Physics 325 – Homework #8 due in 325 homework box by Fri, 1 pm

All solutions must clearly show the steps and/or reasoning you used to arrive at your result. You will lose points for poorly written solutions or incorrect reasoning. Answers given without explanation will not be graded: “NO WORK = NO POINTS”. However you may always use any relation on the 3D-calculus and 1D-math formula sheets without proof. Please write your NAME and DISCUSSION SECTION on your solutions.

Recap of Equilibrium / Small-Oscillation Analysis

An equilibrium point is a particular configuration of a system where, if the system is placed in that configuration at rest, then it will remain at rest. In other words, it is a solution of the system where all coordinates are constants. In class, we discussed two methods for finding a system's equilibrium points and determining their stability and/or their associated small-oscillation frequency :

• Method A: Use the system's EOMs to find the equilibrium points q by seeking solutions of the form q(t) = q = constant. For each q you find, shift coordinates to ε(t) ≡ q(t)− q , then obtain the approximate near-equilibrium solution ε(t) → this solution gives you the stability & small-oscillation frequency at q .

• Method B: Analyze the derivatives of the potential energy U(q).

Method B is much faster, but it doesn't always work. The necessary conditions are:

(1)Energy E must be conserved. The energy in question is T+U in essentially all practical cases.

(2)The conserved energy must have this “normal” form : E(q, !q) =

12 µ(q) !q

2 +Ueff (q) + optional irrelevant constant. The function µ(q) plays the role of an effective mass that may depend on the position or angle of your system. A perfect example is the falling & sliding cube from Homework 7: using the cube's rotation angle φ as our coordinate q, the cube's kinetic energy is

T (φ, !φ) = 1

2!φ 2 ′I + 1

2Mb2 sin2φ( ) . The term in parenthesis

is the effective mass µ(φ) . The function Ueff (q) is usually the actual potential energy U(q) , but it doesn’t have to be; if it’s something else, it is called the effective potential and is written as either Ueff (q) or U *(q) .

(3)The effective mass μ must behave like a normal mass in this way: µ(q) and its derivative ′µ (q) must be finite (no singularities!) at all q that might be equilibrium points. If μ is just a constant, as in T = 1

2 m !x2 or T = 1

2 I !φ2 , this condition is trivially satisfied of course.

If all this is true, then you can use Method B for your equilibrium analysis:

• The equilibrium points, q , are those that satisfy ′Ueff (q ) = 0 .

• The small-oscillation frequency at q is ω = ′′Ueff / µ q. (This notation means ω = ′′Ueff (q ) / µ(q ) .)

• Stable equilibrium exists at q if ω is real.

We proved these results in lecture by applying Method A to the EOM !E = 0 with E = 12 µ(q) !q

2 +Ueff (q) . Since E is very often of this “normal form”, we have already solved the equilibrium / small-oscillation analysis for a large class of problems so we might as well use the results directly. That’s Method B.

PHYS 211 connection : Method B is a juiced-up version of the equilibrium analysis you learned in introductory mechanics: stable equilibrium occurs at potential energy minima and unstable equilibrium occurs at PE maxima. In math: equilibrium requires ′U = 0 (a PE extremum), stable equilibrium requires ′′U > 0 , and unstable equilibrium requires ′′U < 0 . This is all the same as in our Method B except that (1) we have expanded the

analysis to include effective potentials, and (2) if we have an effective mass μ that is negative, the stability conditions on U′′ are reversed since the true stability condition is on U′′ / μ.

Memory Assist : The small-oscillation frequency ω = ′′Ueff / µ q is not studied in introductory mechanics.

Is there any easy way to remember this useful formula? Why yes there is! → Think of the world’s simplest oscillating system: a linear spring k with zero unstretched length that runs along the x axis with one end fixed to the origin and a point mass m fixed to the other end. For this simplest of systems, T = 1

2 m!x2 and U = 1

2 kx2 .

I would guess that you have the oscillation frequency ω = k /m of a spring firmly committed to memory.I would also guess that you have U = 1

2 kx2 committed to memory, so you can see in one second that k = ′′U .

Thus, ω = ′′U /m for a spring. In our generalized analysis, we simply extended this to effective potential and effective mass, both of which can be trivially read off from the normal form E = 1

2 µ(q) !q2 +Ueff (q) by analogy

with the uber-familiar form E = 12 m!x

2 +U(x) . The trickiest thing to remember is that ω = ′′U / µ mustbe evaluated at the equilibrium point you are studying. This is not a concern with our simple spring as ′′U and µ are constants ... but in general, U′′ and μ are functions of your coordinates. If you get a small-oscillationfrequency ω that is not a constant, you know you’ve forgotten this step: ω cannot possibly depend on any coordinates since it is the oscillation frequency at a specific equilibrium point. Makes sense? If not, please ask!

Problem 1 : Rotating Spring

Consider a “standard spring” : it has spring constant k, unstretched length l0, and cannot bend, i.e. it can only stretch or contract. (The spring could be wound around a rigid rod, for example.) One end of this spring is attached to an axle located at the origin; the other end is attached to a point mass m. A motor attached to the axle causes the spring to rotate in the xy-plane with constant angular velocity Ω. Since the spring cannot bend, it always points radially away from the origin. The xy-plane is a frictionless horizontal table so ignore gravity.

For this problem, you must use Method A. (You’ll see why in the later parts). Reminder: you already practiced this method in Discussion 7 and Lecture 8B.

(a) Using cylindrical coordinates (s, φ) for the mass m, find the point s where the mass is in radial equilibrium (i.e. if it is placed at that radius with zero radial speed, it will remain at that radius).

(b) Find the condition on Ω that will make s a point of stable equilibrium and find the frequency ω of small radial oscillations of the mass m around s . Also write down the general solution s(t) for the small-oscillation motion, using the symbols s and ω to simplify your expression.

(c) If the condition on Ω that you found in part (b) is violated, explain why the system actually has no point of equilibrium at all (not even an unstable one). Hint: if you think there is an equilibrium point, try sketching it.

(d) Why couldn’t we use Method B? Construct T and U for the mass and try calculating the equilibrium position s using (1) the familiar equilibrium condition ′U (s ) = 0 from PHYS 211, and (2) our new formula ′Ueff (s ) = 0 . You should find that neither of them gives the correct answer!

FYI: Even though it didn’t work, you should have obtained an effective potential that is not the same as the actual potential. This problem illustrates the typical source of Ueff ≠U : Ueff incorporates a term from the kinetic energy that doesn’t depend on !q (only on q ). In our rotating-spring system, the forced rotation at constant angular velocity Ω is responsible for this term.

(e) What went wrong? Identify which of the three “Method B conditions” on page 1 does not hold and explain the physical origin of the condition’s violation in this system.

Problem 2 : Bead on a Parabolic Wire

A bead of mass m is threaded onto a frictionless wire that is bent into the shape of a parabola: z = x2 / 2b andy = 0, where +z points upward. Uniform gravity g points downward. The bead performs oscillations along the wire, from x = −a to x = +a .

(a) Show that the period τ of these oscillations is given by the integral τ = 4gb

b2 + x2

a2 − x20

a

∫ dx .

Guidance : Conservation Laws & 1st-order ODEs. You can proceed with this problem by finding the EOMvia !F = m!a or !T + !U = 0 , as usual. (About !T + !U = 0 : if T + U is conserved – as it clearly is in this problem –

you can always recover one force-based EOM by taking the time derivative of T +U = constant . This procedure is essentially the inverse of our proof that

!F = m!a for conservative-forces-only leads to !T + !U = 0 .)

If you calculate !T + !U = 0 or Fx = m!!x , you will get an EOM that is second-order in time. However … There is an alternative way to use energy conservation. If the problem supplies enough boundary / initial conditions to determine the total energy E at any point in time, E will have that same value at all other times (it's conserved!). You can then use T +U = E directly as your energy-based EOM. This equation is first-order in time: it depends on coordinates x and velocities !x but not on accelerations !!x . A first-order ODE offers advantages over a second-order ODE: it is usually easier to solve … and even if it's not (which sometimes happens), you can always take its time-derivative to obtain the corresponding second-order ODE. Simply-put, first-order ODEs give you more options than second-order ODEs.

(b) Now consider the small-oscillation case a≪ b . Apply your approximation skills to the integrand to determine τ to lowest non-vanishing order in the small quantity a /b≪1 . (Your answer should have two terms.) The integral table on our 1DMath formula sheet has everything you need.

Problem 3 : A Contraption

A ball of mass m with a hole through it is threaded on a frictionless vertical rod. A massless string of finite length l is attached to the ball and runs over a massless, frictionless pulley that is placed a horizontal distance b from the rod. The other end of the string supports a block of mass M, as shown in the figure. This problem may look multi-dimensional but it’s not: as the positions of both masses are determined by the angle θ, there is only one independent coordinate.

(a) Find the potential energy U(θ) of the two masses under the influence of uniform gravity g, which is in the downward direction.

(b) Calculate all equilibrium positions, θ , of the system and determine what constraint must be imposed on the masses m and M for equilibrium to be possible at all.

(c) Determine the stability (i.e. stable or unstable) of any equilibrium position(s) you found and calculate the small-oscillation frequency for motion around the stable point(s).

Problem 4 : Another Contraption

A massless wheel of radius R is mounted on a frictionless horizontal axle. A point mass M is glued to the edge of the wheel and a mass m hangs from a string wrapped around the perimeter of the wheel.

(a) Determine the total potential energy U(φ) of the two masses under the influence of uniform, downward-pointing gravity g. For definiteness, adjust your expression so that U(φ)=0 when φ=0. (Fixing the overall value of U to some particular value at some reference point makes no physical difference, of course, but the suggested choice will make the ensuing calculations more convenient.)

(b) Determine what constraint must be imposed on the masses m and M so that the system has at least one equilibrium position. Assuming that this constraint holds, find the equilibrium positions, φ , (major hint: there are two of them) and determine their stability (stable or unstable equilibrium).

(c) To generate a warm happy feeling, evaluate the torque on the wheel at the equilibrium angles φ you just found and thereby double-check that the system is indeed at equilibrium.

(d) Use wolframalpha.com or your fancy graphing calculator or phone app to plot U(φ) for these two cases: m = 0.7 M and m = 0.8 M over the range –π < φ < π. Sketch the shape of the curves on your answer sheet then use them to describe the behavior of the system in each case if it is released from rest at φ = 0.Note: You should check that your curves make sense by looking for the two equilibrium points that you found in part (b) ➔ they should appear clearly as two local extrema on both of your U(φ) curves. In addition, the stability-of-equilibrium that you found for these points in part (b) should also be apparent : any point of stable equilibrium will be a local minimum of U(φ) and any point of unstable equilibrium will be a local maximum.

(e) Write down an expression that can be solved for the critical value of the ratio m / M that determines whether the system oscillates or “runs away” (unwinds itself) if it is released from rest at φ = 0. Your expression will not be solvable analytically, but you can use a computer or fancy calculator to solve it numerically; to check your work, you should find a critical value of m / M = 0.725.

HINTS: This question has everything to do with part (d); if you haven’t completed it, this question makes no sense. Once you do have your plots and have interpreted them, you will see that the condition for the critical value of m / M is the situation in between your m / M = 0.7 and m / M = 0.8 plots where the system will just barely maintain an oscillating behavior if released from rest at φ=0. Final hint: You must translate those words into a condition on U(φ) at one of your two equilibrium angles φ . (Which one? … you’ll figure it out. ☺)

Problem 5 : Hemisphere Balanced on a Hemisphere

A solid hemisphere of radius b has its flat surface glued to a horizontal table. A second solid hemisphere of different radius a rests on top of the first one so that the curved surfaces are in contact. The surfaces of the hemispheres are rough (so no slipping occurs between them), and both hemispheres have uniform mass distributions. The two objects are in equilibrium when the top one is “upside down”, i.e. with its flat surface parallel to the table but above the hemispheres' contact point. Show that this equilibrium position is stable if a < 3b / 5 .Suggestion: For this is problem, it is really worth checking the conditions on the introductory page to see if Method B might work. If it does, your stability analysis will be greatly simplified.