PHYS 251 HONOURS CLASSICAL MECHANICS 2014 Homework Set 6 Solutions

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1 PHYS 51 HONOURS CLASSICAL MECHANICS 014 Homework Set 6 Solutions 6.5. Spring on a T Let the l-rod be horizontal (parallel to the x axis at t = 0. Then, as it rotates with constant angular frequency ω, its angle with respect to the x axis will be ωt. Thus, we can write the coordinates of the mass as follows: and so the velocity of the mass is given by So after simplification, We can then write the Lagrangian: x = l cos(ωt r sin(ωt, (1 y = l sin(ωt + r cos(ωt, ( ẋ = ω[ l sin(ωt r cos(ωt] ṙ sin(ωt, (3 ẏ = ω[l cos(ωt r sin(ωt] + ṙ cos(ωt. (4 v = ẋ + ẏ = ω (l + r + ṙ + ωlṙ. (5 L = T V = 1 mv 1 kr = 1 m [ ω (l + r + ṙ + ωlṙ ] 1 kr. (6 Applying the Euler-Lagrange equation, we find d ( = 0 (7 = m r = mω r kr (8 ( k = r + m ω r = 0. (9 We recognize this last equation as the equation of motion of a simple harmonic oscillator. There are three different cases, which yield different solutions: 1. If ω < k/m, then the solution is r(t = A cos( ωt + B sin( ωt, where A and B are constants which k depend on the initial conditions. Note that we defined ω = m ω.. If ω > k/m, then the solution is r(t = Ae ωt + Be ωt, where this time we define ω = B are still constants of integration. 3. If ω = k/m, then the solution is simply r(t = At + B. ω k m. A and The special value of ω is k/m. When it is larger than this value, the centrifugal force in the rotating frame is dominant over the spring force so r grows exponentially. For an angular velocity smaller than k/m, then the usual oscillator motion is dominant. 1

2 Figure 1: Maple code for Problem 6.7. Note that e is some small time interval and that i is some iteration variable Coffee cup and mass The kinetic energy of the coffee cup is simply 1 Mṙ. Since the mass m can also rotate, its kinetic energy is 1 mṙ + 1 mr θ. Their potential energy is then Mgr and mgr sin θ, respectively. Therefore, the Lagrangian is L = T V = 1 Mṙ + 1 mṙ + 1 mr θ Mgr + mgr sin θ, (10 and the equations of motion are and d ( = 0 (11 = (M + m r = mr θ Mg + mg sin θ (1 θ d ( θ = 0 (13 = r θ = ṙ θ + g cos θ. (14 For the second part, see Figure 1 for an example of code written in Maple. We find the ratio of the r at the lowest point to the r at the start to be for m/m = 1/10. We also find that this lowest point is attained after a time of s, and that the angle θ at that moment is rad ẍ dependence With a ẍ dependent term in the Lagrangian, Eq. (6.19 becomes t a S[x a(t] = We then integrate the last term by parts twice: β = β = β [ d ( x a β + ẋ a β + β (15 ( d β (16 ( ( ] d β β. (17

3 Therefore, we obtain S[x a (t] a = t [ β d ( ( ] + d x a ẋ a [ + β d ( ] t + β ẋ a t i t. (18 The boundary term proportional to β is equal to 0, since we assume that β vanishes at the endpoints. However, the boundary proportional to β may not be zero, since we do not assume that the derivative is zero at the endpoints. Therefore, the proposed result is not valid Constraint on a curve The Lagrangian is L = 1 m(ẋ + ẏ V (r where we denote r to be the distance of the bead from the curve. Then, x d ( V (r(x, y = 0 = mẍ = = V ẋ x x, (19 y d ( ẏ V (r(x, y = 0 = mÿ = = V y y, (0 where we note that F (r = V, so mẍ = F (r x and mÿ = F (r y. (1 We are told that the curve is described by a function y = f(x, so we have ẏ = df(x(t = f ẋ = ÿ = f ẍ + f ẋ, ( where denotes a derivative with respect to x. Substituting (1 into (, we obtain F m y = f F m x + f ẋ (3 = F = mf ẋ y f. (4 x Now, to determine the partial derivatives of r with respect to x and y, let θ be the angle between the curve and the x axis at any time. Then, the tangent to the curve at any point on the x axis is simply f (x = tan θ. (5 With the same geometrical description, a bead a distance r from the curve will have Using (5, sin and cos can be re-written in terms of f, so one obtains x = sin θ and = cos θ. (6 y x = f and y = 1. (7 Therefore, the expression for the force found in (4 becomes F = mf ẋ, (8 but we note that since v = ẋˆx + ẏŷ, we have ẋ = v cos θ = v, so in the end 1+f F = mf v. (9 ( 3/ 3

4 6.40. Atwood s machine By conservation of string, the height of the 4m mass is x+y, so the Lagrangian is simply ( x + y L =T V = 1 (ẋ + ẏ (4m + 1 (5mẋ + 1 (3mẏ [ (4mg ] + (5mgx + (3mgy =m(3ẋ + mẋẏ + mẏ mg(3x + y. (31 We note that under the transformation (30 x x + ɛ, (3 y y 3ɛ, (33 the Lagrangian is left invariant (L L. Therefore, we can use Noether s theorem with and the conserved momentum is thus K x = 1 and K y = 3, (34 P = ẋ K x + ẏ K y = m(6ẋ + ẏ 3m(ẋ + yẏ = m(3ẋ 11ẏ, (35 which tells us that m(3ẋ 11ẏ = constant all the times in this system Spring on a spoke Let (x, y be the coordinates of the mass relative to the center of the wheel at t = 0. Also, let r be the length of the spring and θ be the angle through which the wheel has rolled relative to the position where the spring is vertical. Then, x = Rθ r sin θ and y = R r cos θ, (36 so ẋ = R θ r θ cos θ ṙ sin θ and ẏ = r θ sin θ ṙ cos θ (37 = v = ẋ + ẏ = R θ + r θ + ṙ rr θ cos θ ṙr θ sin θ. (38 We assume that θ, θ, ṙ 1, so the velocity becomes ( v = R θ + r θ + ṙ rr θ 1 θ + O(θ4 ṙr (θ θ θ3 6 + O(θ5 = R θ + r θ + ṙ rr θ, (40 where the second equality is valid up to second order in small quantities. The Lagrangian is then L = T V = 1 m(r θ + r θ + ṙ rr θ 1 kr + mgr cos θ. (41 (39 So, θ d ( θ = 0 (4 = m d [ ] (R r θ = mgr sin θ (43 = (R r θ (R rṙ θ = gr sin θ. (44 To first order in small quantities, this EOM simplifies to become (R r θ = grθ. (45 For small oscillations, ṙ 1 = ṙ 0 = r constant. Thus, we have a simple harmonic oscillator, θ + ωθθ = 0, (46 4

5 with ωθ gr (R r. Also, we have d ( = 0 (47 = d (mṙ = m(r R θ kr + mg cos θ (48 = r = kr m + g, (49 up to first order in small quantities. This EOM can be re-written as follows: ρ + ω rρ = 0, (50 with ρ r mg k and ω r k m. From this, we read the equilibrium value of r to be when ρ = 0, i.e. when r = mg k r eq. (51 Thus, we can re-write ω r = g/r eq. Finally, the two frequencies are equal when ω θ = ω r = greq g = = r eq = R R r eq r eq. (5 5