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*From*: brian whatcott <betwys1@sbcglobal.net>*Date*: Mon, 05 Nov 2012 12:38:27 -0600

for a object at rest in some frame, a newly imparted velocity, any velocity, represents an infinite acceleration. You presumably wish to sweep this moment under the carpet. You are left with some acceleration which goes to zero at the moment drag equals unbalanced force. You presumably wish to sweep this limiting velocity under the carpet too.

Brian W

On 11/5/2012 11:49 AM, Paul Nord wrote:

If I've got a blimp inflated to neutral buoyancy and I hang a small mass from it, what will the acceleration of the blimp be immediately after I attach the mass?

Since we're still at zero velocity we can ignore the viscous effects of the air for just a moment. I believe that I need to know the un-inflated mass of the balloon and the payload, the volume of the helium, the mass of the helium, and the mass of the displaced air. Let's assume a very small pressure is held by the balloon so that we can think of it as simply a volume of helium at the ambient pressure. The mathematics of a simple Attwood's Machine would seem to apply.

The total mass going down:

balloon and payload

helium mass

extra ballast weight (call this 'm')

Total mass going up:

mass of displaced air

Let's call the sum of all of the mass except for the ballast weight 'M'.

M = balloon + payload + helium + displaced air

The acceleration of the balloon is then:

a = g * m / (m + M)

Question 1: Of course Pascal's Principle says that air pressure will distribute itself equally on all sides. In the static case I can ignore the effects of pressure and air mass. The net force is zero (ignoring the vertical pressure gradient of the air, yes). However, for a balloon to move down, an identical volume of air needs to move up the same distance. The mass of that air cannot be ignored. Is this a valid assumption?

Question 2: Flow through a vicious fluid is typically modeled with a term which rises as a function of the square of the velocity. There is a force resisting the passage of a moving object. If it use the mass of the displaced fluid in the calculation above, am I already accounting for some of the "drag" force which is normally accounted for in this velocity squared term?

Paul

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**Follow-Ups**:**Re: [Phys-L] Dirigible Flight Question***From:*Paul Nord <paul.nord@valpo.edu>

**References**:**[Phys-L] Dirigible Flight Question***From:*Paul Nord <paul.nord@valpo.edu>

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