Motion of a Particle in Three Dimensions

Transcription

1 Motion of a Particle in Three Dimensions V. Lindberg June 17, General Principles What changes as we move to 2D or 3D motion? We must deal with vectors and components. In Cartesian coordinates Newton s Second Law becomes F x = mẍ (1) F y = mÿ (2) F z = m z (3) Each of the force components can be functions, explicitly or implicitly, of position, x, y, z, velocity, ẋ, ẏ, ż and time, t. The differential equations can quickly become much more complicated than those we saw in 1D, and in general will be partial differential equations. We will begin with a constant force. Cases of force depending solely on position will be treated next, followed by simple cases where the force depends solely on velocity or solely on time. Finally a few words will be said about forces that depend on more than one type of variable. For the moment we will not worry about angular momentum. 1.1 Work in 3D Consider the dot product of Newton s Second Law with velocity, F v = d p d(m v) v = v (4) dt dt Assuming constant mass, we can recognize that v v = 1 d( v v) 2 dt F v = dt dt 1 and hence (5)

2 where T is the kinetic energy of the particle. Now v = d r/dt so we can write F d r = dt (6) dt dt F r = dt (7) and integrating using a line integral between points A and B, W = B A F r = T B T A = T (8) Keep in mind that F is the net force on the particle. Example Suppose that the net force on a particle is F = 3xyzî + 2xĵ + 2xyˆk N. (a) Suppose we start at (0,0,0) move in a straight line to (0,2,0), then in a straight line to (3,2,0), with all the distances in meters. Find the work done in moving along this path. (b) Instead suppose we start at (0,0,0) move in a straight line to (3,0,0), then in a straight line to (3,2,0), with all the distances in meters. Find the work done in moving along this path. (c) Could this force be conservative? Why or why not? 1.2 Force Fields and Conservative Forces In University Physics you discussed potential energy, perhaps using one of the definitions The net work done by a conservative force on a particle moving around any closed path is zero. or The work done by a conservative force on a particle moving between two points does not depend on the path taken by the particle 1. We want to discuss the details of this statement and find a reliable method to decide whether a force is conservative. A pictorial method of showing a force (at least in 2D) is to take a coordinate grid, and at regularly spaced points in the grid to draw short vectors that have the properties that the length of the vector is proportional to the force at that point, and the direction indicates the direction of the force. Examples are shown in the text, Figures and Suppose we have a force defined by F x = by and F y = +bx where b is a positive constant. Figure shows this. Consider going around a closed rectangular path from (x, y) to 1 Fundamentals of Physics, Halliday, Resnick & Walker, 8 th edition. 2

3 (x + x, y) to (x + x, y + y) to (x, y + y) and back to (x, y), and computing the work. (The circle on the integral sign indicates a closed path.) W = F d r = = x+ x x x+ x x y+ y x F x (y)dx + F y (x + x)dy + y [F x (y) F x (y + y)]dx + y+ y = [b(y + y) by] x + [b(x + x) bx] y = 2b x y y x+ x y F x (y + y)dx + F y (x)dy y+ y [F y (x + x) F y (x)]dy For this force, the work done around a closed path is non-zero, therefore this force is non-conservative and has no potential energy. Try a different force, defined by F x = +by and F y = +bx where b is a positive constant. The force field is shown in Figure Evaluating the closed-loop work integral for the same path results in W = 0, meaning that the force is conservative, and that a potential energy does exist. This is still a tedious method (and we need to check that the integral is zero for all closed paths) so we want something better. Stokes Theorem tells us that in general F d r = ( F ) ˆn da (10) surf The closed path of the line integral defines a surface, and the dot product between the curl of the force and the normal to the surface is evaluated over the area. The criterion that we have a conservative force when the closed line integral is zero is thus equivalent to saying that the curl of the force must be zero: For a conservative force, one that has a potential energy defined, F = 0. ConservativeForce (11) (9) 1.3 Computing curl In Cartesian coordinates we can easily write the curl using a determinant. î ĵ ˆk F = det x y z F x F y F z 3

4 ( Fz = î y F ) ( y Fx + ĵ z z F ) ( z + x ˆk Fy x F ) x y (12) Think about this in terms of cyclic permutations: The first term has î, y in the denominator, z in the numerator: think of this as (xyz), x-component is partial with y of F z. The curl terms consist of [(xyz) (xzy)]+[(yzx) (yxz)]+[(zxy) (zyx)]. Getting used to thinking in cyclic permutations takes a little effort, but is very valuable. We may want to find curls in other coordinate systems. Appendix F gives some details, and I will summarize it here. Suppose we have a general orthogonal coordinate system of coordinates (u, v, w) and associated unit vectors (ê 1, ê 2, ê 3 ): for example, spherical polar coordinates (r, θ, φ). It is right handed, meaning ê 1 ê 2 = ê 3. There are associated with the coordinates Lamé factors (or scale factors) (h 1, h 2, h 3 ) that allow us to write volume elements, line elements, gradients, divergences, and curls 2. For spherical coordinates they are (1, r, r sin θ). The volume element in general is The line element is Gradient is Divergence is (note the cyclic nature!) [ Q 1 (h2 h 3 Q 1 ) = h 1 h 2 h 3 u dv = h 1 h 2 h 3 du dv dw (13) d r = ê 1 h 1 du + ê 2 h 2 dv + ê 3 h 3 dw (14) f = ê1 h 1 f u + ê2 h 2 f v + ê3 h 3 f w + (h 3h 1 Q 2 ) v + (h ] 1h 2 Q 3 ) w (15) (16) and curl is Q = 1 det h 1 h 2 h 3 h 1 ê 1 h 2 ê 2 h 3 ê 3 u v w h 1 Q 1 h 2 Q 2 h 3 Q 3 (17) For Cartesian coordinates, h 1 = h 2 = h 3 = 1, for spherical polar, h 1 = 1, h 2 = r, h 3 = r sin θ,and for cylindrical coordinates (R, φ, z), h 1 = h 3 = 1, h 2 = R. Example 1 Suppose F = Ax 2 yî Axy 2 ĵ + Czˆk with A a positive constant. Is this a conservative force? 2 For a description of how to find the Lamè factor, try phys301/classnotes/scalefactorscomplete.pdf 4

5 Example 2 Suppose F = Ar 2 ê 1 in spherical polar coordinates with A a positive constant. Is this a conservative force? Example 3 Suppose F = Ar 2 cos θê 1 + Br 2 sin θê 2 + Cê 3 in spherical polar coordinates. Could this be a conservative force for the proper choice of constants A, B, C? If so what are the values of these constants? Example 4 Suppose F = Ar cos 2 θê 1 + Br 2 sin θ cos θê 2 + Cê 3 in spherical polar coordinates. Could this be a conservative force for the proper choice of constants A, B, C? If so what are the values of these constants? 2 Gradients: Relating Force to Potential Energy We now have an easy way to determine if a force is conservative. When the force is conservative we want to find its potential energy function, V (x, y, z) (or other orthogonal coordinates.) Conversely, if we know the potential energy function, we want to find the vector force. The second task is easier, so we start there. Given V, we can find the force by F = V = V x V y V z (18) where the gradient is given in Cartesian coordinates. It is a simple matter to show that V = 0 verifying that the force is conservative. Example 1 What force is associated with V = 1 2 k(x2 + y 2 + z 2 )? Example 2 What force is associated with V = 1 2 kr2 in spherical polar coordinates? Example 3 What force is associated with V = A 3 r2 cos θ in spherical polar coordinates? So now let s do the reverse, find a potential energy related to a force. F = k 1 xî k 2 yĵ k 3 zˆk. This is an anisotropic spring force. Suppose that We know that V x = F x, so we integrate dv = F x dx to get V = 1 2 k 1x 2 +A(y, z). Likewise from the other force components we get V = 1 2 k 2y 2 + B(x, z) and V = 1 2 k 3z 2 + C(x, y). Combining these we get V (x, y, z) = 1 2 (k 1x 2 + k 2 y 2 + k 3 z 2 ) + V 0 (19) As another example, suppose that in spherical polar coordinates F = Ar 2 cos θê 1 (Ar 2 /3) sin θê 2, with A being a positive constant. Referring back to Equation 15 and using h 1 = 1, h 2 = 5

6 r, h 3 = r sin θ we get the equations F r = Ar 2 cos θ = 1 V 1 r F θ = (Ar 2 /3) sin θ = 1 r F φ = 0 = 1 V r sin θ φ V θ (20) Doing the integrals results in Combining V = Ar3 3 cos θ + A(θ, φ) V = Ar3 cos θ + B(r, φ) 3 V = C(r, θ) (21) V = Ar3 3 cos θ + V 0 (22) where V 0 is a constant (positive or negative). 3 Separable Forces A separable force can be written in Cartesian Coordinates as F = F x (x)î + F y (y)ĵ + F z (z)ˆk (23) Almost by inspection you can see that this is a conservative force. The equations of motion are independent, F x (x) = mẍ F y (y) = mÿ F z (z) = m z (24) Solving these equations by the methods of Chapter 2 gives us r = x(t)î+y(t)ĵ+z(t)ˆk. In chapter 2 we also looked at forces that were functions of velocity as well. In the case where each component depends on the corresponding position and velocity components, the equations of motion look like mẍ = F x (x, ẋ, t) and these can in principle be solved fairly easily. 6

7 3.1 Constant Force: Projectile Motion with No Drag Consider projectile motion in a uniform and constant 3 gravitational field g in a non-rotating coordinate system.. We choose the field to define the z-direction, with up as positive. This is the usual case of two dimensional projectile motion, and we look to see how our formalism gives the well known results. The equation of motion is m d2 r dt 2 = md v dt = mgˆk (25) If we choose the coordinates so that the initial velocity v 0 is in the x-z plane, then y = const and the motion is two dimensional. For simplicity we choose y = 0 so that the location at arbitrary time can be written r = xî + zˆk (26) Integrating the equation of motion once gives Integrating again, with the initial position being r 0 gives v = ˆkgt + v 0 (27) r = ˆk 1 2 gt2 + v 0 t + r 0 (28) The force is conservative with a potential energy V = mgz + V 0 and for convenience we define V 0 = 0 at z = 0. The energy equation, T 0 + V 0 = T + V thus becomes 1 2 m(ẋ2 0 + ż 2 0) + mgz 0 = 1 2 m(ẋ2 + ż 2 ) + mgz = 1 2 mv2 + mgz (29) Frequently we can choose z = 0 at t = 0 so that the energy equation just becomes 1 2 mv2 0 = 1 2 mv2 + mgz v0 2 2gz = v 2 (30) We can write the initial velocity using magnitude and direction as v 0 = (v 0 cos α 0 )î + (v 0 sin α 0 )ˆk where α 0 is the angle from the x axis. (The text uses α.) 3 Recall the meaning of these two terms: uniform means having the same value for all locations in space, but possibly changing in time; constant means unchanging in time, but possibly varying with spatial location. 7

8 Then the solutions to the equation of motion are v = î(v 0 cos α 0 ) + ˆk(v 0 sin α 0 gt) (31) r = î(v 0 cos α 0 )t + [(v ˆk 0 sin α 0 )t 12 ] gt2 (32) The position components are x = (v 0 cos α 0 )t (33) y = 0 (34) z = (v 0 sin α 0 )t 1 2 gt2 (35) and eliminating time we can write ( ) g z = (tan α 0 )x 2v0 2 x 2 (36) cos2 α 0 which is the equation of a parabola. These results should all be familiar to you, and it is easy to use them to get quantities such as the text does: Maximum height where ż = 0, Time to reach maximum height, Time to return to original height, Range = horizontal distance when particle returns to original height. Equation in the text has an error. It uses the trig identity 2 sin α 0 cos α 0 = sin 2α 0 (See Appendix B). The correct result is R = x = v2 0 sin 2α 0 g (37) 3.2 Velocity Dependent Force: Projectile Motion With Linear Drag We have already discussed air drag and found that for most objects the drag is quadratic. With quadratic drag in 2D, the solution is found numerically. We will assume that the drag is linear because then we can solve the result analytically, even if does not apply to most projectiles in air. In Chapter 2 we used F = c 1 v with c 1 = D in SI units. Choose the same orientation of coordinate system with initial velocity in the x-z plane and positive z upwards. For convenience let c 1 = mγ so that the equation of motion becomes m d2 r dt 2 = mγ v ˆkmg d 2 r dt 2 = γ v ˆkg (38) 8

9 In components ẍ = γẋ (39) ÿ = γẏ = 0 (40) z = γż g (41) This shows that the motion is separable and the solutions are variations of those seen in Chapter 2, namely ẋ = ẋ 0 e γt (42) ẏ = 0 (43) ż = ż 0 e γt g γ (1 e γt ) (44) or v = v 0 e γt ˆk g γ (1 e γt ) (45) and integrating again, or x = ẋ0 γ (1 e γt ) (46) y = 0 (47) z = r = (ż0 γ + g γ 2 ) (1 e γt ) gt γ ( v 0 γ + ˆkg ) γ 2 (1 e γt gt ) ˆk γ (48) (49) Notice that as t, ẋ 0 meaning that asymptotically the particle falls vertically, unlike parabolic motion, at a final horizontal location x ẋ 0 /γ. We have seen that the motion is not parabolic. We can again eliminate t to see what the motion actually is, resulting in (ż0 z = γ + g ) γx γ 2 + g ( ẋ 0 γ 2 ln 1 γx ) (50) ẋ 0 If we want to find the horizontal range, R, that is the distance travelled until the particle returns to its original height, we set z = 0 and solve the resulting transcendental equation. 9

10 Approximation methods must be used to solve this equation, starting with a Taylor series expansion of the natural logarithm. The result is R = 2ẋ 0ż 0 g = v2 0 sin 2α 0 g 8ẋ 0ż 2 0 3g 2 γ + (51) 4v3 0 sin 2α 0 sin α 0 3g 2 γ + (52) with the first term being the result with no drag, and the second term being the first order (in γ) correction for drag. Getting this result from Equation 50 is not a trivial project, and is well worth doing. 3.3 Quadratic Drag For projectiles like baseballs or ping-pong balls, the drag in air will be quadratic. For a one dimensional drop we used F = c 2 v v with c 2 = 0.22D 2 for spheres of diameter D in air (all in SI units). In two dimensions we can write F drag = c 2 v v = c 2 v 2 x + v 2 z(îv x + ˆkv z ) (53) This force is not separable, so the solution is much more difficult than that for linear drag, and we will not attempt it analytically. If we have values for all the quantities we can solve it numerically using, for example, EJS. You will do this as an assignment. 3.4 Other Forces on Projectiles In addition to gravitational and drag forces, real balls are subject to buoyant forces and, if they are spinning, Magnus forces. The buoyant force is a constant for motion in a uniform medium. The Magnus force 4 is actually an offshoot of drag and Bernoulli forces. Consider a baseball moving with a velocity v through air. It spins with an angular velocity ω, and at the surface of the ball, air is dragged with the ball. Imagine being in the frame of reference of the ball. The air moves with a translational velocity v, and in addition there is a velocity due to the rotation, of magnitude ωr. The net velocity of the air is the combination of these two, resulting in air having a higher velocity on one side of the ball than on the other, and from Bernoulli s Principle, the pressure is higher when the velocity is lower, resulting in a Magnus force proportional to ω v. 4 See 10

11 4 Position Dependent Forces: Harmonic Oscillator in 3D Suppose we have a particle subject to a linear restoring force F = k r (54) Producing such an isotropic restoring force is NOT easy. The text suggests a set of springs in Figure 4.4.1, but showing whether it really does produce the desired force is not simple 5. This is an example of a central force that, in spherical polar coordinates, depends only the magnitude of the radius vector and not its angular orientation. 4.1 Two Dimensional Case The equation of motion using the x-y plane is separable and is m d2 r dt 2 = k r mẍ = kx (55) mÿ = ky (56) The differential equation should be an old friend and defining ω = k/m, a solution is x = A cos(ωt + α) (57) y = B cos(ωt + β) (58) The four constants A, B, α, β are found from the initial conditions. To find the equation for the path we eliminate time from the equations. Defining = β α, x 2 ( ) 2 cos A 2 xy + y2 AB B 2 = sin2 (59) 4.2 A Primer on Conic Sections Both here, the two-dimensional harmonic oscillator, and later, planetary orbits, we will need to discuss conic sections: ellipses, parabolas, and hyperbolas 6. The term conic section 5 And I suspect that to make it really work that all the springs must initially be slightly stretched. You will solve a related problem in the second quarter of mechanics. 6 Those preferring older spellings will note that the plurals are parabolae and hyperbolae. 11

12 derives from the intersection of a plane surface with a cone, as is illustrated in http: //en.wikipedia.org/wiki/conic_section. There are several ways to describe the conic sections. First consider the geometric view. Two cones meet at an apex, and the z-axis is defined as the symmetry axis. If a plane surface is in the x-y plane, the intersection of the plane with the cone will be a circle, or a point. If the plane is parallel to the edge of the cones, the intersection will be either a parabola or a straight line. As we tilt the plane from the case of the circle to the case of the parabola, we get an ellipse. When the plane is tilted beyond the point of the parabola, we get a pair of hyperbolas. A second method describes the conic sections in terms of a fixed point, the focus, F, a line not through the focus called the directrix, L, and a non-negative real number, the eccentricity. Let us call the separation of focus and directrix q. These are illustrated in Figure 1 and Figure C.1a in Appendix C of the text. The eccentricity is defined as the ratio of lengths ɛ = F C/ĈL c. A circle has eccentricity ɛ = 0, and q =. The focus serves as the center of the circle. All points on the circle are equidistant from F. An ellipse has eccentricity 0 < ɛ < 1, and the distance from a point on the ellipse to F is ɛ times the distance from the point on the ellipse to L. This is shown in Appendix C of the text. Figure 1 has q = 3 and ɛ = 0.5. A parabola has ɛ = 1. Hyperbolas have ɛ > 1. Ellipses have a second symmetrically placed focus and directrix. Likewise hyperbolas have a second symmetrically placed focus and directrix. We can write a general equation for conic sections in terms of ɛ and q using the focal point as the origin. ɛq 1 r = 1 + ɛ cos θ r = 1 + ɛ cos θ (60) r 0 (1 + ɛ) where we have introduced r 0 = ɛq/(1 + ɛ). You can see what r 0 means on Figure 1. The first form of Equation 60 can be written in Cartesian coordinates as 7 (1 ɛ 2 )x 2 + 2ɛ 2 qx + y 2 = ɛ 2 q 2 (61) 7 This is easy to derive if you start down the correct path. From the first of Eq. 60 cross multiply to get r(1 + ɛ cos θ) = ɛq, use x = r cos θ, rearrange to get r on one side, and proceed. 12

13 Figure 1: A conic section symmetric about the x-axis with the origin at the focus, F. The directrix, L, is separated from the focus by a distance q. For a point on a conic section C, the eccentricity ɛ = F C/ĈL c where L c is a point on the directrix opposite C. In polar coordinates, C is at (r, θ), and r 0 is the value at θ = The Circle Using the second form in Equation 60 we see that for a circle with ɛ = 0, r = r 0 for all angles, or equivalently x 2 + y 2 = r The Ellipse Now consider the ellipse with 0 < ɛ < 1. The latus rectum, α, is the y coordinate when x = 0, θ = 90, and it is easy to show that it is α = ɛq. It is also helpful to know the height, 2b and width, 2a, of the ellipse, called the minor and major axes. Half of these are called the semi-minor and semi-major axes. To find the semi-minor axis, write the expression for y = r sin θ and find its maximum value. You will 13

14 find the maximum value occurs at cos θ = ɛ, and has a value b = ɛq 1 ɛ 2 (62) The minor axis occurs at a location x = c relative to the focus where c = ɛ2 q 1 ɛ 2 (63) The semi-major axis is then a = c + r 0 = ɛq 1 ɛ 2 (64) It is more common for ellipses to use the center of the ellipse as the origin. Thus we make a shift of coordinates y y, x x+c. I will now drop the primes. With a bit of work we can show that the equation of the ellipse relative to coordinates located at its center is x 2 a 2 + y2 b 2 = 1 (65) Foci at x = ±c a 2 b 2 = c 2 ɛ = c/a (66) b = a 1 ɛ 2 q = ±b 2 /c r 0 = (1 ɛ)a (67) The ellipse thus written is symmetrical about both x- and y- axes The Parabola For the parabola, ɛ = 1. I find it easier to consider a parabola opening to the left, similar to Figure 1. Calling r 0 c we quickly see that q = 2c. Beginning with the polar expression for the parabola using the focus as the origin, r = 2c/(1 + cos θ), with a bit of work we can get to the relation between the cartesian coordinates, y 2 = 4c(c x ). It is more useful to shift the origin to the apex of the parabola, x = x c and then find the Cartesian equation for a parabola with the origin at the apex, opening to the left y 2 = 4cx (68) The Hyperbola Figure 2 shows an hyperbola 8. Now ɛ > 1. (Note that the figure in Appendix C of the text, page A10, has an error. Can you see what it is?) If we use the center of the the two 8 Some authors write a hyperbola. 14

15 Figure 2: Hyperbola (both branches) with important parameters. branches for our origin, the hyperbola can be written x 2 a 2 y2 b 2 = 1 (69) The apex of the right branch is at x = a. At large magnitudes of x the branches are asymptotic to the lines y = ±bx/a, so b is the distance from the horizontal line to the asymptote at the apex. The triangle formed by a and b has a hypotenuse c = a 2 + b 2, and c is also the distance from the center to the focus. At the focus the latus rectum is α = b 2 /a. Other relations are a 2 + b 2 = c 2 ɛ = c a b = a ɛ 2 1 q = ± b2 c (70) Effect of Translation of the Origin Suppose we translate the origin both in x and y. It is easy to show that this introduces linear terms in x and y so that Ellipse b 2 x 2 + Bx + a 2 y 2 + Dy + E = 0 (71) 15

16 Parabola y 2 + By + E = 4cx (72) Hyperbola b 2 x 2 + Bx a 2 y 2 + Dy + E = 0 (73) Translation and Rotation Suppose the coordinate sytem is rotated and translated, the resulting conic function will be, in general, Ax 2 + Bxy + Cy 2 + Dx + Ey = F (74) To determine which conic section this is, compute the discriminant disc = B 2 4AC. Negative discriminants indicate ellipses, positive discriminants indicate hyperbolas, and zero indicates a parabola. It is a bit of work to show this in the general case, but if we consider just rotation about the origin, and use the rotation matrix from Chapter 1, it is relatively straight forward to show that the discriminant condition is valid. 4.3 Completing the Two Dimensional Isotropic Oscillator In Section 4.1 we obtained the equation of the particles motion, x 2 ( ) 2 cos A 2 xy + y2 AB B 2 = sin2 (75) where = β α is the phase difference between the motion along orthogonal axes, (x = A cos(ωt+α), y = B cos(ωt+β), ω = k/m.) It is quick work to calculate the discriminant and determine that the motion is an ellipse, and a lot more work to show that the major axis is rotated from the x-axis by an amount ψ, tan 2ψ = 2AB cos A 2 B 2 (76) Recall that trig functions are double valued: there are two values of 2ψ that give the same result. You will do an activity to try to clear this up D Isotropic Oscillator The equations of motions are mẍ = kx mÿ = ky m z = kz (77) 16

17 with solutions x = A 1 sin ωt + B 1 cos ωt y = A 2 sin ωt + B 2 cos ωt z = A 3 sin ωt + B 3 cos ωt (78) or in vector notation r = A sin ωt + B cos ωt (79) Vectors A and B define a plane, and the motion is an ellipse located in that plane. Planar motion is characteristic of central forces as we shall see in Chapter Non-Isotropic Oscillator With different elastic constants in three orthogonal directions, we have equations of motion mẍ = k 1 x mÿ = k 2 y m z = k 3 z (80) and solutions x = A cos(ω 1 t + α) y = B cos(ω 2 t + β) z = A cos(ω 3 t + γ) (81) The path of the object is constrained to be in a box of dimensions 2A, 2B, 2C centered on the origin. If A, B and C are commensurate, that is any ratio of the constants is a rario of integers, the path is a closed Lissajous figure. Otherwise the path does not close and will eventually fill the box. An example of a non-isotropic oscillator is an atom in a crystal. 5 Charged Particles in Electric and Magnetic Fields Consider a charged particle of mass m and charge q moving with velocity v in electric and magnetic fields. The Lorentz force is F = q( E + v B) (82) 5.1 Electric Field Only The equation of motion is mẍ = qe x mÿ = qe y m z = qe z or m d2 r dt 2 = q E (83) 17

18 In general the electric field can be a function of location and time (although that will introduce a magnetic field). Consider the simplest case of a constant electric field with axes chosen so that the field is along the z axis. The equations of motion reduce to mẍ = 0 mÿ = 0 z = qe/m = constant (84) and we once again have the equations of motion under a constant acceleration. In E & M you will see that when static charges produce a field, E = 0 and therefore the static electric field can be written in terms of an electric potential with the potential energy being V = qφ. 9 E = Φ (85) 5.2 Static Magnetic Field Only The equation of motion is m d2 r dt 2 = q v B (86) We are here concerned only about a static field (unchanging in time) but it could depend on location. For simplicity, consider the case of a constant magnetic field that we choose to be along the z direction, B = Bˆk. The equation of motion is not separable, but becomes m d2 r dt 2 = q v Bˆk = qb î ĵ ˆk ẋ ẏ ż (87) m(ẍî + ÿĵ + zˆk) = qb(ẏî ẋĵ) (88) Equating components, letting ω = qb/m we get the following differential equations that we can integrate once with respect to time ẍ = ωẏ ÿ = ωẋ z = 0 (89) ẋ = ωy + C 1 ẏ = ωx + C 2 ż = constant = ż 0 (90) Putting the expression for ẏ into the first equation in (89), and defining a = C 2 /ω, we can get ẍ + ω 2 x = ω 2 a (91) 9 Yup notation change here. In Physics III you used U for potential energy, V for electric potential. In this course we use V for potential energy and Φ for electric potential. 18

19 with solution It is then a relatively simple matter to find and x = a + A cos(ωt + θ 0 ) (92) y = b A sin(ωt + θ 0 ) (93) v 2 xy = ẋ 2 + ẏ 2 = A 2 ω 2 = A 2 ( qb m ) 2 (94) This tells us that the speed of the particle is constant, and you should recognize that the motion is circular in the x y plane with motion along z making a helix. Notice that the size of the helix, A depends on the speed in the x y plane, but that the frequency of rotation is constant. This is the basis for the cyclotron, an early (non-relativistic) particle accelerator. Suggested tasks: What is the pitch of the helix? In cylindrical polar coordinates, what are the components of the motion? Motion of charged particles in combined electric and magnetic fields is common. One example is a magnetron used in deposition of material. A non-uniform magnetic field is produced near a target material, and an electric field is perpendicular to the target material. Charged particles (electrons and positive ions) are concentrated in the vicinity of the target and the light electrons have a large mean-free path between collisions. 6 Constraint Forces There are many situations where a particle is constrained in its motion. Consider a bead on a wire: it is constrained to move along the wire and cannot move off the wire. Consider a piece of ice in a bowl: it is constrained to either move on the surface of the bowl or into space above the bowl, but it cannot penetrate the bowl. These are examples of complete and one-sided constraints respectively. Next quarter when we talk about Lagrangians we will term these holonomic and non-holonomic constraints. It is sometimes useful to describe the bead on a wire as a one-dimensional problem, meaning one variable (length along wire) is sufficient to uniquely locate the bead. The ice in the bowl, providing the ice stays on the surface, is two-dimensional problem. This will be useful in Lagrangians, Chapter 10. In the simplest case, the force introduced by the constraint, R, is normal to the surface, and the velocity of the particle, v, is tangent to the surface. If the other forces are a combined value F we write m d v dt = F + R (95) 19

20 m d v dt v = F v + R v (96) ( ) d 1 m v v = F dt 2 v (97) If the other forces are conservative, we can still write conservation of mechanical energy. Eg Period of Ice slipping in a spherical bowl. Consider a frictionless spherical bowl of radius R. A small object of mass m is released from rest at an initial angle θ 0 measured from the vertical. Find an expression for the period of the oscillation. We can write the energy equation Solving for speed as a function of angle, E = 1 2 mv2 mgr cos θ = mgr cos θ 0 (98) v = 2gR(cos θ cos θ 0 ) (99) We can find the time to move from an angle θ to θ + dθ: the distance travelled is Rdθ at a speed v, so Rdθ dt = (100) 2gR(cos θ cos θ0 ) and we integrate to find the period, T θ0 4 = 0 Rdθ 2gR(cos θ cos θ0 ) (101) This integral cannot be done in closed form: it is an elliptic integral, and we will discuss this further in Chapter 8. 20