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Educational Systems® Knowledge Engineering Foundation Assessment and Learning Institute 37 Κoliatsou Street, Corinthos 20100, Greece solomon_antoniou@yahoo.com Mathematical Methods in Physics (Symbolic and Numerical Methods in Undergraduate Physics Courses) Mechanics One-Dimensional Rectilinear Motion Classification of Theory- -Methods and Techniques Dr. Solomon M. Antoniou Copyright© by Educational Systems® © Philosophy This course tries to combine • Traditional mathematical methods with • Symbolic and Numerical techniques For this purpose there is • A short exposition of theory • Emphasis on mathematical techniques • Classification of methods and techniques • Complete step-by-step solutions and proofs 2 Objectives The objectives of this Chapter is the student to be able • To solve problems in one-dimensional Rectilinear Motion • Horizontal or • Vertical using ordinary differential equations • To develop code in any of the symbolic languages in order to solve the differential equations of motion • To develop code in any of the symbolic languages in order to plot the solutions of the equations of motion 3 Contents Chapter 1 One-Dimensional Rectilinear Motion. 1.0. Symbolic Solutions of Ordinary Differential Equations 1.1. Horizontal Motion 1.1.1. First Case. Constant Force 1.1.2. Second Case. Force depends on the instantaneous velocity. 1.1.3. Third Case. Force depends on time. 1.1.4. Fourth Case. Force depends on replacement 1.2. Vertical Motion 1.2.1. Vertical Upward 1.2.2. Vertical Downward 1.2.3. Motion in Earth’s Gravitational Field 1.2.4. Motion of a chain 1.3. Work 1.3.1 Work-Energy Theorem 1.3.2 Conservative Forces 1.4. Momentum 1.4.1 Momentum Conservation 1.5. Impulse 1.5.1 Impulse-Momentum Theorem 4 Learning Outcomes By the time the student completes this chapter, he/she will know • How to write down Newton’s differential equation d2 x dv m = F or m =F dt 2 dt in one of the following cases A. Horizontal Motion 1. When the force is constant 1.1. the force points the direction of motion 1.2. the force opposes the motion 2. When the force depends on the instantaneous velocity 3. When the force depends on time 4. When the force depends on the displacement B. Vertical Motion 1. Upward Vertical Motion 2. Downward Vertical Motion 3. Motion in Earth’s Gravitational Field 4. Motion of a Chain 5 • How to solve the differential equation established in one of the two previous cases and determine 1. The displacement 2. The velocity 3. The acceleration 4. The time taken under some initial conditions imposed as mathematical relations based on the physical principles of the problem and using any one of the two methods a. finding the general solution of the differential equation and determining the arbitrary constant(s) taking into account the initial conditions or b. integrating the differential equation and using the initial conditions as limits in the definite integrals (since the differential equation is, in most of the cases, a first order differential equation). • How to develop code for the solution of any of the differential equations, using any of the Symbolic Languages (Macsyma, Maple, Axiom, Mathematica, Reduce, Derive, Matlab, Maxima, MuPad and Scientific WorkPlace). • How to develop code for plotting the solution of the equations of motion. 6 Chapter 1 One-Dimensional Rectilinear Motion. 1.0. Symbolic Solutions of Ordinary Differential Equations. 1.0.1. Macsyma (Version 2.4.0) Macsyma is the best available package (in the sense of user-friendly and reliability) in the market. On the other hand you may find a lot of add-on packages available, performing a number of very complicated tasks (like Painlevé analysis, Non-linear Partial Differential equations, exploiting Lie Symmetries in solving differential equations, etc). Example 1. 3 d2y dy Step 1. Enter the differential equation 2 + y = 0 by assigning the operator dx dx eqn to the differential equation. Step2. Find the general solution of the equation using the command ode, where gensol is the operator corresponding to the general solution where %k1 and %k 2 are constants. 7 dy Step 3. Enter the initial condition y(0) = 0, = 2 and get the answer dx (0,0) Step 4. Enter the boundary condition y(0) = 1, y(1) = 3 and get the answer Example 2. The system dx =y dt dy m = −k y 2 dt can be solved under the conditions x (0) = 0, y(0) = p using the code References: 1. Macsyma User’s Guide, A Tutorial Introduction Second Edition 1996 (pp.95-107) 1.0.2. Scientific WorkPlace (Version 2.5) From the menu choose Maple and then Solve ODE/Exact, Laplace, Numeric or Series. Example 1. • Step 1. Enter (change from text T mode in mathematics M mode) the expression • Step 2. From the menu bar choose Maple and then Solve ODE/Exact • Step 3. In the appearing ODE Independent Variable box enter the independent variable x and press OK. The program responds with 8 Example 2. • Step 1. From the menu choose Insert and then Display. • Step 2. Enter within the frame the following (pressing Enter after each one of the equations) The reader may easily understand that the first equation is a second order differential equation. The last two equations are the initial conditions. • Step 3. From the menu bar choose Maple and then Solve ODE/Exact • Step 4. In the appearing ODE Independent Variable box enter the independent variable x and press OK. The program responds with Example 3. The system dx =y dt dy m = −k y 2 dt can be solved under the conditions x (0) = 0, y(0) = p using the code References: 1. Doing Mathematics with Scientific WorkPlace™, Darel W. Hardy and Carol L. Walker (pp.233-249) Brookes/Cole 1995 9 1.03. Maple (Version 7.00) Example 1. dy Step 1. We introduce the differential equation x + y = x2 dx Step 2. We use the command dsolve in order to solve the differential equation Step 3. We use initial condition Step 4. We get a special solution of the differential equation by determining the arbitrary constant. Example 2. The system dx =y dt dy 2 m = −k y dt can be solved under the conditions x (0) = 0, y(0) = p using the code 10 References: 1. Maple V Language Reference Manual Bruce W. Char et al., Springer-Verlag 1991 (pp.23-24) 1.0.4. Reduce (Version 3.6) This very popular package written by Anthony C. Hearn is powerful enough and has been used by the scientific community in so many diverse fields (Mathematics, Astronomy, High Energy Physics, etc). To the best of my knowledge, this package can only solve first order differential equations using the user-contributed package ODESOLVE. Step1. Load the package ODESOLVE Step 2. The following command depend says that y is a function of x. dy Step 3. We introduce the differential equation x + y = x2 dx 11 Step 4. We use the command odesolve in order to solve the differential equation and then we get the answer References: 1. Reduce 3.6 Codemist Ltd 1995 Chapter 54, (pp.345-346) 1.0.5. Axiom (Version 2.0a, NAG, Sun SparcStation 4) d2y dy In order to solve the differential equation ( x 2 + 1) 2 + 3x + y( x ) = 0 , dx dx we enter deq : = (x**2+1)* D(y x, x, 2)+3*x*D(y x, x)+y x=0 and then use the command solve(deq, y, x) We get the answer x 3 ( − x / 2) x 3 [particular = 0 , basis = cos %e 2 , %e ( − x / 2) sin 2 ] in a self-explanatory notation. Example 2. The system dx =y dt dy m = −k y 2 dt can be solved under the conditions x (0) = 0, y(0) = p using the code References: 1. Axiom, The Scientific Computation System Richard D. Jenks and Robert S. Sutor Springer-Verlag 1992, (NAG, Oxford) 12 1.06. Matlab (Version 4) Example 1. dy Step1. In order to solve the differential equation = 1 + y 2 with initial condition dx y(0) = 1 , enter Step 2. You get the result Example 2. d2y Step1. In order to solve the differential equation + y = 4 cos( t ) with initial dt 2 π dy conditions y = 2 , = −3 , enter 2 dt ( π / 2) Step 2. You get the result Example 3. The system dx =y dt dy m = −k y 2 dt can be solved under the conditions x (0) = 0, y(0) = p using the code References: 1. The student edition of MATLAB, Version 4 User’s Guide , Prentice Hall 1995 13 1.0.7. Mathematica (Version 3.0) Symbolic solution of an Ordinary Differential Equation can be implemented using the command DSolve[ ]. Example 1. The code for the solution of the differential equation y ′( x ) = k ⋅ y( x ) is The response you get from Mathematica (Shift +Enter) is where C[1] is, as you can easily recognise, an arbitrary constant. In order to determine the constant C[1], we define the function (copy-paste from previous output) and then solve the equation which determines the value of the constant C[1] to be just 1, using the condition y ( 0) = 1 . If you include boundary conditions, for example y(0) = 1 , in the code, the constant C[1] appearing in the general solution will take a specific value. In this case the command is and you get the response dy Example 2. The differential equation = k − x has a solution subject to the dx condition y(0) = 0 which can be found using the code 14 Example 3. The differential equation y ′′ + 2 y ′ + y = 0 has a solution subject to the conditions y(0) = 1, y ′(0) = −2 which can be found using the code Example 4. The differential equation y ′′ − 2 y ′ + y = x 2 e x has a solution subject to the conditions y(0) = 1, y ′(0) = 0 which can be found using the code Example 5. The system dx =y dt dy m = −k y 2 dt can be solved under the conditions x (0) = 0, y(0) = p using the code References: 1. Mathematica, A system for doing mathematics by computer S. Wolfram, Third Edition 2. A Physicist’s guide to MATHEMATICA Patrick T. Tam, AP 1997 3. Mathematica by Example Martha L. Abell, James P. Braselton AP 1997 15 1.0.8. Derive (for Windows) Step 1. From the menu Author, enter y( x ) : = and press Enter. You get the response Step 2. From the menu Author, enter x*dif(y(x),x)+y(x) = x^2 and press Enter. You get the response Step 3. From the menu Calculus, enter Integrate and press Enter. You get Step 4. From the menu press Simplify and Enter. You get Step 4. From the menu Author press F3 and then a. You get the general solution of the differential equation #2 References: 1. Improving Mathematics Teaching with DERIVE, Bernhard Kutzler, Chartwell-Bratt 1996 (pp.93-98) 2. Introduction to DERIVE for Windows, B. Kutzler, 1996 16 1.1. Horizontal Motion. In this case the motion of the particle is being conducted in a horizontal axis (which might be either horizontal or vertical). In the case of Horizontal Motion, we consider previous Figure 1. In this case we have O is the origin of the motion. r i is the unit vector. OA is the vector displacement r v is the instantaneous velocity of the particle at a given point A We usually put OA = x = x i (1) where x is the displacement, which is considered to be a function of time: x = x (t ) (2) r The velocity v at any time t, is the first derivative of the displacement with respect to time: r r dx v= (3) dt The speed v is the magnitude of the velocity (the component of the velocity vector) defined by r r v = v⋅ i (4) r r The acceleration a at time t is defined as the first derivative of the velocity v : r r dv a= (5) dt 17 It is therefore obvious that the acceleration is also the second derivative of the displacement with respect to time: 2r r d x a= 2 (6) dt The basic differential equation which governs the motion is Newton’s Second Law: r d2x r m 2 =F (7) dt or r dv r m =F (8) dt r where F is the total force (resulting force) acting on the particle. It is much easier to solve the differential equation (8) instead of (7), because it is a first order differential equation. Equation (8) can be integrated very easily. It is obvious that the general solution will contain an arbitrary constant, which can be determined using the initial conditions. The initial conditions can be taken into account either • by finding the general solution and then determining the arbitrary constant or • by using definite integrals so as the initial conditions appear as limits of the integrals. Remarks: 1. The origin of the axis is always the starting point of the motion. r 2. The positive direction of the axis (direction of the unit vector i ) is the direction of the motion of the particle. 18 Horizontal Motion --- Solution Strategy Stage 1. Draw a diagram consisting of 1. A horizontal axis 2. A point O as an origin 3. Put an arrow (parallel to the axis and in some reasonable distance from it) showing the direction of motion of the particle 4. Put a unit vector i having as origin the origin O of the axis and direction the positive direction (the direction of the motion) 5. Consider an arbitrary position A on the axis Consider at this point A a. The velocity vector v pointing to the direction of the unit vector b. The resultant Force F Stage 2. Consider the differential equation of the motion (Newton’s Second d2 x Law) m 2 = F dt 1. Split the differential equation into the system dx =v dt (S1) dv m = F dt 2. The above system, since x = x ⋅i , v = v⋅i , F = F⋅i may be written in equivalent form as 19 dx =v dt (S2) dv m = F dt 3. Solve the system (S2) using EITHER a. Symbolic Algebra packages (Macsyma, Mathematica, Maple, Scientific WorkPlace, Matlab, Derive, Reduce, Maxima, MuPad Pro, Axiom, etc) OR dv b. Solve first the 1st order differential equation m = F in dt each one of the cases of Appendix A (and then dx differential equation = v , or a combination between them) dt Stage 3. Plot the various equations – using Symbolic Algebra Packages 1. Displacement vs. Time 2. Velocity vs. Time 3. Acceleration vs. Time 4. Displacement vs. Velocity 5. Acceleration vs. Displacement How to Consider Initial (or Boundary) Conditions. (How to translate physical conditions into mathematical relations) 1. If the initial displacement is x 0 this means that for t = 0 , x = 0 . In other words, x (0) = x 0 . 2. If there is not any initial displacement, then x (0) = 0 3. If the initial velocity is v 0 this means that for t = 0 , v = 0 . In other words, v(0) = v 0 . 20 4. If there is not any initial velocity, then v(0) = 0 . 5. If the particle comes to rest after travelling some distance, this means that the final velocity is zero. In other words, v( t ) = 0 How to implement the Initial Conditions I) By determining the constants in the general solutions of the differential equations II) By considering the limits of the integrals (integrating the differential equations) Appendix A. Solving the system dx =v dt (S2) dv m = F dt in various cases of the force F. First Case. Constant Force. If F = const. , then dv m =F dt is a separable differential equation and may be integrated, m ∫ dv = F∫ dt (1) giving the result mv = F⋅ t + C (2) where C is an arbitrary constant, determined from the initial conditions. dx The first differential equation = v of the system (S2), can be written, dt because of (2) in the form 21 dx m = F⋅t + C (3) dt which is a separable differential equation and may be integrated, m ∫ dx = ∫ (F ⋅ t + C) dt (4) giving the result 1 m ⋅ x = F ⋅ t 2 + C ⋅ t + C1 (5) 2 where C1 is another constant of integration, determined from the initial conditions. If the initial velocity is v 0 , (for t = 0, v = v0 ), then we get from (2): m v 0 = F ⋅ 0 + C and then C = m v0 (6) Eq.(2) gives us m v = F ⋅ t + m v 0 which is equivalent to F v= ⋅ t + v0 (7) m If the initial displacement is zero (i.e. for t = 0, x = 0 ) then 1 we get from (5) m ⋅ 0 = F ⋅ 0 + C ⋅ 0 + C1 and so C1 = 0 . 2 1 Eq.(5) becomes m ⋅ x = F ⋅ t 2 + mv0 ⋅ t which is equivalent to 2 1 F 2 x= ⋅ t + v0 ⋅ t (8) 2m Using Definite Integrals If the initial velocity is v 0 , (for t = 0, v = v0 ), then we get from (1) v t m ∫ dv = F ∫ dt v0 0 which is equivalent to m[v ]v 0 = F[t ]0 ⇔ m( v − v 0 ) = F( t − 0) ⇔ v − v 0 = F v t ⋅t m 22 or F v= ⋅ t + v0 m which is eq.(7). If the initial displacement is zero (i.e. for t = 0, x = 0 ) then we get from (4) x t m ∫ dx = ∫ (F ⋅ t + C)dt 0 0 which is equivalent to t t2 t2 m[x ] x 0 = F ⋅ + C ⋅ t or mx = F ⋅ + C ⋅ t 2 0 2 from which we get 1 F 2 x= ⋅ t + v0 ⋅ t 2m which is eq.(8). Second Case. Force depends on instantaneous velocity. dv If F = F( v) then m = F may be written as dt dv m = F( v) (1) dt which is a separable differential equation and may be integrated giving dv dv m∫ = dt or m ∫ =t+C F( v) ∫ (2) F( v) where C is an arbitrary constant determined from the initial conditions. Equation (2) expresses time in terms of the velocity. dv The same differential equation m = F may be written as dt dv dx dv m = F( v) or m v = F( v) (3) dx dt dx which is a separable differential equation and may be integrated, giving 23 v v m∫ dv = ∫ dx or m ∫ dv = x + C1 (4) F( v) F( v) C1 is a constant determined from the initial conditions. Equation (4) expresses displacement in terms of velocity. Using Definite integrals If the initial velocity is v 0 (i.e. for t = 0, v = v 0 ) we get from (1) after separation of variables and integration v t dv m∫ = dt v0 F( v) ∫ 0 which is equivalent to v dv t=m∫ (5) v0 F( v) If the initial displacement is zero (i.e. for t = 0, x = 0 ), we get from (3) v x v after separation of variables and integration m ∫ dv = ∫ dx , which is v0 F( v) 0 equivalent to v v x=m∫ dv (6) v0 F( v) Third Case. Force depends on time. dv If F = F( t ) the differential equation m = F may be written as dt dv m = F( t ) (1) dt which is a separable differential equation and may be integrated, giving m ∫ dv = ∫ F( t )dt or m ⋅ v + C = ∫ F( t )dt (2) where C is a constant determined from the initial conditions. 24 Equation (2) expresses the velocity in terms of time. The other differential equation dx =v (3) dt is also separable and may be integrated, giving ∫ dx = ∫ v dt or x + C1 = ∫ v dt (4) where v is a function of time, as follows from (2). Equation (2) expresses displacement x in terms of time t . Using Definite integrals If the initial velocity is v 0 (i.e. for t = 0, v = v0 ) we get from (1) after separation of variables and integration v t m ∫ dv = ∫ F( t )dt v0 0 which is equivalent to t m( v − v0 ) = ∫ F( t )dt 0 from which we get t 1 v = v0 + m∫ F( t )dt (5) 0 If the initial displacement is zero (i.e. for t = 0, x = 0 ), we get from (3) after separation of variables and integration x t ∫ dx = ∫ v( t )dt which is equivalent to 0 0 t x = ∫ v( t )dt (6) 0 and v( t ) is given by (5). Second Method. This method refers to § 1.1.3. 25 Fourth Case. Force depends on displacement dv If F = F( x ) then m = F may be written as dt dv dv dx dv m = F( x ) ⇔ m = F( x ) ⇔ m v = F( x ) (1) dt dx dt dx which is a separable differential equation and may be integrated giving m ∫ v dv = ∫ F( x ) dx or 1 mv 2 + C = ∫ F( x ) dx (2) 2 where C is a constant determined from the initial conditions. Equation (2) expresses a relation between the velocity v and the displacement x of the particle. Solving eq.(2) with respect to v we may integrate the first differential equation dx =v (3) dt dx giving ∫ v = ∫ dt or dx ∫ v( x) = t + C1 (4), where v( x ) comes from (1). Equation (2) expresses the time t in terms of the displacement x. Using Definite integrals dv If the initial velocity is v 0 (i.e. for m = F ) and the initial dt displacement is x = 0 , we get from (1) after separation of variables and integration 26 v x m ∫ v dv = ∫ F( x ) dx v0 0 x 1 which is equivalent to m( v 2 − v0 ) = ∫ F( x )dx from which we get 2 2 0 x 2 v 2 = v0 + 2 m∫ F( x )dx (5) 0 If the initial displacement is zero (i.e. for t = 0, x = 0 ), we get from (3) x t dx after separation of variables and integration ∫ v(x ) = ∫ dt which is 0 0 equivalent to x dx t=∫ (6) 0 v( x ) and v( t ) is given by (5). Second Method. We solve the second order differential equation d2x m = F( x ) dt 2 Third Method. This method refers to § 1.1.4. 27 r 1.1.1 First Case. Constant Force, F = const. Example 1. A particle of mass m moves along a straight line under the influence of a constant force of magnitude F (Fig.2). We suppose that the particle has an initial speed v 0 . Find (i) the velocity at any time t. (ii) the displacement x as a function of a) the speed b) of the time. (iii) the speed at any time t. Solution. (i) Using Newton’s second law, we have r dv r dv r r m =F⇔ m i = Fi ⇔ dt dt dv ⇔ m =F (1) dt Indefinite Integrals. Separating the variables, we get m dv = F dt from the previous equation and after integration, m∫ dv = F∫ dt (2) we get mv = F ⋅ t + C (3) where C is a constant. We have now to use the initial conditions in order to determine constant C. 28 Since for t = 0 the speed is v = v 0 , we get from (3): mv 0 = F ⋅ 0 + C from which we get C = mv 0 (4) and so eq.(3) becomes: mv = F ⋅ t + mv 0 which is equivalent to F v = v0 + ⋅t (5) m Definite Integrals. We could get the same result, eq.(5) ,if instead of using indefinite integrals in eq.(2), we would have been using definite integrals instead. In this case, taking into account the initial conditions, eq.(2) gives us v t m ∫ dv = F ∫ dt (6) v0 t0 where there is an obvious correspondence between the limits of integration: t = 0→v = v 0 and t=t → v=v Performing the integration, we get m[v ]v = v 0 = F[t ]0 ⇔ v=v t m( v − v 0 ) = F t ⇔ F v − v0 = t m which is further equivalent to F v = v0 + ⋅t m which is again eq. (5). Using Mathematica, the code for deriving eq. (5) is and after pressing Shift+Enter you get the answer 29 r (ii) The velocity v is given obviously by r F r v = v 0 + t i (7) m (iii) For the calculation of the displacement x , instead of using the second r d2x r order differential equation m 2 =F we can use the first-order equation dt r dv r m =F and then the obvious relation dt dv dv dx dv = = v (8) dt dx dt dx in order to eliminate the time t . dv So, starting from the differential equation m = F of the previous section, dt we get dv dx m =F ⇔ dx dt dv m v=F (9) dx The previous equation can be solved by separation of variables and integration: m v dv = F dx , v x m ∫ v dv = F∫ dx , v0 0 v=v v2 = F [ x] 0 , x m 2 v = v0 v2 v2 m − 0 = F x (10) 2 2 from which we get 2F v2 − v2 = 0 x m which is further equivalent to 30 F v2 = v2 + 2 0 x (11) m and x= m 2 2F ( v − v 20 ) (12) The previous equation expresses the displacement x as a function of the speed. Using Mathematica, the code for deriving eq. (12) is and after pressing Shift+Enter you get the answer Using the value of v from eq.(5) into eq.(12), we get m F 2 x= t + v0 − v2 , 2F m 0 m F2 2 F x= 2 t + 2 t v0 , 2F m m 1 F 2 x= t + t v0 2m which can be written as 1F 2 x = v0 t + t (13) 2m The above equation expresses the displacement x as a function of time. 2nd Method of deriving eq.(13) dx We could get eq.(13) starting from eq.(5). Since v = , eq.(5) is dt equivalent to dx F = v0 + t (14) dt m F and after separation of variables dx = v 0 + t dt m 31 and integration, x t F ∫ dx = ∫ v 0 + m t dt 0 0 we get t F t2 [x ] x 0 = v0 t + m2 0 so as 1F 2 x = v0 t + t . 2m Using (14), we may arrive at eq.(13) by writing the following Mathematica code Important remark: We may derive both equations (5) and (13) by solving the system dx =v dt dv m = F dt under the initial conditions x (0) = 0 , v(0) = v 0 . The corresponding Mathematica code is We may also eliminate the time t between the two previous equations using 32 and then to solve with respect to x : r r r dv (iv) For the acceleration a we have a = and so using eq.(7), we get dt r Fr a= i (15). m Plot. 1. Velocity in terms of time 2. Displacement in terms of time 33 Example 2. A particle of mass m moves in a straight line. A constant resisting force acts on the particle of magnitude F. If v 0 is the initial speed of the particle, find 1. the time taken for the particle to come to rest 2. the distance travelled. Solution. r dv r (i) We start with the differential equation m = F which can be expressed as dt dv r r m i = −F i dt 34 or dv m = −F (1) dt and after separation of variables, mdv = −Fdt (2) Since for t = 0, v = v 0 and for the total time of motion t , v = 0 , we integrate eq.(2) using definite integrals: 0 t m ∫ dv = − F ∫ dt v0 t0 ⇔ m[v]v= v 0 = −F[t ]0 v=0 t ⇔ m(0 − v 0 ) = − F( t − 0) ⇔ mv 0 = Ft mv 0 ⇔t= (3) F Using Mathematica, we may arrive at (3) following the next steps: Step1 Solve differential equation (1) taking into account that v ( 0) = v 0 Ft Step2 Define function v( t ) = − + v0 m Step3 Solve equation v( t ) = 0 with respect to t 35 (ii) We get from (1) dv dx dv m = −F ⇔ m v = −F (4) dx dt dx and after separation of variables, m vdv = − Fdx (5) The previous equation can be integrated using definite integrals: 0 x m ∫ vdv= − F∫ dx , v0 0 v =0 v2 = − F[x ]0 , x m 2 v=v 0 v2 m 0 − 0 = −F(x − 0) 2 from which we get 1 m v0 = F x 2 2 and then 1m 2 x= v0 (6) 2F Using eq.(4), we may arrive at eq.(6) by writing the following Mathematica code and then determining the arbitrary constant C[1] from the initial condition x ( v 0 ) = 0 In a more detailed calculation, we may use the following steps: Step1. Step2. 36 Step3. We may derive the equation of the orbit and the equation for the speed by solving the system dx =v dt dv m = − F dt under the initial conditions x (0) = 0 , v(0) = v 0 . The corresponding Mathematica code is In case of Maple, enter the system and then the initial conditions 37 Enter the following and get the answer The letters in red is the code while the letters in light blue is the answer you get. In Macsyma, enter the two equations as follows and then the initial conditions After that, you enter the code for the solution of the system of the two differential equations under the sided initial conditions and you get the answer 38 In Scientific WorkPlace, from the menu item Insert , choose Display and then enter each one of the equations (by pressing Enter after each one equation), as follows: From the menu item Maple, choose Solve ODE, and then from the appearing submenu, choose Exact. The response you get is Plot We define the function v( t ) = v 0 − at We plot the function v( t ) for v 0 = 10 and a = 3,4,5,6,7 39 1 We define the function x ( t ) = v 0 t − at 2 2 We plot the function for v 0 = 10 and a = 2,3,4,5,6 40 We may also tabulate the functions 41 r 1.1.2. Second Case. The Force F depends on the instantaneous velocity: r r F = F( v ) i . For example: F( v) = − av 2 , F( v) = a + bv 2 , F( v) = F − av 2 where v is the instantaneous speed. Example 1. A particle of mass m moves along a horizontal line such that at t = 0 it has speed v 0 and it is located at x = 0 . The particle is acted upon by a force which opposes the motion and has magnitude proportional to the square of the instantaneous speed : F = − k v 2 , k is the proportionality factor ( k > 0 ). Find the speed, the distance and the acceleration of the particle in terms of time t . Express the speed and the acceleration in terms of the distance travelled. Solution. r dv r Newton’s equation m = F can be written as dt dv r r m i = − kv 2 i dt or dv m = −kv 2 (1) dt 42 Speed in terms of time. After separation of variables, we get from (1) dv m = − k dt (2) v2 Integrating eq.(2) we get v t dv m∫ = − k ∫ dt , v0 v2 0 v 1 m − = −k[ t ]0 , t v v0 1 1 m − + = −kt v v 0 1 1 k − + =− t (3) v v0 m 1 1 k = + t v v0 m 1 m + kv 0 t = v v0m m v= ⋅ v0 . (4) m + kv 0 t Using Mathematica, we may arrive at (4) by finding the general solution of the differential equation (2), and then determining the arbitrary constant C[1] , by t = 0 , v = v 0 . 43 Distance in terms of time. dx Since v = , we get from (4) dt dx mv 0 = (5) dt m + kv 0 t mv 0 and separation variables dx = dt and integration m + kv 0 t x t mv 0 ∫ dx = ∫ 0 0 m + kv 0 t dt we get [ x ]0 = x m [ln(m + kv 0 t )]0t , k x= m {ln(m + kv 0 t ) − ln m} , k m m + kv 0 t x= ln , k m and finally m kv 0 x= ln1 + t (6) k m Using Mathematica, we may arrive at (6) by finding the general solution of the differential equation (5), and then determining the arbitrary constant C[1] , by x (0) = 0 . 44 Acceleration in terms of time. dv Since a = , we get from (4) dt d m − d (m + kv 0 t ) a= dt m + kv t ⋅ v 0 = mv 0 dt (m + kv 0 t )2 = 0 − kv 0 = mv 0 (m + kv 0 t )2 and finally 2 mkv 0 a=− (7) (m + kv 0 t )2 Velocity in terms of displacement In order to express the velocity in terms of the displacement, we write (1) as dv dx m = − kv 2 dx dt dv m v = − kv 2 dx dv m = −kv . (8) dx After separation of variables and integration, we get from equation (8) v x dv m∫ = − k ∫ dx v0 v 0 m[ln v ]v0 = − k[ x ]0 v x m(ln v − ln v 0 ) = − kx v m ln = − kx v0 v k ln =− x v0 m 45 k v − x =e m v0 and finally k − x v = v0 ⋅ e m . (9). Using Mathematica, we may arrive at (9) by finding the general solution of the differential equation (8), and then determining the arbitrary constant C[1] , by v(0) = v 0 . Acceleration in terms of displacement Using equation (9), we get for the acceleration k dv − x d k a= = v0 ⋅ e m − x = dt dt m k dx k = v 0 ⋅ e − kx / m − = − v 0 ⋅ e − kx / m v = m dt m =− k m ( ) k 2 v 0 ⋅ e − kx / m v 0 ⋅ e − kx / m = − v 0 e − 2 kx / m m and finally k 2 − 2 kx / m a=− v0e (10) m Plot We define the velocity function using eq.(4), 46 k where p = . m We substitute v 0 = 10 and p = 0.1, 0.3, 0.5, 0.7, 0.9 We may also tabulate the values of the velocity function We now define the function x ( t ) using eq.(6) 47 and then we plot the function for v 0 = 10 and p = 0.1, 0.3, 0.5, 0.7, 0.9 Finally, we tabulate the values of the function x ( t ) Question 1. Express the time needed by the particle to develop a given velocity v. Answer. We get from (3) k 1 1 t= − m v v0 which is equivalent to m 1 1 t= − v v k 0 Question 2. Express the time needed by the particle to cover a given distance x. Answer. We get from (6) k kv x = ln1 + 0 t m m 48 and after inversion, k x kv 0 em = 1+ t⇔ m k kv 0 x ⇔ t = e m −1 ⇔ m m mx k ⇔t = e − 1 kv 0 Example 2. An object of mass m moves along a straight line under a force of r r constant magnitude F : F = F ⋅ i . We assume that there is a resisting force acting on the object which is proportional to the square of the instantaneous speed : r r FR = − kv 2 i where k is the proportionality factor. Show that the distance traveled in m F − kv1 2 going from a point of speed v1 to a point of speed v 2 is s = ln . 2k F − kv 2 2 r r Solution. The total force acting on the particle is F = ( F − k v 2 ) i . r dv r Newton’s equation m = F can be written as dt dv r r m i = (F − k v 2 ) i dt or 49 dv m = F − kv 2 dt dv dx ⇔ m = F − kv 2 dx dt dv ⇔ m v = F − kv 2 (1) dx Separating the variables and integration we get from the previous equation v2 x2 v m∫ dv = ∫ dx v1 F − kv 2 x1 ⇔ − m 2k [ ln(F − kv 2 ) ] v2 v1 x = [x] x 2 ⇔ 1 ⇔ − m 2k [ ] ln(F − kv 2 ) − ln(F − kv1 ) = x 2 − x1 ⇔ 2 2 m F − kv1 2 ⇔ ln =s (2) 2k F − kv 2 2 Using Mathematica, we may arrive at (2) by finding the general solution of the differential equation (1), and then determining the arbitrary constant C[1] , by x ( v 0 ) = 0 . The code appears in the following 50 Example 3. A small car moves along a horizontal track. There is a resisting r r force acting on the car which is given by FR = − (λ + µ v 2 ) i where λ > 0, µ > 0 . If the initial speed is v 0 calculate (i) The time the car needs to come to rest. r dv r Solution. Newton’s equation m = F , can be written as dt dv r r m i = − (λ + µ v 2 ) i dt or dv m = − (λ + µ v 2 ) (1) dt (i) Separating the variables and integration, we get from equation (1) 0 t dv m∫ = − ∫ dt v0 λ + µv 2 0 0 m dv ⇔ µv∫ λ 2 = −t (2) 0 +v µ Under the change of variables λ λ v= ⋅y, dv = dy , µ µ µ v = v0 → y = v0 , v = 0 → y = 0 . λ 51 equation (2) becomes λ dy m 0 µ ∫ = −t µ λ µ / λ v0 (1 + y 2 ) µ m µ ⇔ [arctan y]0 µ / λ v = − t µ λ 0 m µ arctan 0 − arctan λ v 0 = − t µλ m µ ⇔ t= arctan λ v0 (3) µλ Using Mathematica, we may arrive at (3) by finding the general solution of the differential equation (1), and then determining the arbitrary constant C[1] , by t ( v 0 ) = 0 . In the following, we arrive at (3), using a different procedure Step1. Step2. 52 Step3. Using Reduce, (ii) Equation (1) can be transformed into dv dx m = − (λ + µ ν 2 ) dx dt dv ⇔ m ⋅ v = − (λ + µ ν 2 ) (4) dx After separation of variables and integration, we get 0 x v m∫ dv = − ∫ dx v0 λ + µν 2 0 53 ⇔ m 2µ [ ln(λ + µν 2 ) ]0 v0 = − [ x ]0 x m ⇔ { ln λ − ln(λ + µν 0 )} = − x 2 2µ m λ + µν0 2 ⇔ ⋅ ln =x 2µ λ m µ 2 ⇔x= ln1 + ν 0 (5) 2µ λ Using Mathematica, we may arrive at (5) by finding the general solution of the differential equation (4), and then determining the arbitrary constant C[1] , by x ( v 0 ) = 0 . We may arrive at (5) using the following steps: Step1. Step2. Step3. 54 Example 4. A small object of mass m moves along a horizontal axis. The initial velocity of the object is v 0 . A resisting force acting on the particle is proportional to r r the cube of the instantaneous speed : FR = − k v 3 i where k (k > 0) is the 1 proportionality factor. We suppose that after time τ the speed is v0 . 2 1 (i) Prove that the speed is v0 after time 5 τ . 4 1 (ii) Calculate the distance traveled by the object in reaching the speeds v0 and 2 1 v0 . 4 Solution. r dv r dv r r Newton’s differential equation m = F can be written as m i = − k v3 i dt dt or dv m = − k v3 (1) dt 1 (i) We shall first determine the time τ at which the speed is v 0 . We separate the 2 variables in equation (1) and integrate : v0 / 2 τ dv m ∫ = − k ∫ dt ⇔ v0 v3 0 v /2 1 0 τ m − 2 = − k[ t ]0 ⇔ 2v v 0 55 1 1 m − + 2 = − kτ ⇔ 2(v / 2)2 2 v 0 0 2 1 m − 2 + 2 = − kτ ⇔ v 0 2v 0 3 m 2 = kτ ⇔ 2v0 3m τ= 2 (2) 2kv0 In order to arrive at (2), we use the following steps, using differential equation (1) and the initial condition v(0) = v 0 Step1. Step2. Step3. Integrating equation (1) between the limits v 0 and v 0 / 4 (respectively 0 and τ ′ ) we have v0 / 4 τ′ dv m ∫ = − k ∫ dt , v0 v3 0 56 v /4 1 0 m − 2 = − k [ t ]0′ , τ 2v v0 1 1 m − + 2 = − k τ′ 2 (v 0 / 4 ) 2 v0 8 1 m − 2 + 2 = − k τ′ , v 0 2v 0 15 m 2 = kτ ′ , 2v 0 and finally 3m τ′ = 5 2 (3) 2kv0 A comparison of (2) and (3) gives us τ′ = 5τ (4). In order to arrive at (3), we use the following steps, using differential equation (1) and the initial condition v(0) = v 0 Step1. Step2. Step3. (ii) In order to find the distances travelled by the particle in reaching the speeds 57 1 1 v 0 and v 0 , we start using the differential equation (1) which can be 2 4 transformed as dv dx dv m = −kv 3 ⇔ m v = − kv 3 dx dt dx dv ⇔ m = − kv 2 . (5) dx Separating the variables and integration, we get from equation (5) dv m = −kdx ⇔ v2 v0 / 2 x dv m ∫ v0 v2 = −k ∫ dx ⇔ 0 v0 / 2 1 m − = − k[ x ]0 x v v0 1 1 v / 2 + v = − kx ⇔ m − 0 0 1 m − = − kx ⇔ v 0 m x= ⇔ (6a) kv 0 2 3m x= v0 ⇔ 3 2kv 0 2 2 x= v0 τ . (6) 3 Similarly integrating equation (5) between the limits v 0 and v 0 / 4 we find v0 / 4 x′ dv m ∫ v0 v2 = −k ∫ dx ⇔ 0 58 v /4 1 0 m − = − k[ x ]0 ′ ⇔ x v v0 1 1 v / 4 + v = − kx ′ m − 0 0 3m = kx ′ ⇔ v0 3m x′ = ⇔ (7a) kv 0 3m x ′ = 2v 0 2kv 2 ⇔ 0 x ′ = 2v 0 τ . (7) In order to arrive at (6a) and (7a), we use the following steps, using differential equation (5) and the initial condition v(0) = v 0 Step1. Step2. Step3. Step4. 59 Example 5. A small object moves in a straight path such that the time taken to reach a distance x from the origin is given by t = Ax 2 + Bx + C (1) where A,B,C are constants (A > 0) . Prove that the force acting on the particle is proportional to the cube of the instantaneous speed: F = − k v3 ( k > 0) (2) Solution. The total force acting on the particle is given by d2x F=m (3) dt 2 Therefore, we have to find the second derivative of the displacement with respect to time, using the initial hypothesis (1). We first differentiate eq.(1) with respect to t: d d ( t ) = (Ax 2 + Bx + C) dt dt d dx ⇔1= (Ax 2 + Bx + C) ⋅ dx dt dx ⇔ 1 = (2Ax + B) (4) dt We differentiate further eq.(4) with respect to t: d d dx (1) = (2Ax + B) dt dt dt d dx d dx ⇔ 0 = (2Ax + B) + (2Ax + B) dt dt dt dt d dx dx d2x ⇔ 0 = (2Ax + B) + (2Ax + B) 2 dx dt dt dt 2 dx 2 d x ⇔ 0 = 2A + (2Ax + B) 2 dt dt 60 d2x ⇔ 0 = 2Av 2 + (2Ax + B) (5) dt 2 The factor 2Ax + B appearing in (5) can be expressed in terms of the instantaneous velocity v, using eq.(4): 1 2Ax + B = (6) v We also have from Eq.(3), d2x F = (7) dt 2 m Using (6) and (7), we obtain from (5): 1 F 0 = 2Av 2 + ⋅ v m ⇔ 0 = 2Amv 3 + F ⇔ F = −2Amv 3 ⇔ F = − k v3 , where k = 2AM > 0 . dv 2nd Method. Since m = F , we have dt dv m = dt F v t dv m∫ = ∫ dt = Ax 2 + Bx + C (8) v0 F 0 and differentiating with respect to v, v d dv d ∫ = (Ax + Bx + C) 2 m dv v0 F dv d d dx ⇔m = (Ax 2 + Bx + C) F dx dv m dx ⇔ = (2Ax + B) (9) F dv On the other hand, dv dv dx m =F⇔m =F dt dx dt dv ⇔m v=F dx 61 dx mv ⇔ = (10) dv F Using (9) and (10) we find m mv = (2Ax + B) F F 1 ⇔ 2Ax + B = (11) v 1 1 ⇔ x = − B (12) v 2A Substituting x, as given by Eq.(12), in Eq.(8), we get 2 v dv 1 1 1 1 m∫ = A − B + B v − B 2A + C v0 F v 2A We differentiate with respect to v the previous equation and we get 1 1 1 1 B 1 m =A 2 ⋅ 2 − B − 2 + − 2 F 4A v v 2A v m 1 1 1 B 1 ⇔ = − B − 2 + − 2 F 2A v v 2A v m 1 1 ⇔ =− F 2A v 3 ⇔ F = −2Amv 3 ⇔ F = − kv 3 , where k = 2Am Example 6. A small object moves in a straight path such that the force acting on the particle is given by r r F = −kv 3 i ( k > 0) where v is the instantaneous speed. Prove that the time t taken to reach a distance x is given by t = Ax 2 + Bx + C where A,B,C, are constants (A > 0) . 62 Solution: Since r dv r m =F dt we get dv m = − kv 3 dt dv dx m = −kv 3 dx dt dv m v = − kv 3 dx dv m = − kv 2 (1) dx We integrate eq.(1) and we get v x dv m ∫ 2 = −k ∫ dx v0 v x0 v 1 m − = −k ( x − x 0 ) v v0 1 1 m − + = − k ( x − x 0 ) v v 0 1 1 k = + (x − x 0 ) v v0 m 1 k(x − x 0 ) + m = v mv 0 mv 0 v= (2) kv 0 ( x − x 0 ) + m dx Since v = , eq.(2) is equivalent to dt dx mv 0 = dt kv 0 ( x − x 0 ) + m which can be solved by separation of variables [kv 0 ( x − x 0 ) + m]dx = mv 0 dt and integration x t ∫ [kv 0 (x − x 0 ) + m]dx = mv 0 ∫ dt x0 0 63 x x kv 0 ∫ xdx − (kv 0 x 0 − m) ∫ dx = mv 0 t x0 x0 x x2 kv 0 (kv 0 x 0 − m)( x − x 0 ) = mv 0 t 2 x0 1 kv 0 ( x 2 − x 0 ) − (kv 0 x 0 − m) x + (kv 0 x 0 − m) x 0 = mv 0 t 2 2 1 1 kv 0 x 2 − kv 0 x 0 − (kv 0 x 0 − m) x + (kv 0 x 0 − m) x 0 = mv 0 t 2 2 2 1 1 kv 0 x 2 − (kv 0 x 0 − m) x + kv 0 x 0 − mx 0 = mv 0 t 2 2 2 k 2 m − kv 0 x 0 kv x 2 − 2mx 0 t= x + x+ 0 0 2m mv 0 mv 0 t = Ax 2 + Bx + C where k A= 2m m − kv 0 x 0 B= mv 0 kv 0 x 0 − 2mx 0 2 C= mv 0 r r r 1.1.3. Third Case. The total force F depends on time : F = F( t ) i Consider a particle of mass m moving on a straight line under the action of a force F which depends only on the time t at which the force is measured. The equation of motion becomes m&& = F( t ) . x dx d2t where x = & and && = 2 is Newton’s notation for the time derivatives. x dt dt To integrate this equation you will find it convenient to write it in the form 64 & dx F( t ) = . dt m Integrating with respect to time gives F( t ) x=∫ & dt or x = f ( t ) + c1 & m where f ( t ) is a particular value of the integral and c1 is a constant of integration. Rewriting this equation in the form dx = f ( t ) + c1 dt a second integration with respect to time gives x = ∫ f ( t )dt + c1 t or x = g ( t ) + c1 t + c 2 where g ( t ) is a particular value of the integral ∫ f (t )dt and c 2 is a second constant of integration. With experience it becomes unnecessary to rewrite derivatives explicitly in d the form before integrating with respect to time. dt The displacement of the particle at time t has been found very easily in this case. The function x ( t ) includes two arbitrary constants and describes all possible motions of the particle. To find the actual motion of the particle we require some initial conditions. Suppose that at time t = 0 the particle is observed to pass the point x 0 with velocity v 0 . Here “the point x 0 ” means the point with displacement x 0 relative to the & given origin O. Substituting this information into the equations obtained above for x and x gives v 0 = f (0) + c1 and x 0 = g (0) + c 2 65 These two equations determine the constants c1 and c 2 uniquely. With hindsight it would have been rather clever to choose the particular values f ( t ) and g ( t ) of the integrals to satisfy f (0) = 0 and g (0) = 0 , the constants of integration could then have been interpreted directly as the initial velocity and initial displacement. This can be done by evaluating the two definite integrals t t F( t ) f (t) = ∫ dt and g( t ) = ∫ f ( t )dt . 0 m 0 It is very important to understand that there is no unique choice for the initial conditions. This can be seen quite easily by considering whether the above constants of integration can be determined without knowing both the initial displacement and velocity of the particle explicitly. Suppose that at time t = 0 the particle is observed to pass the point x 1 . Substituting this information into the equation for x gives x 0 = g ( 0) + c 2 and x 1 = g (t 1 ) + c 1 t 1 + c 2 These two equations can be solved algebraically to give c1 and c 2 and hence to determine the actual motion of the particle uniquely. The common property of all the different possible initial conditions is that from each the initial displacement and velocity can be determined either directly or by using one or both of the integrals of the equation of motion. Example 1. A particle of mass m moves on a straight line under the action of a force F( t ) = F0 ⋅ sin(pt ) , where F0 and p are positive constants. The velocity of the particle at time t = 0 , relative to a fixed origin O, is v 0 . Show that the particle F0 will move towards x = +∞ if and only if v 0 > . mp 66 F0 Describe the motion of the particle when v 0 = − . mp Solution. The equation of motion m&& = F0 sin pt x can be integrated twice to give F0 mx = − & cos pt + c1 p and F0 mx = − sin pt + c1 t + c 2 . p2 F0 The term − sin pt in the expression for the displacement of the particle is bounded, p2 in fact it represents an oscillation. As t increases this will become negligible compared to c1 t and so the particle will move towards x = +∞ if and only if c1 > 0 . Now x = v 0 when t = 0 so that & F0 mv 0 = − + c1 p and the inequality c1 > 0 can be written as F0 mv 0 + >0 p F0 from which v 0 > − . mp Similarly the particle will move towards x = −∞ if and only if c1 < 0 which leads to F0 v0 < − . mp F0 When v 0 = − the constant of integration c1 becomes zero so that mp F0 mx = − sin pt + c 2 p2 and the motion is purely oscillatory. Exercises 67 1. The particle of Example 1 is observed to pass the points x 0 and x 1 at times t = 0 2π and t = . Find alternative conditions for the different motions discussed in the p example in terms of x 1 and x 2 . 2. A particle of mass m is moving on a straight line under the action of an exponentially decreasing force F = F0 e − λt , where F0 and λ are positive constants. The particle passes the point x 0 with velocity v 0 at time t = 0 . Obtain an expression for the displacement of the particle at time t and sketch a graph of the F0 displacement x against time for the special case when v 0 = − . mλ − at 3. F = kte 1.1.4. Fourth Case. Force depends on the replacement : F = F( x ) . In this case Newton’s equation can be written as d2x m = F( x ) (1) dt 2 We consider the initial conditions : t = 0 , x = x 0 and v = v 0 . dx Multiply both sides of the equation (1) by 2 . dt We find 2 dx d 2 x dx d dx dx dt dt 2 = 2F( x ) dt ⇔ m dt dt = 2F( x ) dt m 2 and then integrating from 0 to t we find t 2 x d dx m ∫ dt = 2 ∫ F( x )dx 0 dt dt x0 68 1 dx dx 2 2 x ⇔ m − = ∫ F( x )dx 2 dt t = t dt t =0 x 0 1 dx x 2 2 ⇔ m − v 0 = ∫ F( x )dx 2 dt x0 1 dx 1 2 x ⇔ m + − ∫ F( x )dx = mv 0 = const. 2 2 dt x 0 2 The first term of this equation is the kinetic energy, the second term the potential energy of the moving point. From this equation it follows that the sum of the kinetic and potential energy remains constant through the time of motion. Put : x U( x ) = − ∫ F( x )dx x0 from which there follows dU( x ) = − F( x ) dx where U ( x ) = potential function or potential. 1.2. Vertical Motion. In this case we consider two subcases : r a) Vertical upward (necessarily with initial velocity v 0 ≠ 0 ). 69 r r b) Vertical downward (where the initial velocity v 0 = 0 or v 0 ≠ 0 ). 1.2.1. Upward Vertical Motion. Example 1. 70 A small object of mass m is thrown vertically upward with speed v 0 . Assume that air resistance is proportional to the instantaneous speed. Find (i) the maximum height reached by the object. (ii) the time taken to reach the maximum height. Solution. r dv r Newton’s differential equation m = F can be written as dt r dv r r m = w+R. (1) dt r r r where w = − mgk is the weight of the object and R = −avk is air’s resistance (a is the proportionality factor and v is the instantaneous speed). dv r r Equation (1) can so be written as m k = −(mg + av )k dt or = −(mg + av ) dv m (2). dt (i) Equation (2) can be transformed as = −(mg + av ) , dv dx m dx dt v = −(mg + av ) . dv m (3) dx Separation of variables and integration, gives us 71 0 x v m∫ dv = − ∫ dx , v0 mg + av 0 0 m av ∫ mg + avdv = −[ x]0 x a v0 m mg 0 ∫ 1 − mg + av dv = −x , a v0 0 0 m m 2g a ∫ dv − a 2 v∫ mg + av dv = − x a v0 0 m 0 m 2g [ v] v0 − 2 [ln(mg + av)]v0 = − x 0 a a 2 m (0 − v 0 ) − m 2g {ln(mg) − ln(mg + av 0 )} = − x , a a mv 0 m 2 g mg + av 0 − 2 ln =x a a mg and finally mv 0 m 2 g av 0 x= − 2 ln1 + . (4) a a mg Using Mathematica, we may arrive at (4) by finding the general solution of the differential equation (3), and then determining the arbitrary constant C[1] , by x ( v 0 ) = 0 . The steps appear in the following: Step1. 72 Step2. Step3. (ii) From equation (2) we get 0 t m adv ∫ mg + av = −∫ dt , a v0 0 m [ln(mg + av]0 0 = −[ t ]0t v a m {ln(mg) − ln(mg + av 0 )} = − t a m mg + av 0 t= ln a mg m av 0 t= ln1 + (5) a mg Using Mathematica, we may arrive at (5) by finding the general solution of the differential equation (2), and then determining the arbitrary constant C[1] , by t ( v 0 ) = 0 . The steps are the following Step1. 73 Step2. Step3. Example 2. A small stone of mass m is thrown vertically upward from the earth’s surface with initial speed v 0 . Calculate (i) The position of the stone in terms of the time. (ii) The maximum height reached. (iii) The time taken to reach the highest point. 74 (iv) The speed in terms of its distance z from the origin O. r r Solution. The total force acting on the stone is the gravity force : F = −mgk . r dv r Newton’s equation m = F becomes dt dv r r m k = − mgk or dt dv = −g (1). dt (i) Separating variables and integration, we get from (1) : v t ∫ dv = −g ∫ dt , v0 0 [ v] v0 = −g[ t ]0 , v t v − v 0 = −gt v = v 0 − gt . (2) dz dz Since v = , equation (2) gives us = v 0 − gt and after separation dt dt of variables and integration, z t ∫ dz = ∫ (v 0 − gt )dt , 0 0 t t2 [z] = v 0 t − g , z 0 2 0 1 z = v 0 t − gt 2 . (3) 2 We may get eqs.(2) and (3) solving the system dz =v dt dv = −g dt under the initial conditions z(0) = 0 , v(0) = v 0 . 75 The above system can be solved using Mathematica. The code appears in the following (ii) From equation (1) we get dv dz = −g or dz dt dv v = −g (4) dz and after separation of variables and integration 0 h ∫ vdv = −g ∫ dz v0 0 (where it is obvious that at the maximum height h we have v = 0 ) 0 v2 = −g[z]0 z 2 v 0 2 v0 − = −gh 2 and so 2 v0 h= . 2g Using Mathematica, we may arrive at (5) by finding the general solution of the differential equation (4), 76 and then determining the arbitrary constant C[1] , by z( v 0 ) = 0 . We may incorporate the initial condition in the code. In this case we get the result We then define the function z( v) and h is the value z(0) of the function 0 t (iii) From eq.(1) we get ∫ dv = −g ∫ dt where it is obvious that at the highest v0 0 point the velocity is v = 0. We get [v]0v 0 = −g[ t ]0 , t 0 − v 0 = − g ( t − 0) , v 0 = gt and finally v0 t= . g Using Mathematica, we may arrive at (6) by finding the solution of the differential equation (1), 77 (iv) From eq.(4) we get after separation of variables and integration v z ∫ v dv = −g ∫ dz , v0 0 v v2 = −g[z ]0 , z 2 v0 2 v2 v0 − = −gz , 2 2 v 2 − v 0 = −2gz , 2 and finally v 2 = v 0 − 2gz 2 (7) Using Mathematica, we may arrive at (7) by finding the solution of the differential equation (4), under the condition z( v 0 ) = 0 1.2.2. Downward Vertical Motion. 78 Example 1. A parachutist having weight of magnitude mg leaves the plane with initial r velocity v 0 . If the air resistance acting on the parachute is proportional to the instantaneous r r speed, FR = −a v k ( a > 0 ) and a is the proportionality factor, calculate (i) The speed (ii) The distance traveled (iii) The acceleration at any time t > 0 , (iv) The speed as a function of the distance traveled and (v) The limiting speed of the parachutist. r r Solution. The total force acting on the particle is F = ( mg − av ) k . r dv r Newton’s equation m = F becomes dt dv r r m k = (mg − av)k dt or dv m = mg − av (1). dt (i) Separation of variables and integration, equation (1) becomes v t dv m∫ mg − av ∫ = dt , v0 0 − m [ln(mg − av)]v0 = [ t ]0t v a − m [ln(mg − av) ln(mg − av 0 )] = t , a 79 m mg − av − ln =t a mg − av 0 mg − av ln = −at / m , mg − av 0 mg − av = e −at / m mg − av 0 mg − av = (mg − av 0 )e − at / m , av = mg − (mg − av 0 )e − at / m mg mg v= − − v 0 e − at / m (2) a a Using Mathematica, we may arrive at (2) by finding the general solution of the differential dt m equation (1), written in the form = dv mg − av and then determining the arbitrary constant C[1] , by t ( v 0 ) = 0 . We may incorporate the initial condition in the code. In this case we get the result and then solve with respect to v : dz Since v = we get from (2) dt dz mg mg = − − v 0 e −at / m (3) dt a a 80 and after separation of variables and integration, z t mg mg ∫ dz = ∫ − − v 0 e −at / m dt , 0 0 a a t mgt mg m [z] = z 0 − − v 0 − e − at / m a a a 0 z= mgt m mg + ( − v 0 e −at / m − 1 ) a a a and finally z= mg a m t + v0 − a mg a 1− e ( − at / m ) . (4) Using Mathematica, we may arrive at (4) by finding the general solution of the differential equation (3), and then determining the arbitrary constant C[1] , by z(0) = 0 . If we incorporate the initial condition in the code, we get (ii) From equation (1) we get dv dz m = mg − av dz dt 81 dv m v = mg − av . (5) dz After separation of variables and integration, we get from equation (5) : v z v m∫ dv = ∫ dz . (6) v0 mg − av 0 Since v 1 − av =− = mg − av a mg − av 1 (mg − av) − mg 1 mg =− = − 1 − mg − av = a mg − av a 1 mg 1 1 mg − a =− + =− − 2 a a mg − av a a mg − av equation (6) becomes −a v v m mg − ∫ dv − a 2 v∫ mg − av dv = z , a v0 0 [ v] v0 − 2 [ln(mg − av)]v0 = z m v mg − v a a − m (v − v 0 ) − mg ln mg − av = z , a a2 mg − av 0 mg − av 0 m (v 0 − v ) + mg ln =z (7) a a 2 mg − av The previous equation expresses v as a function of z implicitly. Using Mathematica, we may arrive at (7) by finding the general solution of the differential equation (5), 82 and then determining the arbitrary constant C[1] , by z( v 0 ) = 0 . We may also incorporate the initial condition in the code. In this case we find (v) The limiting speed the parachutist acquires can be found by the condition: dv acceleration = 0 ⇔ =0 dt mg and because of equation (1), we get v = which is the limiting velocity. a Example 2. A parachutist falls from rest and acquires a limiting speed of v lim . Assuming that air resistance is proportional to the instantaneous speed, determine how long it takes to reach the speed of v ( v < v lim ). Solution. We follow the same line of reasoning as in previous example in reaching equation (2) with v 0 = 0 : at mg mg − m v= − e (1) a a Since, again, by previous example, the limiting velocity is mg v lim = (2) a we get a system of two equations : (1) and (2). We solve equation (1) with respect to t: We have that at mg − m mg (1) ⇔ e = −v⇔ a a at − a ⇔e m = 1− v⇔ mg at a ⇔− mg v ⇔ = ln1 − m 83 m a ⇔t=− ⋅ ln1 − mg v (3) a From equation (2) we get m v lim = (4) a g Using (2) and (4) in equation (3) we get v lim v t=− ⋅ ln1 − v . g lim Example 3. A small object of mass m is dropped from a height h above the ground, with an initial velocity of magnitude v 0 . Prove that air resistance is negligible, then the v 0 + 2gh − v 0 2 object will reach the ground in time t = with speed g v = v 0 + 2gh . 2 Solution. The total force acting on the object is the gravity force : r r Fw = mgk . r dv r dv r r Newton’s equation m =F becomes m k = mgk dt dt or dv =g (1) dt 84 From equation (1) we get dv dz dv =g ⇔ v=g dz dt dz and after separation of variables and integration, v v z v2 ∫v0 vdv = g ∫ dz ⇔ = g[z]0 ⇔ z 0 2 v0 2 v2 v0 ⇔ − = gz 2 2 ⇔ v 2 − v 0 = 2gz ⇔ 2 ⇔ v 2 = v 0 + 2gz 2 ⇔ v = v 0 + 2gz 2 (2) Equation (2) gives the speed at any point between O and the ground. dz Since v = , we get from (2) dt dz = v 0 + 2gz 2 (3) dt and after separation of variables and integration, h t dz ∫ 0 v 0 + 2gz 2 = ∫ dt 0 h 1 2g ⇔ g0∫ 2 v 2 + 2gz dz = t 0 ⇔ 1 g [v 2 0 + 2gz ] =t h 0 ⇔ 1 g [v 2 0 + 2gh − v 0 = t 2 ] and finally t= 1 g [v 2 0 + 2gh − v 0 ] (4) From equation (2) putting z = h , we obtain 85 v = v 0 + 2gh 2 (5) which is the velocity the object hits the ground. Using Mathematica, we may arrive at (4) by finding the general solution of the dt 1 differential equation (3), written in the form = dz v 0 + 2gz 2 and then determining the arbitrary constant C[1] , by t (0) = 0 . We may also incorporate the initial condition in the code. In this case we find 1.2.3. Motion of a particle in Earth’s Gravitational Field. 86 r The Earth is considered to be motionless. The force F acting on a particle of mass m at a distance r from the center of the Earth is given by Newton’s law of Universal Gravity : r mM r F = −G 2 k r where m : mass of the object M : mass of the Earth. r The origin of the axis is always Earth’s center and the unit vector k shows outward. Example 1. An object is projected vertically upward from the earth’s surface with initial speed v 0 . Neglecting air resistance (a) find the speed at a distance H above the earth’s surface (b) the smallest velocity of projection needed in order that the object never return (escape velocity). Solution. (a) In this case we combine Newton’s Second Law r dv r m =F dt and Newton’s Law of Universal Gravity : r mM r F = −G 2 k r (Another useful relation is MG = gR 2 ). We find dv r mM r m k = −G 2 k dt r dv 1 ⇔ = −GM 2 (1) dt r From equation (1) we get dv dr 1 = −GM 2 dr dt r 87 dv 1 ⇔ v = −GM 2 (2) dr r and after separation of variables and integration, v R+H dr ∫ vdv = −GM v0 ∫ R r2 v R +H v2 1 ⇔ = −GM 2 v0 rR 2 v2 v0 1 1 ⇔ − = −GM − + 2 2 R +H R H ⇔ v 2 − v 0 = −2GM 2 R (R + H ) 2GM H ⇔ v2 = v0 − 2 ⋅ (3) R R+H Since MG = gR 2 we get from (3) : 2gRH v2 = v0 − 2 (4) R+H Using Mathematica, we may arrive at (3) by finding the general solution of the differential equation (2), and then determining the arbitrary constant C[1] , by r ( v 0 ) = R . We may also incorporate the initial condition in the code. In this case we find 88 (b) We suppose that the velocity of the object at an infinite distance (H → ∞ ) from earth’s surface is zero : v ∞ = 0 and the object’s initial velocity is the escape velocity : v 0 = v esc . We then get from equation (4) : H 0 = v esc − 2gR lim 2 H →∞ R + H and since H lim = 1, H →∞ R+H we have finally 0 = v esc − 2gR ⇔ v esc = 2gR 2 (5) Example 2. An object is projected vertically upward from the earth’s surface with initial speed v 0 . (i) Prove that the maximum height H reached above the earth’s surface is 2 v0R H= 2gR − v 0 2 (ii) Discuss the significance of the case where v 0 = 2gR . 2 89 2 v0 (iii) Prove that if H is small, then it is equal to very nearly. 2g Solution. Since the only force acting on the object is Newton’s force, r mM r F = −G 2 k , r Newton’s Second Law r dv r m =F dt becomes dv r mM r m k = −G 2 k dt r or dv 1 = −GM 2 (1) dt r From eq (1) we get dv dr 1 = −GM 2 dr dt r dv 1 ⇔ v = −GM 2 (2) dr r and after separation of variables and integration, 0 R+H dr ∫ vdv = −GM ∫ v0 R r2 0 R +H v2 1 ⇔ = −GM − 2 v0 rR 2 v0 1 1 ⇔− = −GM − + 2 R+H R 2 v0 1 1 ⇔ =− + 2GM R+H R 90 2 1 v0 1 ⇔ =− + R+H 2GM R 2 1 v0 1 ⇔ =− + R+H 2gR 2 R 1 − v 0 + 2gR 2 ⇔ = R+H 2gR 2 2gR 2 ⇔ R+H= 2gR − v 0 2 2gR 2 ⇔H= −R 2gR − v 0 2 2 v0 R ⇔H= (3) 2gR − v 0 2 Using Mathematica, we may arrive at (3) by finding the general solution of the differential equation (2), and then determining the arbitrary constant C[1] , by r ( v 0 ) = R . We may also incorporate the initial condition in the code. In this case we find We may also use the following 91 using from the beginning the fact that GM = gR 2 . We further define the function r[ v] by Equation (3) can be obtained if we solve the equation r[0] = H + R with respect to H : Example 3. An object is projected vertically upward from the Earth’s surface with initial speed v 0 . Prove that the time taken to reach the maximum height H is R+H H R+H R − H + ⋅ arccos 2g R 2R R + H Solution. Since the only force acting on the object is Newton’s force, r mM r F = −G 2 k r r dv r Newton’s Second Law m =F dt becomes dv r mM r m k = −G 2 k dt r 92 or dv 1 = −GM 2 (1) dt r From equation (1) we get dv dr 1 = −GM 2 dr dt r dv 1 ⇔ v = −GM 2 (2) dr r We first calculate the speed v of the object at a distance r from Earth’s center. We separate variables in equation (2) and integrate : v r dr ∫ vdv = −GM ∫ v0 R r 2 v r v2 1 ⇔ = −GM − 2 v0 r R 2 v2 v0 1 1 ⇔ − = −GM − + 2 2 r R 1 1 ⇔ v 2 − v 0 = −2gR 2 − + 2 r R 2 2gR ⇔ v2 = v0 + 2 − 2gR r ⇔ v2 = (v 2 0 ) − 2gR r + 2gR 2 r ⇔ v2 = ( 2gR 2 − 2gR − v 0 r 2 ) (3) r Using Mathematica, we may arrive at (3) by finding the general solution of the differential equation (2), 93 and then determining the arbitrary constant C[1] , by r ( v 0 ) = R . We may also incorporate the initial condition in the code. In this case we find dr Since v = , we get from equation (3) dt dr = 2gR 2 − 2gR − v 0 r 2 2 ( ) dt r from which we obtain dr = ( 2gR 2 − 2gR − v 0 r 2 ) (3a) dt r which after separation of variables and integration, gives us R +H t r ∫ dr = ∫ dt (4) R ( 2gR 2 − 2gR − v 0 r 2 ) 0 The previous equation can be further written as R+H 1 r dr 2gR − v 0 2 ∫ R a2 − r =t (5) where 2gR 2 a = 2 (6) 2gR − v 0 2 It is obvious that since the maximum height H is 2 v0R H= (7) 2gR − v 0 2 (see previous Example), a2 = R + H (8) Making the substitution r = u 2 in the integral (5), we have 94 dr = 2udu and for the new limits of integration we find r=R→u= R r =R+H→u = R+H Integral (5) becomes R+H 1 2u 2 2gR − v 0 2 ∫ a2 − u2 du = t R 1 1 R+H or since = (as follows from (6) and (8)), 2gR − v 0 2 R 2g R +H 1 R+H 2u 2 R 2g ⋅ ∫ a2 − u2 du = t (9) R We also have 2u 2 u ∫ a2 − u2 du = a 2 ⋅ arcsin − u a 2 − u 2 a and so R+H R +H 2u 2 u ∫ a −u 2 2 du = a 2 ⋅ arcsin − u a 2 − u 2 a R R R+H = a 2 ⋅ arcsin − R + H ⋅ a 2 − (R + H ) − a R − a 2 ⋅ arcsin a − R ⋅ a −R 2 = {(R + H) ⋅ arcsin 1 − 0} − (R + H) ⋅ arcsin R − RH = R+H π R = (R + H) − arcsin + RH (10) 2 R+H 95 R If arcsin = ω then as it follows from next figure, R+H π R − ω = φ and since φ = arccos , we have 2 R+H π R R − arcsin = arccos . 2 R+H R+H So (10) gives us R +H 2u 2 R ∫ a −u 2 2 du = (R + H) ⋅ arccos R+H + RH (11) R We finally get from (9) 1 R+H R t= RH + (R + H) arccos R 2g R+H or R+H H R+H R t= + ⋅ arccos (12) 2g R R R+H Using the identity R 1 R−H arccos = arccos R+H 2 R+H (which we leave as an exercise) formula (12) can be written as R+H H R+H R − H t= + ⋅ arccos (13) 2g R 2R R + H 96 Using Mathematica , we may solve differential equation (3a) where C[1] is a constant determined from the initial conditions. Example 4. Prove that if an object is dropped to the Earth’s surface from a height H, then if air resistance is negligible it will hit the Earth with a speed 2g R H v= R+H Calculate also the time taken for the object to reach Earth’s surface. Solution. We start from the differential equation dv 1 v = −GM (1) dr r2 which we have derived in the previous examples. Separation of variables and integration, we get v R dr ∫ vdv = −GM 0 ∫ R +H r 2 v R v2 1 ⇔ = −GM − 2 0 r R+H 97 v2 1 1 ⇔ = −GM − + 2 R R +H v2 −H ⇔ = −GM ⋅ 2 R (R + H ) H ⇔ v 2 = 2GM ⋅ R (R + H) H ⇔ v 2 = 2GM ⋅ R (R + H) H ⇔ v 2 = 2gR 2 ⋅ R (R + H) 2gRH ⇔v= (2) R+H We start again from eq . (1) and integrating, v r dr ∫ vdv = −GM 0 ∫ R +H r 2 v r v2 1 ⇔ = −GM − 2 0 r R+H v2 1 1 ⇔ = −GM − + 2 r R +H 1 1 ⇔ v 2 = 2GM − r R +H dr and since v = , dt 2 dr 21 1 = 2gR − dt r R +H dr R+H−r = − 2g R dt r (R + H ) dr 2g R+H−r = −R ⋅ (3) dt R+H r 98 where we have included a minus sign. Separating variables and integration, we get from eq.(3) R+H R t 1 r − R ∫H R + H − r dr = ∫ dt 2g R + 0 (4) Under the substitution r = u 2 we have dr = 2udu and the new limits are r =R+H→u = R+H r=R→u= R Eq. (4) becomes R+H R 1 2u 2 − R 2g ⋅ ∫ R + H − u2 du = t (5) R +H We also have 2u 2 du = (R + H ) ⋅ arcsin u ∫ R+H−u 2 − u R + H − u2 R+H and so R 2u 2 ∫ R + H − u2 du = R+H R = (R + H ) arcsin u − u R + H − u2 = R+H R+H = (R + H ) arcsin − RH − {(R + H ) arcsin(1) − 0} R R+H = (R + H ) arcsin − RH − (R + H ) R π R+H 2 π = (R + H ) arcsin R − − RH = R + H 2 R = (R + H ) − arccos − RH = R+H R + H R−H = − arccos + RH . 2 R+H 99 We get finally from equation (5) : 1 R+H R+H R − H t= RH + 2 arccos R + H R 2g or R+H H R+H R − H t= + arccos 2g R 2R R + H 1.2.4. Motion of a chain. Example 1. A uniform chain of total length α has a portion b (0 < b < α ) hanging over the edge of a smooth table AB. Prove that the time taken for the chain to slide off the table, if it starts from rest, is α α + α2 − b2 t= ⋅ ln . g b Solution. Suppose that ρ is the linear density of the chain: m ρ= . a It is obvious that for a length ∆l and corresponding mass ∆m we also have 100 ∆m ρ= . ∆l We suppose now that the chain moves a distance x. Then the mass which corresponds to length x is m x = xρ . The total force which is responsible for the motion of the chain is r r r Fx = m x gk = xρgk r dv r Newton’s Second Law m = Fx gives dt dv r r ρα k = xρ gk dt or dv α = xg (1) dt From eq (1) we get dv dx α = xg , dx dt dv α v = xg dx and after separation of variables and integration 0 v a ∫ vdv = g ∫ xdx x b (x > b ) v x v2 x2 ⇔ a = g 2 0 2 b v2 x 2 b2 ⇔a = g − 2 2 2 ( ⇔ av2 = g x 2 − b 2 ) g 2 ⇔v= x − b2 (2) a dx Since v = we get from (2) dt dx g = x 2 − b2 dt a and after separation of variables and integration 101 a t dx g ∫ b x 2 − b2 = a ∫ dt 0 a g ⇔ ln x + x 2 − b 2 = t b a ( ) ⇔ ln a + a 2 − b 2 − ln b = g a t and finally a a + a 2 − b2 t= ⋅ ln g b Example 1a. – Chain Sliding Off Frictionless Table Problem Description: A uniform chain of total length ‘a’ is on a horizontal table so that a length ‘b’ dangles over the side at time zero. Determine how long it will take the chain to slide off the table. Define the density of the chain as σ (mass per unit length) and let x(t) be the length hanging over the side at time t. Therefore, the force on the chain due to gravity is simply the weight of the portion of the chain overhanging the table, or Fg = mg = σxg But the total mass being accelerated is given by dv Ft = M = σxg dt And from Newton’s law we have dv σa = σxg dt Also, we know that position, x(t), and velocity, v(t), are related. In particular, one relationship that is often useful is dv dv dx dv = =v dt dx dt dx Putting this into the above force balance gives the final mathematical model dv g v = x dx a with initial conditions x (0) = b and v(0) = 0 . Solution Method: 1. The above equation is separable. Therefore, it can be written as 102 g v dv = x dx a 2. Integration gives v2 g x 2 = 2 a 2 g + c or v 2 = x 2 + c a ( ) or g 2 v= x +c a 3. Now applying the initial conditions gives g 0= b2 + c a Therefore, c = − b 2 and g 2 v( x , t ) = x − b2 a This is the formal solution for v (x, t). dx 4. To find x(t) explicitly, we note that v( t ) = . Therefore, dt dx g = x 2 − b2 dt a Separating variables again gives dx g = dt x 2 − b2 a and, using standard integral tables, this becomes x g cosh −1 = ( t + c) b a 5. To put this into a more usable form (there is no cosh −1 button on my calculator!), note that 1 x cosh( x ) = (e + e − x ) and cosh −1 (u ) = 1n ( u + u 2 − 1 ) 2 Therefore, x + x 2 − b2 ln = g ( t + c) b a 103 b 6. Now applying the initial conditions again, we have that ln = ln(1) = 0 . b Therefore, c = 0 in the above expression. Finally, the time it takes for the chain to overhang a distance x is given by a x + x 2 − b2 T= ln g b 9 .8 m 7. For example, if a = 10m , b = 1m , and g = , then the time required for the chain s2 to fall off the table is simply (note that x = a for this case) 10 10 + 10 2 − 12 T= ln = 1.01× ln(19.95) = 3.02 sec 9.8 1 Thus, under these conditions, it takes about 3 sec for the chain to side completely off the table. Example 2. A uniform chain of total length a haw a portion b ( 0 < b < a ) hanging over the edge of a smooth table. If the table has coefficient of friction µ prove that the time taken for the chain to slide off the table if it starts from rest is a a+ a 2 − [b (1 + µ) − aµ]2 t= ⋅ ln (1 + µ) g b (1 + µ) − a µ Solution. 104

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