* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Newtonian Mechanics: Rectilinear Motion - RIT
Hooke's law wikipedia , lookup
Specific impulse wikipedia , lookup
Theoretical and experimental justification for the Schrödinger equation wikipedia , lookup
N-body problem wikipedia , lookup
Lagrangian mechanics wikipedia , lookup
Modified Newtonian dynamics wikipedia , lookup
Frame of reference wikipedia , lookup
Routhian mechanics wikipedia , lookup
Relativistic quantum mechanics wikipedia , lookup
Inertial frame of reference wikipedia , lookup
Brownian motion wikipedia , lookup
Velocity-addition formula wikipedia , lookup
Relativistic mechanics wikipedia , lookup
Fictitious force wikipedia , lookup
Hunting oscillation wikipedia , lookup
Centrifugal force wikipedia , lookup
Derivations of the Lorentz transformations wikipedia , lookup
Newton's theorem of revolving orbits wikipedia , lookup
Centripetal force wikipedia , lookup
Seismometer wikipedia , lookup
Classical mechanics wikipedia , lookup
Rigid body dynamics wikipedia , lookup
Equations of motion wikipedia , lookup
Newtonian Mechanics: Rectilinear Motion Vern Lindberg May 25, 2010 1 Newton’s Laws Fowles Chapter 2.1 gives a wonderful historical perspective on the development of what we now call Newton’s Laws. It is worth a read. First Law Every body continues in its state of rest, or of uniform motion in a straight line, unless it is compelled to change that state by forces impressed upon it. This law guides us in the choice of an inertial reference frame, the only frame in which it is true. We can imagine carefully arranging a situation with no known net force, perhaps an air hockey table where the normal and gravitational forces exactly balance. If we give a puck a small push, it will move in a straight line until it hits the boundary of the table. Or will it? I will play part of a video showing a ball rolling on a horizontal table—it will move in a curve, violating the First Law. The reason is that the frame of reference of the camera is in a rotating reference frame in which Newton’s First Law does not work1 . http://www.youtube.com/watch?v=LAX3ALdienQ&feature=related. Motion in an inertial reference frame is usually the simplest to describe. We will discuss motion in a non-inertial reference frame later this quarter in Chapter 5. The concept of an inertial reference frame continues to be important if the areas of special and general relativity. So how do we find an inertial reference frame? In a small lab on earth, the earth is usually sufficient. If we look at long range motion of projectiles, or the motion of air on the earth, the appropriate inertial reference frame is one that uses the center of the earth as an origin, and the direction to the sun to define one of the coordinate axes. 1 Here is an animation http://www.youtube.com/watch?v=49JwbrXcPjc and another video http:// paer.rutgers.edu/pt3/experiment.php?topicid=8&exptid=166. 1 And so on. As we get to larger scale motion, or if we measure more precisely, we will need to be more careful on our choice of reference frame. Second Law The change in motion is proportional to the motive force impressed and is made in the direction of the line in which that force is impressed. Third Law To every action there is always imposed an equal reaction; or, the mutual actions of two bodies upon each other are always equal and directed to contrary parts2 . Newton’s Third Law is a careful definition of forces. Forces are interactions between objects (bodies). If we use F~12 to mean “the force of body 1 acting on body 2”, we can write Newton’s Third Law as F~12 = −F~21 (1) Next we define inertial mass. We take two objects, with one of them having a spring as part of it. We place the objects in an environment where no other forces act, and release the objects from rest with the spring initially compressed. We observe that after the objects have separated, they will have final velocities v~1 and v~2 in opposite directions. We define the ratio of masses as m2 ~v1 = (2) m1 ~v2 We currently choose a reference mass define as 1 kg exactly, made of platinum-iridium alloy and kept in a controlled atmosphere in Paris. Secondary references are kept throughout the world. (And there are problems. The primary reference mass appears to be losing mass relative to its brethren. See http://www.msnbc.msn.com/id/20744160/.) We can write Equation 2 as ∆(m1 v~1 ) = −∆(m2 v~2 ) (3) where ∆(m~v ) = m~v − 0 since the objects start at rest. And defining linear momentum p~ = m~v , ∆p~1 = −∆p~2 (4) Dividing by the time of interaction, ∆t, and taking the limit of small time interval we get dp~1 dp~2 =− (5) dt dt 2 When applied to electric and magnetic forces between two moving charges it may appear that the Third Law is violated. To correctly treat electromagnetic forces requires an electromagnetic field. Most texts in Electricity and Magnetism will discuss this. 2 Newton rigorously defined the “change in motion” in the Second Law as the time rate of change of momentum, so the Second Law can be stated mathematically as d~ p F~ = k dt (6) Here k is a constant that takes care of units. For example, if we wanted to measure force in pounds and momentum in kg· m/s, k = 0.2248. By choosing kg, m, s, N as our units, k = 1. The idea of having k 6= 1 may seem silly in this context, but consider Coulomb’s Law in electricity, F = kq1 q2 /r2 . In SI units k = 1/4π0 , in electrostatic units k = 1. We can look at the dynamics of a system of particles. An isolated system is one in which no external forces exist. By Newton’s Third Law, P the sum of internal forces is zero, therefore we can define a system momentum P~ = p~i and Newton’s Second Law for an isolated system becomes X X X dP~ F~ = F~external + F~internal = 0 = (7) dt Thus by invoking Newton’s Third Law we can extend Newton’s First Law to become the Law of Conservation of Linear Momentum. Conservation of Linear Momentum If the net external force acting on a system is zero, then the linear momentum of the system is constant. 1.1 The view from Landau and Lifshitz “To define the position of a system of N particles in space, it is necessary to specify N radius vectors, i.e. 3N co-ordinates. The number of independent quantities which must be specified in order to define uniquely the position of any system is called the number of degrees of freedom; here this number is 3N .” Landau and Lifshitz3 Degrees of freedom is a powerful concept that is used in a number of fields of physics and statistics. Newton’s Second Law is based on the fact that for a system of N particles, if at some instant we know the positions (3N values) and velocities (another 3N values) of the particles we can uniquely define the accelerations. This implies that the forces can be functions of time, position and velocity only. 3 MECHANICS by L. D. Landau and E. M. Lifschitz, Vol. 1 of Course of Theoretical Physics Second Edition (1965). 3 The equations that arise from Newton’s Second Law are relations between acceleration, velocity, and position, as well as time, and are the equations of motion. 2 Using Newton’s Second Law—Single Particle, Constant Mass We will consider cases of a single particle with a constant mass. Newton’s Second Law for a single particle (no rotational motion for point particle) can be written X d~ p d~v d2~r F~ = =m = m 2 = m~a dt dt dt (8) or using the dot notation X F~ = m~v˙ = m~r¨ (9) Fowles (and I) will begin with rectilinear motion–that is motion along a straight line so that we can drop the vector arrows and use + and − for directions of the component. Let’s call the axis of the motion x. The resulting Newton’s Second Law differential equation is called the Equation of Motion, since we can solve it (analytically or numerically) to determine the motion. Experimentally we find that forces depend on x, v, t only. X Fx (x, ẋ, t) = max = mv˙x = mẍ (10) Some classes of solutions are • Constant net force • Net force a function of position only • Net force a function of time only • Net force a function of velocity only • Combinations of the above We initially look at situations where a first-order differential equation suffices. 2.1 Constant Net Force Equation 10 tells us that if the net force is constant, then the acceleration is constant. We can drop the subscript x since we know we are in 1D, and dv F = dt m 4 (11) which can easily be integrated using initial conditions that at t = 0, v = v0 , x = x0 to get4 Z v Z t F dv = dt (12) v0 0 m F v − v0 = t (13) m Symbols with subscripts (x0 , v0 , t0 ) indicate fixed values while symbols without subscripts are variables. From the definition of velocity we can do another integral to get Z t Z x F v0 + t dt dx = (14) m 0 x0 F 2 (15) t x − x0 = v0 t + 2m These equations are the usual ones for constant acceleration that you have seen before. 2.2 Position Dependent Forces Consider cases where the 1D force depends only on position, F = F (x). Since the force depends only on position, x, we can do the following trick to get rid of the explicit reference to time. dẋ dẋ dx dv ẍ = = =v (16) dt dx dt dx This trick is used a lot, so be sure you understand it! The equation of motion then becomes dv m d(v 2 ) dT F (x) = mv = = (17) dx 2 dx dx where T = 12 mv 2 is the translational kinetic energy of the particle (remember we have no rotation here!) We can rewrite Equation 17, with initial conditions T = T0 , v = v0 when x = x0 , as Z x W = F (x) dx = T − T0 (18) x0 where the left hand integral is defined as the net work done. You should recognize this as the Work-Energy Theorem. 4 I am using somewhat sloppy notation form a mathematical viewpoint, having v as the integration variable and as a limit on the integral. Forgive me. 5 For some types of forces called conservative forces, we can write F (x) = − dV (x) dx (19) where V is the potential energy 5 . Defining V0 = V (x0 ), Equation 18 can be written as − V (x) + V0 = T (v) − T0 (20) Define Mechanical Energy E(x) = T + V and we get the Energy Equation E = T0 + V0 = T + V = const To find the motion we solve Equation 21 for velocity, r dx 2 v= =± [E − V (x)] dt m and integrate to get x, Z x x0 dx q = t − t0 2 ± m [E − V (x)] (21) (22) (23) Notice that for real answers, we must have V (x) ≤ E. This is the classical requirement that gets violated in quantum mechanics when we consider the particle to be represented as a wave. In principle, the integral can be evaluated to give (t − t0 ) = f (x) and this equation can be inverted to give x = x(t). In practice we may not be able either to evaluate the integral in closed form, or to invert the function. In those situations we must rely on numerical techniques. Equation 22 gives v = v(x). If we want to know v(t), we must solve for x(t) and either do a derivative or a substitution. Example Suppose we have a mass m attached to an ideal6 spring of constant k. If the initial conditions are T0 = 0, x0 = A at t0 = 0, Find v(x), t(x), x(t), and v(t). The elastic potential energy can easily be shown to be V (x) = 12 kx2 so we can write the total mechanical energy for this situation as 1 E = T0 + V0 = 0 + kA2 2 5 (24) Why T and V instead of K and U? Just the conventions of the field. Yes, in electricity and magnetism V means potential, not potential energy. We just have to deal with the ambiguity. 6 An ideal spring has no mass, and no length, and a force law of F (x) = −kx. 6 Hence from Equation 22 we get r r dx 2 kp 2 =± A − x2 v= [E − V (x)] = ± dt m m p To make this easier to write, define ω = k/m and we have √ v(x) = ±ω A2 − x2 . (25) Equation 23 becomes Z x A dx √ = t−0 ±ω A2 − x2 x x = ωt ± cos−1 A A (26) (27) The lower limit of evaluation is cos−1 (1) = 0, so we can write x ωt = ± cos−1 A This is easily inverted to give x(t) = A cos(±ωt) = A cos(ωt) since cosine is an even function. A derivative of this gives v(t) = −Aω sin ωt 2.3 Velocity Dependent Forces Forces like air drag are velocity dependent, F = F (v). Using initial conditions v = v0 , x = x0 at t = t0 , we can start with Equation 11 and rearrange to solve for time, F (v) = m dv dt m dv F (v) Z v m dv t − t0 = v0 F (v) dt = (28) (29) (30) We can also use Equation 16 to write Newton’s Second Law as F (v) = mv 7 dv dx (31) and rearranging get Z v m v dv F (v) x − x0 = v0 (32) So we have ways to get t(v) and can in principle invert to get v(t). We also have x(v) and can invert to get v(x). Combining we can get x(t). We will look at the example of drag in detail in Section 3. 2.4 Time dependent forces If F = F (t) we can do direct integration to get Z t v − v0 = t0 F (t)dt m (33) and the rather messy x − x0 = Z t Z v0 + t0 2.5 t t0 F (t0 )dt0 m dt (34) General Conclusions Doing the integrals in closed form (analytically) can be done sometimes, but other times we need to use numerical integration and the results are given as graphs or tables of data, rather than functions. Fortunately there are relatively straightforward methods to do this in Maple, Mathematica, or Easy Java Simulations. It is possible to do numerical integration in software languages such as FORTRAN, C or C++, and even Excel, but we then need to examine the algorithms much more closely. If the force is more general than just F (t), F (x), F (v) we do not have general methods of solutions, but start with Newton’s Second Law and use the methods of differential equations to solve. If the force is separable, e.g. F (v, t) = F1 (v)F2 (t) then the solution is usually easier. 3 Drag force When an object moves through a fluid, a drag force will arise. At low speeds the fluid moves smoothly (laminar flow), but at higher speeds viscosity begins to play an increasingly 8 important role and eventually dominates. We will look at simple situations of drag that can be solved analytically. The Reynold’s Number is a dimensionless parameter that delineates laminar from turbulent flow. For a sphere of diameter D moving with speed v through a fluid of density ρ and dynamic viscosity7 η it is defined as Re = vDρ η (35) The usual model of the drag force for an object moving with velocity ~v is F~drag = −c1~v − c2~v |~v | (36) or in the case of rectilinear (1D) motion, Fdrag = −c1 v − c2 v|v| (37) The linear term is called the viscous drag, Stokes drag, or linear drag, and dominates at low speed, Re < 0.1. There is no turbulence in this case. The coefficient c1 has units kg/s, and for a sphere of diameter D in a fluid having viscosity η, c1 = 3πηD (38) For spheres in air, c1 = 1.55 × 10−4 D kg/(m s). At higher speeds, Re > 1000, the second term called the aerodynamic term, the turbulent term or the quadratic term dominates. For a sphere of cross-sectional area A = πD2 /4 in a medium of density ρ 1 c2 = cd ρA (39) 2 where cd is the drag coefficient, about 0.4 for rough spheres. Coefficient c2 has units kg/m. The equation can be generalized to shapes other than spheres by using the appropriate cross-sectional area. Some drag coefficients are given in Table 1 and on Figure 1. Drag coefficients depend strongly on the shape of the object and on the surface roughness. The extensive field of fluid mechanics deals with such complications. For a rough sphere in air, c2 ≈ 0.22D2 . At low speed viscous drag dominates, while at high speed aerodynamic drag dominates. For 0.1 < Re < 1000 we have a broad transition 7 The SI unit of viscosity is kg/(s·m)≡ (Pa·s). There is a cgs unit, 1 poise = 1 g/(s·cm. 1 kg/(s·m) = 10 poise. The viscosity of water at 25◦ C is 0.00898 poise and 0.000184 poise for air at this temperature. Kinematic viscosity, ν, is the ratio of dynamic viscosity to density, ν = η/ρ. The symbol µ is sometimes used for dynamic viscosity. 9 Table 1: Some drag coefficients (Wikipedia) http://en.wikipedia.org/wiki/Drag_ coefficient smooth sphere 0.1 rough sphere 0.4 2004 Honda Accord 0.29 2003 Hummer H2 0.57 bullet 0.295 flat plate, face on 1.28 Figure 1: Drag coefficients for various shapes (Wikipedia) range of velocities where both the linear and quadratic terms apply. The center of the transition in air will occur when c1 = c2 , or c2 1.55 × 10−5 = vD = 1 c1 0.22 (40) Solving for the speed, in m/s when D is in m, v= 7.14 × 10−4 D (41) Thus for a BB pellet, diameter 4.5 mm, the transition speed is 15 cm/s. So for most of its motion the BB is moving much faster and the quadratic term is the important one. For the BB in air (ρ = 1.204 kg/m3 , η = 1.983 × 10−5 kg/(m s)) at 15 cm/s, Re = 40 which is in the transition region as expected. 10 CAVEAT: Drag is more complicated than Equation 37 suggests, especially in the transition region. 3.1 Horizontal motion with linear drag A block of mass m moves along a horizontal frictionless surface in a fluid at speeds where linear drag dominates. Find the motion subject to initial conditions v = v0 , x0 = 0 at t0 = 0. The only unbalanced force acting is the linear drag, F = −c1 v. dv − c1 v = m dt Z Z t m v dv dt = − c1 v0 v 0 (42) (43) and upon evaluating the integral m ln(v/v0 ) c1 (44) v = v0 exp(−t/τ ) (45) t=− which is easily inverted to where a characteristic time τ = Z m c1 . Then using v = dx/dt we write x Z dx = 0 t v0 exp(−t/τ )dt (46) 0 x = v0 τ (1 − exp(−t/τ )) (47) Now check limiting cases. At t = 0 we have v = v0 and x = 0 as required by the initial conditions. As t → ∞ v → 0 and x → v0 τ . The block moves a finite distance to the right and stops. 3.2 Vertical Motion Near Earth With Linear Drag An object of mass m is falling vertically in a fluid at speeds that provide a linear drag. This occurs near the earth with a constant gravitational field ~g . Initially at time t0 = 0 the object has position x0 = 0 and velocity v0 . 11 Fowles and Cassiday pick up as the positive direction. For variety I will choose down as the positive direction. The equation of motion, Newton’s Second Law is then mg − c1 v = m dv dt (48) The terminal velocity vterminal , occurs when the acceleration is 0, hence vterminal = mg/c1 . Rearrange to get Z v Z t mdv dt = (49) v0 mg − c1 v 0 Evaluate the integral (e.g change variables, u = mg − c1 v) and introduce τ = m/c1 = vterminal /g as before to get mg − c1 v mg − c1 v0 1 − v/gτ = −τ ln 1 − v0 /gτ t = −τ ln (50) (51) With a little bit of algebra—N.B. you should always fill in the steps of algebra— this can be inverted to give v = gτ + (v0 − gτ )e−t/τ (52) Note that at t = 0 this gives v = v0 as it must, and as t → ∞, v ≡ vterminal → gτ , the terminal velocity. We can rewrite Equation 52 as v = v0 e−t/τ + vterminal (1 − e−t/τ ) (53) and note that the initial velocity v0 exponentially disappears while the terminal velocity vterminal exponentially appears. Since v = dx/dt we can integrate the velocity with respect to time to get −t/τ x = vterminal t + (v0 − vterminal )τ 1 − e (54) and for t = 0 this gives x = 0 as required, and as t → ∞, x → vterminal t + (v0 − vterminal )τ ≈ vterminal t 3.3 Horizontal Motion With Quadratic Drag A block moves on a horizontal frictionless surface where the net force is quadratic drag, F = −c2 v|v|. At time t0 = 0, x0 = 0, v = v0 . Find the subsequent motion. 12 We can assume that v0 > 0 so that we can write the equation of motion − c2 v 2 = m Hence Z t t= v Z dt = 0 v0 dv dt −mdv m = c2 v 2 c2 (55) 1 1 − v v0 (56) Defining k = c2 v0 /m and inverting this we get v= v0 1 + kt (57) Note that at t = 0, v = v0 as required, and as t → ∞, v → 0. Integrating again Z x Z t v0 dt v0 dx = = ln(1 + kt) x= 1 + kt k 0 0 (58) At t = 0, x = 0 as required, but as t → ∞, x → ∞. Is this reasonable? If you project a block at high speeds will it continue on indefinitely? If not, what would make it stop? 3.4 Vertical Motion Near Earth With Quadratic Drag For the linear 1D case we write F ∝ −v. The equivalent expression for the quadratic case is F ∝ −v|v|. Having absolute values in calculus expressions is tricky. The usual procedure is to separate the cases of positive and negative velocity components. The text (and I) will look at the case of downward motion, and choose downwards as the positive direction. Then we can write the equation of motion mg − c2 v 2 = m dv dt (59) Suppose the initial conditions are v = v0 ≥ 0, y = 0 at t = t0 . The terminal velocity, vt , occurs when the acceleration is 0, vt2 = mg/c2 . Then dv v2 =g 1− 2 (60) dt vt Rearranging we get the integral Z t (t − t0 ) = Z v dt = t0 v0 13 dv g 1 − v 2 /vt2 (61) Now to evaluate the integral. My CRC Handbook gives Z 1 dx a+x = ln a>x 2 2 a −x 2a a − x but this is not the result given in the text. Going to http://www.quickmath.com/ I get (for x < a) Z dx 1 = tanh−1 (x/a) 2 2 a −x a This is the form used in the text. Hence, evaluating the integral and solving for v, (t − t0 ) −1 v0 v = vt tanh + tanh τ vt (62) When t = t0 , v = v0 and when t → ∞, v → vt , so vt is the terminal velocity. Note that this works providing v0 < vt since tanh−1 v0 /vt is undefined for an argument greater than 1. If this is the case, we need to redo the integral using another form. Using this to determine y from integration yields a very messy equation that will not be used. Instead we will pursue a way to get v as a function of y using the trick dv dv dy dv 1 dv 2 = =v = dt dy dt dy 2 dy Putting this into Equation 60 gives v2 dv 2 = 2g 1 − 2 dy vt Z y= y Z v dy = 0 v0 1 dv 2 vt2 1 − v 2 /vt2 ln = − 2g 1 − v 2 /vt2 g 1 − v02 /vt2 (63) (64) and this can be inverted to give 2 2 v 2 = vt2 1 − e−2gy/vt + v02 e−2gy/vt Calling Lc = vt2 /2g a characteristic length, v 2 = vt2 1 − e−y/Lc + v02 e−y/Lc (65) (66) and we see the initial velocity (squared) fade out and the terminal velocity (squared) fade in with a characteristic length. 14 3.5 Getting Quantitative Suppose we are looking at spheres falling in air. Examples are rain drops and pollen. Given the diameter and mass we want to determine whether to treat the sphere using linear drag, quadratic drag, or worst of all have to consider both and the transition (we would use numeric techniques for this.) We will use the expressions for the drag coefficients of spheres in air, and the expressions for terminal velocity, all using SI units. r mg mg −4 2 c1 = 1.55 × 10 D c2 = 0.22D vt (linear) = vt (quadratic) = c1 c2 So if we know m and D (and assume we are on the earth with known g) then we can calculate the terminal velocities that would result in the linear and quadratic cases. For example a water drop of diameter 10µm has mass 4.2 × 10−12 kg, and we compute linear terminal velocity of 2.65 × 10−7 m/s and a quadratic terminal velocity of 1.37 × 10−5 m/s. If the drop is released from rest, treating it with linear drag is an excellent approximation. If instead we have a water drop of diameter 4 mm with mass 2.7×10−4 kg, and we compute linear terminal velocity of 16.9 m/s and a quadratic terminal velocity of 0.11 m/s. If the drop is released from rest, quadratic drag is a good approximation. If we have a water drop of diameter 0.15 mm with mass 1.4 × 10−8 kg, and we compute linear terminal velocity of 8.94 × 10−8 m/s and a quadratic terminal velocity of 7.9 × 10−4 m/s. If the drop is released from rest, we will need to model more exactly using both linear and quadratic terms. Alternately we can look at the characteristic times τ (linear) = m/c1 and τ (quadratic) = p m/c2 g. When τlinear τquadratic we can use the linear drag. When τlinear τquadratic we can use quadratic drag, and when the two are close we must do more complete modeling. Consider finally a ping-pong ball in air. It has mass 2.7 grams and diameter 40 mm. When released from rest, how should drag be calculated? We find that the linear τ = 17.4 s and the quadratic τ = 0.035 s. Hence we use quadratic drag. Example 2.4.3 gives a third method of deciding. 15