Sign problem when working out a simple pendulum's differential equation












1












$begingroup$


I was trying to work out the differential equation for a simple pendulum, and got a sign mistake along the way (among other things I guess). Here's my reasoning:



assuming that the friction forces are negligible, one may use the conservation of energy principle. When $theta(t)=theta_0$ (position 1), we have



$E_{mech_1}=U_1$ $quad$ (since $K_1=0$)



and that at the equilibrium point (position 2)



$E_{mech_2}=K_2$ $quad$ (since $U_2=0$)



and therefore



$K_2=U_1 iff frac{1}{2}mL^2dot{theta}^2color{red}{boldsymbol{-}}mgL(1-costheta)=0.$



It does however make sense to me that (ref.)



$E_{mech}=U+K$



but I don't really see my mistake. Is it my interpretation of the potential $U$?






classical-mechanics







share|cite|improve this question









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    1












    $begingroup$


    I was trying to work out the differential equation for a simple pendulum, and got a sign mistake along the way (among other things I guess). Here's my reasoning:



    assuming that the friction forces are negligible, one may use the conservation of energy principle. When $theta(t)=theta_0$ (position 1), we have



    $E_{mech_1}=U_1$ $quad$ (since $K_1=0$)



    and that at the equilibrium point (position 2)



    $E_{mech_2}=K_2$ $quad$ (since $U_2=0$)



    and therefore



    $K_2=U_1 iff frac{1}{2}mL^2dot{theta}^2color{red}{boldsymbol{-}}mgL(1-costheta)=0.$



    It does however make sense to me that (ref.)



    $E_{mech}=U+K$



    but I don't really see my mistake. Is it my interpretation of the potential $U$?






    classical-mechanics







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I was trying to work out the differential equation for a simple pendulum, and got a sign mistake along the way (among other things I guess). Here's my reasoning:



      assuming that the friction forces are negligible, one may use the conservation of energy principle. When $theta(t)=theta_0$ (position 1), we have



      $E_{mech_1}=U_1$ $quad$ (since $K_1=0$)



      and that at the equilibrium point (position 2)



      $E_{mech_2}=K_2$ $quad$ (since $U_2=0$)



      and therefore



      $K_2=U_1 iff frac{1}{2}mL^2dot{theta}^2color{red}{boldsymbol{-}}mgL(1-costheta)=0.$



      It does however make sense to me that (ref.)



      $E_{mech}=U+K$



      but I don't really see my mistake. Is it my interpretation of the potential $U$?






      classical-mechanics







      share|cite|improve this question









      $endgroup$




      I was trying to work out the differential equation for a simple pendulum, and got a sign mistake along the way (among other things I guess). Here's my reasoning:



      assuming that the friction forces are negligible, one may use the conservation of energy principle. When $theta(t)=theta_0$ (position 1), we have



      $E_{mech_1}=U_1$ $quad$ (since $K_1=0$)



      and that at the equilibrium point (position 2)



      $E_{mech_2}=K_2$ $quad$ (since $U_2=0$)



      and therefore



      $K_2=U_1 iff frac{1}{2}mL^2dot{theta}^2color{red}{boldsymbol{-}}mgL(1-costheta)=0.$



      It does however make sense to me that (ref.)



      $E_{mech}=U+K$



      but I don't really see my mistake. Is it my interpretation of the potential $U$?






      classical-mechanics




      classical-mechanics






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 12 at 23:06









      ChrisChris

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          $begingroup$

          Write the equation as
          $$
          K_2=U_1 iff frac{1}{2}mL^2dot{theta_2}^2color{red}{boldsymbol{-}}mgL(1-costheta_1)=0,
          $$
          then it is correct. You are computing the relation of maximum angle and maximal velocity, these events happen at different positions, as you also wrote above.






          share|cite|improve this answer









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            $begingroup$

            Write the equation as
            $$
            K_2=U_1 iff frac{1}{2}mL^2dot{theta_2}^2color{red}{boldsymbol{-}}mgL(1-costheta_1)=0,
            $$
            then it is correct. You are computing the relation of maximum angle and maximal velocity, these events happen at different positions, as you also wrote above.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Write the equation as
              $$
              K_2=U_1 iff frac{1}{2}mL^2dot{theta_2}^2color{red}{boldsymbol{-}}mgL(1-costheta_1)=0,
              $$
              then it is correct. You are computing the relation of maximum angle and maximal velocity, these events happen at different positions, as you also wrote above.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Write the equation as
                $$
                K_2=U_1 iff frac{1}{2}mL^2dot{theta_2}^2color{red}{boldsymbol{-}}mgL(1-costheta_1)=0,
                $$
                then it is correct. You are computing the relation of maximum angle and maximal velocity, these events happen at different positions, as you also wrote above.






                share|cite|improve this answer









                $endgroup$



                Write the equation as
                $$
                K_2=U_1 iff frac{1}{2}mL^2dot{theta_2}^2color{red}{boldsymbol{-}}mgL(1-costheta_1)=0,
                $$
                then it is correct. You are computing the relation of maximum angle and maximal velocity, these events happen at different positions, as you also wrote above.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 12 at 23:59









                LutzLLutzL

                58.5k42054




                58.5k42054






























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