21.60 Worksheet 8 - preparation problems - question 1:

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1 Dynamics Worksheet 8 - preparation problems - question 1: A particle of mass m moes under the influence of a conseratie central force F (r) =g(r)r where r = xˆx + yŷ + zẑ and r = x + y + z. A. Show that the angular momentum ector L = mr ṙ is a constant of motion. Guidance: Consider the deriatie of L. Don t expands it in components. B. Assuming that F (r) = kr where k is a positie constant, find the general solution for r(t). C. Assume now that the particle is projected from a point distance a from the force centre along the ˆx direction, with elocity in the ŷ direction. Show that the orbit is gien by: and determine A, B and ω in terms of k, m, and a. r(t) =A cos(ωt)ˆx + B sin(ωt)ŷ D. Sketch the trajectory and show that it is gien by an ellipse of the form: and determine α and β. x α + y β =1. E. Express your solution for the orbit as r(θ) using (plane) polar coordinates, where r = x + y and tan(θ) = y/x. Guidance: Find tan(ωt) in terms of tan(θ), use this to eliminate the dependence on time, expressing the coordinates x and y in terms of tan(θ), and then find r(θ). F. Use your solution to determine the components of the energy: (i) determine the kinetic energy T as a function of time. (ii) determine the potential energy V as a function of time. (iii) summing up the kinetic and potential energy erify that the total energy is consered. Solution: A. The angular momentum is L = mr ṙ To show it is consered we need to show that its time deriatie anished. It is gien by: d L = m(ṙ ṙ + r r) dt where we used the product rule. The first term anishes as a cross product between a ector an itself. In the second we substitute the equation of motion m r = g(r)r, getting: d L = mr r = r g(r)r =0 dt where again we use the fact that a cross product of a ector with itself anishes. We thus erified that with d the gien (central) force law, dtl, so the angular momentum is consered. B. For F (r) = kr the EoM is: m r = kr = r + ω r =0 where ω = k/m. We note that this is a linear equation. In components this looks familiar: ẍ + ω x =0, ÿ + ω y =0, z + ω z =0. For each component we hae a simple harmonic oscillator equation. The general solution is: r = A cos(ωt)+b sin(ωt), where A and B are two constant ectors (yet unknown).

2 Dynamics 191 C. We found aboe the general solution. Its deriatie is: ṙ = ωa sin(ωt)+ωb cos(ωt). The initial conditions are: r(0) = (a, 0, 0) = aˆx, ṙ(0) = (0,,0) = ŷ. Substituting these into out general solution: aˆx = A, ŷ = ωb So the solution is: r = aˆx cos(ωt)+ ω ŷ sin(ωt), so we see that A = a and B = ω, where, as we defined aboe, ω = k/m. D. When t = 0 the position in the (x, y) plane is r =(a, 0), at t = π/ω the position is r =(0, /ω), then at t = πω the position is r =( a, 0) and at t =3π/ω the position is r =(0, /ω). Finally, at π/ω the body returns to the original position (a, 0) with the original elocity. So the motion is periodic. It is conenient to write: x = a cos(ωt), y = ω sin(ωt) So x a = cos(ωt), y ω =sin x (ωt) = a + y ω =1 which is the equation for an ellipse. This is the form in the question with α = a and β = /ω. E. We may conert it to be written as an orbit, namely r as a function of θ. as follows: or tan θ = y x = ωa tan(ωt) tan(ωt) = ωa tan(θ)

3 Dynamics 19 so so and cos(ωt) = tan (ωt) = 1, sin(ωt) = 1+ ω a tan (θ) a x =, y = 1+ ω a tan (θ) tan(ωt) 1 + tan (ωt) = a tan(θ) 1+ ω a tan (θ) 1 + tan (θ) r = a 1+ ω a tan (θ) = a cos (θ)+ ω a sin (θ) F. Use your solution to determine the components of the energy: (i) determine the kinetic energy T as a function of time. T = m (ẋ +ẏ )= m (ii) determine the potential energy V as a function of time. V = F dr = k r = k a ω sin (ωt)+ cos (ωt) a cos (ωt)+ sin (ωt) ω ωa tan(θ) 1+ ω a tan (θ) Using ω = k/m we get: V = m a ω cos (ωt)+ sin (ωt) (iii) summing up the kinetic and potential energy, erify that the total energy is consered. We get: E = T + V = m a ω + = k a + m.

4 Dynamics Worksheet 8 - Workshop problems - question 1: In each of the following cases, is this a conseratie force? If yes, determine the corresponding potential. A. F (r) =xˆx + zŷ + yẑ B. F (r) =xˆx + zxŷ + yẑ C. F (r) = 3e 3x yz ˆx +e 3x z ŷ + e 3x yzẑ Guidance: In each case, assuming it s a conseratie force, write the equations relating V to the components of the (x,y,z) force, example F x (x, y, z) = x. Integrate these relations to find V (x, y, z) or reach a contradiction. Solution: A. The force is: F (r) =xˆx + zŷ + yẑ Thus for it to be a conseratie force the potential V (x, y, z) must admit: Integrating (416a) we get: Differentiating this result with respect to y we get: x =x y = z z = y V (x, y, z) = x + U(y, z) y = U y (416a) (416b) (416c) and by comparing to (416b) we find: U y = z = U(y, z) = yz + W (z) = V (x, y, z) = x yz + W (z) Differentiating this with respect to z we get: z = y dw dz and comparing with (416c) we see that this is consistent and W must be a constant, which we may choose to be zero. So V (x, y, z) = x yz. B. The force is: F (r) =xˆx + zxŷ + yẑ Thus for it to be a conseratie force the potential V (x, y, z) must admit: Integrating (417a) we get: Differentiating this result with respect to y we get: x =x y = xz z = y V (x, y, z) = x + U(y, z) y = U y (417a) (417b) (417c)

5 Dynamics 194 Comparing with (417b) we get: U y = xz (?!) but we know that U is a function of y and z only, so its deriatie cannot depend on x and we conclude that this is not a conseratie force. C. The force is: F (r) = 3e 3x yz ˆx +e 3x z ŷ + e 3x yzẑ Thus for it to be a conseratie force the potential V (x, y, z) must admit: x = 3e3x yz y = e3x z z = e3x yz Integrating (418a) with respect to x we get: V (x, y, z) = 3 dxe 3x yz = e 3x yz + U(y, z) (418a) (418b) (418c) Differentiating this with respect to y we get: and comparing with (418b) we see that U y Finally, differentiating this with respect to z we find: y = e3x z + U (y, z) y =0soU = W (z) and V (x, y, z) = e 3x yz + W (z). z = e3x yz + dw dz so W is a constant which we may choose to be zero. Thus, it is a conseratie force with: V (x, y, z) = e 3x yz.

6 Dynamics Worksheet 8 - Workshop problems - question : Two particles, each of mass m, are connected by a light inflexible string of length l. The string passes through a small smooth hole in the centre of a smooth horizontal table, so that one particle is below the table and the other can moe on the surface of the table. Take the origin of the (plane) polar coordinates to be the hole, and describe the height of the lower particle by the coordinate z, measured downwards from the table surface. A. Show that the equations of motion are: m( r r θ )= T, m(r θ +ṙ θ) =0, m z = mg T where T is the (magnitude of the) tension of the string. B. Impose the condition that the string s length is fixed, and show that m( r r θ + g) =0 where L, the angular momentum, is a constant of motion. mr θ = L, C. By multiplying the equation of motion by ṙ and then integrating, shown that E = 1 (m)ṙ + L + mg(r l), mr the total energy, is a constant of motion. Explain the interpretation of each of the components of the energy. D. Show that the problem is equialent to an effectie one-dimensional system where a particle of mass m moes under the effectie potential: U(r) = + mg(r l) mr Sketch the potential and show that the motion in this potential is oscillatory. L E. Explain why circular orbits of the particle on the table in the original problem corresponds to a particle at the equilibrium point in the effectie one-dimensional problem. Use the equation of motion to find the relation between the angular momentum and the radius of a circular orbit. Use this result to determine the energy corresponding to circular motion in terms of mg, l and r. Solution: A. First let us derie the equations of motion of the upper mass in plane polar coordinates. Defining r = r cos(θ)ˆx + r sin(θ)ŷ and ˆθ = sin(θ)ˆx + cos(θ)ˆx we hae: ṙ =(ṙ cos(θ) r sin(θ) θ)ˆx +(ṙ sin(θ)+r cos(θ) θ)ŷ =ṙˆr + r θˆθ Taking a second deriatie: Substituting r = rˆr +ṙ ˆr +ṙ θˆθ + r θˆθ + r θ ˆθ ˆr = θˆθ, ˆθ = ˆr θ we get: r = rˆr +ṙ θˆθ +ṙ θˆθ + r θˆθ ˆr θ r =ˆr( r θ r)+ˆθ(r θ +ṙ θ) The total force acting on the mass which is on the table is T ˆr, so the equations of motion are: r θ r = T/m, r θ +ṙ θ =0. Consider now the body hanging downwards at the bottom of the string. The forces here (defining ẑ downwards) are: F z = mg T so the equation of motion is: m z = mg T

7 Dynamics 196 B. Imposing the condition that the string length is fixed, l = z + r, we get: ż = ṙ and z = r. We can now determine the tension from the z equation: and substitute that into the r equation: which is the first relation we needed to proe. T/m = g z = g + r r θ r = g r = r θ r = g The second relation, inoling the angular momentum, can be obtained starting with the angular equation of motion: r θ +ṙ θ =0 = r θ +ṙr θ =0 which is equialent to implying that the angular momentum is a constant of motion. C. Consider the equation of motion, times mṙ: Substituting for θ = L mr we get: dl dt = d dt (mr θ) =0 L = mr θ mṙ r m θ rṙ + mgṙ =0 mṙ r L ṙ + mgṙ =0 mr3 This may be written as a time deriatie as follows: d m dt ṙ + L + mg(r l) =0 mr where we identify the constant of motion as the total energy: E = m ṙ + L + mg(r l) mr where the first term is the kinetic energy component due to radial motion (which is associated also with the kinetic energy of the lower body moing up or down along the ẑ direction) the second term corresponds to the angular kinetic energy, which is expressed here ia the angular momentum. Finally the last term is the potential energy due to graity acting on the lower body. D. Based on the form of the consered energy we see that indeed, the problem is equialent to a one dimensional motion of a body of mass m in the coordinate r, where the kinetic energy is and the effectie potential is: U(r) = T = m ṙ L + mg(r l). mr Let us sketch it for L =m and mg = and l =3/. We get the cure in figure 1.6. The motion is periodic because for any gien energy, there are two turning points. E. Circular orbits of the particle on the table correspond to a fixed radial coordinate, namely ṙ = 0 and r = 0 at any time. In the effectie, one-dimensional problem, E = m ṙ + U(r) so the equation of motion (de/dt = 0) is: de dt = m r + du ṙ =0 dr

8 Dynamics 197 so when r = 0 the total force du dr anishes. This is the condition for equilibrium. Let us now look at the equilibrium condition: which yields: Thus for circular motion the energy is: L E = U(r) = mr0 0= du dr = L mr 3 + mg L = m r 3 0g + mg(r 0 l) = mgr mg(r 0 l) =mg r 0 l. Note that this is consistent with the cure aboe: the minimum (there at r 0 = 1, since l =3/) corresponds to the equilibrium point (or a circular motion in the original system).