OK, I accept that an object moving in a circle (a bike on a track, for instance) has a force acting on it to cause the circular motion but that force does no work because it is perpendicular to the direction of motion and work is defined as force through a distance in the direction of motion.
Now, take this scenario. A space ship in deep space traveling in a straight line. For some reason the astronaut’s want to move in a circle. In this instance, they would have to fire a rocket perpendicular to their direction of motion. The stronger the rocket the tighter the circle that would be followed. As soon as the rocket is turned off the space ship would resume moving in a straight line, just a different straight line. Now, no “work” is being done yet substantial energy must be expended in turning the space ship, and the tighter the circle the faster the energy is needed it would seem. How much energy would it take to change the direction of travel 180º?
You’ve got to be F’ing kidding me with this one. No work is done? where the hell did you come up with that? Explain to me how no work is done? You’re thinking (as it usually is) is in a single warped dimension. You forgot about the second and third dimensions. Picture an X and Y coordinate system. The rocket is traveling only along the X axis. Now, the rocket begins turning when its rocket is fired. The Y component of its velocity which was originally 0 is now growing as the rocket accelerates it in that direction. The tangential velocity stays constant though, since no component of force is acting tangentially, but the angular velocity is increasing, thus there is angular acceleration, and thus there must be work done. Work is not soleley defined as W = F x d. This is all assuming that the rocket is placed such that when it fires it will remain constantly pointing perpendicularly to tangential travel.
When traveling in a circle the radius of curvature remains constant
for your space ship example the radius of curvature wouldn’t be constant
also, the center of curvature wouldn’t be constant either
Both the radius of curvature and center or curvature would be constant, as long as the force of the rocket were constant and it were always directed towards perpendicular to the direction of motion the rocket would move in a circle of a specified diameter and center. If you want to presume that the mass of the system changes as the rocket burns fuel, so be it, then let’s change the force of the rocket as the mass “disappears” to ensure the rocket ship follows a circle path. The question is still the same, how are these two concepts, which are similar yet different, reconciled?
You’ve got to be F’ing kidding me with this one. No work is done? where the hell did you come up with that? Explain to me how no work is done? You’re thinking (as it usually is) is in a single warped dimension. You forgot about the second and third dimensions. Picture an X and Y coordinate system. The rocket is traveling only along the X axis. Now, the rocket begins turning when its rocket is fired. The Y component of its velocity which was originally 0 is now growing as the rocket accelerates it in that direction. The tangential velocity stays constant though, since no component of force is acting tangentially, but the angular velocity is increasing, thus there is angular acceleration, and thus there must be work done. Work is not soleley defined as W = F x d. This is all assuming that the rocket is placed such that when it fires it will remain constantly pointing perpendicularly to tangential travel.
Hey, I have been repeatedly lambasted here for thinking that it took any energy to turn a bicycle on the track. In fact, I have been told and shown it takes so little energy that the bicycle actually speeds up during the turn. I am simply asking the question as to why it takes no energy in that situation yet, seemingly plenty of energy in this other situation I have described. Centripetal force is centripetal force isn’t it? How are these two situations reconciled. If it does take energy to turn a bicycle in a turn, how much does it take?
You’ve got to be F’ing kidding me with this one. No work is done? where the hell did you come up with that? Explain to me how no work is done? You’re thinking (as it usually is) is in a single warped dimension. You forgot about the second and third dimensions. Picture an X and Y coordinate system. The rocket is traveling only along the X axis. Now, the rocket begins turning when its rocket is fired. The Y component of its velocity which was originally 0 is now growing as the rocket accelerates it in that direction. The tangential velocity stays constant though, since no component of force is acting tangentially, but the angular velocity is increasing, thus there is angular acceleration, and thus there must be work done. Work is not soleley defined as W = F x d. This is all assuming that the rocket is placed such that when it fires it will remain constantly pointing perpendicularly to tangential travel.
He’s right. Work is being performed. Centripetal force is the force acting along the radius of rotation (normal to the tangential velocity) pointing towards the center of rotation. It’s the force keeping the rocket along it’s circular obit. You are applying a force over a distance. Once you start looking at rotation though you have to adjust your reference frame. tri2doitall is right in looking at in 2 dimensions…As the rocket starts to turn, it is speeding up (a>0) in the y direction and slowing down in the x direction (a<0). F=ma no matter what frame of reference you are in, so to have acceleration, you need force. This force is being applied over the entire distance the rocket is accelerating (a != 0). Even if your angular velocity is constant, you are constantly accelerating in both the x and y directions.
Francois in Montreal http://www.glenbrook.k12.il.us/...s/circles/u6l1c.html
Thus, the work done by the centripetal force in the case of uniform circular motion is 0 Joules. Recall also from Unit 5 of The Physics Classroom that when no work is done upon an object by external forces, the total mechanical energy (potential energy plus kinetic energy) of the object remains constant. So if an object is moving in a horizontal circle at constant speed, the centripetal force does not do work and cannot alter the total mechanical energy of the object. For this reason, the kinetic energy and therefore, the speed of the object will remain constant. The force can indeed accelerate the object - by changing its direction - but it cannot change its speed. In fact, whenever the unbalanced centripetal force acts perpendicular to the direction of motion, the speed of the object will remain constant.
You’ve got to be F’ing kidding me with this one. No work is done? where the hell did you come up with that? Explain to me how no work is done? You’re thinking (as it usually is) is in a single warped dimension. You forgot about the second and third dimensions. Picture an X and Y coordinate system. The rocket is traveling only along the X axis. Now, the rocket begins turning when its rocket is fired. The Y component of its velocity which was originally 0 is now growing as the rocket accelerates it in that direction. The tangential velocity stays constant though, since no component of force is acting tangentially, but the angular velocity is increasing, thus there is angular acceleration, and thus there must be work done. Work is not soleley defined as W = F x d. This is all assuming that the rocket is placed such that when it fires it will remain constantly pointing perpendicularly to tangential travel.
He’s right. Work is being performed. Centripetal force is the force acting along the radius of rotation (normal to the tangential velocity) pointing towards the center of rotation. It’s the force keeping the rocket along it’s circular obit. You are applying a force over a distance. Once you start looking at rotation though you have to adjust your reference frame. tri2doitall is right in looking at in 2 dimensions…As the rocket starts to turn, it is speeding up (a>0) in the y direction and slowing down in the x direction (a<0). F=ma no matter what frame of reference you are in, so to have acceleration, you need force. This force is being applied over the entire distance the rocket is accelerating (a != 0). Even if your angular velocity is constant, you are constantly accelerating in both the x and y directions.
So, how much work is required to turn a bicycle in a semi-circle (like on a track)? How much would you expect that to slow the bicycle down if the power remained constant or how much extra power would it take to keep the bicycle speed constant?
So you are telling me the rocket went from traveling in a straight line to INSTANTANEOUSLY traveling in a circle with a fixed pivot?
I would like to see that happen
Why wouldn’t it if the direction of the force were directed perpendicular to the direction of motion? Seems there might be a small “real world” problem here getting the rocket to get just the right angular momentum to make it “easy” to ensure the force of the rocket is always directed perpendicularly to the direction of travel, but I am sure that is solvable.
Sorry Frank, I was wrong. No work is done by a centripetal force. You know I
it’s time to go home when you start spreading errors about basic physics on a public forum…
Sorry Frank, I was wrong. No work is done by a centripetal force. You know I
it’s time to go home when you start spreading errors about basic physics on a public forum…
Francois in Montreal
Here is the issue I am trying to get my head around. I am sure it is quite simple if I could only see it properly but I am having trouble doing so.
I have no trouble with a spinning disk where there are always two particles of equal mass on opposite sides of the disk chaning momentum in the opposite directions. Momentum is always conserved in that scenario and the disk should spin forever unless an outside force intervenes.
But, that is not the case for a single particle, say a bicycle or a spaceship. There the bicycle or spaceship will tend to go in a straight line unless an outside force acts on it. Since there is no counterbalancing bicycle or spaceship to keep the total momentum of the system constant when the bicycle or spaceship turns it takes energy to turn the bicycle or spaceship even if the total kinetic energy of the bicycle or spaceship remains constant. How do we calculate the energy required when the direction the force must be applied to turn the bicycle is a centripetal force? If we can calculate the energy requirement and if we know the time over which it is applied then we should be able to know the power requirement for such a turn. Or, am I missing something?
You’re mixing and flip flopping between an example where energy is conserved (rocket in space) and one where energy is not conserved (bike on a track). In the space example, yes, no work is done since work is independent of path if energy is conserved (if the spaceship goes all the way around the circle). For the bike on the track, energy is not conserved because there are friction forces which convert the energy into other forms and are lost to the system of the bike. Energy is generally converted into either heat or sound energy: tires, hubs and bottom brackets ‘warm up’, freewheels go ‘clack-clack-clack’, etc. Wind drag will slow you down exponentially versus speed. This is what accounts for the need to keep pedaling whether you’re in a turn or not. In a frictionless void-of-deep-space environment, we could all just pedal until we’re at whatever speed we want to be and then coast ad infinitum.
Oh, the track structure and friction between the track and tire provide the centripetal force to move around the circle.
Stop Nukin’ this stuff, just go ride your bike.
It was fun, however, to revisit some stuff from my undergrad days… been a while since me and Halliday/Resnick’s Fundamentals of Physics 4th Edition had a chat.
OK, I accept that an object moving in a circle (a bike on a track, for instance) has a force acting on it to cause the circular motion but that force does no work because it is perpendicular to the direction of motion and work is defined as force through a distance in the direction of motion.
Now, take this scenario. A space ship in deep space traveling in a straight line. For some reason the astronaut’s want to move in a circle. In this instance, they would have to fire a rocket perpendicular to their direction of motion. The stronger the rocket the tighter the circle that would be followed. As soon as the rocket is turned off the space ship would resume moving in a straight line, just a different straight line. Now, no “work” is being done yet substantial energy must be expended in turning the space ship, and the tighter the circle the faster the energy is needed it would seem. How much energy would it take to change the direction of travel 180º?
How are these two concepts reconciled?
OK, I went to the physics forum and I think I understand this now and can reconcile this “problem”. It is not that anyone gave me the answer (so I might be wrong) but I was able to talk my way through the problem there. Here is what I came up with.
The amount of energy loss in turning an object depends upon the amount of work involved in providing the necessary centripetal force. Work equal to force x distance. In a “perfect” world, using rigid bodies, force does not cause distortion so the “distance” moved in providing the force is zero so the work involved is zero and the energy loss would be zero. But, in the real word all objects react to applied force with some distortion, some more than others, so there would be some distance moved and some work done. So, the amount of work required to turn an object depends entirely on the environment and cannot be calculated without knowing the specifics of the environment. The “stiffer” the better.
In general I would rank the energy required to turn an object of any given mass from lowest energy to highest energy based upon the “rigidity” of the environment thus:
rigid body in completely rigid system (zero energy required, does not exist in real world)
Steel wheels on steel tracks (typical train)
rubber wheels on asphault (typical car, bicycle)
a water environment (typical boat)
an air environment (typical airplane)
“empty” environment (spaceship, particle in a cyclotron)
Does anyone have any correcting thoughts on this? If I have it wrong I would like to know it so I can (eventually) be thinking about this issue “correctly”.
I was thinking about this out on my bike ride today and think I made another conceptual error so i am going to revise my list. However, I think the general principle is correct. What I got wrong, I believe, is the particle in the cyclotron should have the fewest losses of those that I listed, not the most. The reason is the force is coming from magnetic fields that are coming from a very massive and rigid mechanism. Further, it was pointed out on the physics forum that an earth satellite is probably the lowest of all and as I think about it a satellite in geosynchronous orbit would be the lowest of all.
So, my revised rank order of energy cost of turning is:
satellite in geosynchonous orbit (probably no work required as no tidal forces generated, etc. compared to normal satellites).
Sorry Frank, I was wrong. No work is done by a centripetal force. You know I
it’s time to go home when you start spreading errors about basic physics on a public forum…
Francois in Montreal
Here is the issue I am trying to get my head around. I am sure it is quite simple if I could only see it properly but I am having trouble doing so.
I have no trouble with a spinning disk where there are always two particles of equal mass on opposite sides of the disk chaning momentum in the opposite directions. Momentum is always conserved in that scenario and the disk should spin forever unless an outside force intervenes.
But, that is not the case for a single particle, say a bicycle or a spaceship. There the bicycle or spaceship will tend to go in a straight line unless an outside force acts on it. Since there is no counterbalancing bicycle or spaceship to keep the total momentum of the system constant when the bicycle or spaceship turns it takes energy to turn the bicycle or spaceship even if the total kinetic energy of the bicycle or spaceship remains constant. How do we calculate the energy required when the direction the force must be applied to turn the bicycle is a centripetal force? If we can calculate the energy requirement and if we know the time over which it is applied then we should be able to know the power requirement for such a turn. Or, am I missing something?
I don’t know if this pertains or is accurate, but lets say you are traveling in space in a straight line and you have a sudden burst of perpendicular force. Say you have
Vx = 10m/s
Vy = 0m/s
V = sqrt(100)
straight line travel, then you have a burst of perpendicular force your values would be
Vx = 10m/s
Vy = 1m/s (or whatever)
Your final velocity is actually larger than your initial. V = sqrt (101)
The theory revolving around a cyclotron are in a COMPLETE vacuum. As this is the case, once the particle is up to desired speed/rotation of particle (these should not be confused with each other) no work is done. The reason that no work is done, is because W=Fs (Work, F=Force, s=distance). F=ma (m=mass, a=acceleration). Once desired speed is attained, notice that there is NO acceleration. No acceleration means no force. No force means that no work can be done because for work to occur, force must be applied over a distance. In the case of purely centripetal motion (no motion along tangential lines), there is no work being done because although the object may be moving through different spherical coordinates (ranging from 0 to 2Pi), it is not truly moving a distance.
However if a wheel for instance is placed on a surface, work will be done because a distance will be traveled. One thing that must be taken into account when looking at objects undergoing rotational motion of any sort, is their centre of mass. The reason for this, is although when looking strictly at work centre of mass isn’t an issue, it becomes an issue when looking at energy. E=W (E=energy). Also E=PE+KE (PE=Potential energy + Kinetic energy, and yes I am only including these 2 to keep it simple). Lets imagine, that all the E we have is KE, so we can therefore say that for objects involving spherical objects E=(mV^2)/2 + (Iw^2/2) (m=mass, V=velocity, ^=to the power of, I=centre of mass which is different depending on the object, w=V but for circular motion).
So since W=E, we can say that sma=(mV^2)/2 + (Iw^2)/2. It should be noted that friction is not included in these, however if it was, friction is called non-conservative work, so it can be easily added to the equation to see how much work/energy is used/needed to perform a certain amount of work.
