Equation of motion in a disk and slider system












1












$begingroup$


I want to derive the equation of motion in this system: (the slider mass is m and the disk mass is M and the connecting bar is massless)



enter image description here



I have used relative velocity principle to calculate velocity of slider A:
$$vec V_C=Rdottheta hat i $$
$$vec V_B=vec V_C+vec V_{B/C} =Rdottheta (1+sintheta) hat i +Rdotthetacostheta hat j$$
$$vec V_B=vec V_A+vec V_{B/A}=vec V_A+2.5Rdotphisinphi hat i -2.5Rdotphicosphihat j$$
Therfore:
$$vec V_A=[Rdottheta (1+sintheta)-2.5Rdotphisinphi] hat i +[Rdotthetacostheta+2.5Rdotphicosphi ]hat j$$
And as we know the slider has no vertical motion so:
$$Rdotthetacostheta+2.5Rdotphicosphi =0$$
$$dotthetacostheta=-2.5dotphicosphi $$
Therefore:
$$vec V_A=Rdottheta (1+sintheta+frac {costheta}{cosphi})hat i$$
From geometry we know:
$$Rsintheta =2.5RsinphiRightarrow sintheta =2.5sinphi$$
$$cosphi =sqrt{1-sin^2phi}=sqrt{1-frac {1}{2.5^2}sin^2theta}=1+frac{1}{25}cos2theta $$
If we want the acceleration in point A:
$$vec a_A=frac {d}{dt}vec V_A=[Rddottheta (1+sintheta+costheta)+Rdottheta^2 (costheta-sintheta)]hat i$$
So the equation of motion can be derived using newton rule:
$$sum vec F=mvec a $$
$$F (t)= mRddottheta (1+sintheta+costheta)+mRdottheta^2 (costheta-sintheta)$$



Is my solution correct?










share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I want to derive the equation of motion in this system: (the slider mass is m and the disk mass is M and the connecting bar is massless)



    enter image description here



    I have used relative velocity principle to calculate velocity of slider A:
    $$vec V_C=Rdottheta hat i $$
    $$vec V_B=vec V_C+vec V_{B/C} =Rdottheta (1+sintheta) hat i +Rdotthetacostheta hat j$$
    $$vec V_B=vec V_A+vec V_{B/A}=vec V_A+2.5Rdotphisinphi hat i -2.5Rdotphicosphihat j$$
    Therfore:
    $$vec V_A=[Rdottheta (1+sintheta)-2.5Rdotphisinphi] hat i +[Rdotthetacostheta+2.5Rdotphicosphi ]hat j$$
    And as we know the slider has no vertical motion so:
    $$Rdotthetacostheta+2.5Rdotphicosphi =0$$
    $$dotthetacostheta=-2.5dotphicosphi $$
    Therefore:
    $$vec V_A=Rdottheta (1+sintheta+frac {costheta}{cosphi})hat i$$
    From geometry we know:
    $$Rsintheta =2.5RsinphiRightarrow sintheta =2.5sinphi$$
    $$cosphi =sqrt{1-sin^2phi}=sqrt{1-frac {1}{2.5^2}sin^2theta}=1+frac{1}{25}cos2theta $$
    If we want the acceleration in point A:
    $$vec a_A=frac {d}{dt}vec V_A=[Rddottheta (1+sintheta+costheta)+Rdottheta^2 (costheta-sintheta)]hat i$$
    So the equation of motion can be derived using newton rule:
    $$sum vec F=mvec a $$
    $$F (t)= mRddottheta (1+sintheta+costheta)+mRdottheta^2 (costheta-sintheta)$$



    Is my solution correct?










    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      I want to derive the equation of motion in this system: (the slider mass is m and the disk mass is M and the connecting bar is massless)



      enter image description here



      I have used relative velocity principle to calculate velocity of slider A:
      $$vec V_C=Rdottheta hat i $$
      $$vec V_B=vec V_C+vec V_{B/C} =Rdottheta (1+sintheta) hat i +Rdotthetacostheta hat j$$
      $$vec V_B=vec V_A+vec V_{B/A}=vec V_A+2.5Rdotphisinphi hat i -2.5Rdotphicosphihat j$$
      Therfore:
      $$vec V_A=[Rdottheta (1+sintheta)-2.5Rdotphisinphi] hat i +[Rdotthetacostheta+2.5Rdotphicosphi ]hat j$$
      And as we know the slider has no vertical motion so:
      $$Rdotthetacostheta+2.5Rdotphicosphi =0$$
      $$dotthetacostheta=-2.5dotphicosphi $$
      Therefore:
      $$vec V_A=Rdottheta (1+sintheta+frac {costheta}{cosphi})hat i$$
      From geometry we know:
      $$Rsintheta =2.5RsinphiRightarrow sintheta =2.5sinphi$$
      $$cosphi =sqrt{1-sin^2phi}=sqrt{1-frac {1}{2.5^2}sin^2theta}=1+frac{1}{25}cos2theta $$
      If we want the acceleration in point A:
      $$vec a_A=frac {d}{dt}vec V_A=[Rddottheta (1+sintheta+costheta)+Rdottheta^2 (costheta-sintheta)]hat i$$
      So the equation of motion can be derived using newton rule:
      $$sum vec F=mvec a $$
      $$F (t)= mRddottheta (1+sintheta+costheta)+mRdottheta^2 (costheta-sintheta)$$



      Is my solution correct?










      share|cite|improve this question











      $endgroup$




      I want to derive the equation of motion in this system: (the slider mass is m and the disk mass is M and the connecting bar is massless)



      enter image description here



      I have used relative velocity principle to calculate velocity of slider A:
      $$vec V_C=Rdottheta hat i $$
      $$vec V_B=vec V_C+vec V_{B/C} =Rdottheta (1+sintheta) hat i +Rdotthetacostheta hat j$$
      $$vec V_B=vec V_A+vec V_{B/A}=vec V_A+2.5Rdotphisinphi hat i -2.5Rdotphicosphihat j$$
      Therfore:
      $$vec V_A=[Rdottheta (1+sintheta)-2.5Rdotphisinphi] hat i +[Rdotthetacostheta+2.5Rdotphicosphi ]hat j$$
      And as we know the slider has no vertical motion so:
      $$Rdotthetacostheta+2.5Rdotphicosphi =0$$
      $$dotthetacostheta=-2.5dotphicosphi $$
      Therefore:
      $$vec V_A=Rdottheta (1+sintheta+frac {costheta}{cosphi})hat i$$
      From geometry we know:
      $$Rsintheta =2.5RsinphiRightarrow sintheta =2.5sinphi$$
      $$cosphi =sqrt{1-sin^2phi}=sqrt{1-frac {1}{2.5^2}sin^2theta}=1+frac{1}{25}cos2theta $$
      If we want the acceleration in point A:
      $$vec a_A=frac {d}{dt}vec V_A=[Rddottheta (1+sintheta+costheta)+Rdottheta^2 (costheta-sintheta)]hat i$$
      So the equation of motion can be derived using newton rule:
      $$sum vec F=mvec a $$
      $$F (t)= mRddottheta (1+sintheta+costheta)+mRdottheta^2 (costheta-sintheta)$$



      Is my solution correct?







      dynamical-systems






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 6 at 18:29







      H.H

















      asked Jan 6 at 11:48









      H.HH.H

      1466




      1466






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          It seems like a better idea if you try Lagrange method by deriving kinetic and potential energies:
          $$V=0$$
          $$T=frac{1}{2}mV_A^2+frac{1}{2}I_{disk}omega^2$$
          $$I=frac{3}{2}MR^2$$



          if your answer for velocity of the slider is correct we can write:
          $$T=frac{1}{2}m[Rdottheta (1+sintheta+frac {costheta}{cosphi})]^2+frac{1}{2}frac{3}{2}MR^2dottheta^2$$



          so if you use Lagrange equations, you can find the answer:
          $$L=T-V$$



          $${displaystyle {frac {mathrm {d} }{mathrm {d} t}}left({frac {partial L}{partial {dot {q}}_{j}}}right)={frac {partial L}{partial q_{j}}}}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for your suggestion, but can you derive the equation of motion in order to compare two answers?
            $endgroup$
            – H.H
            Jan 6 at 21:57



















          1












          $begingroup$

          The total kinetic energy is given by



          $$
          K = frac 12 M||vec v_C||^2+frac 12 J_C omega^2+frac 12 m ||vec v_A||^2
          $$



          with $omega = dottheta$



          We know also that



          $$
          vec v_B = vec v_C + vecomegatimes(B-C)\
          vec v_B = vec v_A + vec {dotphi} times (B-A)
          $$



          or



          $$
          vec v_C + vecomegatimes(B-C) = vec v_A + vec {dotphi} times (B-A)
          $$



          Here



          $$
          B-C = R(costheta,sintheta)\
          B-A = lambda R(-cosphi,sinphi)\
          vec v_C = R(omega,0)\
          vec v_A = R (omega-(lambdadotphi+omega)sintheta,(omega-lambdadotphi)costheta)
          $$



          then plugin those results into the kinetic energy expression and considering that $T = K - V$ with $V = 0$ we can derive the movement equations as



          $$
          T_{eta}-frac{d}{dt}T_{doteta}=mathcal{F}
          $$



          Here $eta = (theta,phi)$






          share|cite|improve this answer











          $endgroup$













            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3063759%2fequation-of-motion-in-a-disk-and-slider-system%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            It seems like a better idea if you try Lagrange method by deriving kinetic and potential energies:
            $$V=0$$
            $$T=frac{1}{2}mV_A^2+frac{1}{2}I_{disk}omega^2$$
            $$I=frac{3}{2}MR^2$$



            if your answer for velocity of the slider is correct we can write:
            $$T=frac{1}{2}m[Rdottheta (1+sintheta+frac {costheta}{cosphi})]^2+frac{1}{2}frac{3}{2}MR^2dottheta^2$$



            so if you use Lagrange equations, you can find the answer:
            $$L=T-V$$



            $${displaystyle {frac {mathrm {d} }{mathrm {d} t}}left({frac {partial L}{partial {dot {q}}_{j}}}right)={frac {partial L}{partial q_{j}}}}$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thank you for your suggestion, but can you derive the equation of motion in order to compare two answers?
              $endgroup$
              – H.H
              Jan 6 at 21:57
















            1












            $begingroup$

            It seems like a better idea if you try Lagrange method by deriving kinetic and potential energies:
            $$V=0$$
            $$T=frac{1}{2}mV_A^2+frac{1}{2}I_{disk}omega^2$$
            $$I=frac{3}{2}MR^2$$



            if your answer for velocity of the slider is correct we can write:
            $$T=frac{1}{2}m[Rdottheta (1+sintheta+frac {costheta}{cosphi})]^2+frac{1}{2}frac{3}{2}MR^2dottheta^2$$



            so if you use Lagrange equations, you can find the answer:
            $$L=T-V$$



            $${displaystyle {frac {mathrm {d} }{mathrm {d} t}}left({frac {partial L}{partial {dot {q}}_{j}}}right)={frac {partial L}{partial q_{j}}}}$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thank you for your suggestion, but can you derive the equation of motion in order to compare two answers?
              $endgroup$
              – H.H
              Jan 6 at 21:57














            1












            1








            1





            $begingroup$

            It seems like a better idea if you try Lagrange method by deriving kinetic and potential energies:
            $$V=0$$
            $$T=frac{1}{2}mV_A^2+frac{1}{2}I_{disk}omega^2$$
            $$I=frac{3}{2}MR^2$$



            if your answer for velocity of the slider is correct we can write:
            $$T=frac{1}{2}m[Rdottheta (1+sintheta+frac {costheta}{cosphi})]^2+frac{1}{2}frac{3}{2}MR^2dottheta^2$$



            so if you use Lagrange equations, you can find the answer:
            $$L=T-V$$



            $${displaystyle {frac {mathrm {d} }{mathrm {d} t}}left({frac {partial L}{partial {dot {q}}_{j}}}right)={frac {partial L}{partial q_{j}}}}$$






            share|cite|improve this answer









            $endgroup$



            It seems like a better idea if you try Lagrange method by deriving kinetic and potential energies:
            $$V=0$$
            $$T=frac{1}{2}mV_A^2+frac{1}{2}I_{disk}omega^2$$
            $$I=frac{3}{2}MR^2$$



            if your answer for velocity of the slider is correct we can write:
            $$T=frac{1}{2}m[Rdottheta (1+sintheta+frac {costheta}{cosphi})]^2+frac{1}{2}frac{3}{2}MR^2dottheta^2$$



            so if you use Lagrange equations, you can find the answer:
            $$L=T-V$$



            $${displaystyle {frac {mathrm {d} }{mathrm {d} t}}left({frac {partial L}{partial {dot {q}}_{j}}}right)={frac {partial L}{partial q_{j}}}}$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 6 at 21:48









            HarryHarry

            283




            283












            • $begingroup$
              Thank you for your suggestion, but can you derive the equation of motion in order to compare two answers?
              $endgroup$
              – H.H
              Jan 6 at 21:57


















            • $begingroup$
              Thank you for your suggestion, but can you derive the equation of motion in order to compare two answers?
              $endgroup$
              – H.H
              Jan 6 at 21:57
















            $begingroup$
            Thank you for your suggestion, but can you derive the equation of motion in order to compare two answers?
            $endgroup$
            – H.H
            Jan 6 at 21:57




            $begingroup$
            Thank you for your suggestion, but can you derive the equation of motion in order to compare two answers?
            $endgroup$
            – H.H
            Jan 6 at 21:57











            1












            $begingroup$

            The total kinetic energy is given by



            $$
            K = frac 12 M||vec v_C||^2+frac 12 J_C omega^2+frac 12 m ||vec v_A||^2
            $$



            with $omega = dottheta$



            We know also that



            $$
            vec v_B = vec v_C + vecomegatimes(B-C)\
            vec v_B = vec v_A + vec {dotphi} times (B-A)
            $$



            or



            $$
            vec v_C + vecomegatimes(B-C) = vec v_A + vec {dotphi} times (B-A)
            $$



            Here



            $$
            B-C = R(costheta,sintheta)\
            B-A = lambda R(-cosphi,sinphi)\
            vec v_C = R(omega,0)\
            vec v_A = R (omega-(lambdadotphi+omega)sintheta,(omega-lambdadotphi)costheta)
            $$



            then plugin those results into the kinetic energy expression and considering that $T = K - V$ with $V = 0$ we can derive the movement equations as



            $$
            T_{eta}-frac{d}{dt}T_{doteta}=mathcal{F}
            $$



            Here $eta = (theta,phi)$






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              The total kinetic energy is given by



              $$
              K = frac 12 M||vec v_C||^2+frac 12 J_C omega^2+frac 12 m ||vec v_A||^2
              $$



              with $omega = dottheta$



              We know also that



              $$
              vec v_B = vec v_C + vecomegatimes(B-C)\
              vec v_B = vec v_A + vec {dotphi} times (B-A)
              $$



              or



              $$
              vec v_C + vecomegatimes(B-C) = vec v_A + vec {dotphi} times (B-A)
              $$



              Here



              $$
              B-C = R(costheta,sintheta)\
              B-A = lambda R(-cosphi,sinphi)\
              vec v_C = R(omega,0)\
              vec v_A = R (omega-(lambdadotphi+omega)sintheta,(omega-lambdadotphi)costheta)
              $$



              then plugin those results into the kinetic energy expression and considering that $T = K - V$ with $V = 0$ we can derive the movement equations as



              $$
              T_{eta}-frac{d}{dt}T_{doteta}=mathcal{F}
              $$



              Here $eta = (theta,phi)$






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                The total kinetic energy is given by



                $$
                K = frac 12 M||vec v_C||^2+frac 12 J_C omega^2+frac 12 m ||vec v_A||^2
                $$



                with $omega = dottheta$



                We know also that



                $$
                vec v_B = vec v_C + vecomegatimes(B-C)\
                vec v_B = vec v_A + vec {dotphi} times (B-A)
                $$



                or



                $$
                vec v_C + vecomegatimes(B-C) = vec v_A + vec {dotphi} times (B-A)
                $$



                Here



                $$
                B-C = R(costheta,sintheta)\
                B-A = lambda R(-cosphi,sinphi)\
                vec v_C = R(omega,0)\
                vec v_A = R (omega-(lambdadotphi+omega)sintheta,(omega-lambdadotphi)costheta)
                $$



                then plugin those results into the kinetic energy expression and considering that $T = K - V$ with $V = 0$ we can derive the movement equations as



                $$
                T_{eta}-frac{d}{dt}T_{doteta}=mathcal{F}
                $$



                Here $eta = (theta,phi)$






                share|cite|improve this answer











                $endgroup$



                The total kinetic energy is given by



                $$
                K = frac 12 M||vec v_C||^2+frac 12 J_C omega^2+frac 12 m ||vec v_A||^2
                $$



                with $omega = dottheta$



                We know also that



                $$
                vec v_B = vec v_C + vecomegatimes(B-C)\
                vec v_B = vec v_A + vec {dotphi} times (B-A)
                $$



                or



                $$
                vec v_C + vecomegatimes(B-C) = vec v_A + vec {dotphi} times (B-A)
                $$



                Here



                $$
                B-C = R(costheta,sintheta)\
                B-A = lambda R(-cosphi,sinphi)\
                vec v_C = R(omega,0)\
                vec v_A = R (omega-(lambdadotphi+omega)sintheta,(omega-lambdadotphi)costheta)
                $$



                then plugin those results into the kinetic energy expression and considering that $T = K - V$ with $V = 0$ we can derive the movement equations as



                $$
                T_{eta}-frac{d}{dt}T_{doteta}=mathcal{F}
                $$



                Here $eta = (theta,phi)$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 10 at 18:22

























                answered Jan 10 at 14:06









                CesareoCesareo

                8,6733516




                8,6733516






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3063759%2fequation-of-motion-in-a-disk-and-slider-system%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    Popular posts from this blog

                    android studio warns about leanback feature tag usage required on manifest while using Unity exported app?

                    SQL update select statement

                    'app-layout' is not a known element: how to share Component with different Modules