Mixing
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. 
forces rotational-dynamics torque
asked 3 hours ago
Gurbir Singh
306111
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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.
forces rotational-dynamics torque
asked 3 hours ago
Gurbir Singh
306111
1
The net torque about COM describe motion of the center of mass(rotation)
– harambe
3 hours ago
add a comment |Â
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.
forces rotational-dynamics torque
asked 3 hours ago
Gurbir Singh
306111
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.
forces rotational-dynamics torque
forces rotational-dynamics torque
asked 3 hours ago
Gurbir Singh
306111
asked 3 hours ago
Gurbir Singh
306111
asked 3 hours ago
Gurbir Singh
306111
asked 3 hours ago
Gurbir Singh
306111
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
add a comment |Â
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
add a comment |Â
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.
answered 2 hours ago
user7777777
1,8951315
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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.
answered 2 hours ago
V.F.
7,6262621
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- 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.
- 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.
- 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.
answered 1 hour ago
Peter
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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.
answered 2 hours ago
user7777777
1,8951315
add a comment |Â
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.
answered 2 hours ago
user7777777
1,8951315
add a comment |Â
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.
answered 2 hours ago
user7777777
1,8951315
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.
answered 2 hours ago
user7777777
1,8951315
answered 2 hours ago
user7777777
1,8951315
answered 2 hours ago
user7777777
1,8951315
answered 2 hours ago
user7777777
1,8951315
1,8951315
add a comment |Â
add a comment |Â
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.
answered 2 hours ago
V.F.
7,6262621
add a comment |Â
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.
answered 2 hours ago
V.F.
7,6262621
add a comment |Â
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.
answered 2 hours ago
V.F.
7,6262621
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.
answered 2 hours ago
V.F.
7,6262621
answered 2 hours ago
V.F.
7,6262621
answered 2 hours ago
V.F.
7,6262621
answered 2 hours ago
V.F.
7,6262621
7,6262621
add a comment |Â
add a comment |Â
up vote
1
down vote
- 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.
- 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.
- 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.
answered 1 hour ago
Peter
512
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.
add a comment |Â
up vote
1
down vote
- 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.
- 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.
- 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.
answered 1 hour ago
Peter
512
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.
add a comment |Â
up vote
1
down vote
up vote
1
down vote
- 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.
- 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.
- 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.
answered 1 hour ago
Peter
512
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.
- 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.
- 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.
- 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.
answered 1 hour ago
Peter
512
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.
answered 1 hour ago
Peter
512
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.
answered 1 hour ago
Peter
512
answered 1 hour ago
Peter
512
512
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.
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.
Peter 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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add a comment |Â
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1
The net torque about COM describe motion of the center of mass(rotation)
– harambe
3 hours ago