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## Center of mass of infinite cylinder of air

1. The problem statement, all variables and given/known data

The density of air at height z above the Earth’s surface is proportional to e^(−az) , where a is a constant > 0. Find the centre of mass of an infinite cylinder of air above a small flat area on the Earth’s surface. Hint : Consider line density and the identities:

$\frac{d}{dz}e^{-az}=-ae^{-az}$

$\frac{d}{dz}((az+1)e^{-az})=-a^{2}ze^{-az}$

2. Relevant equations

Center of mass = $\frac{1}{M}\sum{m_{i}x_{i}}=\frac{1}{M}\int{xdm}$

3. The attempt at a solution

I have no idea how to get started because I don't know how to use the e^(-az) expression. Could I just write that the density of air at height z = be^(-az) where b is some constant of proportionality? Then I think I would try to find M and dm/dx, plug it into the center of mass equation and integrate from 0 to infinity?
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 Quote by phosgene 1. The problem statement, all variables and given/known data The density of air at height z above the Earth’s surface is proportional to e^(−az) , where a is a constant > 0. Find the centre of mass of an infinite cylinder of air above a small flat area on the Earth’s surface. Hint : Consider line density and the identities: $\frac{d}{dz}e^{-az}=-ae^{-az}$ $\frac{d}{dz}((az+1)e^{-az})=-a^{2}ze^{-az}$ 2. Relevant equations Center of mass = $\frac{1}{M}\sum{m_{i}x_{i}}=\frac{1}{M}\int{xdm}$ 3. The attempt at a solution I have no idea how to get started because I don't know how to use the e^(-az) expression. Could I just write that the density of air at height z = be^(-az) where b is some constant of proportionality? Then I think I would try to find M and dm/dx, plug it into the center of mass equation and integrate from 0 to infinity?
Yes, taking into account that dm=ρ(z)dz, and you integrate with respect to z.

ehild
 Recognitions: Gold Member Thanks :) I did the calculation and got 1/a, is that correct?

Recognitions:
Homework Help

## Center of mass of infinite cylinder of air

 Quote by phosgene Thanks :) I did the calculation and got 1/a, is that correct?
It is correct. Well done!

ehild
 Recognitions: Gold Member Thanks again!