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5.3: More dimensionless groups

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    24109
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    The biggest change in dimensional analysis comes with a second independent dimensionless group. With only one dimensionless group, the most general statement that the universe could make was

    \[\textrm{dimensionless group} \sim 1\]

    With a second group, the universe gains freedom on the right side:

    \[\textrm{group 1} = f (\textrm{group 2})\]

    where f is a dimensionless function: It takes a dimensionless number as its input, and it produces a dimensionless number as its output.

    As an example of two independent dimensionless groups, let’s extend the analysis of Problem 5.10, where you predicted the impact speed of a rock dropped from a height ℎ. The impact speed depends on g and ℎ, so a reasonable choice for the independent dimensionless group is \(v \sqrt{gh}\) Therefore, its value is a universal, dimensionless constant (which turns out to be \(\sqrt{2}\)). To make a second independent dimensionless group, we add a degree of freedom to the problem: that the rock is thrown downward with speed v0 (the previous problem is the case v0 =0).

    With this complication, what are independent dimensionless groups?

    Adding a quantity but no new independent dimension creates one more independent dimensionless group. Therefore, there are two independent groups—for example \(v/\sqrt{gh}\) and \(v_{0}/\sqrt{gh}\). The most general dimensionless statement, which has the form group 1 = f (group 2), is

    \[\frac{v}{\sqrt{gh}} = f(\frac{v_{0}}{\sqrt{gh}}).\]

    This point is as far as dimensional analysis can take us. To go further requires adding physics knowledge (Problem \(\PageIndex{1}\)). But dimensional analysis already tells us that this function f is a universal function: It describes the impact speed of every thrown object in the universe, no matter the launch velocity, the drop height, or the gravitational field strength.

    Exercise \(\PageIndex{1}\): Impact speed of a thrown rock

    For a rock thrown downward with speed v0, use conservation of energy to find the form of the dimensionless function f in \(v/\sqrt{gh} = f(v_{0}/\sqrt{gh}).\)

    Exercise \(\PageIndex{2}\): Nonideal spring

    Imagine a mass connected to a spring with force law \(F \: \propto \: x^{3}\) (instead of the ideal-spring force \(F \: \propto \: x\)) and therefore with potential energy \(V \sim cx^{4}\) (where C is a constant). Which curve shows how the system’s oscillation period T depends on the amplitude x0 ?

    clipboard_e2b9fe99d2fd8beb9fa281732948803e9.png

    Exercise \(\PageIndex{3}\): Rolling down the plane

    In this problem, you use dimensional analysis to simplify finding the acceleration of a ring rolling (without slipping) down an inclined plane.

    clipboard_ef87c8bb0e99156e7d3f01fd9b17a3265.png

    a. List the quantities on which the ring’s acceleration depends. Hint: Include the ring’s moment of inertia and its radius. Do you also need to include its mass?

    b. Form independent dimensionless groups to write the dimensionless statement

    \[\textrm{group proportional to } a = f (\textrm{groups not containing } a)\]

    c. Does a bigger ring roll faster than a smaller ring?

    d. Does a denser ring roll faster than a less dense ring (of the same radius)?

    5.3.1 Bending of starlight by the Sun

    Our next example of two independent dimensionless groups—the deflection of starlight by the Sun—will illustrate how to incorporate physical knowledge into the mathematical results from dimensional analysis.

    Rocks, birds, and people feel the effect of gravity. So why not light? The analysis of its deflection is a triumph of Einstein’s theory of general relativity. However, the theory is based on ten coupled, nonlinear partial-differential equations. Rather than solving these difficult equations, let’s use dimensional analysis.

    Because this problem is more complicated than the previous examples, let’s organize the complexity by making the steps explicit, including the step of incorporating physical knowledge. Divide and conquer!

    1. List the relevant quantities.

    2. Form independent dimensionless groups.

    3. Use the groups to make the most general statement about deflection.

    4. Narrow the possibilities by incorporating physical knowledge.

    Step 1: Listing relevant quantities

    In the first step, we think of and list the quantities that determine the bending. To find them, I often draw a diagram with verbal but without quantitative labels. As the diagram cries out for quantitative labels, it suggests quantities for the list.

    clipboard_ecc5cd4dab4fbd1ed92d0c936449785d2.png

    The bent path means that the star shifts its apparent position. The shift is most naturally measured not as an absolute distance but rather as an angle \(\theta\). For example, if \(\theta\) = 180 (\(\pi\) radians), much larger than the likely deflection, the star would be shifted halfway around the sky.

    \(\theta\) 1 angle
    Gm L3T-2

    Sun's gravity

    r L closest approach

    The labels and the list must include our goal, which is the deflection angle \(\theta\). Because deflection is produced by gravity, we should also include Newton’s gravitational constant G and the Sun’s mass m (we’ll use the more general symbol m rather than MSun, because we may apply the formula to light paths around other stellar objects). These quantities could join the list as two separate quantities. However, the physical consequences of gravity—for example, the gravitational force—depend on G and m only through the product Gm. Therefore, let’s include just the Gm abstraction on the list. (Problem 5.19 shows you how to find its dimensions.)

    The final quantity on the list is based on our knowledge that gravity gets weaker with distance. Therefore, we include the distance from the Sun to the light beam. The phrase “distance from the Sun” is ambiguous, because the beam is at various distances. Our quantity r will be the shortest distance from the center of the Sun to the beam (the distance of closest approach).

    clipboard_e2750dc17b5cf174ffeca659c262fe5ef.png

    Exercise \(\PageIndex{4}\): Dimensions of Gm and G

    Use Newton's law of universal gravitation, \(F = Gm_{1}m_{2}/r^{2}\), to find the dimensions of Gm and G.

    Step 2: Form independent dimensionless groups

    The second step is to form independent dimensionless groups. One group is easy: The angle \(\theta\) is already dimensionless. Unfortunately, three quantities and two independent dimensions (L and T) produce only one independent dimensionless group. With only one dimensionless group, the most general statement is merely \(\theta\) = constant.

    This prediction is absurd. The bending, if it is produced by gravity, has to depend on Gm and r. These quantities have to appear in a second dimensionless group. Creating a second group, we know from the Buckingham Pi theorem, requires at least one more quantity. Its absence indicates that our analysis is missing essential physics.

    What physics is missing?

    No quantity so far distinguishes between the path of light and of, say, a planet. A crucial difference is that light travels far more rapidly than a planet. Let’s represent this important characteristic of light by including the c.

    \(\theta\) 1 angle
    Gm L3T-2

    Sun's gravity

    r L closest approach
    c LT-1 speed of light

    This quantity, a speed, does not introduce a new independent dimension (it uses length and time). Therefore, it increases the number of independent dimensionless groups by one, from one to two. To find the new group, first check whether any dimension appears in only two quantities. If it does, the search simplifies. Time appears in only two quantities: in Gm as T−2 and in c as T−1. To cancel out time, the new dimensionless group must contain Gm/c2. This quotient contains one power of length, so our dimensionless group is Gm/rc2.

    As we hoped, the new group contains Gm and r. Its form illustrates again that quantities with dimensions are not meaningful alone. For knowing just Gm is not enough to decide whether or not gravity is strong. As a quantity with dimensions, Gm must be compared to another relevant quantity with the same dimensions. Here, that quantity is rc2, and the comparison leads to the dimensionless ratio Gm/rc2.

    Can we choose other pairs of independent dimensionless groups?

    Yes. For example, \(\theta\) and \(Gm \theta /rc^{2}\) also make a set of independent dimensionless groups. Mathematically, all pairs of independent dimensionless groups are equivalent, in that any pair can represent any quantitative statement about light bending.

    However, look ahead to the goal: We hope to solve for θ. If θ appears in both groups, we will end up with an implicit equation for θ, where θ appears on both sides of the equals sign. Although such an equation is mathematically legitimate, it is much harder to think about and to solve than is an explicit equation, where θ is on the left side only.

    Therefore, when choosing independent dimensionless groups, place the goal quantity in only one group. This rule of thumb does not remove all our freedom in choosing the groups, but it greatly limits the choices.

    Exercise \(\PageIndex{5}\): Physical interpretation of the new group

    Interpret the dimensionless group Gm/rc2 by multiplying by mlight/mlight and regrouping the quantities until you find physical interpretations for the numerator and denominator.

    Step 3: Make the most general dimensionless statement

    The third step is to use the independent dimensionless groups to write the most general statement about the bending angle. It has the form group 1 = f (group 2). Here,

    \[\theta= f (\frac{gm}{rc^{2}}),\]

    where f is a universal, dimensionless function. Dimensional analysis cannot determine f. However, it has told us that f is a function only of Gm/rc2 and not of the four quantities G, m, r, and c separately. That information is the great simplification.

    Step 4: Use physical knowledge to narrow the possibilities The space of possible functions—here, all nonpathological functions of one variable—is vast. Therefore, the fourth and final step is to narrow the possibilities for f by incorporating physical knowledge. First, imagine increasing the effect of gravity by increasing the mass m—which increases Gm/rc2.

    This change should, based on our physical intuition about gravity, also increase the bending angle. Therefore, f should be a monotonically increasing function of Gm/rc2. Second, imagine an antigravity world, where the gravitational constant G is negative. Then the Sun would deflect light away from it, making the bending angle negative. In terms of \(x \equiv Gm/rc^{2}\), this constraint eliminates even functions of x, such as \(f(x) \sim x^{2}\), which produce the same sign for the bending angle independent of the sign of x.

    The simplest function meeting both the monotonicity and sign constraints is f(x) ~ x. In terms of the second group GM/rc2, the form f(x) ~ x is the dimensionless statement about bending

    \[\theta \sim \frac{GM}{rc^{2}}.\]

    All reasonable theories of gravity will predict this relation, because it is almost entirely a mathematical requirement. The theories differ only in the dimensionless factor hidden in the single approximation sign ~:

    \[\theta = \frac{Gm}{rc^{2}} \times \left \{ \begin{array} {ll} 1 \: \textrm{(simplest guess);} \\ 2 \: \textrm{(Newtonian gravity);} \\ 4 \: \textrm{(general relativity).} \end{array} \right. \]

    The factor of 2 for Newtonian gravity results from solving for the trajectory of a rock roving past the Sun with speed c. The factor of 4 for generalrelativity, double the Newtonian value, results from solving the ten partial-differential equations in the limit that the gravitational field is weak.

    How large are these angles?

    Let’s first estimate the bending angles closer to home—produced by the Earth’s gravity. For a light ray just grazing the surface of the Earth, the bending angle (in radians!) is roughly 10−9:

    \[\theta_{Earth} \sim \frac{\overbrace{6 \times 10^{-11} \textrm{kg}^{-1} \textrm{ m}^{3} \textrm{ s}^{-2}}^{G} \times \overbrace{6 \times 10^{24}}^{m_{Earth}}}{\underbrace{6.4 \times 10^{6} \textrm{m}}_{R_{Earth}} \times \underbrace{10^{17} \textrm{m}^{2} \textrm{s}^{-2}}_{c^{2}}} \approx 0.7 \times 10^{-9}.\]

    Can we observe this angle?

    The bending angle is the angular shift in the position of the star making the starlight. To observe these shifts, astronomers compare a telescope picture of the star and surrounding sky, with and without the deflection. A telescope with lens of diameter D can resolve angles roughly as small as \(\lambda /D\), where \(\lambda\) is the wavelength of light. As a result, a lens that can resolve 0.7 ×10−9 radians has a diameter of at least 700 meters:

    \[D \sim \frac{\lambda}{\theta_{Earth}} \sim \frac{0.5 \times 10^{-6} m}{0.7 \times 10^{-9}} \approx 700 m.\]

    On its own, this length does not mean much. However, the largest telescope lens has a diameter of roughly 1 meter; the largest telescope mirror, in a different telescope design, is still only 10 meters. No practical lens or mirror can be 700 meters in diameter. Thus, there is no way to see the deflection produced by the Earth’s gravitational field.

    Physicists therefore searched for a stronger source of light bending. The bending angle is proportional to m/r. The largest mass in the solar system is the Sun. For a light ray grazing the surface of the Sun, r = RSun and m = MSun. So the ratio of Sun-to-Earth-produced bending angles is

    \[\frac{\theta_{Sun}}{\theta_{Earth}} = \frac{M_{Sun}}{m_{Earth}} \times (\frac{R_{sun}}{R_{Earth}})^{-1}.\]

    The mass ratio is roughly 3×105; the radius ratio is roughly 100. The ratio of deflection angles is therefore roughly 3000. The required lens diameter, which is inversely proportional to \(\theta\), is correspondingly smaller than 700 meters by a factor of 3000: roughly 25 centimeters or 10 inches. That lens size is plausible, and the deflection might just be measurable.

    Between 1909 and 1916, Einstein believed that a correct theory of gravity would predict the Newtonian value of 4.2×10−6 radians or 0.87 arcseconds:

    \[\underbrace{0.7 \times 10^{-9} \textrm{rad}}_{\sim \theta_{Earth}} \times \underbrace{2}_{\textrm{Newtonian gravity}} \times \underbrace{3000}_{\theta_{Sun}/\theta_{Earth}} \sim 4.2 \times 10^{-6} \textrm{ rad}.\]

    The German astronomer Soldner had derived the same result in 1803. The eclipse expeditions to test his (and Soldner’s) prediction got rained out or clouded out. By the time an expedition got lucky with the weather, in 1919, Einstein had invented a new theory of gravity—general relativity—and it predicted a deflection twice as large, or 1.75 arcseconds.

    The eclipse expedition of 1919, led by Arthur Eddington of Cambridge University and using a 13-inch lens, measured the deflection. The measurements are difficult, and the results were not accurate enough to decide clearly which theory was right. But 1919 was the first year after World War One, in which Germany and Britain had fought each other almost to oblivion. A theory invented by a German, confirmed by an Englishman (from Newton’s university, no less)—such a picture was welcome after the war. The world press and scientific community declared Einstein vindicated.

    A proper confirmation of Einstein’s prediction came only with the advent of radio astronomy, which allowed small deflections to be measured accurately (Problem 5.21). The results, described as the dimensionless factor multiplying Gm/rc2, were around 4 ± 0.2—definitely different from the Newtonian prediction of 2 and consistent with general relativity.

    Exercise \(\PageIndex{6}\): Accuracy of radio telescope measurements

    The angular resolution of a telescope is \(\lambda/D\), where \(\lambda\) is the wavelength and D is the telescope’s diameter. But radio waves have a much longer wavelength than light. How can measurements of the bending angle that are made with radio telescopes be so much more accurate than measurements made with optical telescopes?

    Exercise \(\PageIndex{7}\): Another theory of gravity

    An alternative to Newtonian gravity and to general relativity is the Brans–Dicke theory of gravitation [3]. Look up what it predicts for the angle that the Sun would deflect starlight. Express your answer as the dimensionless prefactor in \(\theta \sim Gm/rc^{2}\).

    5.3.2 Drag

    Even more difficult than the equations of general relativity, which at least have been solved for realistic problems, are the Navier–Stokes equations of fluid mechanics. We first met them in Section 3.5, where their complexity pushed us to find an alternative to solving them directly. We used a conservation argument, which we tested using a falling cone, and concluded that the drag force on an object traveling through a fluid is

    \[F_{drag} \sim \rho v^{2}A_{cs},\]

    where \(\rho\) is the density of the fluid, v is the speed of the object, and Acs is its cross-sectional area. This result is the starting point for our dimensional analysis.

    With Fdrag dependent on \(\rho\), v, and Acs, there are four quantities and three independent dimensions. Therefore, by the Buckingham Pi theorem, there is one independent dimensionless group. Our expression for the drag force already gives one such group

    \[\frac{F_{drag}}{\rho v^{2}A_{cs}}.\]

    Because the \(\rho v^{2}\) in the denominator looks like mv2 in kinetic energy, a factor of one-half is traditionally included in the denominator:

    \[\textrm{group 1} \equiv \frac{F_{drag}}{\frac{1}{2} \rho v^{2} A_{cs}}.\]

    The resulting dimensionless group is called the drag coefficient cd.

    \[c_{d} \equiv \frac{F_{drag}}{\frac{1}{2} \rho v^{2} A_{cs}}\]

    As the only dimensionless group, it has to be a constant: There is no other group on which it could depend. Therefore,

    \[\frac{F_{drag}}{\frac{1}{2} \rho v^{2} A_{cs}} = \textrm{dimensionless constant}.\]

    This result is equivalent to our prediction based on conservation of energy, that \(F_{drag} \sim \rho v^{2} A_{cs}\). However, trouble happens when we wonder about the dimensionless constant’s value.

    Dimensional analysis, as a mathematical technique, cannot predict the constant. Doing so requires physical reasoning, which starts with the observation that drag consumes energy. Here, no quantity on which Fdrag depends—namely, \(rho\), v, or Acs—represents a mechanism of energy loss. Therefore, the drag coefficient should be zero. Although this value is consistent with the prediction that the drag coefficient is constant, it contradicts all our experience of fluids!

    Our analysis needs to include a mechanism of energy loss. In fluids, loss is due to viscosity (for an object moving at a constant speed and generating no waves). Its physics is the topic of Section 7.3.2, as an example of a diffusion constant. For our purposes here, we need just its dimensions in order to incorporate viscosity into a dimensionless group: The dimensions of kinematic viscosity \(\nu\) are L2T−1. (Unfortunately, in almost every font, the standard symbols for velocity and kinematic viscosity, \(v\) and \(\nu\), look similar; however, even more confusion would result from selecting new symbols.)

    Thus, the kinematic viscosity joins our list. Adding one quantity without adding a new independent dimension creates a new independent dimensionless group. Similarly, in our analysis of the deflection of starlight (Section ), adding the speed of light c created a second independent dimensionless group.

    What is this new group containing viscosity?

    Before finding the group, look ahead to how it will be used. With a second group, the most general statement has the form

    \[\underbrace{\textrm{group 1}}_{c_{d}} = f(\textrm{group 2}).\]

    Our goal is Fdrag, which is already part of group 1. Including Fdrag also in group 2 would be a bad tactic, because it would produce an equation with Fdrag on both sides and, additionally, wrapped on the right side in the unknown function f. Keep Fdrag out of group 2.

    \(v\) LT-1 speed
    Acs L2 area
    \(rho\) ML-3 fluid density
    \(\nu\) L2T-1 viscosity

    To make this group from the other quantities \(v\), Acs, \(\rho\), and \(\nu\), look for dimensions that appear in only one or two quantities. Mass appears only in the density; thus, the density cannot appear in group 2: If it were part of group 2, there would be no way to cancel the dimensions of mass, and group 2 could not be dimensionless.

    Time appears in two quantities: speed \(v\) and viscosity \(\nu\). Because each quantity contains the same power of time (T−1), group 2 has to contain the quotient \(v\)/\(\nu\)—otherwise the dimensions of time would not cancel.

    Here's what we know so far about this dimensionless group: So that mass disappears, it cannot contain the denisty \(\rho\); so that time disappears, it must contian \(v\) and \(\nu\) as their quotient \(v/\nu\). The remaining task is to make length disappear--for which purpose we use the object's cross-sectional area Acs. The quotient \(v/\nu\) has dimensions of L-1 ,so \(\sqrt{A_{cs}}v/\nu\) is a new, independent dimensionless group.

    Instead of \(\sqrt{A_{cs}}\), usually one uses the diameter D. With that choice, our group 2 is called the Reynolds number Re.

    \[\textrm{Re} \equiv \frac{vD}{\nu}.\]

    The conclusion from dimensional analysis is then

    \[\underbrace{\textrm{drag coefficient}}_{c_{d}} = \underbrace{f(\textrm{Reynolds number}).}_{\textrm{Re}}\]

    For each shape, the dimensionless function f is a universal function; it depends on the shape of the object, but not on its size. For example, a sphere, a cylinder, and a cone have different functions fsphere, fcylinder, and fcone. Similarly, a narrow-angle cone and a wide-angle cone have different functions.

    But a small and a large sphere are described by the same function, as are a large and a small cone of the same opening angle. Any difference in the drag coefficients for different-sized objects of the same shape results only from the difference in Reynolds numbers (different because the sizes are different).

    clipboard_e180a036e332e8c1a4b02ed4a98f825fa.png

    Here we see the power of dimensional analysis. It shows us that the universe doesn’t care about the size, speed, or viscosity individually. The universe cares about them only through the abstraction known as the Reynolds number. It is the only information needed to determine the drag coefficient (for a given shape).

    Now let’s use this dimensionless framework to analyze the cone experiment: The experimental data showed that the small and large cones fell at the same speed—roughly 1 meter per second.

    What are the corresponding Reynolds numbers?

    The small cone has a diameter of 0.75×7 centimeters (0.75 because one-quarter of the original circumference was removed), which is roughly 5.3 centimeters. The viscosity of air is roughly 1.5×10−5 square meters per second (an estimate that we will make in Section 7.3.2). The resulting Reynolds number is roughly 3500:

    \[\textrm{RE} = \frac{\overbrace{1 \textrm{ m s}^{-1}}^{v} \times \overbrace{0.053 \textrm{ m}}^{D}}{\underbrace{1.5 \times 10^{-5} \textrm{ m}^{2} \textrm{ s}^{-1}}_{v}} \approx 3500.\]

    What is the Reynolds number for the large cone?

    Never make a calculation from scratch when you can use proportional reasoning. Both cones experience the same viscosity and share the same fall speed. Therefore, we can consider \(\nu\) and \(v\) to be constants, leaving the Reynolds number \(vD/\nu\) proportional just to the diameter D.

    Because the large cone has twice the diameter of the small cone, it has twice the Reynolds number:

    \[\textrm{RE}_{\textrm{large}} = 2 \times \textrm{Re}_{\textrm{small}} \approx 7000.\]

    At that Reynolds number, what is the cone’s drag coefficient?

    The drag coefficient is

    \[c_{d} \equiv \frac{F_{drag}}{\frac{1}{2} \rho_{air} v^{2} A_{cs}}\]

    The cone falls at its terminal speed, so the drag force is also its weight W:

    \[F_{drag} = W = A_{paper}\sigma_{paper}g,\]

    where \(\sigma_{paper}\) is the areal density (mass per area) of paper and Apaper is the area of the cone template. The drag coefficient is then

    \[c_{d} = \frac{\overbrace{A_{paper} \sigma_{paper} g}^{F_{drag}}}{\frac{1}{2} \rho_{air} A_{cs} v^{2}}.\]

    As we showed in Section 3.5.2,

    \[A_{cs} = \frac{3}{4} A_{paper}.\]

    This proportionality means that the areas cancel out of the drag coefficient:

    \[c_{d} = \frac{\sigma_{paper}g}{\frac{1}{2} \rho_{air} \times \frac{3}{4} v^{2}}.\]

    To compute cd, plug in the areal density \(\sigma_{paper} \approx 80\) grams per square meter and the measured speed \(v \approx 1\) meter per second:

    \[c_{d} \approx \frac{\overbrace{8 \times 10^{-2} \textrm{ kg m}^{-2}}^{\sigma_{paper}} \times \overbrace{10 \textrm{ m s}^{-2}}^{g}}{\frac{1}{2} \times \underbrace{1.2 \textrm{ kg m}^{-3}}_{\rho_{air}} \times \frac{3}{4} \times \underbrace{1 \textrm{ m}^{2} \textrm{ s}^{-2}}_{v^{2}}} \approx 1.8.\]

    Because no quantity in this calculation depends on the cone’s size, both cones have the same drag coefficient. (Our estimated drag coefficient is significantly largerthan the canonical drag coefficient for a solid cone, roughly 0.7, and is approximately the drag coefficient for a wedge.)

    Thus, the drag coefficient is independent of Reynolds number—at least, for Reynolds numbers between 3500 and 7000. The giant-cone experiment of Problem 4.16 shows that the independence holds even to Re ∼ 14 000. Within this range, the dimensionless function f in

    \[\textrm{drag coefficient} = f_{cone}\textrm{(Reynolds number)}\]

    is a constant. What a simple description of the complexity of fluid flow!

    This conclusion for fcone is valid for most shapes. The most extensive drag data is for a sphere—plotted below on log–log axes (data adapted from Fluid-Dynamic Drag: Practical Information on Aerodynamic Drag and Hydrodynamic Resistance [23]):

    clipboard_e7a7d507863804fbe7107809125ca7476.png

    Like the drag coefficient for the cone, the drag coefficient for a sphere is almost constant as the Reynolds number varies from 3500 to 7000. The drag coefficient holds constant even throughout the wider range 2000…3 × 105. Around Re ≈ 3 × 105, the drag coefficient falls by a factor of 5, from 0.5 to roughly 0.1. This drop is the reason that ggolf balls have dimples (you will explain the connection in Problem 7.24).

    At low Reynolds numbers, the drag coefficient becomes large. This behavior, which represents a small object oozing through honey, will be explained in Section 8.3.1.2 using the tool of easy cases. The main point here is that, for most everyday flows, where the Reynolds number is in the “few thousand to few hundred thousand” range, the drag coefficient is a constant that depends only on the shape of the object.

    Exercise \(\PageIndex{8}\): Giant cone

    How large, as measured by the diameter of its cross section, would a paper cone need to be in order for its fall to have a Reynolds number of roughly 3 × 105 (the Reynolds number at which the drag coefficient of a sphere drops significantly)?

    Exercise \(\PageIndex{9}\): Reynolds numbers

    Estimate the Reynolds number for (a) a falling raindrop, (b) a flying mosquito, (c) a person walking, and (d) a jumbo jet flying at its cruising speed

    Exercise \(\PageIndex{10}\): Compound pendulum

    Use dimensional analysis to deduce as much as you can about the period T of a compound pendulum—that is, a pendulum where the bob is not a point mass but is an extended object of mass m. The lightrod (no longer a string) of length l is fixed to the center of mass of the object, which has a moment of inertia ICM about the point of attachment. (Assume that the oscillation amplitude is small and therefore doesn’t affect the period.)

    clipboard_e01bc9964a070a533887ec483ac9750c8.png

    Exercise \(\PageIndex{11}\): Terminal velocity of a raindrop

    In Problem 3.37, you estimated the terminal speed of a raindrop using

    \[F_{drag} \sim \rho_{air} A_{cs}v^{2}.\]

    Redo the calculation using the more precise information that cd ≈ 0.5 and retaining the factor of 4\(\pi\)/3 in the volume of a spherical raindrop


    This page titled 5.3: More dimensionless groups 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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