Physics 106b/196b – Problem Set 9 – Due Jan 19, 2007Version 3: January 18, 2007This problem set focuses on dynamics in rotating coordinate systems (Section 5.2), with someadditional early material on dynamics of rigid bodies (Section 5.3). Again, the length of theproblem set is misleading; much of it is in expository material. Problems 1 and 2 are for 106bstudents only, 3 though 5 for 106b an 196b students, and 6 and 7 are for 196b students only. It issuggested you start thinking about Problem 5 early in the week; it may require a bit of time foryou to get your mind around the problem.Version 2: Now provide expected result for position of CM in Problem 2, and explicitly ask youto calculate it. Even though the problem did not ask for it in v. 1, you need to calculate it to getthe right principal moments. Hint added for Problem 4. Problem 5: Additional details given onhow to set up coordinate system. Corrected the expected result for kinetic energy (it didn’t makesense dimensionally).Version 3: A couple typos in Problem 5 still – the y components of ~ω and~L were m issing a minussign and there was a 1/2 missing from T . The first one had no impact on the rest of the problem.The second one would make you calculate the incorrect Lagrangian and oscillation frequency inProblem 5b. Also, the explanation of the axis orientation, “At t = 0, the xz plane is normal to thex0y0-plane with the x axis in the first quadrant of the x0z0-plane” is incorrect in an obvious way: thex axis is in the third quadrant of the x0z0-plane at t = 0. Given the late date of these corrections,we will be lenient in grading on any errors that may have resulted from these typos.1. (106b) Two glorified plug-and-chug problems:(a) A wagon wheel with spokes is mounted on a vertical axis s o it is free to rotate in thehorizontal plane. The wheel is rotating with an angular speed of ω = 3.0 radians/sec.A bug crawls out on one of the spokes of the wheel with a velocity of 0.5 cm/s holdingon to the s poke with a coefficient of friction µ = 0.30. How far can the bug crawl outalong the spoke before it starts to slip?(b) A carousel (counter-clockwise merry-go-round) starts from rest and accelerates at aconstant angular accelereration of 0.02 revolutions/s2. A girl sitting on a bench on theplatform 7.0 m from the center is holding a 3.0 kg ball. Calculate the magnitude anddirection of the force she must exert to hold the ball 6.0 s after the c arousel starts tomove. Give the direction with respect to the line from the center of rotation to the girl.2. (106b) Show that the position of the center-of-mass and the values of the principal mome ntsof inertia of a right circular cone of height h, half-angle a, mas s M, and uniform density are~R =0, 0,h4I1= I2=320M h214+ tan2αI3=310M h2tan2α1where the center-of-mass is calculated in a coordinate system in which the base of the conesits in the xy-plane with its center at the origin. Of course, the principal moments mustalways be calculated in a system in which the origin sits at the center of mass, so you haveto move origins to then calculate the principal moments.3. (106b/196b) In Section 5.2.3 of the lecture notes, we derived the Coriolis force for the Fou-cault’s pendulum problem via Lagrangian methods. We assumed ω was constant and we ne-glected terms of order ω2, resulting in the loss of the centrifugal and Euler force terms. Let’snow generalize this to allow ~ω to point in an arbitrary direction and to be time-dep endent.Write the Lagrangian for this more general case, keeping the ω2terms. Find the Euler-Lagrange equations – you should obtain a set of equations equivalent to Equation 5.4 fromthe lecture notes, which is the equation of motion in the rotating frame with all the fictitiousforces included. The canonical momentum you obtain along the way will not be m ~vbody;give a physical interpretation of the canonical momentum. Calculate the Hamiltonian (butdon’t bother to calculate Hamilton’s equations). Show that H = Hω=0− ~ω ·~lbodywhere~lbody= ~r × ~pbodyand ~pbody=∂∂~vbodyL(~r, ~vbody, t).If you have any confusion as to why we use andbodyhere in ~vbodyand ~pbody, you shouldconsult someone to clear it up.You will find it most convenient to do this problem using index notation for the vectorarithmetic. In particular, you will obtain cross products in the kinetic energy like ~vbody·(~ω × ~r)and (~ω× ~r)·(~ω× ~r); leave them in cross-product form, and then write them in index notationwhen you want to take the usual derivatives to get the Euler-Lagrange equations.4. (106b/196b) Water being diverted during a flood in Helsinki, Finland (latitude 60◦N) flowsalong a diversion channel of width 47 m in the south direction at a speed of 3.4 m/s. Onwhich side is the water highest (from the standpoint of a noninertial system) and by howmuch? Hint: What relation defines the surface of the water? Consider a mass element ofwater on the water surface. In what direction must the net force on that element point?5. (106b/196b) Fun with rolling cones. Apologies for the length of the exposition, hopefully itwill prevent you from heading off in a wrong direction. Two parts:(a) The right circular cone in Problem 2 above rolls on its side without slipping on a hor-izontal plane. It completes a circle in time τ . Set up your body and space frames asfollows:• The space and body frame origins are at the (motionless) apex of the cone; the bodyframe origin is not at the CM of the cone.• The body frame z-axis is along the axis of the cone with the +z direction pointingfrom the base to the apex.• The space frame z0-axis is perpendicular to the horizontal plane.• At t = 0, the cone’s line of contact with the horizontal plane coincides with thex0-axis.• At t = 0, the xz plane is normal to the x0y0-plane with the x axis in the thirdquadrant of the x0z0-plane, or, equivalently, the yz plane is inclined at an angle αto the x0y0-plane with the y and y0axes coincident at t = 0.2Defineωp≡2πτΩ ≡2πτ1sin α˜I1= I1+ M34h2=320M h24 + tan2αShow that the angular velocity, angular momentum, and kinetic energy of the cone inthe space and body frames as a function of time are as follows:~ω =ωpcos α cos Ωt−ωpcos α sin ΩtΩ − ωpsin α~ω0=−Ω cos α cos ωpt−Ω cos α sin ωpt0~L =˜I1ωpcos α cos Ωt−˜I1ωpcos α sin ΩtI3(Ω − ωpsin