Does it take energy to move something in a circle?
$begingroup$
Work can be calculated by $W = Fdcostheta$ where $theta$ is the angle between the force $F$ and the displacement $d$
Let's say there's a ball and a (physically ideal - no friction etc.) robotic arm situated in otherwise empty space. The arm takes the ball, moves it around in a circle, and then returns it exactly to where it started. The arm also returns to its starting position.
In this case, there is no displacement overall. The ball and the robot arm are in the exact same positions as when they began.
Thus, $W = Fcdot 0 cdot costheta = 0$ So no energy was required to move the ball in a circle.
However, this disagrees with my intuition, because if I were to make a robot arm that would do this, I feel like I would need to give it an energy source (for example, a battery), and that by the time the robot was done, I would have lost energy from that battery.
Does the movement require energy? If so, where did I mess up in my calculation? If not, why would my robot need a battery pack or some energy source in order to make this movement? (If it doesn't, please explain.)
newtonian-mechanics energy rotational-dynamics energy-conservation work
New contributor
$endgroup$
add a comment |
$begingroup$
Work can be calculated by $W = Fdcostheta$ where $theta$ is the angle between the force $F$ and the displacement $d$
Let's say there's a ball and a (physically ideal - no friction etc.) robotic arm situated in otherwise empty space. The arm takes the ball, moves it around in a circle, and then returns it exactly to where it started. The arm also returns to its starting position.
In this case, there is no displacement overall. The ball and the robot arm are in the exact same positions as when they began.
Thus, $W = Fcdot 0 cdot costheta = 0$ So no energy was required to move the ball in a circle.
However, this disagrees with my intuition, because if I were to make a robot arm that would do this, I feel like I would need to give it an energy source (for example, a battery), and that by the time the robot was done, I would have lost energy from that battery.
Does the movement require energy? If so, where did I mess up in my calculation? If not, why would my robot need a battery pack or some energy source in order to make this movement? (If it doesn't, please explain.)
newtonian-mechanics energy rotational-dynamics energy-conservation work
New contributor
$endgroup$
add a comment |
$begingroup$
Work can be calculated by $W = Fdcostheta$ where $theta$ is the angle between the force $F$ and the displacement $d$
Let's say there's a ball and a (physically ideal - no friction etc.) robotic arm situated in otherwise empty space. The arm takes the ball, moves it around in a circle, and then returns it exactly to where it started. The arm also returns to its starting position.
In this case, there is no displacement overall. The ball and the robot arm are in the exact same positions as when they began.
Thus, $W = Fcdot 0 cdot costheta = 0$ So no energy was required to move the ball in a circle.
However, this disagrees with my intuition, because if I were to make a robot arm that would do this, I feel like I would need to give it an energy source (for example, a battery), and that by the time the robot was done, I would have lost energy from that battery.
Does the movement require energy? If so, where did I mess up in my calculation? If not, why would my robot need a battery pack or some energy source in order to make this movement? (If it doesn't, please explain.)
newtonian-mechanics energy rotational-dynamics energy-conservation work
New contributor
$endgroup$
Work can be calculated by $W = Fdcostheta$ where $theta$ is the angle between the force $F$ and the displacement $d$
Let's say there's a ball and a (physically ideal - no friction etc.) robotic arm situated in otherwise empty space. The arm takes the ball, moves it around in a circle, and then returns it exactly to where it started. The arm also returns to its starting position.
In this case, there is no displacement overall. The ball and the robot arm are in the exact same positions as when they began.
Thus, $W = Fcdot 0 cdot costheta = 0$ So no energy was required to move the ball in a circle.
However, this disagrees with my intuition, because if I were to make a robot arm that would do this, I feel like I would need to give it an energy source (for example, a battery), and that by the time the robot was done, I would have lost energy from that battery.
Does the movement require energy? If so, where did I mess up in my calculation? If not, why would my robot need a battery pack or some energy source in order to make this movement? (If it doesn't, please explain.)
newtonian-mechanics energy rotational-dynamics energy-conservation work
newtonian-mechanics energy rotational-dynamics energy-conservation work
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edited 4 hours ago
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asked 5 hours ago
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2 Answers
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oldest
votes
$begingroup$
First, you are not equating the work done correctly. This is a good physics lesson. Please understand your equations before you use them. Blindly plugging in numbers will not work out. The equation you give is only true for motion in one dimension and with a constant force. Plugging in $0$ for displacement is not correct here. In general you need to look at infinitesimal displacements $text dmathbf x$ and calculate the work $text d W=mathbf Fcdottext dmathbf x$, then integrate (add up) the total work.
Now, I am assuming the ball starts and stops at rest. Therefore, the arm does work to increase the ball's speed, and then it does the same amount of negative work to bring it to rest. So the net work is $0$, but it is because the total change in kinetic energy is $0$ (since $W=Delta K$), not because the displacement is $0$ around the circle.
Now, this is not the same thing as the robot using something like a battery. The robot (neglecting friction) has to apply forces to change the speed, and this requires power from the power supply. Just imagine yourself doing the action of the robot. You will need to exert effort to get the ball (and yourself) moving, and you will need to exert effort to get the ball (and yourself) to stop rotating.
$endgroup$
$begingroup$
Does that "exerting effort" for the ideal robot require energy? I noticed that you said the net work is 0, but you never said the net energy required to make the motion was 0.
$endgroup$
– Pro Q
5 hours ago
$begingroup$
@ProQ The "exerting effort" will require energy from the power supply, but this isn't the same energy of the motion of the ball because, as I said, the net work is $0$. An even more drastic example: imagine you holding a weight up in place. No motion means no work, but you are going to get tired eventually.
$endgroup$
– Aaron Stevens
4 hours ago
$begingroup$
Does this mean that if (in the ideal situation) I use some energy from the battery to make the robot move the ball in a circle, I can recover back that same amount of energy over the course of moving of the ball (for example with something like perfect regenerative braking)?
$endgroup$
– Pro Q
4 hours ago
$begingroup$
@ProQ I'm not an engineer, but I think your motor has the be designed a specific way to have regenerative braking. It certainly isn't always possible in general. The energy will be lost. Even in an ideal system the battery will "lose power" during this process.
$endgroup$
– Aaron Stevens
4 hours ago
$begingroup$
Negating friction the robot does not need to draw power. The robot can just be a bar connecting to a heavy object in the center. (It needs to be heavy because the whole system will rotate around the systems COM)
$endgroup$
– Taemyr
44 mins ago
|
show 5 more comments
$begingroup$
The definition of work is energy = force * displacement * cos(theta)
This definition is only strictly accurate for movement on a straight line.
In general, you need to solve an integral
$$E = int_a^bvec{F}cdot dvec{ell}$$
Does the movement require energy?
Two reasons a real robot will require energy to do this operation:
Real robots have friction in their joints, so there will always be a force opposing the direction the robot is trying to move, and so the integral of force along the path of motion won't go to zero
If you are considering cases where the motion includes some components downward in relation to gravity and some components upward, not every robot will be designed to recover the energy gained by moving downward and use it to move the object back upward, so some energy will be used lifting the object on the upward part of the path.
$endgroup$
$begingroup$
I'm assuming an ideal robot in space (so no gravity towards the earth). Are you saying that an ideal robot would not require any energy to do this motion?
$endgroup$
– Pro Q
5 hours ago
1
$begingroup$
@ProQ, not if the robot had magical frictionless joints.
$endgroup$
– The Photon
5 hours ago
1
$begingroup$
Also, if there is no gradient of potential energy (i.e. no gravity and the ball doesn't have electric charge, etc) then the movement doesn't even have to be in a circle for it to require no energy.
$endgroup$
– The Photon
5 hours ago
$begingroup$
@ThePhoton I don't think friction is the only place energy will go into.
$endgroup$
– Aaron Stevens
4 hours ago
add a comment |
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2 Answers
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2 Answers
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$begingroup$
First, you are not equating the work done correctly. This is a good physics lesson. Please understand your equations before you use them. Blindly plugging in numbers will not work out. The equation you give is only true for motion in one dimension and with a constant force. Plugging in $0$ for displacement is not correct here. In general you need to look at infinitesimal displacements $text dmathbf x$ and calculate the work $text d W=mathbf Fcdottext dmathbf x$, then integrate (add up) the total work.
Now, I am assuming the ball starts and stops at rest. Therefore, the arm does work to increase the ball's speed, and then it does the same amount of negative work to bring it to rest. So the net work is $0$, but it is because the total change in kinetic energy is $0$ (since $W=Delta K$), not because the displacement is $0$ around the circle.
Now, this is not the same thing as the robot using something like a battery. The robot (neglecting friction) has to apply forces to change the speed, and this requires power from the power supply. Just imagine yourself doing the action of the robot. You will need to exert effort to get the ball (and yourself) moving, and you will need to exert effort to get the ball (and yourself) to stop rotating.
$endgroup$
$begingroup$
Does that "exerting effort" for the ideal robot require energy? I noticed that you said the net work is 0, but you never said the net energy required to make the motion was 0.
$endgroup$
– Pro Q
5 hours ago
$begingroup$
@ProQ The "exerting effort" will require energy from the power supply, but this isn't the same energy of the motion of the ball because, as I said, the net work is $0$. An even more drastic example: imagine you holding a weight up in place. No motion means no work, but you are going to get tired eventually.
$endgroup$
– Aaron Stevens
4 hours ago
$begingroup$
Does this mean that if (in the ideal situation) I use some energy from the battery to make the robot move the ball in a circle, I can recover back that same amount of energy over the course of moving of the ball (for example with something like perfect regenerative braking)?
$endgroup$
– Pro Q
4 hours ago
$begingroup$
@ProQ I'm not an engineer, but I think your motor has the be designed a specific way to have regenerative braking. It certainly isn't always possible in general. The energy will be lost. Even in an ideal system the battery will "lose power" during this process.
$endgroup$
– Aaron Stevens
4 hours ago
$begingroup$
Negating friction the robot does not need to draw power. The robot can just be a bar connecting to a heavy object in the center. (It needs to be heavy because the whole system will rotate around the systems COM)
$endgroup$
– Taemyr
44 mins ago
|
show 5 more comments
$begingroup$
First, you are not equating the work done correctly. This is a good physics lesson. Please understand your equations before you use them. Blindly plugging in numbers will not work out. The equation you give is only true for motion in one dimension and with a constant force. Plugging in $0$ for displacement is not correct here. In general you need to look at infinitesimal displacements $text dmathbf x$ and calculate the work $text d W=mathbf Fcdottext dmathbf x$, then integrate (add up) the total work.
Now, I am assuming the ball starts and stops at rest. Therefore, the arm does work to increase the ball's speed, and then it does the same amount of negative work to bring it to rest. So the net work is $0$, but it is because the total change in kinetic energy is $0$ (since $W=Delta K$), not because the displacement is $0$ around the circle.
Now, this is not the same thing as the robot using something like a battery. The robot (neglecting friction) has to apply forces to change the speed, and this requires power from the power supply. Just imagine yourself doing the action of the robot. You will need to exert effort to get the ball (and yourself) moving, and you will need to exert effort to get the ball (and yourself) to stop rotating.
$endgroup$
$begingroup$
Does that "exerting effort" for the ideal robot require energy? I noticed that you said the net work is 0, but you never said the net energy required to make the motion was 0.
$endgroup$
– Pro Q
5 hours ago
$begingroup$
@ProQ The "exerting effort" will require energy from the power supply, but this isn't the same energy of the motion of the ball because, as I said, the net work is $0$. An even more drastic example: imagine you holding a weight up in place. No motion means no work, but you are going to get tired eventually.
$endgroup$
– Aaron Stevens
4 hours ago
$begingroup$
Does this mean that if (in the ideal situation) I use some energy from the battery to make the robot move the ball in a circle, I can recover back that same amount of energy over the course of moving of the ball (for example with something like perfect regenerative braking)?
$endgroup$
– Pro Q
4 hours ago
$begingroup$
@ProQ I'm not an engineer, but I think your motor has the be designed a specific way to have regenerative braking. It certainly isn't always possible in general. The energy will be lost. Even in an ideal system the battery will "lose power" during this process.
$endgroup$
– Aaron Stevens
4 hours ago
$begingroup$
Negating friction the robot does not need to draw power. The robot can just be a bar connecting to a heavy object in the center. (It needs to be heavy because the whole system will rotate around the systems COM)
$endgroup$
– Taemyr
44 mins ago
|
show 5 more comments
$begingroup$
First, you are not equating the work done correctly. This is a good physics lesson. Please understand your equations before you use them. Blindly plugging in numbers will not work out. The equation you give is only true for motion in one dimension and with a constant force. Plugging in $0$ for displacement is not correct here. In general you need to look at infinitesimal displacements $text dmathbf x$ and calculate the work $text d W=mathbf Fcdottext dmathbf x$, then integrate (add up) the total work.
Now, I am assuming the ball starts and stops at rest. Therefore, the arm does work to increase the ball's speed, and then it does the same amount of negative work to bring it to rest. So the net work is $0$, but it is because the total change in kinetic energy is $0$ (since $W=Delta K$), not because the displacement is $0$ around the circle.
Now, this is not the same thing as the robot using something like a battery. The robot (neglecting friction) has to apply forces to change the speed, and this requires power from the power supply. Just imagine yourself doing the action of the robot. You will need to exert effort to get the ball (and yourself) moving, and you will need to exert effort to get the ball (and yourself) to stop rotating.
$endgroup$
First, you are not equating the work done correctly. This is a good physics lesson. Please understand your equations before you use them. Blindly plugging in numbers will not work out. The equation you give is only true for motion in one dimension and with a constant force. Plugging in $0$ for displacement is not correct here. In general you need to look at infinitesimal displacements $text dmathbf x$ and calculate the work $text d W=mathbf Fcdottext dmathbf x$, then integrate (add up) the total work.
Now, I am assuming the ball starts and stops at rest. Therefore, the arm does work to increase the ball's speed, and then it does the same amount of negative work to bring it to rest. So the net work is $0$, but it is because the total change in kinetic energy is $0$ (since $W=Delta K$), not because the displacement is $0$ around the circle.
Now, this is not the same thing as the robot using something like a battery. The robot (neglecting friction) has to apply forces to change the speed, and this requires power from the power supply. Just imagine yourself doing the action of the robot. You will need to exert effort to get the ball (and yourself) moving, and you will need to exert effort to get the ball (and yourself) to stop rotating.
answered 5 hours ago
Aaron StevensAaron Stevens
11.1k31743
11.1k31743
$begingroup$
Does that "exerting effort" for the ideal robot require energy? I noticed that you said the net work is 0, but you never said the net energy required to make the motion was 0.
$endgroup$
– Pro Q
5 hours ago
$begingroup$
@ProQ The "exerting effort" will require energy from the power supply, but this isn't the same energy of the motion of the ball because, as I said, the net work is $0$. An even more drastic example: imagine you holding a weight up in place. No motion means no work, but you are going to get tired eventually.
$endgroup$
– Aaron Stevens
4 hours ago
$begingroup$
Does this mean that if (in the ideal situation) I use some energy from the battery to make the robot move the ball in a circle, I can recover back that same amount of energy over the course of moving of the ball (for example with something like perfect regenerative braking)?
$endgroup$
– Pro Q
4 hours ago
$begingroup$
@ProQ I'm not an engineer, but I think your motor has the be designed a specific way to have regenerative braking. It certainly isn't always possible in general. The energy will be lost. Even in an ideal system the battery will "lose power" during this process.
$endgroup$
– Aaron Stevens
4 hours ago
$begingroup$
Negating friction the robot does not need to draw power. The robot can just be a bar connecting to a heavy object in the center. (It needs to be heavy because the whole system will rotate around the systems COM)
$endgroup$
– Taemyr
44 mins ago
|
show 5 more comments
$begingroup$
Does that "exerting effort" for the ideal robot require energy? I noticed that you said the net work is 0, but you never said the net energy required to make the motion was 0.
$endgroup$
– Pro Q
5 hours ago
$begingroup$
@ProQ The "exerting effort" will require energy from the power supply, but this isn't the same energy of the motion of the ball because, as I said, the net work is $0$. An even more drastic example: imagine you holding a weight up in place. No motion means no work, but you are going to get tired eventually.
$endgroup$
– Aaron Stevens
4 hours ago
$begingroup$
Does this mean that if (in the ideal situation) I use some energy from the battery to make the robot move the ball in a circle, I can recover back that same amount of energy over the course of moving of the ball (for example with something like perfect regenerative braking)?
$endgroup$
– Pro Q
4 hours ago
$begingroup$
@ProQ I'm not an engineer, but I think your motor has the be designed a specific way to have regenerative braking. It certainly isn't always possible in general. The energy will be lost. Even in an ideal system the battery will "lose power" during this process.
$endgroup$
– Aaron Stevens
4 hours ago
$begingroup$
Negating friction the robot does not need to draw power. The robot can just be a bar connecting to a heavy object in the center. (It needs to be heavy because the whole system will rotate around the systems COM)
$endgroup$
– Taemyr
44 mins ago
$begingroup$
Does that "exerting effort" for the ideal robot require energy? I noticed that you said the net work is 0, but you never said the net energy required to make the motion was 0.
$endgroup$
– Pro Q
5 hours ago
$begingroup$
Does that "exerting effort" for the ideal robot require energy? I noticed that you said the net work is 0, but you never said the net energy required to make the motion was 0.
$endgroup$
– Pro Q
5 hours ago
$begingroup$
@ProQ The "exerting effort" will require energy from the power supply, but this isn't the same energy of the motion of the ball because, as I said, the net work is $0$. An even more drastic example: imagine you holding a weight up in place. No motion means no work, but you are going to get tired eventually.
$endgroup$
– Aaron Stevens
4 hours ago
$begingroup$
@ProQ The "exerting effort" will require energy from the power supply, but this isn't the same energy of the motion of the ball because, as I said, the net work is $0$. An even more drastic example: imagine you holding a weight up in place. No motion means no work, but you are going to get tired eventually.
$endgroup$
– Aaron Stevens
4 hours ago
$begingroup$
Does this mean that if (in the ideal situation) I use some energy from the battery to make the robot move the ball in a circle, I can recover back that same amount of energy over the course of moving of the ball (for example with something like perfect regenerative braking)?
$endgroup$
– Pro Q
4 hours ago
$begingroup$
Does this mean that if (in the ideal situation) I use some energy from the battery to make the robot move the ball in a circle, I can recover back that same amount of energy over the course of moving of the ball (for example with something like perfect regenerative braking)?
$endgroup$
– Pro Q
4 hours ago
$begingroup$
@ProQ I'm not an engineer, but I think your motor has the be designed a specific way to have regenerative braking. It certainly isn't always possible in general. The energy will be lost. Even in an ideal system the battery will "lose power" during this process.
$endgroup$
– Aaron Stevens
4 hours ago
$begingroup$
@ProQ I'm not an engineer, but I think your motor has the be designed a specific way to have regenerative braking. It certainly isn't always possible in general. The energy will be lost. Even in an ideal system the battery will "lose power" during this process.
$endgroup$
– Aaron Stevens
4 hours ago
$begingroup$
Negating friction the robot does not need to draw power. The robot can just be a bar connecting to a heavy object in the center. (It needs to be heavy because the whole system will rotate around the systems COM)
$endgroup$
– Taemyr
44 mins ago
$begingroup$
Negating friction the robot does not need to draw power. The robot can just be a bar connecting to a heavy object in the center. (It needs to be heavy because the whole system will rotate around the systems COM)
$endgroup$
– Taemyr
44 mins ago
|
show 5 more comments
$begingroup$
The definition of work is energy = force * displacement * cos(theta)
This definition is only strictly accurate for movement on a straight line.
In general, you need to solve an integral
$$E = int_a^bvec{F}cdot dvec{ell}$$
Does the movement require energy?
Two reasons a real robot will require energy to do this operation:
Real robots have friction in their joints, so there will always be a force opposing the direction the robot is trying to move, and so the integral of force along the path of motion won't go to zero
If you are considering cases where the motion includes some components downward in relation to gravity and some components upward, not every robot will be designed to recover the energy gained by moving downward and use it to move the object back upward, so some energy will be used lifting the object on the upward part of the path.
$endgroup$
$begingroup$
I'm assuming an ideal robot in space (so no gravity towards the earth). Are you saying that an ideal robot would not require any energy to do this motion?
$endgroup$
– Pro Q
5 hours ago
1
$begingroup$
@ProQ, not if the robot had magical frictionless joints.
$endgroup$
– The Photon
5 hours ago
1
$begingroup$
Also, if there is no gradient of potential energy (i.e. no gravity and the ball doesn't have electric charge, etc) then the movement doesn't even have to be in a circle for it to require no energy.
$endgroup$
– The Photon
5 hours ago
$begingroup$
@ThePhoton I don't think friction is the only place energy will go into.
$endgroup$
– Aaron Stevens
4 hours ago
add a comment |
$begingroup$
The definition of work is energy = force * displacement * cos(theta)
This definition is only strictly accurate for movement on a straight line.
In general, you need to solve an integral
$$E = int_a^bvec{F}cdot dvec{ell}$$
Does the movement require energy?
Two reasons a real robot will require energy to do this operation:
Real robots have friction in their joints, so there will always be a force opposing the direction the robot is trying to move, and so the integral of force along the path of motion won't go to zero
If you are considering cases where the motion includes some components downward in relation to gravity and some components upward, not every robot will be designed to recover the energy gained by moving downward and use it to move the object back upward, so some energy will be used lifting the object on the upward part of the path.
$endgroup$
$begingroup$
I'm assuming an ideal robot in space (so no gravity towards the earth). Are you saying that an ideal robot would not require any energy to do this motion?
$endgroup$
– Pro Q
5 hours ago
1
$begingroup$
@ProQ, not if the robot had magical frictionless joints.
$endgroup$
– The Photon
5 hours ago
1
$begingroup$
Also, if there is no gradient of potential energy (i.e. no gravity and the ball doesn't have electric charge, etc) then the movement doesn't even have to be in a circle for it to require no energy.
$endgroup$
– The Photon
5 hours ago
$begingroup$
@ThePhoton I don't think friction is the only place energy will go into.
$endgroup$
– Aaron Stevens
4 hours ago
add a comment |
$begingroup$
The definition of work is energy = force * displacement * cos(theta)
This definition is only strictly accurate for movement on a straight line.
In general, you need to solve an integral
$$E = int_a^bvec{F}cdot dvec{ell}$$
Does the movement require energy?
Two reasons a real robot will require energy to do this operation:
Real robots have friction in their joints, so there will always be a force opposing the direction the robot is trying to move, and so the integral of force along the path of motion won't go to zero
If you are considering cases where the motion includes some components downward in relation to gravity and some components upward, not every robot will be designed to recover the energy gained by moving downward and use it to move the object back upward, so some energy will be used lifting the object on the upward part of the path.
$endgroup$
The definition of work is energy = force * displacement * cos(theta)
This definition is only strictly accurate for movement on a straight line.
In general, you need to solve an integral
$$E = int_a^bvec{F}cdot dvec{ell}$$
Does the movement require energy?
Two reasons a real robot will require energy to do this operation:
Real robots have friction in their joints, so there will always be a force opposing the direction the robot is trying to move, and so the integral of force along the path of motion won't go to zero
If you are considering cases where the motion includes some components downward in relation to gravity and some components upward, not every robot will be designed to recover the energy gained by moving downward and use it to move the object back upward, so some energy will be used lifting the object on the upward part of the path.
edited 5 hours ago
answered 5 hours ago
The PhotonThe Photon
9,44311831
9,44311831
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I'm assuming an ideal robot in space (so no gravity towards the earth). Are you saying that an ideal robot would not require any energy to do this motion?
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– Pro Q
5 hours ago
1
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@ProQ, not if the robot had magical frictionless joints.
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– The Photon
5 hours ago
1
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Also, if there is no gradient of potential energy (i.e. no gravity and the ball doesn't have electric charge, etc) then the movement doesn't even have to be in a circle for it to require no energy.
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– The Photon
5 hours ago
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@ThePhoton I don't think friction is the only place energy will go into.
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– Aaron Stevens
4 hours ago
add a comment |
$begingroup$
I'm assuming an ideal robot in space (so no gravity towards the earth). Are you saying that an ideal robot would not require any energy to do this motion?
$endgroup$
– Pro Q
5 hours ago
1
$begingroup$
@ProQ, not if the robot had magical frictionless joints.
$endgroup$
– The Photon
5 hours ago
1
$begingroup$
Also, if there is no gradient of potential energy (i.e. no gravity and the ball doesn't have electric charge, etc) then the movement doesn't even have to be in a circle for it to require no energy.
$endgroup$
– The Photon
5 hours ago
$begingroup$
@ThePhoton I don't think friction is the only place energy will go into.
$endgroup$
– Aaron Stevens
4 hours ago
$begingroup$
I'm assuming an ideal robot in space (so no gravity towards the earth). Are you saying that an ideal robot would not require any energy to do this motion?
$endgroup$
– Pro Q
5 hours ago
$begingroup$
I'm assuming an ideal robot in space (so no gravity towards the earth). Are you saying that an ideal robot would not require any energy to do this motion?
$endgroup$
– Pro Q
5 hours ago
1
1
$begingroup$
@ProQ, not if the robot had magical frictionless joints.
$endgroup$
– The Photon
5 hours ago
$begingroup$
@ProQ, not if the robot had magical frictionless joints.
$endgroup$
– The Photon
5 hours ago
1
1
$begingroup$
Also, if there is no gradient of potential energy (i.e. no gravity and the ball doesn't have electric charge, etc) then the movement doesn't even have to be in a circle for it to require no energy.
$endgroup$
– The Photon
5 hours ago
$begingroup$
Also, if there is no gradient of potential energy (i.e. no gravity and the ball doesn't have electric charge, etc) then the movement doesn't even have to be in a circle for it to require no energy.
$endgroup$
– The Photon
5 hours ago
$begingroup$
@ThePhoton I don't think friction is the only place energy will go into.
$endgroup$
– Aaron Stevens
4 hours ago
$begingroup$
@ThePhoton I don't think friction is the only place energy will go into.
$endgroup$
– Aaron Stevens
4 hours ago
add a comment |
Pro Q is a new contributor. Be nice, and check out our Code of Conduct.
Pro Q is a new contributor. Be nice, and check out our Code of Conduct.
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Pro Q is a new contributor. Be nice, and check out our Code of Conduct.
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