Equation of motion in a disk and slider system












1














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







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    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



























      1












      1








      1


      1





      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















      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






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      share|cite|improve this question













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      edited Jan 6 at 18:29







      H.H

















      asked Jan 6 at 11:48









      H.HH.H

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          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





















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











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          1 Answer
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          active

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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          1














          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





















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
















          1














          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





















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














          1












          1








          1






          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












          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












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


















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
















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




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


















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