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?






dynamical-systems







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?






    dynamical-systems







    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?






      dynamical-systems







      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




      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

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          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











          $endgroup$













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            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

                9,4713517




                9,4713517






























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                    .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
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