Does it take energy to move something in a circle?
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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
edited Feb 25 at 3:47
Chair
4,39872241
asked Feb 25 at 2:20
Pro QPro Q
18118
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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
edited Feb 25 at 3:47
Chair
4,39872241
asked Feb 25 at 2:20
Pro QPro Q
18118
$endgroup$
4
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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.
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– David White
Feb 26 at 3:15
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A better example of this is to consider a puck sliding on a surface on without any drag or other losses, attached to a massless string that goes through a hole in the surface. Assume the puck is moving in a circle, then a downwards force on the string below the hole equals the tension in the string above the hole which is equal to the centripetal force exerted by the string onto the puck. Since there's no vertical motion of the string below the hole, then the work done is force x zero distance = zero.
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– rcgldr
Feb 26 at 16:13
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Continuing, now consider what happens if the string is pulled downwards, the puck's path is no longer a circle and a component of tension in the string will be in the direction of the puck's path, increasing the speed and mechanical energy of the puck (angular momentum will be conserved). If the string is allowed to move upwards, then "negative" work is done, as energy is extracted from the puck, and the puck's speed and energy decrease.
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– rcgldr
Feb 26 at 16:16
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You get the same circular motion if you replace the robot arm with a (well-lubricated) rod that rotates around a pole. Does this setup need a battery?
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– Rahul
Feb 26 at 22:14
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Sorry, I thought the question was specifically about the robot.
$endgroup$
– Aaron Stevens
Feb 27 at 10:08
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
edited Feb 25 at 3:47
Chair
4,39872241
asked Feb 25 at 2:20
Pro QPro Q
18118
$endgroup$
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
edited Feb 25 at 3:47
Chair
4,39872241
asked Feb 25 at 2:20
Pro QPro Q
18118
edited Feb 25 at 3:47
Chair
4,39872241
asked Feb 25 at 2:20
Pro QPro Q
18118
edited Feb 25 at 3:47
Chair
4,39872241
edited Feb 25 at 3:47
Chair
4,39872241
edited Feb 25 at 3:47
Chair
4,39872241
4,39872241
asked Feb 25 at 2:20
Pro QPro Q
18118
asked Feb 25 at 2:20
Pro QPro Q
18118
asked Feb 25 at 2:20
Pro QPro Q
18118
18118
4
$begingroup$
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.
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– David White
Feb 26 at 3:15
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A better example of this is to consider a puck sliding on a surface on without any drag or other losses, attached to a massless string that goes through a hole in the surface. Assume the puck is moving in a circle, then a downwards force on the string below the hole equals the tension in the string above the hole which is equal to the centripetal force exerted by the string onto the puck. Since there's no vertical motion of the string below the hole, then the work done is force x zero distance = zero.
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– rcgldr
Feb 26 at 16:13
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Continuing, now consider what happens if the string is pulled downwards, the puck's path is no longer a circle and a component of tension in the string will be in the direction of the puck's path, increasing the speed and mechanical energy of the puck (angular momentum will be conserved). If the string is allowed to move upwards, then "negative" work is done, as energy is extracted from the puck, and the puck's speed and energy decrease.
$endgroup$
– rcgldr
Feb 26 at 16:16
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You get the same circular motion if you replace the robot arm with a (well-lubricated) rod that rotates around a pole. Does this setup need a battery?
$endgroup$
– Rahul
Feb 26 at 22:14
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Sorry, I thought the question was specifically about the robot.
$endgroup$
– Aaron Stevens
Feb 27 at 10:08
add a comment |
4
$begingroup$
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
Feb 26 at 3:15
$begingroup$
A better example of this is to consider a puck sliding on a surface on without any drag or other losses, attached to a massless string that goes through a hole in the surface. Assume the puck is moving in a circle, then a downwards force on the string below the hole equals the tension in the string above the hole which is equal to the centripetal force exerted by the string onto the puck. Since there's no vertical motion of the string below the hole, then the work done is force x zero distance = zero.
$endgroup$
– rcgldr
Feb 26 at 16:13
$begingroup$
Continuing, now consider what happens if the string is pulled downwards, the puck's path is no longer a circle and a component of tension in the string will be in the direction of the puck's path, increasing the speed and mechanical energy of the puck (angular momentum will be conserved). If the string is allowed to move upwards, then "negative" work is done, as energy is extracted from the puck, and the puck's speed and energy decrease.
$endgroup$
– rcgldr
Feb 26 at 16:16
$begingroup$
You get the same circular motion if you replace the robot arm with a (well-lubricated) rod that rotates around a pole. Does this setup need a battery?
$endgroup$
– Rahul
Feb 26 at 22:14
$begingroup$
Sorry, I thought the question was specifically about the robot.
$endgroup$
– Aaron Stevens
Feb 27 at 10:08
4
4
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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
Feb 26 at 3:15
$begingroup$
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
Feb 26 at 3:15
$begingroup$
A better example of this is to consider a puck sliding on a surface on without any drag or other losses, attached to a massless string that goes through a hole in the surface. Assume the puck is moving in a circle, then a downwards force on the string below the hole equals the tension in the string above the hole which is equal to the centripetal force exerted by the string onto the puck. Since there's no vertical motion of the string below the hole, then the work done is force x zero distance = zero.
$endgroup$
– rcgldr
Feb 26 at 16:13
$begingroup$
A better example of this is to consider a puck sliding on a surface on without any drag or other losses, attached to a massless string that goes through a hole in the surface. Assume the puck is moving in a circle, then a downwards force on the string below the hole equals the tension in the string above the hole which is equal to the centripetal force exerted by the string onto the puck. Since there's no vertical motion of the string below the hole, then the work done is force x zero distance = zero.
$endgroup$
– rcgldr
Feb 26 at 16:13
$begingroup$
Continuing, now consider what happens if the string is pulled downwards, the puck's path is no longer a circle and a component of tension in the string will be in the direction of the puck's path, increasing the speed and mechanical energy of the puck (angular momentum will be conserved). If the string is allowed to move upwards, then "negative" work is done, as energy is extracted from the puck, and the puck's speed and energy decrease.
$endgroup$
– rcgldr
Feb 26 at 16:16
$begingroup$
Continuing, now consider what happens if the string is pulled downwards, the puck's path is no longer a circle and a component of tension in the string will be in the direction of the puck's path, increasing the speed and mechanical energy of the puck (angular momentum will be conserved). If the string is allowed to move upwards, then "negative" work is done, as energy is extracted from the puck, and the puck's speed and energy decrease.
$endgroup$
– rcgldr
Feb 26 at 16:16
$begingroup$
You get the same circular motion if you replace the robot arm with a (well-lubricated) rod that rotates around a pole. Does this setup need a battery?
$endgroup$
– Rahul
Feb 26 at 22:14
$begingroup$
You get the same circular motion if you replace the robot arm with a (well-lubricated) rod that rotates around a pole. Does this setup need a battery?
$endgroup$
– Rahul
Feb 26 at 22:14
$begingroup$
Sorry, I thought the question was specifically about the robot.
$endgroup$
– Aaron Stevens
Feb 27 at 10:08
$begingroup$
Sorry, I thought the question was specifically about the robot.
$endgroup$
– Aaron Stevens
Feb 27 at 10:08
add a comment |
8 Answers
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Many objects move in circles without needing any energy.
This starts with stuff like geostationary satellites, continues with moons that orbit planets, goes on to planets orbiting stars, stars orbiting other stars or black holes, and stars orbiting their galaxies' center of mass. All of these perform thousands, millions, and billions of rotations without any need for energy.
Of course, these objects all need to have sufficient kinetic energy to be able to orbit in the first place. An object that's stationary with respect to earth, will just fall down. To get it to orbit earth, it must be accelerated first, and that needs energy. And if you want to get it stationary again after it has completed an orbit you have to decelerated it again. However, while the object is orbiting on a circle, its kinetic energy remains the same all the time, no energy needs to be put in or removed.
The trick about saying that moving the object in a circle does not take any energy is that energy for acceleration and deceleration are added. Acceleration increases the kinetic energy of the object, the orbit does not change that kinetic energy, and deceleration removes the same amount of kinetic energy again. If you can perfectly recover the kinetic energy during deceleration, you get out with a black zero. (Of course, if you have losses, as you always do, you loose energy overall.)
However, that's not the entire story. Because, when an electrically charged object goes in circles, it emits electromagnetic radiation. A circling charge induces a magnetic field. That is what happens in any electric motor, including the one that spins your computer's fan right now. The reverse process works as well, a changing magnetic field accelerates charges round in circles. That's the working principle of any electrical generator, including your bike's dynamo.
Even more curiously, what can be said about the electro-magnetic field can be said about the gravitational field as well: Rotating masses emit gravitational waves, which carry away a tiny amount of energy. Usually we can totally ignore this kind of effect, but when really heavy objects (like black holes many times as heavy as our sun) dance around each other really closely, and really fast (like going round ten times per second or more), that tiny bit of energy loss can amount to the equivalent of three times the mass of our sun radiated away as pure gravitational wave energy... (https://en.wikipedia.org/wiki/First_observation_of_gravitational_waves)
TL;DR:
You can move stuff around in circles without needing energy, as long as the object is not electrically charged, and losses to gravitational waves are negligible. Which is pretty much always the case.
However, when electro-magnetic or gravitational effects enter the stage, that's where the real fun starts...
And, at least for the electro-magnetic effects, you personally rely on them every single day.
answered Feb 26 at 21:57
cmastercmaster
6009
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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.
answered Feb 25 at 2:44
Aaron StevensAaron Stevens
13.5k42250
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Comments are not for extended discussion; this conversation has been moved to chat.
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– ACuriousMind♦
Feb 26 at 19:38
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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
$$E = int_a^bvecFcdot dvecell$$
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.
answered Feb 25 at 2:28
The PhotonThe Photon
9,68711833
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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
Feb 25 at 2:33
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@ProQ, not if the robot had magical frictionless joints.
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– The Photon
Feb 25 at 2:39
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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
Feb 25 at 2:41
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@ThePhoton I don't think friction is the only place energy will go into.
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– Aaron Stevens
Feb 25 at 3:14
5
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@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
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– Neil Slater
Feb 25 at 14:28
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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.
answered Feb 25 at 10:57
gandalf61gandalf61
41628
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NO, an ideal robot can move the load in a circle (actually in any closed curve, not just a circle), producing zero total work done and hence no loss of 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 loss. A simple electric engine can do this, but obviously not without loss.
- Move the load without friction (well, you have stated in the question that your robot can do this, but even spacecraft do hit atoms on their way through the vacuum).
The robot will need energy in general, especially if the circle stands vertically like a Ferris wheel. Only an ideal robot would 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.
edited Feb 26 at 13:29
Captain Lepton
1033
answered Feb 25 at 20:04
h22h22
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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.
answered Feb 25 at 17:32
Chris PercivalChris Percival
411
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Consider using MathJax for writing equations.
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– exp ikx
Feb 25 at 18:18
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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.
answered Feb 25 at 18:17
AcccumulationAcccumulation
2,696312
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Interesting question and a simple answer here. Newtonian first law claims, that:
$$fractext d mathbf vtext d t = 0$$ 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:
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.
answered Feb 26 at 11:12
Agnius VasiliauskasAgnius Vasiliauskas
1264
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Well, that's interesting. So who's spending the energy in keeping the planets in orbit? Angels pushing on spheres?
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– AtmosphericPrisonEscape
Feb 26 at 11:22
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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 inconstant falltowards 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.
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– Agnius Vasiliauskas
Feb 26 at 11:31
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You're mixing center-of-mass system and co-rotating system. Your argumentation is plain wrong. Compute $oint vec F cdot dvecl $ for the gravitational force in the CMS frame for a perfectly circular orbit. You will see that, because of the free-fall you've quoted but misunderstood, no work is done onto the body. For eccentric orbits the net work is still zero, because gravity is a conservative force field.
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– AtmosphericPrisonEscape
Feb 26 at 16:25
add a comment |
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8 Answers
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8 Answers
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$begingroup$
Many objects move in circles without needing any energy.
This starts with stuff like geostationary satellites, continues with moons that orbit planets, goes on to planets orbiting stars, stars orbiting other stars or black holes, and stars orbiting their galaxies' center of mass. All of these perform thousands, millions, and billions of rotations without any need for energy.
Of course, these objects all need to have sufficient kinetic energy to be able to orbit in the first place. An object that's stationary with respect to earth, will just fall down. To get it to orbit earth, it must be accelerated first, and that needs energy. And if you want to get it stationary again after it has completed an orbit you have to decelerated it again. However, while the object is orbiting on a circle, its kinetic energy remains the same all the time, no energy needs to be put in or removed.
The trick about saying that moving the object in a circle does not take any energy is that energy for acceleration and deceleration are added. Acceleration increases the kinetic energy of the object, the orbit does not change that kinetic energy, and deceleration removes the same amount of kinetic energy again. If you can perfectly recover the kinetic energy during deceleration, you get out with a black zero. (Of course, if you have losses, as you always do, you loose energy overall.)
However, that's not the entire story. Because, when an electrically charged object goes in circles, it emits electromagnetic radiation. A circling charge induces a magnetic field. That is what happens in any electric motor, including the one that spins your computer's fan right now. The reverse process works as well, a changing magnetic field accelerates charges round in circles. That's the working principle of any electrical generator, including your bike's dynamo.
Even more curiously, what can be said about the electro-magnetic field can be said about the gravitational field as well: Rotating masses emit gravitational waves, which carry away a tiny amount of energy. Usually we can totally ignore this kind of effect, but when really heavy objects (like black holes many times as heavy as our sun) dance around each other really closely, and really fast (like going round ten times per second or more), that tiny bit of energy loss can amount to the equivalent of three times the mass of our sun radiated away as pure gravitational wave energy... (https://en.wikipedia.org/wiki/First_observation_of_gravitational_waves)
TL;DR:
You can move stuff around in circles without needing energy, as long as the object is not electrically charged, and losses to gravitational waves are negligible. Which is pretty much always the case.
However, when electro-magnetic or gravitational effects enter the stage, that's where the real fun starts...
And, at least for the electro-magnetic effects, you personally rely on them every single day.
answered Feb 26 at 21:57
cmastercmaster
6009
$endgroup$
add a comment |
$begingroup$
Many objects move in circles without needing any energy.
This starts with stuff like geostationary satellites, continues with moons that orbit planets, goes on to planets orbiting stars, stars orbiting other stars or black holes, and stars orbiting their galaxies' center of mass. All of these perform thousands, millions, and billions of rotations without any need for energy.
Of course, these objects all need to have sufficient kinetic energy to be able to orbit in the first place. An object that's stationary with respect to earth, will just fall down. To get it to orbit earth, it must be accelerated first, and that needs energy. And if you want to get it stationary again after it has completed an orbit you have to decelerated it again. However, while the object is orbiting on a circle, its kinetic energy remains the same all the time, no energy needs to be put in or removed.
The trick about saying that moving the object in a circle does not take any energy is that energy for acceleration and deceleration are added. Acceleration increases the kinetic energy of the object, the orbit does not change that kinetic energy, and deceleration removes the same amount of kinetic energy again. If you can perfectly recover the kinetic energy during deceleration, you get out with a black zero. (Of course, if you have losses, as you always do, you loose energy overall.)
However, that's not the entire story. Because, when an electrically charged object goes in circles, it emits electromagnetic radiation. A circling charge induces a magnetic field. That is what happens in any electric motor, including the one that spins your computer's fan right now. The reverse process works as well, a changing magnetic field accelerates charges round in circles. That's the working principle of any electrical generator, including your bike's dynamo.
Even more curiously, what can be said about the electro-magnetic field can be said about the gravitational field as well: Rotating masses emit gravitational waves, which carry away a tiny amount of energy. Usually we can totally ignore this kind of effect, but when really heavy objects (like black holes many times as heavy as our sun) dance around each other really closely, and really fast (like going round ten times per second or more), that tiny bit of energy loss can amount to the equivalent of three times the mass of our sun radiated away as pure gravitational wave energy... (https://en.wikipedia.org/wiki/First_observation_of_gravitational_waves)
TL;DR:
You can move stuff around in circles without needing energy, as long as the object is not electrically charged, and losses to gravitational waves are negligible. Which is pretty much always the case.
However, when electro-magnetic or gravitational effects enter the stage, that's where the real fun starts...
And, at least for the electro-magnetic effects, you personally rely on them every single day.
answered Feb 26 at 21:57
cmastercmaster
6009
$endgroup$
add a comment |
$begingroup$
Many objects move in circles without needing any energy.
This starts with stuff like geostationary satellites, continues with moons that orbit planets, goes on to planets orbiting stars, stars orbiting other stars or black holes, and stars orbiting their galaxies' center of mass. All of these perform thousands, millions, and billions of rotations without any need for energy.
Of course, these objects all need to have sufficient kinetic energy to be able to orbit in the first place. An object that's stationary with respect to earth, will just fall down. To get it to orbit earth, it must be accelerated first, and that needs energy. And if you want to get it stationary again after it has completed an orbit you have to decelerated it again. However, while the object is orbiting on a circle, its kinetic energy remains the same all the time, no energy needs to be put in or removed.
The trick about saying that moving the object in a circle does not take any energy is that energy for acceleration and deceleration are added. Acceleration increases the kinetic energy of the object, the orbit does not change that kinetic energy, and deceleration removes the same amount of kinetic energy again. If you can perfectly recover the kinetic energy during deceleration, you get out with a black zero. (Of course, if you have losses, as you always do, you loose energy overall.)
However, that's not the entire story. Because, when an electrically charged object goes in circles, it emits electromagnetic radiation. A circling charge induces a magnetic field. That is what happens in any electric motor, including the one that spins your computer's fan right now. The reverse process works as well, a changing magnetic field accelerates charges round in circles. That's the working principle of any electrical generator, including your bike's dynamo.
Even more curiously, what can be said about the electro-magnetic field can be said about the gravitational field as well: Rotating masses emit gravitational waves, which carry away a tiny amount of energy. Usually we can totally ignore this kind of effect, but when really heavy objects (like black holes many times as heavy as our sun) dance around each other really closely, and really fast (like going round ten times per second or more), that tiny bit of energy loss can amount to the equivalent of three times the mass of our sun radiated away as pure gravitational wave energy... (https://en.wikipedia.org/wiki/First_observation_of_gravitational_waves)
TL;DR:
You can move stuff around in circles without needing energy, as long as the object is not electrically charged, and losses to gravitational waves are negligible. Which is pretty much always the case.
However, when electro-magnetic or gravitational effects enter the stage, that's where the real fun starts...
And, at least for the electro-magnetic effects, you personally rely on them every single day.
answered Feb 26 at 21:57
cmastercmaster
6009
$endgroup$
Many objects move in circles without needing any energy.
This starts with stuff like geostationary satellites, continues with moons that orbit planets, goes on to planets orbiting stars, stars orbiting other stars or black holes, and stars orbiting their galaxies' center of mass. All of these perform thousands, millions, and billions of rotations without any need for energy.
Of course, these objects all need to have sufficient kinetic energy to be able to orbit in the first place. An object that's stationary with respect to earth, will just fall down. To get it to orbit earth, it must be accelerated first, and that needs energy. And if you want to get it stationary again after it has completed an orbit you have to decelerated it again. However, while the object is orbiting on a circle, its kinetic energy remains the same all the time, no energy needs to be put in or removed.
The trick about saying that moving the object in a circle does not take any energy is that energy for acceleration and deceleration are added. Acceleration increases the kinetic energy of the object, the orbit does not change that kinetic energy, and deceleration removes the same amount of kinetic energy again. If you can perfectly recover the kinetic energy during deceleration, you get out with a black zero. (Of course, if you have losses, as you always do, you loose energy overall.)
However, that's not the entire story. Because, when an electrically charged object goes in circles, it emits electromagnetic radiation. A circling charge induces a magnetic field. That is what happens in any electric motor, including the one that spins your computer's fan right now. The reverse process works as well, a changing magnetic field accelerates charges round in circles. That's the working principle of any electrical generator, including your bike's dynamo.
Even more curiously, what can be said about the electro-magnetic field can be said about the gravitational field as well: Rotating masses emit gravitational waves, which carry away a tiny amount of energy. Usually we can totally ignore this kind of effect, but when really heavy objects (like black holes many times as heavy as our sun) dance around each other really closely, and really fast (like going round ten times per second or more), that tiny bit of energy loss can amount to the equivalent of three times the mass of our sun radiated away as pure gravitational wave energy... (https://en.wikipedia.org/wiki/First_observation_of_gravitational_waves)
TL;DR:
You can move stuff around in circles without needing energy, as long as the object is not electrically charged, and losses to gravitational waves are negligible. Which is pretty much always the case.
However, when electro-magnetic or gravitational effects enter the stage, that's where the real fun starts...
And, at least for the electro-magnetic effects, you personally rely on them every single day.
answered Feb 26 at 21:57
cmastercmaster
6009
edited Feb 26 at 22:07
answered Feb 26 at 21:57
cmastercmaster
6009
answered Feb 26 at 21:57
cmastercmaster
6009
answered Feb 26 at 21:57
cmastercmaster
6009
6009
add a comment |
add a comment |
$begingroup$
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.
answered Feb 25 at 2:44
Aaron StevensAaron Stevens
13.5k42250
$endgroup$
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– ACuriousMind♦
Feb 26 at 19:38
add a comment |
$begingroup$
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.
answered Feb 25 at 2:44
Aaron StevensAaron Stevens
13.5k42250
$endgroup$
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– ACuriousMind♦
Feb 26 at 19:38
add a comment |
$begingroup$
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.
answered Feb 25 at 2:44
Aaron StevensAaron Stevens
13.5k42250
$endgroup$
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 Feb 25 at 2:44
Aaron StevensAaron Stevens
13.5k42250
answered Feb 25 at 2:44
Aaron StevensAaron Stevens
13.5k42250
answered Feb 25 at 2:44
Aaron StevensAaron Stevens
13.5k42250
answered Feb 25 at 2:44
Aaron StevensAaron Stevens
13.5k42250
13.5k42250
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– ACuriousMind♦
Feb 26 at 19:38
add a comment |
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– ACuriousMind♦
Feb 26 at 19:38
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– ACuriousMind♦
Feb 26 at 19:38
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– ACuriousMind♦
Feb 26 at 19:38
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
$$E = int_a^bvecFcdot dvecell$$
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.
answered Feb 25 at 2:28
The PhotonThe Photon
9,68711833
$endgroup$
$begingroup$
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
Feb 25 at 2:33
2
$begingroup$
@ProQ, not if the robot had magical frictionless joints.
$endgroup$
– The Photon
Feb 25 at 2:39
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
Feb 25 at 2:41
4
$begingroup$
@ThePhoton I don't think friction is the only place energy will go into.
$endgroup$
– Aaron Stevens
Feb 25 at 3:14
5
$begingroup$
@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
Feb 25 at 14:28
|
show 1 more 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
$$E = int_a^bvecFcdot dvecell$$
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.
answered Feb 25 at 2:28
The PhotonThe Photon
9,68711833
$endgroup$
$begingroup$
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
Feb 25 at 2:33
2
$begingroup$
@ProQ, not if the robot had magical frictionless joints.
$endgroup$
– The Photon
Feb 25 at 2:39
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
Feb 25 at 2:41
4
$begingroup$
@ThePhoton I don't think friction is the only place energy will go into.
$endgroup$
– Aaron Stevens
Feb 25 at 3:14
5
$begingroup$
@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
Feb 25 at 14:28
|
show 1 more 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
$$E = int_a^bvecFcdot dvecell$$
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.
answered Feb 25 at 2:28
The PhotonThe Photon
9,68711833
$endgroup$
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
$$E = int_a^bvecFcdot dvecell$$
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.
answered Feb 25 at 2:28
The PhotonThe Photon
9,68711833
edited Feb 25 at 2:38
answered Feb 25 at 2:28
The PhotonThe Photon
9,68711833
answered Feb 25 at 2:28
The PhotonThe Photon
9,68711833
answered Feb 25 at 2:28
The PhotonThe Photon
9,68711833
9,68711833
$begingroup$
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
Feb 25 at 2:33
2
$begingroup$
@ProQ, not if the robot had magical frictionless joints.
$endgroup$
– The Photon
Feb 25 at 2:39
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
Feb 25 at 2:41
4
$begingroup$
@ThePhoton I don't think friction is the only place energy will go into.
$endgroup$
– Aaron Stevens
Feb 25 at 3:14
5
$begingroup$
@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
Feb 25 at 14:28
|
show 1 more 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
Feb 25 at 2:33
2
$begingroup$
@ProQ, not if the robot had magical frictionless joints.
$endgroup$
– The Photon
Feb 25 at 2:39
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
Feb 25 at 2:41
4
$begingroup$
@ThePhoton I don't think friction is the only place energy will go into.
$endgroup$
– Aaron Stevens
Feb 25 at 3:14
5
$begingroup$
@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
Feb 25 at 14:28
$begingroup$
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
Feb 25 at 2:33
$begingroup$
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
Feb 25 at 2:33
2
2
$begingroup$
@ProQ, not if the robot had magical frictionless joints.
$endgroup$
– The Photon
Feb 25 at 2:39
$begingroup$
@ProQ, not if the robot had magical frictionless joints.
$endgroup$
– The Photon
Feb 25 at 2:39
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
Feb 25 at 2:41
$begingroup$
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
Feb 25 at 2:41
4
4
$begingroup$
@ThePhoton I don't think friction is the only place energy will go into.
$endgroup$
– Aaron Stevens
Feb 25 at 3:14
$begingroup$
@ThePhoton I don't think friction is the only place energy will go into.
$endgroup$
– Aaron Stevens
Feb 25 at 3:14
5
5
$begingroup$
@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
Feb 25 at 14:28
$begingroup$
@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
Feb 25 at 14:28
|
show 1 more comment
$begingroup$
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.
answered Feb 25 at 10:57
gandalf61gandalf61
41628
$endgroup$
add a comment |
$begingroup$
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.
answered Feb 25 at 10:57
gandalf61gandalf61
41628
$endgroup$
add a comment |
$begingroup$
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.
answered Feb 25 at 10:57
gandalf61gandalf61
41628
$endgroup$
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.
answered Feb 25 at 10:57
gandalf61gandalf61
41628
answered Feb 25 at 10:57
gandalf61gandalf61
41628
answered Feb 25 at 10:57
gandalf61gandalf61
41628
answered Feb 25 at 10:57
gandalf61gandalf61
41628
41628
add a comment |
add a comment |
$begingroup$
NO, an ideal robot can move the load in a circle (actually in any closed curve, not just a circle), producing zero total work done and hence no loss of 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 loss. A simple electric engine can do this, but obviously not without loss.
- Move the load without friction (well, you have stated in the question that your robot can do this, but even spacecraft do hit atoms on their way through the vacuum).
The robot will need energy in general, especially if the circle stands vertically like a Ferris wheel. Only an ideal robot would 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.
edited Feb 26 at 13:29
Captain Lepton
1033
answered Feb 25 at 20:04
h22h22
25328
$endgroup$
add a comment |
$begingroup$
NO, an ideal robot can move the load in a circle (actually in any closed curve, not just a circle), producing zero total work done and hence no loss of 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 loss. A simple electric engine can do this, but obviously not without loss.
- Move the load without friction (well, you have stated in the question that your robot can do this, but even spacecraft do hit atoms on their way through the vacuum).
The robot will need energy in general, especially if the circle stands vertically like a Ferris wheel. Only an ideal robot would 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.
edited Feb 26 at 13:29
Captain Lepton
1033
answered Feb 25 at 20:04
h22h22
25328
$endgroup$
add a comment |
$begingroup$
NO, an ideal robot can move the load in a circle (actually in any closed curve, not just a circle), producing zero total work done and hence no loss of 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 loss. A simple electric engine can do this, but obviously not without loss.
- Move the load without friction (well, you have stated in the question that your robot can do this, but even spacecraft do hit atoms on their way through the vacuum).
The robot will need energy in general, especially if the circle stands vertically like a Ferris wheel. Only an ideal robot would 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.
edited Feb 26 at 13:29
Captain Lepton
1033
answered Feb 25 at 20:04
h22h22
25328
$endgroup$
NO, an ideal robot can move the load in a circle (actually in any closed curve, not just a circle), producing zero total work done and hence no loss of 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 loss. A simple electric engine can do this, but obviously not without loss.
- Move the load without friction (well, you have stated in the question that your robot can do this, but even spacecraft do hit atoms on their way through the vacuum).
The robot will need energy in general, especially if the circle stands vertically like a Ferris wheel. Only an ideal robot would 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.
edited Feb 26 at 13:29
Captain Lepton
1033
answered Feb 25 at 20:04
h22h22
25328
edited Feb 26 at 13:29
Captain Lepton
1033
edited Feb 26 at 13:29
Captain Lepton
1033
edited Feb 26 at 13:29
Captain Lepton
1033
1033
answered Feb 25 at 20:04
h22h22
25328
answered Feb 25 at 20:04
h22h22
25328
answered Feb 25 at 20:04
h22h22
25328
25328
add a comment |
add a comment |
$begingroup$
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.
answered Feb 25 at 17:32
Chris PercivalChris Percival
411
$endgroup$
$begingroup$
Consider using MathJax for writing equations.
$endgroup$
– exp ikx
Feb 25 at 18:18
add a comment |
$begingroup$
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.
answered Feb 25 at 17:32
Chris PercivalChris Percival
411
$endgroup$
$begingroup$
Consider using MathJax for writing equations.
$endgroup$
– exp ikx
Feb 25 at 18:18
add a comment |
$begingroup$
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.
answered Feb 25 at 17:32
Chris PercivalChris Percival
411
$endgroup$
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.
answered Feb 25 at 17:32
Chris PercivalChris Percival
411
answered Feb 25 at 17:32
Chris PercivalChris Percival
411
answered Feb 25 at 17:32
Chris PercivalChris Percival
411
answered Feb 25 at 17:32
Chris PercivalChris Percival
411
411
$begingroup$
Consider using MathJax for writing equations.
$endgroup$
– exp ikx
Feb 25 at 18:18
add a comment |
$begingroup$
Consider using MathJax for writing equations.
$endgroup$
– exp ikx
Feb 25 at 18:18
$begingroup$
Consider using MathJax for writing equations.
$endgroup$
– exp ikx
Feb 25 at 18:18
$begingroup$
Consider using MathJax for writing equations.
$endgroup$
– exp ikx
Feb 25 at 18:18
add a comment |
$begingroup$
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.
answered Feb 25 at 18:17
AcccumulationAcccumulation
2,696312
$endgroup$
add a comment |
$begingroup$
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.
answered Feb 25 at 18:17
AcccumulationAcccumulation
2,696312
$endgroup$
add a comment |
$begingroup$
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.
answered Feb 25 at 18:17
AcccumulationAcccumulation
2,696312
$endgroup$
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.
answered Feb 25 at 18:17
AcccumulationAcccumulation
2,696312
answered Feb 25 at 18:17
AcccumulationAcccumulation
2,696312
answered Feb 25 at 18:17
AcccumulationAcccumulation
2,696312
answered Feb 25 at 18:17
AcccumulationAcccumulation
2,696312
2,696312
add a comment |
add a comment |
$begingroup$
Interesting question and a simple answer here. Newtonian first law claims, that:
$$fractext d mathbf vtext d t = 0$$ 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:
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.
answered Feb 26 at 11:12
Agnius VasiliauskasAgnius Vasiliauskas
1264
$endgroup$
1
$begingroup$
Well, that's interesting. So who's spending the energy in keeping the planets in orbit? Angels pushing on spheres?
$endgroup$
– AtmosphericPrisonEscape
Feb 26 at 11:22
$begingroup$
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 inconstant falltowards 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
Feb 26 at 11:31
$begingroup$
You're mixing center-of-mass system and co-rotating system. Your argumentation is plain wrong. Compute $oint vec F cdot dvecl $ for the gravitational force in the CMS frame for a perfectly circular orbit. You will see that, because of the free-fall you've quoted but misunderstood, no work is done onto the body. For eccentric orbits the net work is still zero, because gravity is a conservative force field.
$endgroup$
– AtmosphericPrisonEscape
Feb 26 at 16:25
add a comment |
$begingroup$
Interesting question and a simple answer here. Newtonian first law claims, that:
$$fractext d mathbf vtext d t = 0$$ 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:
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.
answered Feb 26 at 11:12
Agnius VasiliauskasAgnius Vasiliauskas
1264
$endgroup$
1
$begingroup$
Well, that's interesting. So who's spending the energy in keeping the planets in orbit? Angels pushing on spheres?
$endgroup$
– AtmosphericPrisonEscape
Feb 26 at 11:22
$begingroup$
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 inconstant falltowards 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
Feb 26 at 11:31
$begingroup$
You're mixing center-of-mass system and co-rotating system. Your argumentation is plain wrong. Compute $oint vec F cdot dvecl $ for the gravitational force in the CMS frame for a perfectly circular orbit. You will see that, because of the free-fall you've quoted but misunderstood, no work is done onto the body. For eccentric orbits the net work is still zero, because gravity is a conservative force field.
$endgroup$
– AtmosphericPrisonEscape
Feb 26 at 16:25
add a comment |
$begingroup$
Interesting question and a simple answer here. Newtonian first law claims, that:
$$fractext d mathbf vtext d t = 0$$ 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:
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.
answered Feb 26 at 11:12
Agnius VasiliauskasAgnius Vasiliauskas
1264
$endgroup$
Interesting question and a simple answer here. Newtonian first law claims, that:
$$fractext d mathbf vtext d t = 0$$ 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:
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.
answered Feb 26 at 11:12
Agnius VasiliauskasAgnius Vasiliauskas
1264
answered Feb 26 at 11:12
Agnius VasiliauskasAgnius Vasiliauskas
1264
answered Feb 26 at 11:12
Agnius VasiliauskasAgnius Vasiliauskas
1264
answered Feb 26 at 11:12
Agnius VasiliauskasAgnius Vasiliauskas
1264
1264
1
$begingroup$
Well, that's interesting. So who's spending the energy in keeping the planets in orbit? Angels pushing on spheres?
$endgroup$
– AtmosphericPrisonEscape
Feb 26 at 11:22
$begingroup$
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 inconstant falltowards 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
Feb 26 at 11:31
$begingroup$
You're mixing center-of-mass system and co-rotating system. Your argumentation is plain wrong. Compute $oint vec F cdot dvecl $ for the gravitational force in the CMS frame for a perfectly circular orbit. You will see that, because of the free-fall you've quoted but misunderstood, no work is done onto the body. For eccentric orbits the net work is still zero, because gravity is a conservative force field.
$endgroup$
– AtmosphericPrisonEscape
Feb 26 at 16:25
add a comment |
1
$begingroup$
Well, that's interesting. So who's spending the energy in keeping the planets in orbit? Angels pushing on spheres?
$endgroup$
– AtmosphericPrisonEscape
Feb 26 at 11:22
$begingroup$
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 inconstant falltowards 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
Feb 26 at 11:31
$begingroup$
You're mixing center-of-mass system and co-rotating system. Your argumentation is plain wrong. Compute $oint vec F cdot dvecl $ for the gravitational force in the CMS frame for a perfectly circular orbit. You will see that, because of the free-fall you've quoted but misunderstood, no work is done onto the body. For eccentric orbits the net work is still zero, because gravity is a conservative force field.
$endgroup$
– AtmosphericPrisonEscape
Feb 26 at 16:25
1
1
$begingroup$
Well, that's interesting. So who's spending the energy in keeping the planets in orbit? Angels pushing on spheres?
$endgroup$
– AtmosphericPrisonEscape
Feb 26 at 11:22
$begingroup$
Well, that's interesting. So who's spending the energy in keeping the planets in orbit? Angels pushing on spheres?
$endgroup$
– AtmosphericPrisonEscape
Feb 26 at 11:22
$begingroup$
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
Feb 26 at 11:31
$begingroup$
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
Feb 26 at 11:31
$begingroup$
You're mixing center-of-mass system and co-rotating system. Your argumentation is plain wrong. Compute $oint vec F cdot dvecl $ for the gravitational force in the CMS frame for a perfectly circular orbit. You will see that, because of the free-fall you've quoted but misunderstood, no work is done onto the body. For eccentric orbits the net work is still zero, because gravity is a conservative force field.
$endgroup$
– AtmosphericPrisonEscape
Feb 26 at 16:25
$begingroup$
You're mixing center-of-mass system and co-rotating system. Your argumentation is plain wrong. Compute $oint vec F cdot dvecl $ for the gravitational force in the CMS frame for a perfectly circular orbit. You will see that, because of the free-fall you've quoted but misunderstood, no work is done onto the body. For eccentric orbits the net work is still zero, because gravity is a conservative force field.
$endgroup$
– AtmosphericPrisonEscape
Feb 26 at 16:25
add a comment |
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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
Feb 26 at 3:15
$begingroup$
A better example of this is to consider a puck sliding on a surface on without any drag or other losses, attached to a massless string that goes through a hole in the surface. Assume the puck is moving in a circle, then a downwards force on the string below the hole equals the tension in the string above the hole which is equal to the centripetal force exerted by the string onto the puck. Since there's no vertical motion of the string below the hole, then the work done is force x zero distance = zero.
$endgroup$
– rcgldr
Feb 26 at 16:13
$begingroup$
Continuing, now consider what happens if the string is pulled downwards, the puck's path is no longer a circle and a component of tension in the string will be in the direction of the puck's path, increasing the speed and mechanical energy of the puck (angular momentum will be conserved). If the string is allowed to move upwards, then "negative" work is done, as energy is extracted from the puck, and the puck's speed and energy decrease.
$endgroup$
– rcgldr
Feb 26 at 16:16
$begingroup$
You get the same circular motion if you replace the robot arm with a (well-lubricated) rod that rotates around a pole. Does this setup need a battery?
$endgroup$
– Rahul
Feb 26 at 22:14
$begingroup$
Sorry, I thought the question was specifically about the robot.
$endgroup$
– Aaron Stevens
Feb 27 at 10:08