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8.5: Summary and Further Problems

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    24132
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    When the going gets tough, the tough lower their standards. In this chapter, you learned how to do that by studying the easy cases of a problem. This tool is based on the idea that a correct solution works in all cases, including the easy cases. Therefore, look at the easy cases first. Often, we can completely solve a problem simply by understanding the easy cases.

    Exercise \(\PageIndex{1}\): Easy cases for the period of a pendulum

    Does the period of a pendulum increase, decrease, or remain constant as the amplitude is increased? Decide by selecting an amplitude for which you can easily predict the period.

    Exercise \(\PageIndex{2}\): Pyramid volume

    Use easy cases to find the dimensionless prefactor in the volume of a pyramid with height h and a square (b × b) base:

    clipboard_ea506d66b70234dc5ec6ae85f8bbe07a1.png

    \[V = \textrm{dimensionless prefactor} \times b^{2}h.\]

    In particular, choose the easy case of a pyramid that, with a few more copies of itself, can be assembled into a cube.

    Exercise \(\PageIndex{3}\): Power means

    The arithmetic and geometric means are easy cases of a higher-level abstraction: the power mean. The kth power mean of two positive numbers a and b is defined by

    \[M_{k} (a,b) \equiv (\frac{a^{k} + b^{k}}{2})^{1/k}.\]

    You raise the numbers to the kth power, take the (regular) mean, and then undo the exponentiation by taking the kth root.

    a. What is k for an arithmetic mean?

    b. What is k for the rms (root mean square)?

    c. The harmonic mean of a and b is sometimes written as 2(ab), where ∥ denotes the parallel combination of a and b (introduced in Section 2.4.3). What is k for the harmonic mean?

    d. (Surprising!) What is k for the geometric mean?

    Exercise \(\PageIndex{4}\): Easy case of the compound pendulum

    For the compound pendulum of Problem 5.25, what easy case produces an ordinary, noncompound pendulum? Check that, in this limit, your formula for the period from Problem 5.25 behaves correctly.

    Exercise \(\PageIndex{5}\): Means in an elliptical orbit

    A planetary orbit (an ellipse) has two important radii: rmin and rmax. The other lengths in the ellipse are power means of these radii (see Problem 8.22 about power means).

    clipboard_e89cb1e2d8cb97e9ac577daea19f75084.png

    Match the three power means—arithmetic, geometric, and harmonic—to the three lengths: the semimajor axis a (which is related to the orbital period), the semiminor axis b, and the semilatus rectum l (which is related to the orbital angular momentum)

    Hint: The power-mean theorem says that Mm(a, b) < Mn(a, b) if and only if m < n.

    Exercise \(\PageIndex{6}\): Four regimes in orbital motion

    Once in a while, there are four interesting easy-cases regimes. An example is orbits. The dimensionless parameter \(\beta\) that characterizes the type of an orbit is

    \[\beta \equiv \frac{\textrm{kinetic energy}}{\vert \textrm{gravitational potential energy} \vert},\]

    where the absolute value handles a possibly negative potential energy.

    A related dimensionless parameter is the orbit eccentricity \(\epsilon\). In terms of the eccentricity, a planet’s orbit in polar coordinates is

    \[r (\theta) = \frac{l}{1 + \epsilon cos \theta},\]

    where the Sun is at the origin, and \(l\) is the length scale of the orbit (\(l\) is diagrammed in Problem 8.24). Sketch and classify the four orbit shapes according to their values of \(\beta\) and \(\epsilon\) (giving a point value or a range, as appropriate): (a) circle, (b) ellipse, (c) parabola, and (d) hyperbola.

    Exercise \(\PageIndex{7}\): Superfluid Helium

    Helium, when cold, turns into a liquid. When very cold, the liquid turns into a superfluid—a quantum liquid. Here is a dimensionless ratio determining how quantum the liquid is

    \[\beta \equiv \frac{\textrm{quantum uncertainty in the position of a helium atom}}{\textrm{separation between atoms}}.\]

    a. Estimate \(\beta\) in terms of the quantum constant \(\hbar\), helium’s density \(\rho\) (as a liquid), the thermal energy kBT, and the atomic mass mHe.

    b. In the \(\beta\) ∼ 1 regime, helium becomes a quantum liquid (a superfluid). Thus, estimate the superfluid transition temperature.

    Exercise \(\PageIndex{8}\): Adiabatic atmosphere

    The simplest model of the atmosphere is isothermal: The atmosphere has one temperature throughout it. A better approximation, the adiabatic atmosphere, relaxes this assumption and incorporates the adiabatic gas law:

    \[p V^{\gamma} \propto 1,\]

    where p is atmospheric pressure, V is the volume of a parcel of air, and \(\gamma \equiv c_{p}/c_{v}\) is the ratio of the two specific heats in the gas. (For dry air, \(\gamma\) = 1.4.) Imagine an air parcel rising up a mountain. As the parcel rises into air with a lower pressure, it expands, and its volume and temperature change according to a combination of the adiabatic and ideal gas laws.

    a. What easy case of \(\gamma\) reproduces the isothermal atmosphere?

    b. For \(\gamma\) = 1.4 (dry air), will air temperature decrease with, increase with, or be independent of height?

    Exercise \(\PageIndex{9}\): Loan payments

    A fixed-term, fixed-interest-rate loan has four important parameters: the principal P (the amount borrowed), the interest rate r, the repayment interval \(\tau\), and the number of payments n. The loan is repaid in n equal payments over the loan term \(n \tau\). Each payment consists of a principal and an interest portion. The interest portion is the interest accumulated on the principal outstanding during the term; the principal portion reduces the outstanding principal.

    The dimensionless quantity determining the type of loan is \(\beta \equiv n \tau r.\).

    a. Estimate the payment (the amount per term) in the easy case \(\beta\) = 0, in terms of P, n, and \(\tau\). (The term \(n \tau\) and the repayment interval \(\tau\) don’t vary that much—\(\tau\) is usually 1 month and \(n \tau\) is somewhere between 3 to 30 years—so \(\beta\) ≪ 1 is usually reached by lowering the interest rate r.)

    b. Estimate the payment in the slightly harder case where \(\beta\) ≪ 1 (which includes the \(\beta\) = 0 case). In this regime, the loan is called an installment loan.

    c. Estimate the payment in the easy case \(\beta\) ≫ 1. In this regime, the loan is called an annuity. (This regime is usually reached by increasing n.)

    Exercise \(\PageIndex{10}\): Heavy nuclei

    In this problem, you study the innermost electron in an atom such as uranium that has many protons, and analyze a surprising physical consequence of its binding energy. Imagine a nucleus with Z protons around which orbits one electron. Let E(Z) be the binding energy (the hydrogen binding energy is the case Z = 1).

    a. Show that the ratio E(Z)/E(1) is Z2.

    b. In Problem 5.36, you showed that E(1) is the kinetic energy of an electron moving with speed \(\alpha c\) where \(\alpha\) is the fine-structure constant (roughly 10−2). How fast does the innermost electron move around a heavy nucleus with charge Z?

    c. When that speed is comparable to the speed of light, the electron has a kinetic energy comparable to its (relativistic) rest energy. One consequence of such a high kinetic energy is that the electron has enough kinetic energy to produce a positron (an anti-electron) out of nowhere; this process is called pair creation. That positron leaves the nucleus, turning a proton into a neutron as it exits: The atomic number Z decreases by one. The nucleus is unstable! Relativity therefore places an upper limit to Z. Estimate this maximum Z and compare it with the Z for the heaviest stable nucleus (uranium).

    Exercise \(\PageIndex{11}\): Minimum wave speed

    For deep-water waves, estimate the minimum wave speed in terms of \(\rho\), \(g\), and \(\gamma\) (the surface tension). Test your prediction in two different ways. (1) Drop a pebble into water, and observe how fast the slowest ripples move outward. (2) Move a toothpick through a pan of water, and look for the fastest speed at which the toothpick generates no waves.

    Exercise \(\PageIndex{12}\): Surface tension and the size of raindrops

    A liquid’s surface tension, usually denoted \(\gamma\), is the energy required to create new surface divided by the area of the new surface. For a falling raindrop, surface tension and drag compete: the drag force flattens the raindrop, and surface tension keeps it spherical. If the drop gets too flat, it can lower its surface energy by breaking into smaller and more spherical droplets. The fluid dynamics is complicated, but we don’t need to know it. Instead, use competition reasoning (easy cases) to estimate the maximum size of raindrops.

    Exercise \(\PageIndex{13}\): Waves driven by surface tension

    Imagine a wave with reduced wavelength \(\cancel{\lambda}\) on the surface of a fluid.

    a. Show that the dimensionless ratio

    \[R \equiv \frac{\textrm{potential energy due to gravity}}{\textrm{potential energy due to surface tension}}\]

    is, after ignoring dimensionless constants, the dimensionless group \(\rho g \cancel{\lambda}^{2}/\gamma\) that we used in Section 8.4.1 to distinguish waves driven by gravity from waves driven by surface tension.

    b. For water, estimate the critical wavelength \(\lambda\) at which R ∼ 1.

    Exercise \(\PageIndex{14}\): Including buoyancy

    The terminal speed \(v\) of a raindrop with radius r can be written in the following dimensionless form:

    \[\frac{v^{2}}{rg} = f(\frac{\rho_{water}}{\rho_{air}}).\]

    In this problem, you use easy cases of \(x \equiv \rho_{water}/\rho_{air}\) to guess how buoyancy affects this result. (Imagine that you may vary the density of air or water as needed.) In dimensional analysis, including the buoyant force requires including \(\rho_{air}\), g, and r in order to compute the weight of the displaced fluid (which is the buoyant force)—but those variables are already included in the dimensional analysis. Therefore, including buoyancy doesn’t require a new dimensionless group. So it must change the form of the dimensionless function \(f\).

    a. Before you account for buoyancy: What is the dimensionless function \(f (x)\)? Assume spherical raindrops and that cd ≈ 0.5.

    b. What would be the effect of buoyancy if \(\rho_{water}\) were equal to \(rho_{air}\)? This thought experiment is the easy case x = 1. Therefore, find \(f(1)\).

    c. Guess the general form of \(f\) with buoyancy, and thereby find \(v\) including the effect of buoyancy.

    d. Explain physically the difference between \(v\) without and with buoyancy. Hint: How does buoyancy affect g?


    This page titled 8.5: Summary and Further Problems is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Sanjoy Mahajan (MIT OpenCourseWare) .

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