Mixing
Why can't we assume the positive direction is upwards when solving the DE motion of air drag?
$begingroup$
In essence my question is: Why is it that in differential equations that motion of the object has to be assumed as the positive direction? is there a way to write DEs with the upwards direction as positive which will lead to the right answer?
For instance: Almost all reference material lists:
m(dv/dt) = mg - kv
but what if I wanted to interpret the positive direction as upwards:
m(dv/dt) = kv - mg
when solving the Differential eq this way, there is no transient term (terminal velocity cannot be achieved).
ordinary-differential-equations
asked Feb 1 at 15:25
PrandalsPrandals
858
$endgroup$
add a comment |
$begingroup$
In essence my question is: Why is it that in differential equations that motion of the object has to be assumed as the positive direction? is there a way to write DEs with the upwards direction as positive which will lead to the right answer?
For instance: Almost all reference material lists:
m(dv/dt) = mg - kv
but what if I wanted to interpret the positive direction as upwards:
m(dv/dt) = kv - mg
when solving the Differential eq this way, there is no transient term (terminal velocity cannot be achieved).
ordinary-differential-equations
asked Feb 1 at 15:25
PrandalsPrandals
858
$endgroup$
$begingroup$
The laws of physics are the same in all frames of reference! I'd just carefully define your coordinate system, including origin and positive directions of all relevant dimensions, and go from there.
$endgroup$
– Adrian Keister
Feb 1 at 15:40
add a comment |
$begingroup$
In essence my question is: Why is it that in differential equations that motion of the object has to be assumed as the positive direction? is there a way to write DEs with the upwards direction as positive which will lead to the right answer?
For instance: Almost all reference material lists:
m(dv/dt) = mg - kv
but what if I wanted to interpret the positive direction as upwards:
m(dv/dt) = kv - mg
when solving the Differential eq this way, there is no transient term (terminal velocity cannot be achieved).
ordinary-differential-equations
asked Feb 1 at 15:25
PrandalsPrandals
858
$endgroup$
In essence my question is: Why is it that in differential equations that motion of the object has to be assumed as the positive direction? is there a way to write DEs with the upwards direction as positive which will lead to the right answer?
For instance: Almost all reference material lists:
m(dv/dt) = mg - kv
but what if I wanted to interpret the positive direction as upwards:
m(dv/dt) = kv - mg
when solving the Differential eq this way, there is no transient term (terminal velocity cannot be achieved).
ordinary-differential-equations
ordinary-differential-equations
asked Feb 1 at 15:25
PrandalsPrandals
858
asked Feb 1 at 15:25
PrandalsPrandals
858
asked Feb 1 at 15:25
PrandalsPrandals
858
asked Feb 1 at 15:25
PrandalsPrandals
858
asked Feb 1 at 15:25
PrandalsPrandals
858
858
$begingroup$
The laws of physics are the same in all frames of reference! I'd just carefully define your coordinate system, including origin and positive directions of all relevant dimensions, and go from there.
$endgroup$
– Adrian Keister
Feb 1 at 15:40
add a comment |
$begingroup$
The laws of physics are the same in all frames of reference! I'd just carefully define your coordinate system, including origin and positive directions of all relevant dimensions, and go from there.
$endgroup$
– Adrian Keister
Feb 1 at 15:40
$begingroup$
The laws of physics are the same in all frames of reference! I'd just carefully define your coordinate system, including origin and positive directions of all relevant dimensions, and go from there.
$endgroup$
– Adrian Keister
Feb 1 at 15:40
$begingroup$
The laws of physics are the same in all frames of reference! I'd just carefully define your coordinate system, including origin and positive directions of all relevant dimensions, and go from there.
$endgroup$
– Adrian Keister
Feb 1 at 15:40
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You can define the axes whichever direction you want as long as you are consistent. You are correct that the first form has the axis positive downward, which makes $g$ be positive and the velocity positive. If you define the axis as positive upward and take $g$ to be a positive number the equation should be $$mfrac {dv}{dt}=-mg-kv$$
because the change in velocity is opposite to the velocity. You can then compue terminal velocity by setting the derivative to zero, getting $$v=-frac {mg}k$$ This is negative as it should be because the object is moving downward.
answered Feb 1 at 15:44
Ross MillikanRoss Millikan
301k24200375
$endgroup$
$begingroup$
ohhh. So the way that I have it set up the terminal velocity is "positive" downwards when I referenced upwards as positive; so this value wouldn't have made sense. Thank you.
$endgroup$
– Prandals
Feb 1 at 15:50
$begingroup$
The terminal velocity is negative because downward velocity is negative in your coordinates. That is shown in my answer.
$endgroup$
– Ross Millikan
Feb 1 at 15:52
$begingroup$
So in my initial attempt: m(dv/dt) = kv - mg the v here would be (-v) thus m(dv/dt) = -kv - mg. Is that a correct interpretation?
$endgroup$
– Prandals
Feb 1 at 15:54
$begingroup$
No, you don't change the equation because $v$ is negative. The equation has to be $m(dv/dt)=-kv$ because the friction makes a force opposite to the velocity, whatever the sign is. In some problems the sign of $v$ changes during the problem, like if you throw a ball upward. The sign on $g$ changes if you change the coordinate system, but that is all.
$endgroup$
– Ross Millikan
Feb 1 at 17:56
$begingroup$
So the negative sign is to indicate the opposite nature of the air drag to the velocity. In essence it makes it so that whatever the sign of v is the resulting kv term is opposite?
$endgroup$
– Prandals
Feb 3 at 3:53
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
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$begingroup$
You can define the axes whichever direction you want as long as you are consistent. You are correct that the first form has the axis positive downward, which makes $g$ be positive and the velocity positive. If you define the axis as positive upward and take $g$ to be a positive number the equation should be $$mfrac {dv}{dt}=-mg-kv$$
because the change in velocity is opposite to the velocity. You can then compue terminal velocity by setting the derivative to zero, getting $$v=-frac {mg}k$$ This is negative as it should be because the object is moving downward.
answered Feb 1 at 15:44
Ross MillikanRoss Millikan
301k24200375
$endgroup$
$begingroup$
ohhh. So the way that I have it set up the terminal velocity is "positive" downwards when I referenced upwards as positive; so this value wouldn't have made sense. Thank you.
$endgroup$
– Prandals
Feb 1 at 15:50
$begingroup$
The terminal velocity is negative because downward velocity is negative in your coordinates. That is shown in my answer.
$endgroup$
– Ross Millikan
Feb 1 at 15:52
$begingroup$
So in my initial attempt: m(dv/dt) = kv - mg the v here would be (-v) thus m(dv/dt) = -kv - mg. Is that a correct interpretation?
$endgroup$
– Prandals
Feb 1 at 15:54
$begingroup$
No, you don't change the equation because $v$ is negative. The equation has to be $m(dv/dt)=-kv$ because the friction makes a force opposite to the velocity, whatever the sign is. In some problems the sign of $v$ changes during the problem, like if you throw a ball upward. The sign on $g$ changes if you change the coordinate system, but that is all.
$endgroup$
– Ross Millikan
Feb 1 at 17:56
$begingroup$
So the negative sign is to indicate the opposite nature of the air drag to the velocity. In essence it makes it so that whatever the sign of v is the resulting kv term is opposite?
$endgroup$
– Prandals
Feb 3 at 3:53
add a comment |
$begingroup$
You can define the axes whichever direction you want as long as you are consistent. You are correct that the first form has the axis positive downward, which makes $g$ be positive and the velocity positive. If you define the axis as positive upward and take $g$ to be a positive number the equation should be $$mfrac {dv}{dt}=-mg-kv$$
because the change in velocity is opposite to the velocity. You can then compue terminal velocity by setting the derivative to zero, getting $$v=-frac {mg}k$$ This is negative as it should be because the object is moving downward.
answered Feb 1 at 15:44
Ross MillikanRoss Millikan
301k24200375
$endgroup$
$begingroup$
ohhh. So the way that I have it set up the terminal velocity is "positive" downwards when I referenced upwards as positive; so this value wouldn't have made sense. Thank you.
$endgroup$
– Prandals
Feb 1 at 15:50
$begingroup$
The terminal velocity is negative because downward velocity is negative in your coordinates. That is shown in my answer.
$endgroup$
– Ross Millikan
Feb 1 at 15:52
$begingroup$
So in my initial attempt: m(dv/dt) = kv - mg the v here would be (-v) thus m(dv/dt) = -kv - mg. Is that a correct interpretation?
$endgroup$
– Prandals
Feb 1 at 15:54
$begingroup$
No, you don't change the equation because $v$ is negative. The equation has to be $m(dv/dt)=-kv$ because the friction makes a force opposite to the velocity, whatever the sign is. In some problems the sign of $v$ changes during the problem, like if you throw a ball upward. The sign on $g$ changes if you change the coordinate system, but that is all.
$endgroup$
– Ross Millikan
Feb 1 at 17:56
$begingroup$
So the negative sign is to indicate the opposite nature of the air drag to the velocity. In essence it makes it so that whatever the sign of v is the resulting kv term is opposite?
$endgroup$
– Prandals
Feb 3 at 3:53
add a comment |
$begingroup$
You can define the axes whichever direction you want as long as you are consistent. You are correct that the first form has the axis positive downward, which makes $g$ be positive and the velocity positive. If you define the axis as positive upward and take $g$ to be a positive number the equation should be $$mfrac {dv}{dt}=-mg-kv$$
because the change in velocity is opposite to the velocity. You can then compue terminal velocity by setting the derivative to zero, getting $$v=-frac {mg}k$$ This is negative as it should be because the object is moving downward.
answered Feb 1 at 15:44
Ross MillikanRoss Millikan
301k24200375
$endgroup$
You can define the axes whichever direction you want as long as you are consistent. You are correct that the first form has the axis positive downward, which makes $g$ be positive and the velocity positive. If you define the axis as positive upward and take $g$ to be a positive number the equation should be $$mfrac {dv}{dt}=-mg-kv$$
because the change in velocity is opposite to the velocity. You can then compue terminal velocity by setting the derivative to zero, getting $$v=-frac {mg}k$$ This is negative as it should be because the object is moving downward.
answered Feb 1 at 15:44
Ross MillikanRoss Millikan
301k24200375
edited Feb 1 at 17:50
answered Feb 1 at 15:44
Ross MillikanRoss Millikan
301k24200375
answered Feb 1 at 15:44
Ross MillikanRoss Millikan
301k24200375
answered Feb 1 at 15:44
Ross MillikanRoss Millikan
301k24200375
301k24200375
$begingroup$
ohhh. So the way that I have it set up the terminal velocity is "positive" downwards when I referenced upwards as positive; so this value wouldn't have made sense. Thank you.
$endgroup$
– Prandals
Feb 1 at 15:50
$begingroup$
The terminal velocity is negative because downward velocity is negative in your coordinates. That is shown in my answer.
$endgroup$
– Ross Millikan
Feb 1 at 15:52
$begingroup$
So in my initial attempt: m(dv/dt) = kv - mg the v here would be (-v) thus m(dv/dt) = -kv - mg. Is that a correct interpretation?
$endgroup$
– Prandals
Feb 1 at 15:54
$begingroup$
No, you don't change the equation because $v$ is negative. The equation has to be $m(dv/dt)=-kv$ because the friction makes a force opposite to the velocity, whatever the sign is. In some problems the sign of $v$ changes during the problem, like if you throw a ball upward. The sign on $g$ changes if you change the coordinate system, but that is all.
$endgroup$
– Ross Millikan
Feb 1 at 17:56
$begingroup$
So the negative sign is to indicate the opposite nature of the air drag to the velocity. In essence it makes it so that whatever the sign of v is the resulting kv term is opposite?
$endgroup$
– Prandals
Feb 3 at 3:53
add a comment |
$begingroup$
ohhh. So the way that I have it set up the terminal velocity is "positive" downwards when I referenced upwards as positive; so this value wouldn't have made sense. Thank you.
$endgroup$
– Prandals
Feb 1 at 15:50
$begingroup$
The terminal velocity is negative because downward velocity is negative in your coordinates. That is shown in my answer.
$endgroup$
– Ross Millikan
Feb 1 at 15:52
$begingroup$
So in my initial attempt: m(dv/dt) = kv - mg the v here would be (-v) thus m(dv/dt) = -kv - mg. Is that a correct interpretation?
$endgroup$
– Prandals
Feb 1 at 15:54
$begingroup$
No, you don't change the equation because $v$ is negative. The equation has to be $m(dv/dt)=-kv$ because the friction makes a force opposite to the velocity, whatever the sign is. In some problems the sign of $v$ changes during the problem, like if you throw a ball upward. The sign on $g$ changes if you change the coordinate system, but that is all.
$endgroup$
– Ross Millikan
Feb 1 at 17:56
$begingroup$
So the negative sign is to indicate the opposite nature of the air drag to the velocity. In essence it makes it so that whatever the sign of v is the resulting kv term is opposite?
$endgroup$
– Prandals
Feb 3 at 3:53
$begingroup$
ohhh. So the way that I have it set up the terminal velocity is "positive" downwards when I referenced upwards as positive; so this value wouldn't have made sense. Thank you.
$endgroup$
– Prandals
Feb 1 at 15:50
$begingroup$
ohhh. So the way that I have it set up the terminal velocity is "positive" downwards when I referenced upwards as positive; so this value wouldn't have made sense. Thank you.
$endgroup$
– Prandals
Feb 1 at 15:50
$begingroup$
The terminal velocity is negative because downward velocity is negative in your coordinates. That is shown in my answer.
$endgroup$
– Ross Millikan
Feb 1 at 15:52
$begingroup$
The terminal velocity is negative because downward velocity is negative in your coordinates. That is shown in my answer.
$endgroup$
– Ross Millikan
Feb 1 at 15:52
$begingroup$
So in my initial attempt: m(dv/dt) = kv - mg the v here would be (-v) thus m(dv/dt) = -kv - mg. Is that a correct interpretation?
$endgroup$
– Prandals
Feb 1 at 15:54
$begingroup$
So in my initial attempt: m(dv/dt) = kv - mg the v here would be (-v) thus m(dv/dt) = -kv - mg. Is that a correct interpretation?
$endgroup$
– Prandals
Feb 1 at 15:54
$begingroup$
No, you don't change the equation because $v$ is negative. The equation has to be $m(dv/dt)=-kv$ because the friction makes a force opposite to the velocity, whatever the sign is. In some problems the sign of $v$ changes during the problem, like if you throw a ball upward. The sign on $g$ changes if you change the coordinate system, but that is all.
$endgroup$
– Ross Millikan
Feb 1 at 17:56
$begingroup$
No, you don't change the equation because $v$ is negative. The equation has to be $m(dv/dt)=-kv$ because the friction makes a force opposite to the velocity, whatever the sign is. In some problems the sign of $v$ changes during the problem, like if you throw a ball upward. The sign on $g$ changes if you change the coordinate system, but that is all.
$endgroup$
– Ross Millikan
Feb 1 at 17:56
$begingroup$
So the negative sign is to indicate the opposite nature of the air drag to the velocity. In essence it makes it so that whatever the sign of v is the resulting kv term is opposite?
$endgroup$
– Prandals
Feb 3 at 3:53
$begingroup$
So the negative sign is to indicate the opposite nature of the air drag to the velocity. In essence it makes it so that whatever the sign of v is the resulting kv term is opposite?
$endgroup$
– Prandals
Feb 3 at 3:53
add a comment |
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$begingroup$
The laws of physics are the same in all frames of reference! I'd just carefully define your coordinate system, including origin and positive directions of all relevant dimensions, and go from there.
$endgroup$
– Adrian Keister
Feb 1 at 15:40