Pendulum Rotation












3












$begingroup$


A simple pendulum has a bob with a mass of $0.50$ 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.




$$omega^2_f = 2 alphaDeltatheta$$




I get the following expression:



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



Is this expression correct to solve the question?






calculus







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    A simple pendulum has a bob with a mass of $0.50$ 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.




    $$omega^2_f = 2 alphaDeltatheta$$




    I get the following expression:



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



    Is this expression correct to solve the question?






    calculus







    share|cite|improve this question











    $endgroup$















      3












      3








      3


      1



      $begingroup$


      A simple pendulum has a bob with a mass of $0.50$ 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.




      $$omega^2_f = 2 alphaDeltatheta$$




      I get the following expression:



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



      Is this expression correct to solve the question?






      calculus







      share|cite|improve this question











      $endgroup$




      A simple pendulum has a bob with a mass of $0.50$ 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.




      $$omega^2_f = 2 alphaDeltatheta$$




      I get the following expression:



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



      Is this expression correct to solve the question?






      calculus




      calculus






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 6 at 19:41









      Andrews

      1,2641420




      1,2641420










      asked Mar 3 at 1:40









      EnlightenedFunkyEnlightenedFunky

      82411022




      82411022






















          2 Answers
          2






          active

          oldest

          votes


















          10












          $begingroup$

          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
            Mar 3 at 2:10










          • $begingroup$
            Secondly, everything was given to you.in the detail.
            $endgroup$
            – EnlightenedFunky
            Mar 3 at 2:11






          • 4




            $begingroup$
            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
            Mar 3 at 2:15










          • $begingroup$
            Isn't the force is $mgsintheta$
            $endgroup$
            – EnlightenedFunky
            Mar 3 at 2:19












          • $begingroup$
            No, the force is $mg$ always pointing downwards. The torque is $mgLsintheta$
            $endgroup$
            – Andrei
            Mar 3 at 2:22



















          3












          $begingroup$

          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









          $endgroup$













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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            10












            $begingroup$

            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
              Mar 3 at 2:10










            • $begingroup$
              Secondly, everything was given to you.in the detail.
              $endgroup$
              – EnlightenedFunky
              Mar 3 at 2:11






            • 4




              $begingroup$
              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
              Mar 3 at 2:15










            • $begingroup$
              Isn't the force is $mgsintheta$
              $endgroup$
              – EnlightenedFunky
              Mar 3 at 2:19












            • $begingroup$
              No, the force is $mg$ always pointing downwards. The torque is $mgLsintheta$
              $endgroup$
              – Andrei
              Mar 3 at 2:22
















            10












            $begingroup$

            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
              Mar 3 at 2:10










            • $begingroup$
              Secondly, everything was given to you.in the detail.
              $endgroup$
              – EnlightenedFunky
              Mar 3 at 2:11






            • 4




              $begingroup$
              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
              Mar 3 at 2:15










            • $begingroup$
              Isn't the force is $mgsintheta$
              $endgroup$
              – EnlightenedFunky
              Mar 3 at 2:19












            • $begingroup$
              No, the force is $mg$ always pointing downwards. The torque is $mgLsintheta$
              $endgroup$
              – Andrei
              Mar 3 at 2:22














            10












            10








            10





            $begingroup$

            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 Mar 3 at 2:05









            AndreiAndrei

            13.1k21230




            13.1k21230












            • $begingroup$
              But, I would like to solve this particular manner the professor solved it that way already.
              $endgroup$
              – EnlightenedFunky
              Mar 3 at 2:10










            • $begingroup$
              Secondly, everything was given to you.in the detail.
              $endgroup$
              – EnlightenedFunky
              Mar 3 at 2:11






            • 4




              $begingroup$
              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
              Mar 3 at 2:15










            • $begingroup$
              Isn't the force is $mgsintheta$
              $endgroup$
              – EnlightenedFunky
              Mar 3 at 2:19












            • $begingroup$
              No, the force is $mg$ always pointing downwards. The torque is $mgLsintheta$
              $endgroup$
              – Andrei
              Mar 3 at 2:22


















            • $begingroup$
              But, I would like to solve this particular manner the professor solved it that way already.
              $endgroup$
              – EnlightenedFunky
              Mar 3 at 2:10










            • $begingroup$
              Secondly, everything was given to you.in the detail.
              $endgroup$
              – EnlightenedFunky
              Mar 3 at 2:11






            • 4




              $begingroup$
              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
              Mar 3 at 2:15










            • $begingroup$
              Isn't the force is $mgsintheta$
              $endgroup$
              – EnlightenedFunky
              Mar 3 at 2:19












            • $begingroup$
              No, the force is $mg$ always pointing downwards. The torque is $mgLsintheta$
              $endgroup$
              – Andrei
              Mar 3 at 2:22
















            $begingroup$
            But, I would like to solve this particular manner the professor solved it that way already.
            $endgroup$
            – EnlightenedFunky
            Mar 3 at 2:10




            $begingroup$
            But, I would like to solve this particular manner the professor solved it that way already.
            $endgroup$
            – EnlightenedFunky
            Mar 3 at 2:10












            $begingroup$
            Secondly, everything was given to you.in the detail.
            $endgroup$
            – EnlightenedFunky
            Mar 3 at 2:11




            $begingroup$
            Secondly, everything was given to you.in the detail.
            $endgroup$
            – EnlightenedFunky
            Mar 3 at 2:11




            4




            4




            $begingroup$
            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
            Mar 3 at 2:15




            $begingroup$
            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
            Mar 3 at 2:15












            $begingroup$
            Isn't the force is $mgsintheta$
            $endgroup$
            – EnlightenedFunky
            Mar 3 at 2:19






            $begingroup$
            Isn't the force is $mgsintheta$
            $endgroup$
            – EnlightenedFunky
            Mar 3 at 2:19














            $begingroup$
            No, the force is $mg$ always pointing downwards. The torque is $mgLsintheta$
            $endgroup$
            – Andrei
            Mar 3 at 2:22




            $begingroup$
            No, the force is $mg$ always pointing downwards. The torque is $mgLsintheta$
            $endgroup$
            – Andrei
            Mar 3 at 2:22











            3












            $begingroup$

            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









            $endgroup$


















              3












              $begingroup$

              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









              $endgroup$
















                3












                3








                3





                $begingroup$

                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









                $endgroup$



                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












                share|cite|improve this answer



                share|cite|improve this answer










                answered Mar 3 at 2:18









                Yuzheng LinYuzheng Lin

                512




                512






























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