Axis of rotation of the body

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A couple is applied to a rod such that the forces are at equal distance from the centre of mass of body. The body will rotate about an axis through its centre of mass.
However, if the couple is applied to some distance away from the centre of mass, will the body rotate about an axis which will pass through its centre of mass? Please explain.
The fig. Is shown here .






forces rotational-dynamics torque







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    The net torque about COM describe motion of the center of mass(rotation)
    – harambe
    3 hours ago














up vote
2
down vote

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A couple is applied to a rod such that the forces are at equal distance from the centre of mass of body. The body will rotate about an axis through its centre of mass.
However, if the couple is applied to some distance away from the centre of mass, will the body rotate about an axis which will pass through its centre of mass? Please explain.
The fig. Is shown here .






forces rotational-dynamics torque







share|cite|improve this question

















  • 1




    The net torque about COM describe motion of the center of mass(rotation)
    – harambe
    3 hours ago












up vote
2
down vote

favorite









up vote
2
down vote

favorite











A couple is applied to a rod such that the forces are at equal distance from the centre of mass of body. The body will rotate about an axis through its centre of mass.
However, if the couple is applied to some distance away from the centre of mass, will the body rotate about an axis which will pass through its centre of mass? Please explain.
The fig. Is shown here .






forces rotational-dynamics torque







share|cite|improve this question













A couple is applied to a rod such that the forces are at equal distance from the centre of mass of body. The body will rotate about an axis through its centre of mass.
However, if the couple is applied to some distance away from the centre of mass, will the body rotate about an axis which will pass through its centre of mass? Please explain.
The fig. Is shown here .






forces rotational-dynamics torque




forces rotational-dynamics torque






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asked 3 hours ago









Gurbir Singh

306111




306111







  • 1




    The net torque about COM describe motion of the center of mass(rotation)
    – harambe
    3 hours ago












  • 1




    The net torque about COM describe motion of the center of mass(rotation)
    – harambe
    3 hours ago







1




1




The net torque about COM describe motion of the center of mass(rotation)
– harambe
3 hours ago




The net torque about COM describe motion of the center of mass(rotation)
– harambe
3 hours ago










3 Answers
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The rod accelerates as if all the forces acting on it are acting upon the center of mass. Thus, if the rod is free to move on a frictionless surface, the rod will still obey the equation $$vec F = mvec a$$ Again, if the rod is free to move (not constrained to rotate about a certain axis, and no other force acting on it), it will always rotate about its center of mass. That is, if the force is applied at a distance $r$ from the center, we have $$Fr = Ialpha$$ where $I = frac112mL^2$ is the moment of inertia about the center of mass, and $alpha$ is the angular acceleration.



The translational motion and the rotational motion can be separated. Imagine the rod as a point mass (that is, a dot) accelerating with $vec F = mvec a$, and rotating about it.






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    The effect of a couple on a rod shouldn't not depend on its application point - it is a free vector.



    We can also observe that the net force of a couple is zero, which means that the COM shouldn't accelerate.






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      1. For the pair of equal forces shown, we know there is no net translation force on the rod, so its centre of mass will not move.

      2. If you make the centre of mass a fixed point we have equal forces at different distances from the fixed point, so the rod will rotate (anticlockwise). The net torque (radius times force) is the same wherever the forces are applied, as long as they stay the same distance apart.

      3. The centre of mass does not move (because of 1.) so 2. applies.

      If you tried this with a rod on a smooth table, and fingers applying the force I think you would automatically make the two forces unequal, so the centre of rotation could be between them.



      While it is easy enough to construct examples in space or on water, I find it hard to imagine a situation like this, where the rod is free to move in any direction and the two equal forces are the only relevant forces acting, and which is commonplace enough to be obvious.






      share|cite|improve this answer








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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        1
        down vote



        accepted










        The rod accelerates as if all the forces acting on it are acting upon the center of mass. Thus, if the rod is free to move on a frictionless surface, the rod will still obey the equation $$vec F = mvec a$$ Again, if the rod is free to move (not constrained to rotate about a certain axis, and no other force acting on it), it will always rotate about its center of mass. That is, if the force is applied at a distance $r$ from the center, we have $$Fr = Ialpha$$ where $I = frac112mL^2$ is the moment of inertia about the center of mass, and $alpha$ is the angular acceleration.



        The translational motion and the rotational motion can be separated. Imagine the rod as a point mass (that is, a dot) accelerating with $vec F = mvec a$, and rotating about it.






        share|cite|improve this answer
























          up vote
          1
          down vote



          accepted










          The rod accelerates as if all the forces acting on it are acting upon the center of mass. Thus, if the rod is free to move on a frictionless surface, the rod will still obey the equation $$vec F = mvec a$$ Again, if the rod is free to move (not constrained to rotate about a certain axis, and no other force acting on it), it will always rotate about its center of mass. That is, if the force is applied at a distance $r$ from the center, we have $$Fr = Ialpha$$ where $I = frac112mL^2$ is the moment of inertia about the center of mass, and $alpha$ is the angular acceleration.



          The translational motion and the rotational motion can be separated. Imagine the rod as a point mass (that is, a dot) accelerating with $vec F = mvec a$, and rotating about it.






          share|cite|improve this answer






















            up vote
            1
            down vote



            accepted







            up vote
            1
            down vote



            accepted






            The rod accelerates as if all the forces acting on it are acting upon the center of mass. Thus, if the rod is free to move on a frictionless surface, the rod will still obey the equation $$vec F = mvec a$$ Again, if the rod is free to move (not constrained to rotate about a certain axis, and no other force acting on it), it will always rotate about its center of mass. That is, if the force is applied at a distance $r$ from the center, we have $$Fr = Ialpha$$ where $I = frac112mL^2$ is the moment of inertia about the center of mass, and $alpha$ is the angular acceleration.



            The translational motion and the rotational motion can be separated. Imagine the rod as a point mass (that is, a dot) accelerating with $vec F = mvec a$, and rotating about it.






            share|cite|improve this answer












            The rod accelerates as if all the forces acting on it are acting upon the center of mass. Thus, if the rod is free to move on a frictionless surface, the rod will still obey the equation $$vec F = mvec a$$ Again, if the rod is free to move (not constrained to rotate about a certain axis, and no other force acting on it), it will always rotate about its center of mass. That is, if the force is applied at a distance $r$ from the center, we have $$Fr = Ialpha$$ where $I = frac112mL^2$ is the moment of inertia about the center of mass, and $alpha$ is the angular acceleration.



            The translational motion and the rotational motion can be separated. Imagine the rod as a point mass (that is, a dot) accelerating with $vec F = mvec a$, and rotating about it.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 2 hours ago









            user7777777

            1,8951315




            1,8951315




















                up vote
                1
                down vote













                The effect of a couple on a rod shouldn't not depend on its application point - it is a free vector.



                We can also observe that the net force of a couple is zero, which means that the COM shouldn't accelerate.






                share|cite|improve this answer
























                  up vote
                  1
                  down vote













                  The effect of a couple on a rod shouldn't not depend on its application point - it is a free vector.



                  We can also observe that the net force of a couple is zero, which means that the COM shouldn't accelerate.






                  share|cite|improve this answer






















                    up vote
                    1
                    down vote










                    up vote
                    1
                    down vote









                    The effect of a couple on a rod shouldn't not depend on its application point - it is a free vector.



                    We can also observe that the net force of a couple is zero, which means that the COM shouldn't accelerate.






                    share|cite|improve this answer












                    The effect of a couple on a rod shouldn't not depend on its application point - it is a free vector.



                    We can also observe that the net force of a couple is zero, which means that the COM shouldn't accelerate.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 2 hours ago









                    V.F.

                    7,6262621




                    7,6262621




















                        up vote
                        1
                        down vote













                        1. For the pair of equal forces shown, we know there is no net translation force on the rod, so its centre of mass will not move.

                        2. If you make the centre of mass a fixed point we have equal forces at different distances from the fixed point, so the rod will rotate (anticlockwise). The net torque (radius times force) is the same wherever the forces are applied, as long as they stay the same distance apart.

                        3. The centre of mass does not move (because of 1.) so 2. applies.

                        If you tried this with a rod on a smooth table, and fingers applying the force I think you would automatically make the two forces unequal, so the centre of rotation could be between them.



                        While it is easy enough to construct examples in space or on water, I find it hard to imagine a situation like this, where the rod is free to move in any direction and the two equal forces are the only relevant forces acting, and which is commonplace enough to be obvious.






                        share|cite|improve this answer








                        New contributor




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





















                          up vote
                          1
                          down vote













                          1. For the pair of equal forces shown, we know there is no net translation force on the rod, so its centre of mass will not move.

                          2. If you make the centre of mass a fixed point we have equal forces at different distances from the fixed point, so the rod will rotate (anticlockwise). The net torque (radius times force) is the same wherever the forces are applied, as long as they stay the same distance apart.

                          3. The centre of mass does not move (because of 1.) so 2. applies.

                          If you tried this with a rod on a smooth table, and fingers applying the force I think you would automatically make the two forces unequal, so the centre of rotation could be between them.



                          While it is easy enough to construct examples in space or on water, I find it hard to imagine a situation like this, where the rod is free to move in any direction and the two equal forces are the only relevant forces acting, and which is commonplace enough to be obvious.






                          share|cite|improve this answer








                          New contributor




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



















                            up vote
                            1
                            down vote










                            up vote
                            1
                            down vote









                            1. For the pair of equal forces shown, we know there is no net translation force on the rod, so its centre of mass will not move.

                            2. If you make the centre of mass a fixed point we have equal forces at different distances from the fixed point, so the rod will rotate (anticlockwise). The net torque (radius times force) is the same wherever the forces are applied, as long as they stay the same distance apart.

                            3. The centre of mass does not move (because of 1.) so 2. applies.

                            If you tried this with a rod on a smooth table, and fingers applying the force I think you would automatically make the two forces unequal, so the centre of rotation could be between them.



                            While it is easy enough to construct examples in space or on water, I find it hard to imagine a situation like this, where the rod is free to move in any direction and the two equal forces are the only relevant forces acting, and which is commonplace enough to be obvious.






                            share|cite|improve this answer








                            New contributor




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









                            1. For the pair of equal forces shown, we know there is no net translation force on the rod, so its centre of mass will not move.

                            2. If you make the centre of mass a fixed point we have equal forces at different distances from the fixed point, so the rod will rotate (anticlockwise). The net torque (radius times force) is the same wherever the forces are applied, as long as they stay the same distance apart.

                            3. The centre of mass does not move (because of 1.) so 2. applies.

                            If you tried this with a rod on a smooth table, and fingers applying the force I think you would automatically make the two forces unequal, so the centre of rotation could be between them.



                            While it is easy enough to construct examples in space or on water, I find it hard to imagine a situation like this, where the rod is free to move in any direction and the two equal forces are the only relevant forces acting, and which is commonplace enough to be obvious.







                            share|cite|improve this answer








                            New contributor




                            Peter 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






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                            Peter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                            answered 1 hour ago









                            Peter

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                            512




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                            Peter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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