Pendulum Rotation












1












$begingroup$


A simple pendulum has a bob with a mass of .50 kg. The cord has a length of 1.5 m, and the bob is displaced $20^circ$.



I am trying to use this expression to find the maximum velocity of the bob.




$$omega^2_f=2 alphatriangletheta$$




I get the following expression:



$$frac{v^2}{L^2}=2Lfrac{mgsintheta}{I}triangletheta$$



Is this expression correct to solve the question?






mathematical-physics







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    A simple pendulum has a bob with a mass of .50 kg. The cord has a length of 1.5 m, and the bob is displaced $20^circ$.



    I am trying to use this expression to find the maximum velocity of the bob.




    $$omega^2_f=2 alphatriangletheta$$




    I get the following expression:



    $$frac{v^2}{L^2}=2Lfrac{mgsintheta}{I}triangletheta$$



    Is this expression correct to solve the question?






    mathematical-physics







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      A simple pendulum has a bob with a mass of .50 kg. The cord has a length of 1.5 m, and the bob is displaced $20^circ$.



      I am trying to use this expression to find the maximum velocity of the bob.




      $$omega^2_f=2 alphatriangletheta$$




      I get the following expression:



      $$frac{v^2}{L^2}=2Lfrac{mgsintheta}{I}triangletheta$$



      Is this expression correct to solve the question?






      mathematical-physics







      share|cite|improve this question











      $endgroup$




      A simple pendulum has a bob with a mass of .50 kg. The cord has a length of 1.5 m, and the bob is displaced $20^circ$.



      I am trying to use this expression to find the maximum velocity of the bob.




      $$omega^2_f=2 alphatriangletheta$$




      I get the following expression:



      $$frac{v^2}{L^2}=2Lfrac{mgsintheta}{I}triangletheta$$



      Is this expression correct to solve the question?






      mathematical-physics




      mathematical-physics






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 1 hour ago







      EnlightenedFunky

















      asked 1 hour ago









      EnlightenedFunkyEnlightenedFunky

      81211022




      81211022






















          2 Answers
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          3












          $begingroup$

          First of all, you should explain what your parameters are. But you seem to over complicate the problem. Write conservation of energy. At the bottom of the trajectory, we choose the potential energy to be $0$. Then you have only kinetic energy $frac 12 mv^2$. At the top of the trajectory, the bob is at rest, so kinetic energy is zero, and you have only potential energy. The height of the bob is $L(1-costheta)$, so $$mgL(1-costheta)=frac 12 mv^2$$
          Notice that the velocity is independent of mass






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            But, I would like to solve this particular manner the professor solved it that way already.
            $endgroup$
            – EnlightenedFunky
            1 hour ago










          • $begingroup$
            Secondly, everything was given to you.in the detail.
            $endgroup$
            – EnlightenedFunky
            1 hour ago










          • $begingroup$
            You did not explain what the $alpha$ or $I$ or $omega_f$ are. I guessed that you they represent angular acceleration, moment of inertia, and angular velocity. However, the formula that you wrote cannot be used in this case. That formula is valid for constant $alpha$, which is not the case here. The torque depends on the angle between gravity and the cord.
            $endgroup$
            – Andrei
            1 hour ago










          • $begingroup$
            Isn't the force is $mgsintheta$
            $endgroup$
            – EnlightenedFunky
            1 hour ago












          • $begingroup$
            No, the force is $mg$ always pointing downwards. The torque is $mgLsintheta$
            $endgroup$
            – Andrei
            1 hour ago



















          2












          $begingroup$

          The equation only holds when angular acceleration $alpha$ is a constant. However in this case $alpha=frac{gsintheta}{l}$, which depends on $theta$






          share|cite|improve this answer








          New contributor




          Yuzheng Lin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$

            First of all, you should explain what your parameters are. But you seem to over complicate the problem. Write conservation of energy. At the bottom of the trajectory, we choose the potential energy to be $0$. Then you have only kinetic energy $frac 12 mv^2$. At the top of the trajectory, the bob is at rest, so kinetic energy is zero, and you have only potential energy. The height of the bob is $L(1-costheta)$, so $$mgL(1-costheta)=frac 12 mv^2$$
            Notice that the velocity is independent of mass






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              But, I would like to solve this particular manner the professor solved it that way already.
              $endgroup$
              – EnlightenedFunky
              1 hour ago










            • $begingroup$
              Secondly, everything was given to you.in the detail.
              $endgroup$
              – EnlightenedFunky
              1 hour ago










            • $begingroup$
              You did not explain what the $alpha$ or $I$ or $omega_f$ are. I guessed that you they represent angular acceleration, moment of inertia, and angular velocity. However, the formula that you wrote cannot be used in this case. That formula is valid for constant $alpha$, which is not the case here. The torque depends on the angle between gravity and the cord.
              $endgroup$
              – Andrei
              1 hour ago










            • $begingroup$
              Isn't the force is $mgsintheta$
              $endgroup$
              – EnlightenedFunky
              1 hour ago












            • $begingroup$
              No, the force is $mg$ always pointing downwards. The torque is $mgLsintheta$
              $endgroup$
              – Andrei
              1 hour ago
















            3












            $begingroup$

            First of all, you should explain what your parameters are. But you seem to over complicate the problem. Write conservation of energy. At the bottom of the trajectory, we choose the potential energy to be $0$. Then you have only kinetic energy $frac 12 mv^2$. At the top of the trajectory, the bob is at rest, so kinetic energy is zero, and you have only potential energy. The height of the bob is $L(1-costheta)$, so $$mgL(1-costheta)=frac 12 mv^2$$
            Notice that the velocity is independent of mass






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              But, I would like to solve this particular manner the professor solved it that way already.
              $endgroup$
              – EnlightenedFunky
              1 hour ago










            • $begingroup$
              Secondly, everything was given to you.in the detail.
              $endgroup$
              – EnlightenedFunky
              1 hour ago










            • $begingroup$
              You did not explain what the $alpha$ or $I$ or $omega_f$ are. I guessed that you they represent angular acceleration, moment of inertia, and angular velocity. However, the formula that you wrote cannot be used in this case. That formula is valid for constant $alpha$, which is not the case here. The torque depends on the angle between gravity and the cord.
              $endgroup$
              – Andrei
              1 hour ago










            • $begingroup$
              Isn't the force is $mgsintheta$
              $endgroup$
              – EnlightenedFunky
              1 hour ago












            • $begingroup$
              No, the force is $mg$ always pointing downwards. The torque is $mgLsintheta$
              $endgroup$
              – Andrei
              1 hour ago














            3












            3








            3





            $begingroup$

            First of all, you should explain what your parameters are. But you seem to over complicate the problem. Write conservation of energy. At the bottom of the trajectory, we choose the potential energy to be $0$. Then you have only kinetic energy $frac 12 mv^2$. At the top of the trajectory, the bob is at rest, so kinetic energy is zero, and you have only potential energy. The height of the bob is $L(1-costheta)$, so $$mgL(1-costheta)=frac 12 mv^2$$
            Notice that the velocity is independent of mass






            share|cite|improve this answer









            $endgroup$



            First of all, you should explain what your parameters are. But you seem to over complicate the problem. Write conservation of energy. At the bottom of the trajectory, we choose the potential energy to be $0$. Then you have only kinetic energy $frac 12 mv^2$. At the top of the trajectory, the bob is at rest, so kinetic energy is zero, and you have only potential energy. The height of the bob is $L(1-costheta)$, so $$mgL(1-costheta)=frac 12 mv^2$$
            Notice that the velocity is independent of mass







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 1 hour ago









            AndreiAndrei

            12.5k21128




            12.5k21128












            • $begingroup$
              But, I would like to solve this particular manner the professor solved it that way already.
              $endgroup$
              – EnlightenedFunky
              1 hour ago










            • $begingroup$
              Secondly, everything was given to you.in the detail.
              $endgroup$
              – EnlightenedFunky
              1 hour ago










            • $begingroup$
              You did not explain what the $alpha$ or $I$ or $omega_f$ are. I guessed that you they represent angular acceleration, moment of inertia, and angular velocity. However, the formula that you wrote cannot be used in this case. That formula is valid for constant $alpha$, which is not the case here. The torque depends on the angle between gravity and the cord.
              $endgroup$
              – Andrei
              1 hour ago










            • $begingroup$
              Isn't the force is $mgsintheta$
              $endgroup$
              – EnlightenedFunky
              1 hour ago












            • $begingroup$
              No, the force is $mg$ always pointing downwards. The torque is $mgLsintheta$
              $endgroup$
              – Andrei
              1 hour ago


















            • $begingroup$
              But, I would like to solve this particular manner the professor solved it that way already.
              $endgroup$
              – EnlightenedFunky
              1 hour ago










            • $begingroup$
              Secondly, everything was given to you.in the detail.
              $endgroup$
              – EnlightenedFunky
              1 hour ago










            • $begingroup$
              You did not explain what the $alpha$ or $I$ or $omega_f$ are. I guessed that you they represent angular acceleration, moment of inertia, and angular velocity. However, the formula that you wrote cannot be used in this case. That formula is valid for constant $alpha$, which is not the case here. The torque depends on the angle between gravity and the cord.
              $endgroup$
              – Andrei
              1 hour ago










            • $begingroup$
              Isn't the force is $mgsintheta$
              $endgroup$
              – EnlightenedFunky
              1 hour ago












            • $begingroup$
              No, the force is $mg$ always pointing downwards. The torque is $mgLsintheta$
              $endgroup$
              – Andrei
              1 hour ago
















            $begingroup$
            But, I would like to solve this particular manner the professor solved it that way already.
            $endgroup$
            – EnlightenedFunky
            1 hour ago




            $begingroup$
            But, I would like to solve this particular manner the professor solved it that way already.
            $endgroup$
            – EnlightenedFunky
            1 hour ago












            $begingroup$
            Secondly, everything was given to you.in the detail.
            $endgroup$
            – EnlightenedFunky
            1 hour ago




            $begingroup$
            Secondly, everything was given to you.in the detail.
            $endgroup$
            – EnlightenedFunky
            1 hour ago












            $begingroup$
            You did not explain what the $alpha$ or $I$ or $omega_f$ are. I guessed that you they represent angular acceleration, moment of inertia, and angular velocity. However, the formula that you wrote cannot be used in this case. That formula is valid for constant $alpha$, which is not the case here. The torque depends on the angle between gravity and the cord.
            $endgroup$
            – Andrei
            1 hour ago




            $begingroup$
            You did not explain what the $alpha$ or $I$ or $omega_f$ are. I guessed that you they represent angular acceleration, moment of inertia, and angular velocity. However, the formula that you wrote cannot be used in this case. That formula is valid for constant $alpha$, which is not the case here. The torque depends on the angle between gravity and the cord.
            $endgroup$
            – Andrei
            1 hour ago












            $begingroup$
            Isn't the force is $mgsintheta$
            $endgroup$
            – EnlightenedFunky
            1 hour ago






            $begingroup$
            Isn't the force is $mgsintheta$
            $endgroup$
            – EnlightenedFunky
            1 hour ago














            $begingroup$
            No, the force is $mg$ always pointing downwards. The torque is $mgLsintheta$
            $endgroup$
            – Andrei
            1 hour ago




            $begingroup$
            No, the force is $mg$ always pointing downwards. The torque is $mgLsintheta$
            $endgroup$
            – Andrei
            1 hour ago











            2












            $begingroup$

            The equation only holds when angular acceleration $alpha$ is a constant. However in this case $alpha=frac{gsintheta}{l}$, which depends on $theta$






            share|cite|improve this answer








            New contributor




            Yuzheng Lin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.






            $endgroup$


















              2












              $begingroup$

              The equation only holds when angular acceleration $alpha$ is a constant. However in this case $alpha=frac{gsintheta}{l}$, which depends on $theta$






              share|cite|improve this answer








              New contributor




              Yuzheng Lin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.






              $endgroup$
















                2












                2








                2





                $begingroup$

                The equation only holds when angular acceleration $alpha$ is a constant. However in this case $alpha=frac{gsintheta}{l}$, which depends on $theta$






                share|cite|improve this answer








                New contributor




                Yuzheng Lin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.






                $endgroup$



                The equation only holds when angular acceleration $alpha$ is a constant. However in this case $alpha=frac{gsintheta}{l}$, which depends on $theta$







                share|cite|improve this answer








                New contributor




                Yuzheng Lin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.









                share|cite|improve this answer



                share|cite|improve this answer






                New contributor




                Yuzheng Lin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.









                answered 1 hour ago









                Yuzheng LinYuzheng Lin

                411




                411




                New contributor




                Yuzheng Lin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.





                New contributor





                Yuzheng Lin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.






                Yuzheng Lin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.






























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