien the spring constant k and the amplitude o we can calculate the alues of restoring force, acceleration, elocity and position by using the equations: F z (t+ = k (t ) a(t+ = F z (t+ /m V (t+ = V (+a (t + t ) t (t+ = (t )+V (t + t We will get simple pendulum by attaching a tiny ball of mass m to a weightless, non-stretchy thread of the length of a line L. Initial displacement angle is α (Fig. 4.). Fig. 4.. Simple pendulum. Calculation of the period of pendulum T: F = mg sinα Fz sinα = L Fz = k mg sinα k = mg L mg mω = ω = L T = π L g g L
Damped harmonic motion occurs when a block of mass m is attached to a spring of spring constant k, the other end of the spring is fied in a constant position and while moing on the surface the block is affected by the sliding friction force F t. The alues of the forces acting upon the object: F s = k, force of stress in a spring F t = µmg, friction force opposite to F t = F s, counterbalances the strength of a spring, when an object is stationary and F t µ mg Fig 4.3. Damped harmonic motion. In case in resistance forces, position of the block (see Fig. 4.1) is defined by the equation: ( = oe ( = oe ω 1 = βt βt k b m 4m sin( t + ω1 φ ) where drag coefficient is b = mβ and β is damping coefficient. (4.1) We will get isualization of the motion by calculating position and changing the time abruptly according to: (t+ = ( + o e βt sin (ω 1 In order to calculate the period of oscillation T we will use formula 4.1by substituting π ω 1 =. T Problem 1 a) Perform a simulation of the motion of a harmonic oscillator by calculating its current parameters. b) Check if the mechanical energy is consered. c) Perform isualization of the motion.
d) Compare the results of the simulation of the object position ( with theoretical alues. e) Compare calculated alue of period T with period T s based on the simulation results. Problem Perform a simulation of the motion of a simple pendulum for the gien by superisor alue of the length of a thread and many different initial inclination angles in order to find a maimum angle α ma for which the actual period of oscillation T s will differ by less than 1% from the theoretical alue T. Problem 3 a) Perform a simulation of the damping harmonic motion, on the eample of a harmonic oscillator with friction. b) Perform isualization of the motion. c) Compare the results of the simulation of the object position ( with theoretical alues. d) Predict precisely the time of the motion till its final stopping.
Eercise V. Simulation of the motion in the force field Theory Two-dimensional motion of an object in the field of graity Two spherical objects act upon each other with a force F 1 = F 1 = G M 1 M R where M 1 and M are masses of the objects, R distances between their centers and G = 6.67 1-11 Nm /kg is the graitational constant. Fig.5.1 Graitational force acting between two spherical objects. I Cosmic speed The force of graity acts as the centripetal force Mm G R = m R Thus I cosmic speed = GM R Satellites moe around great masses in elliptic orbits with the period defined by Kepler s third law: 4π R T = GM 3 where M is the mass of a heaenly body (i.e. the Sun) around which a satellite orbits, and R is the semi-major ais of the orbit. Rotational motion of an object attached to a spring. In this case the centripetal force is the spring force (see eercise IV) F = -k. Motion of a charged particle in the magnetic field. The particle will moe around the circumference of a circle, the radius of which is defined by:
R = M/(Bq) Fig. 5. proides an eample of two-dimensional motion of this type where is perpendicular to B. Fig. 5.. Motion of a charged particle in the magnetic field where B induction of magnetic field, q charge of a particle, elocity of a particle. Problem 1 a) Perform a simulation of the motion of two heaenly bodies of gien masses M 1 and M. b) Perform a simulation of the motion of a body in the planet orbit. c) Perform isualization of the motion. d) Find I cosmic speed. e) Proe the rightness of Kepler s Third Law. Problem a) Perform a simulation of circular motion of a body attached to a spring. b) Perform isualization of the motion. Problem 3 a) Perform a simulation of the two-dimensional motion of a particle of charge q and elocity in the homogeneous magnetic field with induction B. b) Perform isualization of the motion.
Eercise VI. Statics and dynamics of a many-body system Theory A many-body system may be ery different in terms of size and number of objects N; the solar system with the planets and the Sun constitute a ery large system but there are relatiely few objects whereas the atom system of nanometer size can contain many atoms. In human scale the closest to the physical points N are balls of mass m i connected by the springs of spring constant k. In each such case a simulation of time eolution of a system is possible if we use numerical solutions for the equations of motion, which is fundamental for the molecular dynamics but also requires selecting the time step t appropriate to the analyzed phenomenon. In case of atoms the step must be small enough so that during this period of time the atom can make the distance of a fraction of the atom radius and for planets the step should be bigger so that they moe according to the scale of cosmic distances. Fig. 6.1. The resultant force F wi acting upon i-th physical point is the ector sum of the forces of interaction (green colour). In each case we must calculate the resultant force acting upon each of N bodies (Fig. 6.1). In case of two-body interaction in the distance (graitational, interatomic) for the point, it is defined by: j = N j =, j i F wi = F ij 1 (6.1) therefore we add up forces for each couple of bodies i and j. The force of their interaction is defined by the general formula:
rij F ij = F ( rij ) rij (6.) The components of this force are F ij = F ( rij ) / r ij and F ijy = F ( rij ) y / when the motion is twodimensional. For the graitational interaction of two masses M 1 and M it is: M M r 1 F ( rij ) = G ij, r ij whereas interaction of atoms, here for inert gases, can be described by the formula proposed by Lennard-Jones: F ( r ij 4ε σ ) = σ rij 13 7 σ rij (6.3) where the alues of parameters σ and ε depend on the type of the chosen gas. We calculate the resultant force for N bodies connected by springs by adding up ectors of forces of interaction of directly connected bodies. When such a system is made up by the balls connected by springs in one row and numbered from 1 to N then the resultant force is defined by the formula: F wi = k ( r l r i 1, i i 1, i ) + ( i, i+ 1 ) ri 1, i k r l r i, i+ 1 r i, i+ 1 (6.4) where r i, i+ 1 means the ector connecting i and (i + 1) ball, and l is a free length of a spring when it is neither tight nor stretched. Simulation of the motion of N bodies on XY plane can be performed with the molecular dynamics method een by using the simplest arithmetically Euler s method. For i-th body it is defined by the following equations: a a i y i i yi = F = F ( t + = ( t + = / m / m + a + a ( t + = + i y ( t + = y + i w i wy i i i i yi i i i yi i yi t t t t (6.5)
Problem 1 The system is composed of two masses M 1 and M and there is graitational interaction between them and the initial distance between them is Earth-Moon. Mass M 1 was gien elocity 9 km/s at angle α = 45 o to the straight line joining the centres of the masses. Perform a simulation of the motion of eery mass for the time step t = 1h and masses: a) M 1 = M = M Z, b) M 1 =,5 M Z and M = M Z. Check if the momentum, the angular momentum and mechanical energy of the system are consered. Problem The system is composed of two masses m 1 = m = 1 kg connected by a spring of spring constant k = 75 N/m and free length l = 5 cm. Mass m1 was gien elocity 5 m/s: a) in the direction of m, b) at angle α = 9 o to the straight line joining the centres of the masses. Perform a simulation of the motion of eery mass for the time step t =,1 s. Check if the momentum, the moment of momentum and mechanical energy of the system are consered. Problem 3 The system comprises N = 1 physical points of the same masses m = 1 g connected by springs of spring constant k = 5 N/m and free length l = 1 cm. One end of the string of balls is fied in the wall and the other is free. During the period of time t imp =, s we generate an impulse by using harmonic motion to moe the last ball transersely to the chain oer a distance A = cm and then returning to the point of departure. Perform a simulation of the motion of the chain of balls and obsere the displacement and rebound against the wall of the impulse. What is the elocity of the impulse and how does it depend on the mass of the balls and how on the stress of the springs? Problem 4 The system comprises N = 31 identical point masses connected by springs of spring constant k = 1 5 N/m and free length l = 5 cm spread between two walls L = 1 m away. Two outermost masses/balls are fied in the wall. Assuming that the initial shape of chain is a downward-pointing isosceles triangle, perform a simulation of the displacement of the balls in order to find their equilibrium position. To find the equilibrium of such chain in the net Nk = 4 steps, masses must be moed along the resultant force by ector T i = A F wi, where A = 1 1-6 m/n. Compare the final position of the balls with theoretical predictions based on the chain cure gien by the equation: F ρ l g y( ) = cosh ρ l g F
where ρ l is the linear density of the chain, g graitational acceleration, and F tension of the chain in its lowest horizontal part. Eercise VII. Simulation of the motion of a rocket Theory A rocket consists of the body with the system stabilizing direction of its flight i.e. stabilizers, and the substance which thrown with high elocity from the body of the rocket causes the rocket motion. Most frequently the substance is solid or liquid fuel which burns and changes into gas of high temperature and pressure and is epelled from the nozzle. The epulsion of combustion products with high elocity s relatie to the rocket body results in thrust F c. Its alue can be calculated using Newton s second and third laws for the system of two interacting bodies (Fig. 7.1): substance of mass m epelled at time t from the rocket, and the other part of the rocket of mass M. Fig. 7.1. Idea of deeloping thrust of the rocket of mass M when mass m is epelled from the rocket with elocity s. Notation of Newton s second law for the epelled substance, showing equality between a change of momentum of the substance and the impulse of a force, is as follows proides the alue of an aerage force acting upon mass m: F c s m s = Fc t. This dependency m = (7.1) t which proides the mass m with elocity s at time t. According to Newton s third law the force of identical alue will also act upon the remaining part of the rocket of mass M. The thrust gies the rocket acceleration of alue:
Fc a =, (7.) M ( where notation M( informs that the rocket mass is not constant but it decreases due to using up of the fuel. By introducing an important parameter of the rocket, the speed of change of the rocket mass defined as S M = M/ t, we can define the change of mass as follows: M( = M S m t, if this speed is constant. This dependency and equation (7.) proides the equation of the rocket motion: d dt Fc = M S t m where according to equation (7.1) F = S mass of the rocket body and mass of the whole fuel. c m s (7.3) and M is the rocket s initial mass equal of the sum of Equation (7.3) is a differential equation and its result proides us with an interesting dependency of the alue of the rocket elocity on time (. In order to achiee this result we must perform the following operations: Fc d = dt M S t d = Fc dt M S t = Sm ln( M S m m m + C. The alue of constant C is based on the condition () = m/s resulting from the assumption that at the initial moment the rocket was not moing. In this way we get alue C = S m lnm. Final ersion of the equation of the rocket s motion is: ( = S m M ln M S or in different but equialent notation: M = Sm ln. M m t The aboe equations show that the rocket elocity is directly proportional to the elocity of epulsion of mass/fuel from the rocket and to the natural logarithm of the relation of the initial mass to the rocket current mass. Figure 7.. shows dependency of the rocket elocity on time calculated with equation 7.4. We can note that the rocket s elocity increases up to the moment when the whole fuel is used up t w = 7, s, when it stays and thrust from F c = 75 N decreases abruptly to zero. The rocket s critical elocity reaches almost 7 m/s therefore it is oer two times higher than the elocity of epulsion of (7.4) (7.5)
combustion products of s = 3 m/s. According to equation 7.5 it would be een higher if the ratio of the initial mass to the final mass, or the mass of the body M k, was lower. Fig. 7.. Dependency on time of the rocket elocity of parameters: M = kg (including mass of the body M k =, kg), S m =,5 kg/s and s. = 3 m/s. Finding information about the rocket position for a gien moment t of its motion requires integration of equation 7.4. The easier way is to get the result through the numerical method i.e. Euler s method for equations: a( = F w / M, ( t + = ( + a( t, ( t + = ( + ( t. Symbol of thrust F c replaced with symbol of resultant force F w in order to use some eternal forces in the simulation for eample perform the simulation a) in ertical flight in the force of graity (F w = F c Mg) or b) in horizontal but it would be hold back by air resistance (F w = F c Mg k ). (7.6) Problem 1 Perform a simulation of the motion of a rocket moing horizontally only under the influence of thrust for parameters gien by the person performing the simulation. Initial alues of the motion shall be: () = and () = m/s. a) Find the rocket s elocity in time function and its maimum alue. Compare the results with theoretical alues. b) Calculate the rocket s position at regular time interals. c) Perform isualization of the rocket s motion.
Problem Perform a simulation of the motion of a rocket moing ertically in the force of graity for parameters gien by the person performing the simulation. Initial alues of the motion shall be: y() = and () = m/s. a) Calculate the resultant force acting upon the rocket and find whether it will be able to take off. b) Calculate the rocket elocity in time function and its maimum alue. c) Calculate the rocket position, maimum height and time necessary to reach this height. d) Perform isualization of the rocket motion. BIBLIOGRAPHY 1. D. Holliday, R. Resnick, J. Walker, Fundamental of Physics. PWN, Warszawa 5.. H.D. Young, R.A. Freedman, Uniersity Physics, 1 th ed., Pearson Education, Addison Wesley 8. 3. I. M. Matyka, Symulacje komputerowe w fizyce, Helion,. 4. T. Pang, Metody obliczeniowe w fizyce, PWN, Warszawa 1.