Differential equation thought question [duplicate]
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This question already has an answer here:
First order differential equations with 2 varriables
2 answers
Does anybody know how to solve the following differential equation?
$$w^2cdot r+frac{dw}{dt}cdot r+frac{dr}{dt}cdot w=3g$$
where w and r are a function of time so $w(t)$ and $r(t)$.
$g$ is a constant
This equation came from a thought problem of mine which I will now explain since I still do not know how to approach the problem.
Basically I want to design a spiral track in which a vehicle will accelerate. However the total acceleration needs to be constant.
The vehicle will have 2 acceleration components, a radial component ($a_r$) and a change in speed component ($a_s$). Since we want constant acceleration:
$a_r + a_s = constant$
From circular motion we know that $a_r = omega ^2 r$
where $r$ is the radius, and $omega$
Further we know that $v=omega*r$, therefore:
$$a_s=frac{d}{dt}(omega r)$$
For constant acceleration $a_{total} = a_r + a_s = omega ^2 r + frac{d}{dt}(omega r) =constant$
Further usefull information:
Length of the track covered at a certain time
$$l=vcdot t=omega r t$$
With all this the radius will be a function of time, $r(t)$
and the angular velocity will be a function of time, $omega (t)$
Boundary conditions:
$$r(0) = 0$$
$$omega (0) = 0$$
Now I would like to find a possible set of solutions to create a track by implementing end boundary conditions that i choose e.g. $v(t_{end})=100 [m/s]$ and $l(t_{end})=5000 [m]$
Here is a diagram for clarity:
Constant acceleration spiral track
ordinary-differential-equations
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marked as duplicate by quid♦ Jan 1 at 2:28
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
First order differential equations with 2 varriables
2 answers
Does anybody know how to solve the following differential equation?
$$w^2cdot r+frac{dw}{dt}cdot r+frac{dr}{dt}cdot w=3g$$
where w and r are a function of time so $w(t)$ and $r(t)$.
$g$ is a constant
This equation came from a thought problem of mine which I will now explain since I still do not know how to approach the problem.
Basically I want to design a spiral track in which a vehicle will accelerate. However the total acceleration needs to be constant.
The vehicle will have 2 acceleration components, a radial component ($a_r$) and a change in speed component ($a_s$). Since we want constant acceleration:
$a_r + a_s = constant$
From circular motion we know that $a_r = omega ^2 r$
where $r$ is the radius, and $omega$
Further we know that $v=omega*r$, therefore:
$$a_s=frac{d}{dt}(omega r)$$
For constant acceleration $a_{total} = a_r + a_s = omega ^2 r + frac{d}{dt}(omega r) =constant$
Further usefull information:
Length of the track covered at a certain time
$$l=vcdot t=omega r t$$
With all this the radius will be a function of time, $r(t)$
and the angular velocity will be a function of time, $omega (t)$
Boundary conditions:
$$r(0) = 0$$
$$omega (0) = 0$$
Now I would like to find a possible set of solutions to create a track by implementing end boundary conditions that i choose e.g. $v(t_{end})=100 [m/s]$ and $l(t_{end})=5000 [m]$
Here is a diagram for clarity:
Constant acceleration spiral track
ordinary-differential-equations
$endgroup$
marked as duplicate by quid♦ Jan 1 at 2:28
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
First order differential equations with 2 varriables
2 answers
Does anybody know how to solve the following differential equation?
$$w^2cdot r+frac{dw}{dt}cdot r+frac{dr}{dt}cdot w=3g$$
where w and r are a function of time so $w(t)$ and $r(t)$.
$g$ is a constant
This equation came from a thought problem of mine which I will now explain since I still do not know how to approach the problem.
Basically I want to design a spiral track in which a vehicle will accelerate. However the total acceleration needs to be constant.
The vehicle will have 2 acceleration components, a radial component ($a_r$) and a change in speed component ($a_s$). Since we want constant acceleration:
$a_r + a_s = constant$
From circular motion we know that $a_r = omega ^2 r$
where $r$ is the radius, and $omega$
Further we know that $v=omega*r$, therefore:
$$a_s=frac{d}{dt}(omega r)$$
For constant acceleration $a_{total} = a_r + a_s = omega ^2 r + frac{d}{dt}(omega r) =constant$
Further usefull information:
Length of the track covered at a certain time
$$l=vcdot t=omega r t$$
With all this the radius will be a function of time, $r(t)$
and the angular velocity will be a function of time, $omega (t)$
Boundary conditions:
$$r(0) = 0$$
$$omega (0) = 0$$
Now I would like to find a possible set of solutions to create a track by implementing end boundary conditions that i choose e.g. $v(t_{end})=100 [m/s]$ and $l(t_{end})=5000 [m]$
Here is a diagram for clarity:
Constant acceleration spiral track
ordinary-differential-equations
$endgroup$
This question already has an answer here:
First order differential equations with 2 varriables
2 answers
Does anybody know how to solve the following differential equation?
$$w^2cdot r+frac{dw}{dt}cdot r+frac{dr}{dt}cdot w=3g$$
where w and r are a function of time so $w(t)$ and $r(t)$.
$g$ is a constant
This equation came from a thought problem of mine which I will now explain since I still do not know how to approach the problem.
Basically I want to design a spiral track in which a vehicle will accelerate. However the total acceleration needs to be constant.
The vehicle will have 2 acceleration components, a radial component ($a_r$) and a change in speed component ($a_s$). Since we want constant acceleration:
$a_r + a_s = constant$
From circular motion we know that $a_r = omega ^2 r$
where $r$ is the radius, and $omega$
Further we know that $v=omega*r$, therefore:
$$a_s=frac{d}{dt}(omega r)$$
For constant acceleration $a_{total} = a_r + a_s = omega ^2 r + frac{d}{dt}(omega r) =constant$
Further usefull information:
Length of the track covered at a certain time
$$l=vcdot t=omega r t$$
With all this the radius will be a function of time, $r(t)$
and the angular velocity will be a function of time, $omega (t)$
Boundary conditions:
$$r(0) = 0$$
$$omega (0) = 0$$
Now I would like to find a possible set of solutions to create a track by implementing end boundary conditions that i choose e.g. $v(t_{end})=100 [m/s]$ and $l(t_{end})=5000 [m]$
Here is a diagram for clarity:
Constant acceleration spiral track
This question already has an answer here:
First order differential equations with 2 varriables
2 answers
ordinary-differential-equations
ordinary-differential-equations
edited Dec 30 '18 at 13:28
Larry
2,0602826
2,0602826
asked Dec 30 '18 at 12:09
Alessandro SimonelliAlessandro Simonelli
113
113
marked as duplicate by quid♦ Jan 1 at 2:28
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by quid♦ Jan 1 at 2:28
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
add a comment |
1 Answer
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oldest
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You will not be able to solve this equation since it has two unknowns, you need to impose an additional constraint (and appropriate initial conditions) to solve it. One possible approach would be to choose a function $r=r(t)$, in which case
$$
frac{dv}{dt} = c - frac{v^2} {r}
$$
is the equation to solve, where $c=3g$ frpm the question.
Edit:
If $omega$ is constant then
$$
frac{dr} {dt} = frac{c -omega^2 r} {omega}
$$
which can be solved by seperation of variables.
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And what if I were to say that for example, $omega$ is constant?
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– Alessandro Simonelli
Jan 2 at 15:32
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What if $omega$ is constant?
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– Alessandro Simonelli
Jan 2 at 15:33
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You will not be able to solve this equation since it has two unknowns, you need to impose an additional constraint (and appropriate initial conditions) to solve it. One possible approach would be to choose a function $r=r(t)$, in which case
$$
frac{dv}{dt} = c - frac{v^2} {r}
$$
is the equation to solve, where $c=3g$ frpm the question.
Edit:
If $omega$ is constant then
$$
frac{dr} {dt} = frac{c -omega^2 r} {omega}
$$
which can be solved by seperation of variables.
$endgroup$
$begingroup$
And what if I were to say that for example, $omega$ is constant?
$endgroup$
– Alessandro Simonelli
Jan 2 at 15:32
$begingroup$
What if $omega$ is constant?
$endgroup$
– Alessandro Simonelli
Jan 2 at 15:33
add a comment |
$begingroup$
You will not be able to solve this equation since it has two unknowns, you need to impose an additional constraint (and appropriate initial conditions) to solve it. One possible approach would be to choose a function $r=r(t)$, in which case
$$
frac{dv}{dt} = c - frac{v^2} {r}
$$
is the equation to solve, where $c=3g$ frpm the question.
Edit:
If $omega$ is constant then
$$
frac{dr} {dt} = frac{c -omega^2 r} {omega}
$$
which can be solved by seperation of variables.
$endgroup$
$begingroup$
And what if I were to say that for example, $omega$ is constant?
$endgroup$
– Alessandro Simonelli
Jan 2 at 15:32
$begingroup$
What if $omega$ is constant?
$endgroup$
– Alessandro Simonelli
Jan 2 at 15:33
add a comment |
$begingroup$
You will not be able to solve this equation since it has two unknowns, you need to impose an additional constraint (and appropriate initial conditions) to solve it. One possible approach would be to choose a function $r=r(t)$, in which case
$$
frac{dv}{dt} = c - frac{v^2} {r}
$$
is the equation to solve, where $c=3g$ frpm the question.
Edit:
If $omega$ is constant then
$$
frac{dr} {dt} = frac{c -omega^2 r} {omega}
$$
which can be solved by seperation of variables.
$endgroup$
You will not be able to solve this equation since it has two unknowns, you need to impose an additional constraint (and appropriate initial conditions) to solve it. One possible approach would be to choose a function $r=r(t)$, in which case
$$
frac{dv}{dt} = c - frac{v^2} {r}
$$
is the equation to solve, where $c=3g$ frpm the question.
Edit:
If $omega$ is constant then
$$
frac{dr} {dt} = frac{c -omega^2 r} {omega}
$$
which can be solved by seperation of variables.
edited Jan 2 at 17:32
answered Dec 30 '18 at 13:14
EddyEddy
869612
869612
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And what if I were to say that for example, $omega$ is constant?
$endgroup$
– Alessandro Simonelli
Jan 2 at 15:32
$begingroup$
What if $omega$ is constant?
$endgroup$
– Alessandro Simonelli
Jan 2 at 15:33
add a comment |
$begingroup$
And what if I were to say that for example, $omega$ is constant?
$endgroup$
– Alessandro Simonelli
Jan 2 at 15:32
$begingroup$
What if $omega$ is constant?
$endgroup$
– Alessandro Simonelli
Jan 2 at 15:33
$begingroup$
And what if I were to say that for example, $omega$ is constant?
$endgroup$
– Alessandro Simonelli
Jan 2 at 15:32
$begingroup$
And what if I were to say that for example, $omega$ is constant?
$endgroup$
– Alessandro Simonelli
Jan 2 at 15:32
$begingroup$
What if $omega$ is constant?
$endgroup$
– Alessandro Simonelli
Jan 2 at 15:33
$begingroup$
What if $omega$ is constant?
$endgroup$
– Alessandro Simonelli
Jan 2 at 15:33
add a comment |