Solutions 9
Transcription
Solutions 9
PHY 3003 SPRING 2015: WEEK OF MARCH 30 Reading: Taylor, Chapter 7.10, Chapter 9.1, 9.3, 9.4 NOTE EXAM in class Wednesday April 1. Covers material up through Taylor Chapter 8 except for Lagrange multipliers Homework: Due in class April 6. All problems count equally towards your grade. (Numbers refer to problems in Classical Mechanics, J. R. Talyor, 2005 Edition). (1) 7.50 The Lagrangian is (note y increases downwards so the potential energy is −mgy): L= m1 2 m2 2 x˙ + y˙ + m2 gy + λ (x + y − L) 2 2 so we have m1 x ¨ = λ m2 y¨ = m2 g + λ x+y = L From the third equation we have x ¨ = −¨ y so λ = m1 x ¨ = −m1 y¨ and (m2 + m1 ) y¨ = m2 g or y(t) = y0 + vy t + and λ=− m2 g t2 2 (m1 + m2 ) m1 m2 g m1 + m2 (2) 7.52 L= so m 2 I ˙2 x˙ + φ + mgx + λ (x + Rφ) 2 2 m¨ x = mg + λ I φ¨ = λR x = −Rφ The last equation implies so x ¨ φ¨ = − R λ=− 1 Ix ¨ R2 2 PHY 3003 SPRING 2015: WEEK OF MARCH 30 so x ¨= mg m + RI2 and mg I 2 R m + RI2 Solving Newton’s equations in the standard way shows that λR is the torque exerted by the tension in the rope on the wheel. (3) 9.2 λ=− We need ω such that ω2R = g i.e. r g 10 ω= = = 0.5/sec R 40 Changing R from 40 to 38 is about 2 parts out of 40 or 5% change, so δgef f = ω 2 δR so δg/g = δR/R = 0.05. (4) 9.12 (a) If nothing is moving (in the rotating frame) then one simply has the extra force ω 2 R (eq. 9.36) (b) From the point of view of an observer in the rotating frame there is a radial force F = ω 2 R which must be balanced by the force of static friction µmg so µmg Rmax = 2 ω r (5) 9.14 The easy way to approach this is to note that in the rotating frame the centripetal force is equivalent to a potential energy 1 Vcentripetal = − Ω2 R2 2 That the surface of the water is an equipotential implies that 1 V = mgz(R) − Ω2 R2 = const 2 or Ω2 2 z=C+ R 2mg which is the equation for a parabola.