# 2- Units, Dimensional Analysis, Estimation - (Mostly) Classical Mechanics

Note by Hope S, updated more than 1 year ago Created by Hope S about 4 years ago
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### Description

Units, Dimensional Analysis, Estimation. The math behind the kilogram prototype (problem solving).

## Resource summary

### Page 1

Units, Dimensional Analysis and Estimation (MIT) Absolute time is an abstraction, things can only be measured in terms of other things. We are, however, sure of the speed of light, which will give us a way to measure both time and distance. (t*v = d)System Internationale: Mass (which can be related to the speed of light) is measured in kilograms, which are based on a platinum-iridium reference kilogram. Because of the intrinsic flaws in using an artifact as a standard, something which compares mass and the number of atoms would be preferable, but currently error in the reference kilo and the error in measurement for a silicon crystal (the most likely current alternative) are about the same. Time we can measure more accurately, with the number of atomic oscillations per second. With an accurate measure for time we have a good definition for a meter: the distance that light travels in a fraction of a second. (Very large distances can be measured more manageably in light-years.) Radians relate the arc of a circle to its radius (angles can also be used to measure very large distances, that is, with parsecs). Steradians are similar units of solid angle (a steradian cuts of the surface area of a sphere equal to the area of a square with sides equal to the sphere's radius). This is useful in measuring radiant intensity. ________________________ Dimensions can often be used to estimate the order of magnitude of a physical equation. If a given equation is described by length (L) time (T), and mass (M), the chance for confusion about terms is less, and it may give a better intuition about what's actually going on.(Why does this work? See first chapter of Street-fighting Mathematics and constraint propagation)________________________Likewise, using known information and reasonable guesses to arrive at an estimation gives a general idea of the answer, shows whether a solution is reasonable (or worth further effort) and allows you to run through the entire process in your head. (Ideally.)

### Page 2

The Logistics of Defining Mass (MIT texbook) Mass is currently measured with the reference kilogram, made of platinum-iridium alloy. To reduce the effects of corrosion, this prototype is a right cylinder with a radius that minimizes the surface area for the given volume. We know the density of the alloy (21.56 grams per cubic centimeter) and can find the optimal radius to height ratio, which will give us the volume of the prototype kilogram:The surface area can be described in terms of the radius and volume. SA = 2πr^2 + 2πr*(V/πr^2) SA = 2πr^2 + 2V/r dSA = 4πr - 2 V/r^2 when the derivative is zero r = (V/2π)^(1/3) giving a ratio r:h of 1/2 Mass/Volume = Density, so density times the mass gives a volume of 46.38 cubic centimeters, with an optimal radius of (V/2π)^(1/3), this gives us a radius of about 1.95 centimeters. However, due to the inherent risks of relying on a physical prototype, and the unavoidable wear and tear that results in a loss of mass of about 1 μg per year, we'd like a better solution....______________________________Being able to measure mass in terms of discrete, unchanging tiny masses (i.e. atoms) seems like a good idea. Silicon grows in large uniform crystals, so we'd hope that we could come up with a definition that describes a kilogram mass in terms of silicon atoms. How do we practically measure this? mass = density*volume moles = density*volume/molar_mass (that is, the number of moles in a given mass is the product of the density and the volume, divided by the molar mass) atoms = (N_a)*density*volume/molar_mass (that is, the number of atoms in a given mass is the product of Avogadro's number and the number of moles in the mass) ρ = m / V . (the density of the unit cell is the unit mass divided by the unit volume) m = M_mol*V*N_0/(N_a*V_0) or, the density of the unit cell is the molar mass times the number of atoms in a sample divided by avogadro's constant times the sample volume) (or, the unit mass is the mass of a single atom (molar_mass/avogadro's constant) times the number of atoms in the sample (sample atoms times unit volume divided by sample volume).

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