Lagrange Equation and Motion in Rotating Frames: Two-Particle Systems and Central Forces, Exams of Mechanics

Download Lagrange Equation and Motion in Rotating Frames: Two-Particle Systems and Central Forces and more Exams Mechanics in PDF only on Docsity! Phys 3610, Fall 2008 Exam #3 1. Define: a) Cyclic (ignorable) coordinate The generalized coordinate qi is cyclic if the lagrangian L does not depend on it. In that case, ∂L/∂q̇i is constant which is also expressed by saying that the momentum conjugate to q i is conserved. b) Reduced mass For a two-particle system, the reduced mass µ is µ = m1m2 m1 +m2 In the equivalent one--particle problem µ plays the role of the (single) particle's mass. c) Coriolis force This is a velocity-dependent (phony) force used in a rotating frame when we express Newton's 2nd law in that frame. It is given by 2mΩ × ṙ d) Foucault pendulum. A fine physics demo found a quality science museums everywhere. It is a long simple pendulum which is set to swing in a plane but it is found that over time the plane of motion slowly rotates. The rate of this rotation depends on the latitude of the location. This drift is caused by the Coriolis force. e) Principal axes. This are axes associated with a rigid body for which the total angular momentum L is parallel to ω if ω lies along one of these axes. q m1 m 2 2. Massm1 is suspended from a string which passes over an ideal massless pulley; the other end of the string is attached to a mass m2 which slides on a frictionless inclined surface, inclined at angle θ. (The string pulls parallel to the slope.) You can assume m1 > m2. Find the acceleration of the masses using the Lagrange approach. Note: There is one degree of freedom; the total length of the string is constant. You only need to write down an expression for the potential energy up to a constant term. (Does the answer match what you know from elementary physics?) 1 If we let x be the distance of m1 below the pulley (that is, the negative of its y coordinate) then if the length of the string is L, thenm2 is a distance L−x downward along the slope from the pulley. (This ignores the part of string that goes around the pulley, but that just adds a constant to the coordinates.) Then the kinetic energy of the system is T = 1 2 m1 ( d dt x )2 + 1 2 m2 ( d dt (L − x) )2 = 1 2 m1ẋ 2 + 1 2 m2ẋ 2 = 1 2 (m2 +m2)ẋ 2 Of course, both masses have the same speed . The potential energy of the system is U = −m1gx−m2g(L− x) sin θ = −m1gx+m2gx sin θ = gx(−m1 +m2 sin θ) so the Lagrangian is L = T − U = 1 2 (m1 +m2)ẋ 2 + gx(m1 −m2 sin θ) Set up the Lagrange equation: ∂L ∂x = g(m1 −m2 sin θ) ∂L ∂ẋ = (m1 +m2)ẋ Then the Lagrange equation gives us g(m1 −m2 sin θ) − d dt (m1 +m2)ẋ = (m1 +m2)ẍ which then gives ẍ = g(m1 −m2 sin θ) (m1 +m2) This is what we get from Phys 2110 physics; if the string has tension T then the free-body diagrams give the equations m1g − T = m2ẍ T −m2g sin θ = m2ẍ which, on eliminating T (add the two equations) gives m1g −m2g sin θ = (m1 +m2)ẍ =⇒ ẍ = g(m1 −m2 sin θ) (m1 +m2) as before. I’ve decided to make #3 extra-credit! Finish the rest first and do as much as you have time for. 2 so that ẍ = −g sin θ − a sin θ = −(g + a) sin θ We note that in the usual answer of ``g sin θ down the slope'' we have g + a in place of g. This is as we expect; the upward acceleration of the reference frame gives a new effective value of g. Car accelerating to the right Use the degree of freedom x in the same way as the case above but have the X axis go to the right. Then the coordinates of the mass are X = −x cos θ + 1 2 at2 Y = x sin θ and the kinetic energy is T = m 2 (Ẋ2 + Ẏ 2) = m 2 ((−ẋ cos θ + at)2 + (ẋ sin θ)2) = m 2 (ẋ2 − 2ẋat cos θ + a2t2) and the potential energy is U = mgx sin θ so L = T − U = m 2 (ẋ2 − 2ẋat cos θ + a2t2) −mgx sin θ Set up the Lagrange equation: ∂L ∂x = −mg sin θ ∂L ∂ẋ = m 2 (2ẋ− 2at cos θ) = m(ẋ− at cos θ) Then the lagrange equation gives −mg sin θ = m(ẍ− a cos θ) =⇒ ẍ = −g sin θ + a cos θ Remembering that x points up the slope (to the left), the effect of gravity is add a component of acceleration a cos θ up the slope. Ueff r 5. What’s the deal with the “effective potential” for the Ke- pler problem? (Graph of Ueff shown at right.) Why does Ueff(r) goes to +∞ at r = 0? (Doesn’t the gravitational potential go to negative infinity there?) The effective radial potential is a combination of the potential from the true central force plus the ``centrifugal potential''. The latter is proportional to +1/r2 so it blows up at the origin and overpowers the real gravitational potential which only blows up as −1/r. Explain how we would use this graph the find minimum and maximum distance of a Kepler orbit for a given total energy (and `). 5 At the values of r where a horizontal line at E crosses the curve of Ueff the radial velocity ṙ is zero, so these are the maximum and minimum radii. 6. Pluto’s closest approach to sun is 30.17 AU and its farthest distance from the sun is 48.48 AU. (The “AU” is the average distance of the earth from the sun.) Find the eccen- tricity  of Pluto’s orbit. (Hint: You may want to use the equation for r(φ) and take ratios for the two distances. Then solve for .) Using the formula for a Kepler elliptical orbit ( < 1) relating r and φ, the minimum r occurs where cosφ = 1; this gives rmin = 30.17 AU = c 1 +  and the maximum is where cos φ = −1, which gives rmax = 48.48 AU = c 1 −  Divide the second equation by the first and get 48.48 30.17 = 1.6069 = 1 +  1 −  =⇒ 1 +  = 1.6069(1 − ) Solve for : 2.6069 = 0.6069 =⇒  = 0.233 7. Suppose the central potential for a mass m moving around a center of force is U(r) = −A e−ar r a) What is F (r)? (I.e. radial component of the central force.) For the given central potential, U(r) = −Ae−αr/r (which approaches the Kepler problem as α→ 0, we have F (r) = − dU dr = +A [ −αe−αr r − e−αr r2 ] = −Ae−αr (αr + 1 r2 b) Write out the differential equation that would in principle allow you to find the trajectory r(φ) if the mass has angular momentum `. (You don’t need to solve it!) With u = 1/r we have a differential equation for u(φ): u′′ = −u− µ `2u2 F We need to write F as a function of u: F = −Au2[α/u+ 1]e−α/u = −A[αu+ u2]e−α/u 6 so with this we write u′′(φ) = −u(φ) + A µ `2u(φ)2 [αu+ u2]e−α/u which is probably very challenging to solve and certainly will not give closed orbits. 8. Recall the equation from the text, ( dQ dt ) S0 = ( dQ dt ) S + Ω × Q a) What does this equation tell us? The equation relates the rate of change of any vector Q in the inertial and the rotating frames (S0 and S , respectively). b) Describe how it was used to derive “Newton’s 2nd law in a rotating frame”. Newton's 2nd law is first written down for an inertial frame; it gives r̈ in terms of the true forces acting on a particle. This equation involves two d/dt operations acting on a vector r in the inertial frame; we use the above equation to relate this to two d/dt operating on r in the rotating frame. The result is that in addition to the true forces we get a couple terms which act like forces: The centrifugal ``force'' and the Coriolis ``force''. 9. In solving problems showing how the Coriolis force affects free-fall, we grouped the centrifugal force together with gravity and didn’t consider its separate effect. Why did we do that? Since it is independent of the (rotating frame) velocity and proportional to the mass, for a particular location on the earth the centrifugal force is a constant force like gravity. A plumb line points along the direction of the sum of gravitational and centrifugal forces and so we can separate them (experimentally) only with great difficulty. Rather we punt and call this direction the ``vertical''. 10. Suppose we fire a bullet southward at 36 deg latitude at a speed of 1000 m/s. Find the magnitude and direction of the acceleration due to the Coriolis force. (A picture may help you. . . ) W v 36 o The situation is illustrated at the right. From geometry, the angle between Ω and v is 90◦ + 54◦ = 144◦ . The value of Ω is Ω = 2π (24)(3600) s = 7.27 × 10−5 s−1 and since the Coriolis force if 2mṙ × Ω, its direction is out of the page for the diagram here, which means it points to the West and its magnitude is |FCor = 2mΩṙ sin 144 ◦ = m(2)(7.27 × 10−5 s−1)(1000 m s ) sin 144◦ = m(8.55 × 10−2 m s2 ) 7