Mechanics 1: Motion of a Particle in 1 Dimension

Transcription

1 Mechanics 1: Motion of a Particle in 1 Diension Motion of a Particle in One Diension eans that the particle is restricted to lie on a curve. If you think hard about this stateent you ight think that the otion of all particles is one diensional since the otion traces out a curve in space. Actually, this is true. However, in order to take advantage of this you ust know the curve, i.e., know the otion. In practice, this is usually what we are trying to find out. The situations we have in ind here are when the forces on a particle are always acting so that at each point of a known curve they are tangential to that curve. Initially, the curve of interest will be a straight line. In this case, the forces act only in the direction of this straight line. We will also consider particles constrained to ove in a circle, as well as other plane curves. We will denote the generic coordinate for the particle by s. For a particle oving horizontally in the x direction, s would just be x. For a particle falling under the influence of gravity s would be z. For a siple pendulu the particle would be constrained to ove in a circle and s would be an angular coordinate, where the angle is easure fro soe fixed (say vertical) line. If we assue constant ass, Newton s second axio of otion becoes: = F. (1) dt2 This is an exaple of an ordinary differential equation (ODE) and all we now need to do is solve it to obtain the otion, s(t). However, we are getting a bit ahead of ourselves, because there is the proble of specifying the right-hand-side of (1), the F. This requires an understanding of the physics of the syste under consideration. But first, we want to consider soe strictly atheatical issues associated with (1). First soe terinology. Newton s equations are second order ODEs. This eans that the highest derivatives appearing in the equation are two. The other point to ake is that solutions to second order ODEs require the specification of two constants (which have physical eaning). Why is this? Here is an arguent that is not generally found in echanics textbooks. Suppose s(t) is a solution of (1) and we are interested in the solution near t =. So we could consider a Taylor expansion near : s(t + ) = s( ) + dt ()t + 1 d 2 s 2 dt 2 ()t d 3 s 6 dt 3 ()t 3 +, (2) and the Taylor coefficients are just the derivatives of s(t) evaluated at. Now in what sense does the equation (1) provide the inforation to copletely specify (2)? Here, copletely specify would ean to copletely specify all of the Taylor coefficients. Equation (1) provides us with the second derivative of s(t), we can repeatedly differentiate it to obtain all the higher order derivatives. However, it tells us nothing about s( ) and dt (). These we ust specify separately. If is the initial tie we say that we ust specify initial conditions, the initial position, s( ), and the initial velocity, ṡ( ). Now for what type of forces can we actually solve (1) analytically (as opposed to using a coputer)? This depen on the functional for of the force, i.e., how it depen on s, ṡ and t. We suarize the results in Table 1, and then provide soe detailed calculations backing up the clais. Force Function F = constant F = F(t) F = F(ṡ) F = F(s) F = F(ṡ, t) F = F(s, t) F = F(s, ṡ, t) Solvability of Newton s Equations Table 1: Everything was going fine until the last three rows of Table 1, and that requires soe explanation. 1

2 There is an iportant distinction between linear and nonlinear ODEs. The distinction being that solutions of linear ODEs are fairly siple, while the solutions of nonlinear ODEs ay be extreely coplicated (in ways that can be ade atheatically precise). For these reasons it is iportant to understand fro the outset whether you are dealing with a linear or nonlinear ODE. First, let s define what we ean by a linear ODE. This eans that F(s, ṡ, t) is a linear function of s and ṡ, plus a function solely of t (which could be constant), i.e., F(s, ṡ, t) = (a 0 + a 1 (t))s + (b 0 + b 1 (t))ṡ + c 0 + c 1 (t), (3) where a 0, b 0, c 0 are constants, and a 1 (t), b 1 (t), c 1 (t) are functions of t. So are Newton s equations solvable with a force of the for of (3). No. The proble coes fro the coefficients on the s and ṡ ters. If they are constant (i.e., a 1 (t) = b 1 (t) = 0), then Newton s equations can always be solved. So what is a nonlinear ODE? It is one that is not linear, according to our definition above. Now let s turn to justifying Table 1. F = constant. Newton s equations are: dt 2 = F = constant, s() = s 0, ṡ( ) = v 0, and this is about the easiest differential equation that you could be given to solve. To solve it, you just integrate twice, as we now show. ( ) d dτ dτ (τ) dτ = Fdτ, Perforing the integrals gives: Integrating this equation gives: dt (t) = v 0 + F (t ). dτ (τ)dτ = v 0 dτ + F (τ )dτ. Perforing the integrals gives: F = F(t). Newton s equations are: s(t) = s 0 + v 0 (t ) + F 2 (t ) 2. dt 2 = F(t), s() = s 0, ṡ( ) = v 0, and this is about the second easiest differential equation that you could be given to solve. To solve it, you also integrate twice. ( ) d dτ dτ (τ) dτ = F(τ)dτ, Perforing the integrals gives: 2

3 dt (t) = v F(τ)dτ. Integrating this expression gives: or dτ (τ )dτ = v 0 dτ + 1 τ F(τ)dτdτ, F = F(ṡ). Newton s equations are: To solve this equation let s(t) = s 0 + v 0 (t ) + 1 dt 2 = F τ F(τ)dτdτ. ( ), s( ) = s 0, ṡ( ) = v 0. dt then Newton s equations becoe: u = dt, (4) du = F(u). (5) dt We can solve this equation for u(t) (provided we can do the integrals that arise), and then integrate u(t) with respect to t to get s(t). More precisely, we solve for u(t) by rewriting (5) in the following for: u(t) du t u()=v 0 F(u) = dτ. If this integral can be perfored (which will depend on F(u)), then we ay be able to obtain u(t) = dt (t). We integrate this expression fro to t to obtain s(t). F = F(s). Newton s equations are given by: dt 2 = F(s), s() = s 0, ṡ( ) = v 0. (6) Solving this equation requires a certain insight that, fortunately, others have had earlier. Define the function: V (s) = where c is any constant. Then Eq. (6) becoes: s c F(s ), (7) Notice that: dt 2 = dv (s), s() = s 0, ṡ( ) = v 0. (8) d ( ( dt 2 ṡ2 + V (s)) = ṡ s + dv ) (s) = 0. (9) 3

4 Now it is iportant to interpret this equation correctly. It says that a solution of s + dv (s) = 0 ust also satisfy: 2 ṡ2 + V (s) = constant. (10) The constant is deterined by the initial conditions. This is a very iportant expression and later we will see how it is related to energy. F = F(ṡ,t). If we let ṡ = u, then Newton s equations becoe first order equations for the velocity: u = F(u, t), u( ) = v 0. If these equations could be solved then the velocity could be integrated to give the position. Unfortunately, even though they are first order, the general equation cannot be solved explicitly (although any special cases are known that can be solved). However, it is true that all linear first order equations can be solved, i.e., equations of the for: u = a(t)u + b(t), u( ) = v 0, (11) where a(t) and b(t) are continuous functions of t. Before deonstrating how this can be solved, let s first siplify this equation by getting rid of the annoying ass ter (we can restore it later). We do this by defining: Then (11) becoes: ā(t) a(t), b(t) b(t). u = ā(t)u + b(t), u( ) = v 0, (12) Now the trick to solving this equation is due to Johann Bernoulli. He proposed to write the solution of (12) in the for u(t) = n(t)(t). Substituting this into (12) gives: ṅ + ṁn = ā(t)n + b(t), u( ) = n( )( ) = v 0. (13) The solution of this equation can be obtained by breaking it into two pieces, and solving each piece separately (why?): The solution of (14) is given by: ṅ = ā(t)n, solve for n, (14) ṁ = b(t), integrate to get, with n substituted in fro above. (15) n This is then substituted into (15), and integrated, to obtain: To obtain u(t), we ultiply n(t) and (t) to obtain: R t t n(t) = n( )e ā(τ)dτ 0. (16) (t) = ( ) + 1 R τ ā(τ b(τ)e )dτ dτ. (17) n( ) 4

5 and reeber that R t t u(t) = n(t)(t) = n( )( )e ā(τ)dτ R t t R ā(τ)dτ τ ā(τ 0 t + e 0 b(τ)e )dτ dτ, (18) u( ) = n( )( ). This expression illustrates the fact that the general solution of (12) is the su of a solution to the hoogeneous equation: u = ā(t)u, which is the first ter in the solution (18), and a particular solution (the second ter in the solution (18). A particular solution is just any solution of: u = ā(t)u + b(t), where you don t worry about the initial conditions. The initial conditions are then satisfied for the su of the hoogeneous plus particular solution. F = F(s, t). Newton s equations are: dt 2 = F(s, t), s() = s 0, ṡ( ) = v 0. Equations of this type cannot generally be solved analytically, even if they are linear. However, there is one class of systes of this for which always have a solution: the linear, constant coefficient systes, i.e., systes of the for: = as + b(t), dt2 where a is a real nuber and b(t) is a continuous function of t. F = F(s, ṡ, t). Newton s equations are: dt 2 = F(s, ṡ, t), s() = s 0, ṡ( ) = v 0. Equations of this type cannot generally be solved analytically, even if they are linear. However, there is one class of systes of this for which always have a solution: the linear, constant coefficient systes, i.e., systes of the for: = aṡ + bs + c(t), dt2 where a and b are real nubers and c(t) is a continuous function of t. 5