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Does it take energy to move something in a circle?


Block on cart, equation of motionDoes an opposing force cause loss/ waste of energy?What's wrong with this idea for recover energy from pressure?Conservation of energy in a solenoidAttitude control with a Geared reaction wheelConceptual Difficulty About WorkRegarding gravity force during circular motionKinetic energy for generalized coordinates?Finding the moment about a PointAbout the work of the friction force













9












$begingroup$


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.)










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  • 1




    $begingroup$
    A centripetal force always acts at right angles to displacement, and does no work as a consequence. This is why satellites remain in orbit around the earth in the absence of air friction ... the earth's gravity is applying a force to the satellite to cause it to orbit, but that force is doing no work on the satellite, so the satellite's speed remains constant.
    $endgroup$
    – David White
    10 hours ago
















9












$begingroup$


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.)










share|cite|improve this question









New contributor




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







$endgroup$








  • 1




    $begingroup$
    A centripetal force always acts at right angles to displacement, and does no work as a consequence. This is why satellites remain in orbit around the earth in the absence of air friction ... the earth's gravity is applying a force to the satellite to cause it to orbit, but that force is doing no work on the satellite, so the satellite's speed remains constant.
    $endgroup$
    – David White
    10 hours ago














9












9








9


3



$begingroup$


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.)










share|cite|improve this question









New contributor




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







$endgroup$




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






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edited yesterday









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  • 1




    $begingroup$
    A centripetal force always acts at right angles to displacement, and does no work as a consequence. This is why satellites remain in orbit around the earth in the absence of air friction ... the earth's gravity is applying a force to the satellite to cause it to orbit, but that force is doing no work on the satellite, so the satellite's speed remains constant.
    $endgroup$
    – David White
    10 hours ago














  • 1




    $begingroup$
    A centripetal force always acts at right angles to displacement, and does no work as a consequence. This is why satellites remain in orbit around the earth in the absence of air friction ... the earth's gravity is applying a force to the satellite to cause it to orbit, but that force is doing no work on the satellite, so the satellite's speed remains constant.
    $endgroup$
    – David White
    10 hours ago








1




1




$begingroup$
A centripetal force always acts at right angles to displacement, and does no work as a consequence. This is why satellites remain in orbit around the earth in the absence of air friction ... the earth's gravity is applying a force to the satellite to cause it to orbit, but that force is doing no work on the satellite, so the satellite's speed remains constant.
$endgroup$
– David White
10 hours ago




$begingroup$
A centripetal force always acts at right angles to displacement, and does no work as a consequence. This is why satellites remain in orbit around the earth in the absence of air friction ... the earth's gravity is applying a force to the satellite to cause it to orbit, but that force is doing no work on the satellite, so the satellite's speed remains constant.
$endgroup$
– David White
10 hours ago










7 Answers
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19












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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.






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$endgroup$









  • 2




    $begingroup$
    @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
    yesterday








  • 2




    $begingroup$
    @Taemyr So the robot can accelerate and decelerate itself and the ball without any power needed? How do you suppose these forces arise?
    $endgroup$
    – Aaron Stevens
    yesterday






  • 2




    $begingroup$
    @Taemyr Then we are in agreement, and you should read my answer more carefully.
    $endgroup$
    – Aaron Stevens
    yesterday






  • 3




    $begingroup$
    @Taemyr I agree. Nowhere in my answer do I say power is needed to keep the ball rotating at a constant speed. I only say power is needed to speed up or slow down the ball. Once again, please read my answer more carefully. You are objecting to things I didn't even say.
    $endgroup$
    – Aaron Stevens
    yesterday








  • 3




    $begingroup$
    @AaronStevens yes, in general. All energy can be recovered from a process that does no (net) work. This is what it means for a process to do no work
    $endgroup$
    – DreamConspiracy
    17 hours ago



















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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.







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$endgroup$













  • $begingroup$
    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
    yesterday








  • 2




    $begingroup$
    @ProQ, not if the robot had magical frictionless joints.
    $endgroup$
    – The Photon
    yesterday






  • 2




    $begingroup$
    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
    yesterday






  • 4




    $begingroup$
    @ThePhoton I don't think friction is the only place energy will go into.
    $endgroup$
    – Aaron Stevens
    yesterday






  • 4




    $begingroup$
    @ProQ: As well as magical frictionless joints, the robot would need to use some kind of energy recovery mechanism when slowing down the ball to convert the kinetic energy back into electrical energy in the store - even in space. The idea would be similar to energy recovery used in some electric vehicles, and in reality has losses, similar to friction. Finally, there may still be a small unrecoverable energy loss due to the information processing required for the robot to perform the action - but that is probably off-topic for your original question
    $endgroup$
    – Neil Slater
    22 hours ago





















9












$begingroup$

If we can neglect external gravitational or electromagnetic fields then we only need to consider the kinetic energy of the ball. This is zero at the beginning of the motion and zero at the end, so the net change of energy in the ball is zero.



If we assume an ideal robot arm (no friction, perfect conductors, no air resistance etc.) then the energy that the robot arm puts into the ball to accelerate it at the start of the motion can be 100% recovered when the ball decelerates at the end of motion. So the net loss of energy from the robot arm is also zero.



In practice the robot arm will lose energy due to friction, resistive heating, air resistance etc.






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$endgroup$





















    3












    $begingroup$

    If the robot arm is extended holding the ball and rotated horizontally, then the energy required would be calculated using torque, using the distance of the centre of mass and the total mass being moved by the motor.



    Let us say that with the arm extended the centre of mass is 0.5m (distance from axis) and the total mass of arm + ball = 5kg, the torque rating is 10 Nm. Because torque is a rotational equivalent of linear motion then power = torque x 2 x pi x rotational speed (revs per second) (https://en.wikipedia.org/wiki/Torque)



    Therefore, if we say the arm takes 1 second to complete the circle, the power required is 10 x 2 x pi x 1 = 62.83 Watts (= joules/sec). A typical stepper motor for a robot arm with a reach about 1m would draw 12V, therefore would have to be rated at 6A which in a perfect system would be capable of delivering 72 Watts.






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    • $begingroup$
      Consider using MathJax for writing equations.
      $endgroup$
      – exp ikx
      18 hours ago



















    3












    $begingroup$

    NO, an ideal robot can move the load in a circle (actually in an any closed curve, not just a circle), producing zero total job and hence without any loss of the energy. However in order to do that, the robot must:




    • Recover back all potential (lifting up the load) and kinetic (accelerating the load from zero initial velocity so it can be moved) energy without the loss. A simple electric engine can do this, but obviously not without the loss.

    • Move the load without friction (well, you have stated in the question your robot can do this but even spacecraft do hit one another atom on they way through the vacuum).


    The robot will need the energy in general, especially if the circle stands vertically like a Ferris wheel. Just an ideal robot will be able to recover all the energy back at the end of the loop.



    A great example of such a machine is the Buzz Aldrin cycler. It is a "space bus" that travels between Earth and Mars essentially for free because it travels in a closed loop. In space, it is relatively easy to satisfy the two conditions above in a significant degree.






    share|cite











    $endgroup$





















      1












      $begingroup$

      The infinitesimal work on an object is $Fdcosθ$. If any of those quantities aren't constant, then we have to take the integral over some path. However, all paths will the same starting and ending conditions will yield the same answer. If the ball ends up in the same condition that its started in, then it hasn't had any work done on it. So this is not the central issue.



      What is the central issue is that this is the formula for doing work on an object. Just because the robot arm has done zero work on the ball, that does not mean that the robot arm has expended no energy. It just means that none of the energy expended by the robot arm has gone towards permanently increasing the kinetic energy of the ball. The robot arm could have expended energy otherwise, such as overcoming friction in its internal mechanisms. If the robot arm accelerated to move the ball, and then decelerated to stop the ball, then it took energy to accelerate, and then the energy went somewhere when the arm decelerated. If the deceleration happened through friction, then the energy dissipated into heat. But the robot arm could have regenerative braking, in which case some of the energy went back into the battery.



      In any real world system, there will be some loss of energy to heat. No motor and no regenerative braking system operates with perfect efficiency. But in an ideal system with no friction or other inefficiency, the robot arm could indeed move the ball and end up with the same amount of energy in its battery, and thus this would not use up any energy. However, we would still have to have some energy to start with to power the system, even if that energy isn't used up.






      share|cite|improve this answer









      $endgroup$





















        1












        $begingroup$

        Interesting question and a simple answer here. Newtonian first law claims, that:
        $$frac{text d mathbf v}{text d t} = 0$$ In words it would be:
        If net force applied to an object is zero, then object's speed must be constant



        Your object moves in a circle and that means, that it changes speed, because speed is a vector which has magnitude and direction:



        enter image description here



        So if direction of movement changes, speed changes also. And if speed changes - there MUST be a force acting upon an object. And for keeping force, you need an energy of course.






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        $endgroup$













        • $begingroup$
          Well, that's interesting. So who's spending the energy in keeping the planets in orbit? Angels pushing on spheres?
          $endgroup$
          – AtmosphericPrisonEscape
          1 hour ago










        • $begingroup$
          Easy to answer. Planets in orbits are kept by a gravity force of a more massive body - such as star, a bigger planet (in case of satellite planet orbiting around other planet), etc. In other words - orbiting planet is in constant fall towards the massive body (actually towards the center of gravity, because massive body will rotate too a bit). But planet in orbit can't fall to center of gravity, because it is also exposed to a centrifugal force.
          $endgroup$
          – Agnius Vasiliauskas
          1 hour ago













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        7 Answers
        7






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        7 Answers
        7






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        active

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        active

        oldest

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        19












        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$









        • 2




          $begingroup$
          @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
          yesterday








        • 2




          $begingroup$
          @Taemyr So the robot can accelerate and decelerate itself and the ball without any power needed? How do you suppose these forces arise?
          $endgroup$
          – Aaron Stevens
          yesterday






        • 2




          $begingroup$
          @Taemyr Then we are in agreement, and you should read my answer more carefully.
          $endgroup$
          – Aaron Stevens
          yesterday






        • 3




          $begingroup$
          @Taemyr I agree. Nowhere in my answer do I say power is needed to keep the ball rotating at a constant speed. I only say power is needed to speed up or slow down the ball. Once again, please read my answer more carefully. You are objecting to things I didn't even say.
          $endgroup$
          – Aaron Stevens
          yesterday








        • 3




          $begingroup$
          @AaronStevens yes, in general. All energy can be recovered from a process that does no (net) work. This is what it means for a process to do no work
          $endgroup$
          – DreamConspiracy
          17 hours ago
















        19












        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$









        • 2




          $begingroup$
          @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
          yesterday








        • 2




          $begingroup$
          @Taemyr So the robot can accelerate and decelerate itself and the ball without any power needed? How do you suppose these forces arise?
          $endgroup$
          – Aaron Stevens
          yesterday






        • 2




          $begingroup$
          @Taemyr Then we are in agreement, and you should read my answer more carefully.
          $endgroup$
          – Aaron Stevens
          yesterday






        • 3




          $begingroup$
          @Taemyr I agree. Nowhere in my answer do I say power is needed to keep the ball rotating at a constant speed. I only say power is needed to speed up or slow down the ball. Once again, please read my answer more carefully. You are objecting to things I didn't even say.
          $endgroup$
          – Aaron Stevens
          yesterday








        • 3




          $begingroup$
          @AaronStevens yes, in general. All energy can be recovered from a process that does no (net) work. This is what it means for a process to do no work
          $endgroup$
          – DreamConspiracy
          17 hours ago














        19












        19








        19





        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered yesterday









        Aaron StevensAaron Stevens

        11.3k31744




        11.3k31744








        • 2




          $begingroup$
          @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
          yesterday








        • 2




          $begingroup$
          @Taemyr So the robot can accelerate and decelerate itself and the ball without any power needed? How do you suppose these forces arise?
          $endgroup$
          – Aaron Stevens
          yesterday






        • 2




          $begingroup$
          @Taemyr Then we are in agreement, and you should read my answer more carefully.
          $endgroup$
          – Aaron Stevens
          yesterday






        • 3




          $begingroup$
          @Taemyr I agree. Nowhere in my answer do I say power is needed to keep the ball rotating at a constant speed. I only say power is needed to speed up or slow down the ball. Once again, please read my answer more carefully. You are objecting to things I didn't even say.
          $endgroup$
          – Aaron Stevens
          yesterday








        • 3




          $begingroup$
          @AaronStevens yes, in general. All energy can be recovered from a process that does no (net) work. This is what it means for a process to do no work
          $endgroup$
          – DreamConspiracy
          17 hours ago














        • 2




          $begingroup$
          @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
          yesterday








        • 2




          $begingroup$
          @Taemyr So the robot can accelerate and decelerate itself and the ball without any power needed? How do you suppose these forces arise?
          $endgroup$
          – Aaron Stevens
          yesterday






        • 2




          $begingroup$
          @Taemyr Then we are in agreement, and you should read my answer more carefully.
          $endgroup$
          – Aaron Stevens
          yesterday






        • 3




          $begingroup$
          @Taemyr I agree. Nowhere in my answer do I say power is needed to keep the ball rotating at a constant speed. I only say power is needed to speed up or slow down the ball. Once again, please read my answer more carefully. You are objecting to things I didn't even say.
          $endgroup$
          – Aaron Stevens
          yesterday








        • 3




          $begingroup$
          @AaronStevens yes, in general. All energy can be recovered from a process that does no (net) work. This is what it means for a process to do no work
          $endgroup$
          – DreamConspiracy
          17 hours ago








        2




        2




        $begingroup$
        @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
        yesterday






        $begingroup$
        @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
        yesterday






        2




        2




        $begingroup$
        @Taemyr So the robot can accelerate and decelerate itself and the ball without any power needed? How do you suppose these forces arise?
        $endgroup$
        – Aaron Stevens
        yesterday




        $begingroup$
        @Taemyr So the robot can accelerate and decelerate itself and the ball without any power needed? How do you suppose these forces arise?
        $endgroup$
        – Aaron Stevens
        yesterday




        2




        2




        $begingroup$
        @Taemyr Then we are in agreement, and you should read my answer more carefully.
        $endgroup$
        – Aaron Stevens
        yesterday




        $begingroup$
        @Taemyr Then we are in agreement, and you should read my answer more carefully.
        $endgroup$
        – Aaron Stevens
        yesterday




        3




        3




        $begingroup$
        @Taemyr I agree. Nowhere in my answer do I say power is needed to keep the ball rotating at a constant speed. I only say power is needed to speed up or slow down the ball. Once again, please read my answer more carefully. You are objecting to things I didn't even say.
        $endgroup$
        – Aaron Stevens
        yesterday






        $begingroup$
        @Taemyr I agree. Nowhere in my answer do I say power is needed to keep the ball rotating at a constant speed. I only say power is needed to speed up or slow down the ball. Once again, please read my answer more carefully. You are objecting to things I didn't even say.
        $endgroup$
        – Aaron Stevens
        yesterday






        3




        3




        $begingroup$
        @AaronStevens yes, in general. All energy can be recovered from a process that does no (net) work. This is what it means for a process to do no work
        $endgroup$
        – DreamConspiracy
        17 hours ago




        $begingroup$
        @AaronStevens yes, in general. All energy can be recovered from a process that does no (net) work. This is what it means for a process to do no work
        $endgroup$
        – DreamConspiracy
        17 hours ago











        13












        $begingroup$


        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.







        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          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
          yesterday








        • 2




          $begingroup$
          @ProQ, not if the robot had magical frictionless joints.
          $endgroup$
          – The Photon
          yesterday






        • 2




          $begingroup$
          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
          yesterday






        • 4




          $begingroup$
          @ThePhoton I don't think friction is the only place energy will go into.
          $endgroup$
          – Aaron Stevens
          yesterday






        • 4




          $begingroup$
          @ProQ: As well as magical frictionless joints, the robot would need to use some kind of energy recovery mechanism when slowing down the ball to convert the kinetic energy back into electrical energy in the store - even in space. The idea would be similar to energy recovery used in some electric vehicles, and in reality has losses, similar to friction. Finally, there may still be a small unrecoverable energy loss due to the information processing required for the robot to perform the action - but that is probably off-topic for your original question
          $endgroup$
          – Neil Slater
          22 hours ago


















        13












        $begingroup$


        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.







        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          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
          yesterday








        • 2




          $begingroup$
          @ProQ, not if the robot had magical frictionless joints.
          $endgroup$
          – The Photon
          yesterday






        • 2




          $begingroup$
          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
          yesterday






        • 4




          $begingroup$
          @ThePhoton I don't think friction is the only place energy will go into.
          $endgroup$
          – Aaron Stevens
          yesterday






        • 4




          $begingroup$
          @ProQ: As well as magical frictionless joints, the robot would need to use some kind of energy recovery mechanism when slowing down the ball to convert the kinetic energy back into electrical energy in the store - even in space. The idea would be similar to energy recovery used in some electric vehicles, and in reality has losses, similar to friction. Finally, there may still be a small unrecoverable energy loss due to the information processing required for the robot to perform the action - but that is probably off-topic for your original question
          $endgroup$
          – Neil Slater
          22 hours ago
















        13












        13








        13





        $begingroup$


        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.







        share|cite|improve this answer











        $endgroup$




        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.








        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited yesterday

























        answered yesterday









        The PhotonThe Photon

        9,56311833




        9,56311833












        • $begingroup$
          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
          yesterday








        • 2




          $begingroup$
          @ProQ, not if the robot had magical frictionless joints.
          $endgroup$
          – The Photon
          yesterday






        • 2




          $begingroup$
          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
          yesterday






        • 4




          $begingroup$
          @ThePhoton I don't think friction is the only place energy will go into.
          $endgroup$
          – Aaron Stevens
          yesterday






        • 4




          $begingroup$
          @ProQ: As well as magical frictionless joints, the robot would need to use some kind of energy recovery mechanism when slowing down the ball to convert the kinetic energy back into electrical energy in the store - even in space. The idea would be similar to energy recovery used in some electric vehicles, and in reality has losses, similar to friction. Finally, there may still be a small unrecoverable energy loss due to the information processing required for the robot to perform the action - but that is probably off-topic for your original question
          $endgroup$
          – Neil Slater
          22 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
          yesterday








        • 2




          $begingroup$
          @ProQ, not if the robot had magical frictionless joints.
          $endgroup$
          – The Photon
          yesterday






        • 2




          $begingroup$
          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
          yesterday






        • 4




          $begingroup$
          @ThePhoton I don't think friction is the only place energy will go into.
          $endgroup$
          – Aaron Stevens
          yesterday






        • 4




          $begingroup$
          @ProQ: As well as magical frictionless joints, the robot would need to use some kind of energy recovery mechanism when slowing down the ball to convert the kinetic energy back into electrical energy in the store - even in space. The idea would be similar to energy recovery used in some electric vehicles, and in reality has losses, similar to friction. Finally, there may still be a small unrecoverable energy loss due to the information processing required for the robot to perform the action - but that is probably off-topic for your original question
          $endgroup$
          – Neil Slater
          22 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
        yesterday






        $begingroup$
        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
        yesterday






        2




        2




        $begingroup$
        @ProQ, not if the robot had magical frictionless joints.
        $endgroup$
        – The Photon
        yesterday




        $begingroup$
        @ProQ, not if the robot had magical frictionless joints.
        $endgroup$
        – The Photon
        yesterday




        2




        2




        $begingroup$
        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
        yesterday




        $begingroup$
        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
        yesterday




        4




        4




        $begingroup$
        @ThePhoton I don't think friction is the only place energy will go into.
        $endgroup$
        – Aaron Stevens
        yesterday




        $begingroup$
        @ThePhoton I don't think friction is the only place energy will go into.
        $endgroup$
        – Aaron Stevens
        yesterday




        4




        4




        $begingroup$
        @ProQ: As well as magical frictionless joints, the robot would need to use some kind of energy recovery mechanism when slowing down the ball to convert the kinetic energy back into electrical energy in the store - even in space. The idea would be similar to energy recovery used in some electric vehicles, and in reality has losses, similar to friction. Finally, there may still be a small unrecoverable energy loss due to the information processing required for the robot to perform the action - but that is probably off-topic for your original question
        $endgroup$
        – Neil Slater
        22 hours ago






        $begingroup$
        @ProQ: As well as magical frictionless joints, the robot would need to use some kind of energy recovery mechanism when slowing down the ball to convert the kinetic energy back into electrical energy in the store - even in space. The idea would be similar to energy recovery used in some electric vehicles, and in reality has losses, similar to friction. Finally, there may still be a small unrecoverable energy loss due to the information processing required for the robot to perform the action - but that is probably off-topic for your original question
        $endgroup$
        – Neil Slater
        22 hours ago













        9












        $begingroup$

        If we can neglect external gravitational or electromagnetic fields then we only need to consider the kinetic energy of the ball. This is zero at the beginning of the motion and zero at the end, so the net change of energy in the ball is zero.



        If we assume an ideal robot arm (no friction, perfect conductors, no air resistance etc.) then the energy that the robot arm puts into the ball to accelerate it at the start of the motion can be 100% recovered when the ball decelerates at the end of motion. So the net loss of energy from the robot arm is also zero.



        In practice the robot arm will lose energy due to friction, resistive heating, air resistance etc.






        share|cite|improve this answer









        $endgroup$


















          9












          $begingroup$

          If we can neglect external gravitational or electromagnetic fields then we only need to consider the kinetic energy of the ball. This is zero at the beginning of the motion and zero at the end, so the net change of energy in the ball is zero.



          If we assume an ideal robot arm (no friction, perfect conductors, no air resistance etc.) then the energy that the robot arm puts into the ball to accelerate it at the start of the motion can be 100% recovered when the ball decelerates at the end of motion. So the net loss of energy from the robot arm is also zero.



          In practice the robot arm will lose energy due to friction, resistive heating, air resistance etc.






          share|cite|improve this answer









          $endgroup$
















            9












            9








            9





            $begingroup$

            If we can neglect external gravitational or electromagnetic fields then we only need to consider the kinetic energy of the ball. This is zero at the beginning of the motion and zero at the end, so the net change of energy in the ball is zero.



            If we assume an ideal robot arm (no friction, perfect conductors, no air resistance etc.) then the energy that the robot arm puts into the ball to accelerate it at the start of the motion can be 100% recovered when the ball decelerates at the end of motion. So the net loss of energy from the robot arm is also zero.



            In practice the robot arm will lose energy due to friction, resistive heating, air resistance etc.






            share|cite|improve this answer









            $endgroup$



            If we can neglect external gravitational or electromagnetic fields then we only need to consider the kinetic energy of the ball. This is zero at the beginning of the motion and zero at the end, so the net change of energy in the ball is zero.



            If we assume an ideal robot arm (no friction, perfect conductors, no air resistance etc.) then the energy that the robot arm puts into the ball to accelerate it at the start of the motion can be 100% recovered when the ball decelerates at the end of motion. So the net loss of energy from the robot arm is also zero.



            In practice the robot arm will lose energy due to friction, resistive heating, air resistance etc.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered yesterday









            gandalf61gandalf61

            26914




            26914























                3












                $begingroup$

                If the robot arm is extended holding the ball and rotated horizontally, then the energy required would be calculated using torque, using the distance of the centre of mass and the total mass being moved by the motor.



                Let us say that with the arm extended the centre of mass is 0.5m (distance from axis) and the total mass of arm + ball = 5kg, the torque rating is 10 Nm. Because torque is a rotational equivalent of linear motion then power = torque x 2 x pi x rotational speed (revs per second) (https://en.wikipedia.org/wiki/Torque)



                Therefore, if we say the arm takes 1 second to complete the circle, the power required is 10 x 2 x pi x 1 = 62.83 Watts (= joules/sec). A typical stepper motor for a robot arm with a reach about 1m would draw 12V, therefore would have to be rated at 6A which in a perfect system would be capable of delivering 72 Watts.






                share|cite|improve this answer








                New contributor




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






                $endgroup$













                • $begingroup$
                  Consider using MathJax for writing equations.
                  $endgroup$
                  – exp ikx
                  18 hours ago
















                3












                $begingroup$

                If the robot arm is extended holding the ball and rotated horizontally, then the energy required would be calculated using torque, using the distance of the centre of mass and the total mass being moved by the motor.



                Let us say that with the arm extended the centre of mass is 0.5m (distance from axis) and the total mass of arm + ball = 5kg, the torque rating is 10 Nm. Because torque is a rotational equivalent of linear motion then power = torque x 2 x pi x rotational speed (revs per second) (https://en.wikipedia.org/wiki/Torque)



                Therefore, if we say the arm takes 1 second to complete the circle, the power required is 10 x 2 x pi x 1 = 62.83 Watts (= joules/sec). A typical stepper motor for a robot arm with a reach about 1m would draw 12V, therefore would have to be rated at 6A which in a perfect system would be capable of delivering 72 Watts.






                share|cite|improve this answer








                New contributor




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






                $endgroup$













                • $begingroup$
                  Consider using MathJax for writing equations.
                  $endgroup$
                  – exp ikx
                  18 hours ago














                3












                3








                3





                $begingroup$

                If the robot arm is extended holding the ball and rotated horizontally, then the energy required would be calculated using torque, using the distance of the centre of mass and the total mass being moved by the motor.



                Let us say that with the arm extended the centre of mass is 0.5m (distance from axis) and the total mass of arm + ball = 5kg, the torque rating is 10 Nm. Because torque is a rotational equivalent of linear motion then power = torque x 2 x pi x rotational speed (revs per second) (https://en.wikipedia.org/wiki/Torque)



                Therefore, if we say the arm takes 1 second to complete the circle, the power required is 10 x 2 x pi x 1 = 62.83 Watts (= joules/sec). A typical stepper motor for a robot arm with a reach about 1m would draw 12V, therefore would have to be rated at 6A which in a perfect system would be capable of delivering 72 Watts.






                share|cite|improve this answer








                New contributor




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






                $endgroup$



                If the robot arm is extended holding the ball and rotated horizontally, then the energy required would be calculated using torque, using the distance of the centre of mass and the total mass being moved by the motor.



                Let us say that with the arm extended the centre of mass is 0.5m (distance from axis) and the total mass of arm + ball = 5kg, the torque rating is 10 Nm. Because torque is a rotational equivalent of linear motion then power = torque x 2 x pi x rotational speed (revs per second) (https://en.wikipedia.org/wiki/Torque)



                Therefore, if we say the arm takes 1 second to complete the circle, the power required is 10 x 2 x pi x 1 = 62.83 Watts (= joules/sec). A typical stepper motor for a robot arm with a reach about 1m would draw 12V, therefore would have to be rated at 6A which in a perfect system would be capable of delivering 72 Watts.







                share|cite|improve this answer








                New contributor




                Chris Percival 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




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









                answered 19 hours ago









                Chris PercivalChris Percival

                311




                311




                New contributor




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





                New contributor





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






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












                • $begingroup$
                  Consider using MathJax for writing equations.
                  $endgroup$
                  – exp ikx
                  18 hours ago


















                • $begingroup$
                  Consider using MathJax for writing equations.
                  $endgroup$
                  – exp ikx
                  18 hours ago
















                $begingroup$
                Consider using MathJax for writing equations.
                $endgroup$
                – exp ikx
                18 hours ago




                $begingroup$
                Consider using MathJax for writing equations.
                $endgroup$
                – exp ikx
                18 hours ago











                3












                $begingroup$

                NO, an ideal robot can move the load in a circle (actually in an any closed curve, not just a circle), producing zero total job and hence without any loss of the energy. However in order to do that, the robot must:




                • Recover back all potential (lifting up the load) and kinetic (accelerating the load from zero initial velocity so it can be moved) energy without the loss. A simple electric engine can do this, but obviously not without the loss.

                • Move the load without friction (well, you have stated in the question your robot can do this but even spacecraft do hit one another atom on they way through the vacuum).


                The robot will need the energy in general, especially if the circle stands vertically like a Ferris wheel. Just an ideal robot will be able to recover all the energy back at the end of the loop.



                A great example of such a machine is the Buzz Aldrin cycler. It is a "space bus" that travels between Earth and Mars essentially for free because it travels in a closed loop. In space, it is relatively easy to satisfy the two conditions above in a significant degree.






                share|cite











                $endgroup$


















                  3












                  $begingroup$

                  NO, an ideal robot can move the load in a circle (actually in an any closed curve, not just a circle), producing zero total job and hence without any loss of the energy. However in order to do that, the robot must:




                  • Recover back all potential (lifting up the load) and kinetic (accelerating the load from zero initial velocity so it can be moved) energy without the loss. A simple electric engine can do this, but obviously not without the loss.

                  • Move the load without friction (well, you have stated in the question your robot can do this but even spacecraft do hit one another atom on they way through the vacuum).


                  The robot will need the energy in general, especially if the circle stands vertically like a Ferris wheel. Just an ideal robot will be able to recover all the energy back at the end of the loop.



                  A great example of such a machine is the Buzz Aldrin cycler. It is a "space bus" that travels between Earth and Mars essentially for free because it travels in a closed loop. In space, it is relatively easy to satisfy the two conditions above in a significant degree.






                  share|cite











                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    NO, an ideal robot can move the load in a circle (actually in an any closed curve, not just a circle), producing zero total job and hence without any loss of the energy. However in order to do that, the robot must:




                    • Recover back all potential (lifting up the load) and kinetic (accelerating the load from zero initial velocity so it can be moved) energy without the loss. A simple electric engine can do this, but obviously not without the loss.

                    • Move the load without friction (well, you have stated in the question your robot can do this but even spacecraft do hit one another atom on they way through the vacuum).


                    The robot will need the energy in general, especially if the circle stands vertically like a Ferris wheel. Just an ideal robot will be able to recover all the energy back at the end of the loop.



                    A great example of such a machine is the Buzz Aldrin cycler. It is a "space bus" that travels between Earth and Mars essentially for free because it travels in a closed loop. In space, it is relatively easy to satisfy the two conditions above in a significant degree.






                    share|cite











                    $endgroup$



                    NO, an ideal robot can move the load in a circle (actually in an any closed curve, not just a circle), producing zero total job and hence without any loss of the energy. However in order to do that, the robot must:




                    • Recover back all potential (lifting up the load) and kinetic (accelerating the load from zero initial velocity so it can be moved) energy without the loss. A simple electric engine can do this, but obviously not without the loss.

                    • Move the load without friction (well, you have stated in the question your robot can do this but even spacecraft do hit one another atom on they way through the vacuum).


                    The robot will need the energy in general, especially if the circle stands vertically like a Ferris wheel. Just an ideal robot will be able to recover all the energy back at the end of the loop.



                    A great example of such a machine is the Buzz Aldrin cycler. It is a "space bus" that travels between Earth and Mars essentially for free because it travels in a closed loop. In space, it is relatively easy to satisfy the two conditions above in a significant degree.







                    share|cite














                    share|cite



                    share|cite








                    edited 17 hours ago

























                    answered 17 hours ago









                    h22h22

                    23318




                    23318























                        1












                        $begingroup$

                        The infinitesimal work on an object is $Fdcosθ$. If any of those quantities aren't constant, then we have to take the integral over some path. However, all paths will the same starting and ending conditions will yield the same answer. If the ball ends up in the same condition that its started in, then it hasn't had any work done on it. So this is not the central issue.



                        What is the central issue is that this is the formula for doing work on an object. Just because the robot arm has done zero work on the ball, that does not mean that the robot arm has expended no energy. It just means that none of the energy expended by the robot arm has gone towards permanently increasing the kinetic energy of the ball. The robot arm could have expended energy otherwise, such as overcoming friction in its internal mechanisms. If the robot arm accelerated to move the ball, and then decelerated to stop the ball, then it took energy to accelerate, and then the energy went somewhere when the arm decelerated. If the deceleration happened through friction, then the energy dissipated into heat. But the robot arm could have regenerative braking, in which case some of the energy went back into the battery.



                        In any real world system, there will be some loss of energy to heat. No motor and no regenerative braking system operates with perfect efficiency. But in an ideal system with no friction or other inefficiency, the robot arm could indeed move the ball and end up with the same amount of energy in its battery, and thus this would not use up any energy. However, we would still have to have some energy to start with to power the system, even if that energy isn't used up.






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          The infinitesimal work on an object is $Fdcosθ$. If any of those quantities aren't constant, then we have to take the integral over some path. However, all paths will the same starting and ending conditions will yield the same answer. If the ball ends up in the same condition that its started in, then it hasn't had any work done on it. So this is not the central issue.



                          What is the central issue is that this is the formula for doing work on an object. Just because the robot arm has done zero work on the ball, that does not mean that the robot arm has expended no energy. It just means that none of the energy expended by the robot arm has gone towards permanently increasing the kinetic energy of the ball. The robot arm could have expended energy otherwise, such as overcoming friction in its internal mechanisms. If the robot arm accelerated to move the ball, and then decelerated to stop the ball, then it took energy to accelerate, and then the energy went somewhere when the arm decelerated. If the deceleration happened through friction, then the energy dissipated into heat. But the robot arm could have regenerative braking, in which case some of the energy went back into the battery.



                          In any real world system, there will be some loss of energy to heat. No motor and no regenerative braking system operates with perfect efficiency. But in an ideal system with no friction or other inefficiency, the robot arm could indeed move the ball and end up with the same amount of energy in its battery, and thus this would not use up any energy. However, we would still have to have some energy to start with to power the system, even if that energy isn't used up.






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            The infinitesimal work on an object is $Fdcosθ$. If any of those quantities aren't constant, then we have to take the integral over some path. However, all paths will the same starting and ending conditions will yield the same answer. If the ball ends up in the same condition that its started in, then it hasn't had any work done on it. So this is not the central issue.



                            What is the central issue is that this is the formula for doing work on an object. Just because the robot arm has done zero work on the ball, that does not mean that the robot arm has expended no energy. It just means that none of the energy expended by the robot arm has gone towards permanently increasing the kinetic energy of the ball. The robot arm could have expended energy otherwise, such as overcoming friction in its internal mechanisms. If the robot arm accelerated to move the ball, and then decelerated to stop the ball, then it took energy to accelerate, and then the energy went somewhere when the arm decelerated. If the deceleration happened through friction, then the energy dissipated into heat. But the robot arm could have regenerative braking, in which case some of the energy went back into the battery.



                            In any real world system, there will be some loss of energy to heat. No motor and no regenerative braking system operates with perfect efficiency. But in an ideal system with no friction or other inefficiency, the robot arm could indeed move the ball and end up with the same amount of energy in its battery, and thus this would not use up any energy. However, we would still have to have some energy to start with to power the system, even if that energy isn't used up.






                            share|cite|improve this answer









                            $endgroup$



                            The infinitesimal work on an object is $Fdcosθ$. If any of those quantities aren't constant, then we have to take the integral over some path. However, all paths will the same starting and ending conditions will yield the same answer. If the ball ends up in the same condition that its started in, then it hasn't had any work done on it. So this is not the central issue.



                            What is the central issue is that this is the formula for doing work on an object. Just because the robot arm has done zero work on the ball, that does not mean that the robot arm has expended no energy. It just means that none of the energy expended by the robot arm has gone towards permanently increasing the kinetic energy of the ball. The robot arm could have expended energy otherwise, such as overcoming friction in its internal mechanisms. If the robot arm accelerated to move the ball, and then decelerated to stop the ball, then it took energy to accelerate, and then the energy went somewhere when the arm decelerated. If the deceleration happened through friction, then the energy dissipated into heat. But the robot arm could have regenerative braking, in which case some of the energy went back into the battery.



                            In any real world system, there will be some loss of energy to heat. No motor and no regenerative braking system operates with perfect efficiency. But in an ideal system with no friction or other inefficiency, the robot arm could indeed move the ball and end up with the same amount of energy in its battery, and thus this would not use up any energy. However, we would still have to have some energy to start with to power the system, even if that energy isn't used up.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 19 hours ago









                            AcccumulationAcccumulation

                            2,526312




                            2,526312























                                1












                                $begingroup$

                                Interesting question and a simple answer here. Newtonian first law claims, that:
                                $$frac{text d mathbf v}{text d t} = 0$$ In words it would be:
                                If net force applied to an object is zero, then object's speed must be constant



                                Your object moves in a circle and that means, that it changes speed, because speed is a vector which has magnitude and direction:



                                enter image description here



                                So if direction of movement changes, speed changes also. And if speed changes - there MUST be a force acting upon an object. And for keeping force, you need an energy of course.






                                share|cite|improve this answer








                                New contributor




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






                                $endgroup$













                                • $begingroup$
                                  Well, that's interesting. So who's spending the energy in keeping the planets in orbit? Angels pushing on spheres?
                                  $endgroup$
                                  – AtmosphericPrisonEscape
                                  1 hour ago










                                • $begingroup$
                                  Easy to answer. Planets in orbits are kept by a gravity force of a more massive body - such as star, a bigger planet (in case of satellite planet orbiting around other planet), etc. In other words - orbiting planet is in constant fall towards the massive body (actually towards the center of gravity, because massive body will rotate too a bit). But planet in orbit can't fall to center of gravity, because it is also exposed to a centrifugal force.
                                  $endgroup$
                                  – Agnius Vasiliauskas
                                  1 hour ago


















                                1












                                $begingroup$

                                Interesting question and a simple answer here. Newtonian first law claims, that:
                                $$frac{text d mathbf v}{text d t} = 0$$ In words it would be:
                                If net force applied to an object is zero, then object's speed must be constant



                                Your object moves in a circle and that means, that it changes speed, because speed is a vector which has magnitude and direction:



                                enter image description here



                                So if direction of movement changes, speed changes also. And if speed changes - there MUST be a force acting upon an object. And for keeping force, you need an energy of course.






                                share|cite|improve this answer








                                New contributor




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






                                $endgroup$













                                • $begingroup$
                                  Well, that's interesting. So who's spending the energy in keeping the planets in orbit? Angels pushing on spheres?
                                  $endgroup$
                                  – AtmosphericPrisonEscape
                                  1 hour ago










                                • $begingroup$
                                  Easy to answer. Planets in orbits are kept by a gravity force of a more massive body - such as star, a bigger planet (in case of satellite planet orbiting around other planet), etc. In other words - orbiting planet is in constant fall towards the massive body (actually towards the center of gravity, because massive body will rotate too a bit). But planet in orbit can't fall to center of gravity, because it is also exposed to a centrifugal force.
                                  $endgroup$
                                  – Agnius Vasiliauskas
                                  1 hour ago
















                                1












                                1








                                1





                                $begingroup$

                                Interesting question and a simple answer here. Newtonian first law claims, that:
                                $$frac{text d mathbf v}{text d t} = 0$$ In words it would be:
                                If net force applied to an object is zero, then object's speed must be constant



                                Your object moves in a circle and that means, that it changes speed, because speed is a vector which has magnitude and direction:



                                enter image description here



                                So if direction of movement changes, speed changes also. And if speed changes - there MUST be a force acting upon an object. And for keeping force, you need an energy of course.






                                share|cite|improve this answer








                                New contributor




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






                                $endgroup$



                                Interesting question and a simple answer here. Newtonian first law claims, that:
                                $$frac{text d mathbf v}{text d t} = 0$$ In words it would be:
                                If net force applied to an object is zero, then object's speed must be constant



                                Your object moves in a circle and that means, that it changes speed, because speed is a vector which has magnitude and direction:



                                enter image description here



                                So if direction of movement changes, speed changes also. And if speed changes - there MUST be a force acting upon an object. And for keeping force, you need an energy of course.







                                share|cite|improve this answer








                                New contributor




                                Agnius Vasiliauskas 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




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









                                answered 2 hours ago









                                Agnius VasiliauskasAgnius Vasiliauskas

                                1113




                                1113




                                New contributor




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





                                New contributor





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






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












                                • $begingroup$
                                  Well, that's interesting. So who's spending the energy in keeping the planets in orbit? Angels pushing on spheres?
                                  $endgroup$
                                  – AtmosphericPrisonEscape
                                  1 hour ago










                                • $begingroup$
                                  Easy to answer. Planets in orbits are kept by a gravity force of a more massive body - such as star, a bigger planet (in case of satellite planet orbiting around other planet), etc. In other words - orbiting planet is in constant fall towards the massive body (actually towards the center of gravity, because massive body will rotate too a bit). But planet in orbit can't fall to center of gravity, because it is also exposed to a centrifugal force.
                                  $endgroup$
                                  – Agnius Vasiliauskas
                                  1 hour ago




















                                • $begingroup$
                                  Well, that's interesting. So who's spending the energy in keeping the planets in orbit? Angels pushing on spheres?
                                  $endgroup$
                                  – AtmosphericPrisonEscape
                                  1 hour ago










                                • $begingroup$
                                  Easy to answer. Planets in orbits are kept by a gravity force of a more massive body - such as star, a bigger planet (in case of satellite planet orbiting around other planet), etc. In other words - orbiting planet is in constant fall towards the massive body (actually towards the center of gravity, because massive body will rotate too a bit). But planet in orbit can't fall to center of gravity, because it is also exposed to a centrifugal force.
                                  $endgroup$
                                  – Agnius Vasiliauskas
                                  1 hour ago


















                                $begingroup$
                                Well, that's interesting. So who's spending the energy in keeping the planets in orbit? Angels pushing on spheres?
                                $endgroup$
                                – AtmosphericPrisonEscape
                                1 hour ago




                                $begingroup$
                                Well, that's interesting. So who's spending the energy in keeping the planets in orbit? Angels pushing on spheres?
                                $endgroup$
                                – AtmosphericPrisonEscape
                                1 hour ago












                                $begingroup$
                                Easy to answer. Planets in orbits are kept by a gravity force of a more massive body - such as star, a bigger planet (in case of satellite planet orbiting around other planet), etc. In other words - orbiting planet is in constant fall towards the massive body (actually towards the center of gravity, because massive body will rotate too a bit). But planet in orbit can't fall to center of gravity, because it is also exposed to a centrifugal force.
                                $endgroup$
                                – Agnius Vasiliauskas
                                1 hour ago






                                $begingroup$
                                Easy to answer. Planets in orbits are kept by a gravity force of a more massive body - such as star, a bigger planet (in case of satellite planet orbiting around other planet), etc. In other words - orbiting planet is in constant fall towards the massive body (actually towards the center of gravity, because massive body will rotate too a bit). But planet in orbit can't fall to center of gravity, because it is also exposed to a centrifugal force.
                                $endgroup$
                                – Agnius Vasiliauskas
                                1 hour ago












                                Pro Q is a new contributor. Be nice, and check out our Code of Conduct.










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