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



enter image description here










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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.




















    1












    $begingroup$



    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



    enter image description here










    share|cite|improve this question











    $endgroup$



    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.


















      1












      1








      1


      1



      $begingroup$



      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



      enter image description here










      share|cite|improve this question











      $endgroup$





      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



      enter image description here





      This question already has an answer here:




      • First order differential equations with 2 varriables

        2 answers








      ordinary-differential-equations






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      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.
























          1 Answer
          1






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          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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




















          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          0












          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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


















          0












          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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
















          0












          0








          0





          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$



          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.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 2 at 17:32

























          answered Dec 30 '18 at 13:14









          EddyEddy

          869612




          869612












          • $begingroup$
            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










          • $begingroup$
            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





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