PHY W3003: Practice Final solutions You may bring to the exam on sheet (both sides) of paper. Use of calculators or other electronic devices is not allowed during the examination. Where numerical answers are requested, answers to one significant figure are acceptable. 1 A pendulum is constructed by putting a bob of mass m at the end of a massless spring which can stretch along its length but does not bend (see Fig). The spring constant is k and the relaxed length of the spring is 0. In addition to the spring the mass is subject to the force of gravity which points vertically down. The mass can move in x θ (shown on fig) and radially along the length of the spring but not in the azimuthal direction, so motion is confined to a plane. (a) Please write the Lagrangian of the system, using as generalized coordinates θ and the length x of the spring. Call the two coordinates z (vertical) and y (horizontal). Then z = −xcosθ y = xsinθ so the kinetic and potential energies are m 2 m 2 z˙ + y˙ 2 = x˙ + x2 θ˙2 2 2 k 2 = −mgxcosθ + x 2 K = V so the Lagrangian is L= k m 2 x˙ + x2 θ˙2 + mgxcosθ − x2 2 2 (b) Please write (but do not solve!) the Euler-Lagrange equations m¨ x = mxθ˙2 + mgcosθ − kx m d 2˙ x θ = −mgxsinθ dt (c) Please find the equilibrium values of the generalized coordinates. We minimize V with respect to the coordinate values, obtaining θ=0 x= mg k (d) Please find the frequencies of the normal modes describing small oscillations about the equilibrium position 1 Near the minimum we have mg 2 (mg)2 m k 2 L= x˙ + θ˙2 − θ 2 − x2 2 k 2k 2 so we have one normal mode describing oscillations of x with frequency Ω2x = k describing oscillations in θ also with frequency Ω2θ = m k m and one 2 A particle of mass m = 1 moves in one dimension subject to a harmonic potential and an additional force linearly proportional to time such that its equation of motion is x ¨ + 4x˙ + 4x = t If the particle is released from rest at position x = 0 at time t = 0 please find its position x(t) for all t > 0. The solution of the homogeneous equation is xh (t) = Ae−2t + Bte−2t A solution of the inhomogeneous equation is xp (t) = 0.25t − 0.25 At time t = 0 we have x(t = 0) = A − 0.25 = 0 x(t ˙ = 0) = 0.25 − 2A + B = 0 => A = 0.25 B = 0.25 so the full solution is x(t) = 0.25 e−2t − 1 + 0.25t e−2t + 1 3 Three k x=0 particles equal k k x1 of k x2 x3 η1 = η2 = η3 = mass m are coupled by springs as shown in the figure. At time t = 0 particles 1 and 3 are at rest in their equilibrium positions while particle 2 is at rest but with a small displacement q2 in the +x direction. Please find the position of particle 3 for all t > 0. The three normal modes are x=4a √ √1 2− 2 k 2 2 Ω1 = m 1 1 k 0 Ω22 = m −1 √ 1 √ 2+ 2 k 2 − 2 Ω3 = m 1 Thus if the normal mode amplitudes are A1,2,3 then the time dependence is x1 (t) = A1 cos (Ω1 t + φ1 ) + A2 cos (Ω2 t + φ2 ) + A3 cos (Ω3 t + φ3 ) √ √ x2 (t) = 2A1 cos (Ω1 t + φ1 ) − 2A3 cos (Ω3 t + φ3 ) x3 (t) = A1 cos (Ω1 t + φ1 ) − A2 cos (Ω2 t + φ2 ) + A3 cos (Ω3 t + φ3 ) If x1 (t = 0) = x3 (t = 0) = 0 then subtracting the first and last equation implies A2 = 0. Adding the two equations and doing the analogous thing for the velocities gives 0 = A1 cosφ1 + A3 cosφ3 0 = −Ω1 A1 sinφ1 − Ω3 A3 sinφ3 so we set φ1 = φ3 = 0 and A1 = −A3 so √ implying A1 = q2 /(2 2) so √ x2 (t = 0) = 2 2A1 q2 x3 (t) = √ (cos (Ω1 t) − cos (Ω3 t)) 2 2 4 Origin Consider a rectangular object of dimensions a, b and c as shown in the Figure, with mass M at each of the 8 corners. Take the origin of coordinates as shown to be the front lower left corner of the object, and choose x to be along b, y to be along a and z to be along c (a) Please find the inertia tensor for rotations about the origin. Ixx = 4M a2 + c2 Iyy = 4M b2 + c2 Izz = 4M a2 + b2 Ixy = −2M ab Ixz = −2M bc Iyz = −2M ac (b) If the object rotates with angular velocity Ω = 1radian/sec about the axis shown, the masses M are each 0.5kG and a = 1, b = 2 and c = 0.5 with distances measured in meters, please find the three components of the angular momentum vector (numerical values, to one significant figure). 4 − √4.25 0 −2 Lx 2.5 −2 −1 2 16.75 Ly = −2 8.5 −0.5 √4.25 = √4.25 ≈ 8 0.5 √ 2 Lz −1 −0.5 10 √4 4.25 4.25 5 Consider a particle of mass m moving in two dimensions and subject to the central potential k V (r) = − 2 r + b2 Please (a) find (in terms of m, k and b) the values of the angular momentum L (measured about an axis perpendicular to the plane of motion and passing through the origin) for which stable circular motion is possible, and for these values, the radius of the circular motion (terms of m, k and b and L) The effective potential is k L2 − Vef f (r) = 2mr2 r2 + b2 and is minimized at 2kr L2 0=− 3 + mr (r2 + b2 )2 or 2 2kmr4 = L2 r2 + b2 or √ 2kmr2 = L r2 + b2 or Lb2 r2 = √ 2km − L The second derivative is 8kr2 d2 V 3L2 2k − = + dr2 mr4 (r2 + b2 )2 (r2 + b2 )3 Simplifying using the results above √ 4L2 L d2 V 3L2 L2 2 2L3 = 1− √ = + − √ dr2 mr4 mr4 m mkr4 mr4 2mk √ What is inside the parens is positive so the minimum is stable if L > 2mk, so all physically allowed circular orbits are stable, with r given by the square root of the expression above (b) for values of L such that stable circular motion is possible please sketch the effective potential and the phase space orbits (in the half plane r > 0, r˙ arbitrary) indicating the bound unbound and separatrix orbits (if any) and give the energy of the separatrix orbit (if any) (c) Find the value of L for which the stable circular orbit has radius b. From above r km L= 2 Consider a particle moving on the orbit found in (c). Suppose that at time t = 0 the particle receives a tangential impulse such that its angular momentum is instantaneously doubled (L → 2L) but immediately after the impulse the radial velocity remains zero. (d) After the impulse the orbit is no longer circular. Please determine the smallest and largest values of r reached by the particle. After the impulse L2 = 2km and the energy is k k E= 2 − 2 >0 b 2b so the particle is unbound; the largest r = ∞; the smallest distance is b, the time taken to reach infinity must be infinite.
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