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This is a continuation of this question.

http://ocw.mit.edu/courses/physics/8-01-physics-i-classical-mechanics-fall-1999/video-lectures/lecture-1/ skip this lecture to around 25:50.

After doing dimensional analysis on $t\propto h^\alpha m^\beta g^\gamma$ Lewin concludes that:

$$\alpha = \frac{1}{2}, \gamma = -1/2, \beta = 0$$

This is all fully understood, but he then goes to conclude from this that:

$$t = C \sqrt{\frac{h}{g}}$$

How did he get to this? And why is he allowed to just assume that there is a constant, C, there when he doesn't even know its value or what it is?

Keep things as simple as possible please, I'm 16.

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ODP
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4 Answers4

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As requested, but please don't mark this as the answer because it's a rehash of Manishearth's post:

When physicists say for example "time is proportional to $\sqrt{h}$" what they mean is that:

$$t = C \times \sqrt{h}$$

where C is just a number, called the constant of proportionality, and it doesn't change. The only way to find out the value of the constant $C$ is to measure $t$ for a variety of values of $\sqrt{h}$. Typically you would draw a graph of $t$ against $\sqrt{h}$ and you should get a straight line. The gradient of the line gives you the value of $C$. If you don't get a straight line then you've made a mistake somewhere and $C$ isn't constant.

So the lecturer in the video starts by guessing that $t$ is proportional to $h^\alpha m^\beta g^\gamma$, and what he means by this is:

$$t = C \times h^\alpha m^\beta g^\gamma$$

where $C$ is the constant of proportionality. Dimensional analysis can't tell you what the constant of proportionality is. The only way to find the value of $C$ is to do the experiment.

John Rennie
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The next answer is wrong and should not be considered, please.

Why a constant like the one $C$ in: $t = C \sqrt{\frac{h}{g}}$

You can express time in hours, or minutes or years, ...
You can express length in meters, miles or light-years, or ...

Clearly the constant we have to use is dependent on the choice of the units. We can also choose a system of units (existent or created ad-hoc because it is a convention) in such a way that $C=1$.

Edit add, after the correct remark of Mr Mark.
Indeed C is dimensionless. I beleive that it is tied to the geometry of the problem and, in this case, I think that its value is $2\pi$, normally associated to a planar problem (2D) involving angles. The value of $2\pi$ is not only a number but the expression of the measure of an angle (360°) (the length of the circle with radius=1 u is 2$\pi\cdot u$, it have length units!) .
I will show with an example why the above answer is wrong.
The formula in question is the pendulum equation for small apertures and gives its period in function of the lenght $h$ and the $g$ gravitational acceleration.
For instance using $100\:\rm{m}$ and $9.8 \:\rm{m/s^2}$, it is $ t=2\pi\sqrt{\frac{100}{9.8}}, t=20.07s$ the period of such pendulum (in the usual units meter and second).
Now if we adopt a unit of measure of time ($ut$) with the double of the extension of one second, say $2s=ut$, and keep measuring in meters. The period in this units has the value $10.035\:\rm{ut}$ (half of the $20.07$ but it has the same extension of time) . Clearly we can not use the value $9.8\:\rm{m/s^2}$ for the acceleration $g$. We have to express the acceleration in the units $\:\rm{m/ut^2} $ which is $g = 39.20 \:\rm{m/ut^2}$.

Manishearth
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Helder Velez
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Remember there's a proportionality sign $\propto$ $$t\propto h^\alpha m^\beta g^\gamma$$ Whenever we have $A\propto B$, then $A=CB$, where C is some constant.For example, gravitational force on the surface of the earth $\propto m$, but it's actually $mg$, where $g$ is (sort of ) constant.

For a pendulum, in SI units, the $c$ is $2\pi$.

The point is, while using dimensional analysis, you know which quantities your variable depends on, but not how. Generally, a dependency is of the form $kA^\alpha$, where A is the variable, and the other two are constants that you need to determine. You can determine $\alpha$ via dimensional analysis, but $k$ requires experimental data. The $c$ comes directly from the product of the $k$s.

Manishearth
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Have not watched your lesson but your question was:

$$t= C\sqrt{\frac{h}g}$$

How did he get to this?

$h^{1/2} = \sqrt{h}$

$m^0 = 1$ (not used)

$g^{-1/2} = \sqrt\frac1g$

$t$ could therefore be said, to be proportional to $\sqrt{h}$, since, $\sqrt{h}$ is the only variable besides $t$.

$\therefore t= C\sqrt{h}$ where $C = \sqrt\frac1g$ (a constant)

Not sure why g was left in the equation, rather than including it in with constant C, but it was, so

$t= C\sqrt\frac{h}{g} $

This is the standard formula for,

Time $t$,taken for an object to fall distance h.

In this formula $C = \sqrt2$

$g$ is a constant, if distance $h$, is not excessive.

Hope that makes sense to you.

---------------------------

Left this as an answer, to your other similar question.

Yes $g$ is a constant, so can be dropped from the equation.

So $t$ is proportional to $\sqrt h$

or $h$ is proportional to $t^2$

I.E. Distance $h$,is proportional to the square of the elapsed fall time, $t$.

Clive Ballard
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