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Changing mass |
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| Jan31-13, 01:15 PM | #1 |
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Changing mass
hi,
I want to calculate how is acceleration changing if I have changing mass, but constant trust i.e. : T = m*a a = T / m (I know it has to be calculus). Then again I also wan't to be able to calculate displacement and velocities etc.. Trying to find somewhere on the internet a tutorial on equations of motion when the acceleration is varying.. but most of the time I find equations for constant-acceleration. Do you have a good tutorial ? (don't point me to wikipedia, it is good as reference but not as tutorial) I would like also to have some simple Exersises, so I can figure out how it is done in general. thank you |
| Jan31-13, 02:06 PM | #2 |
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You are on the right track. a = T/m and it takes calculus.
The derivitive of 1/m with respect to m is -1/m2 So for constant T, the derivitive of T/m with respect to m is -T/m2 The minus sign indicates that as m increases the quotient T/m decreases. |
| Jan31-13, 04:14 PM | #3 |
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Nice.. ok now how can I calculate displacement or time taken to cross specific distans having this acceleration...
I suppose I can't use : d = x + v*t + 1/2 a*t^2 because this is only valid for constant acceleration ? |
| Feb1-13, 07:02 AM | #4 |
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Changing mass
You are correct. For a non-constant acceleration instead of computing the change in velocity by simply multiplying acceleration by time, you have to compute it by integrating acceleration over time using calculus.
Similarly, for a non-constant velocity you compute change in position by integrating velocity over time rather than simply multiplying velocity by time. You end up with a double integral. The first integration to compute velocity as a function of time results in: http://en.wikipedia.org/wiki/Tsiolko...ocket_equation |
| Feb1-13, 08:36 AM | #5 |
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Conservation of momentum, mv, leads to (mv)'= m'v+ mv'= 0 (the ' indicates the derivative) if there is no external force. If there is a force, then we do not have conservartion of momentum but have (mv)'= m'v+ mv'= F.
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