DrP Posted May 7, 2015 Posted May 7, 2015 When you fall off the edge of the roundabout you fly off at a tangent as discussed. As the roundabout rotates a further 90 degrees it now looks like you are just flying off of it radially outward from the centre, but you are not, you are just going in a straight line still on what was the tangent a quarter spin ago and your frame of reference has moved... maybe that is where the confusion lies for some people. I really don't want to get into it as it has all been said before, I am just trying to see why some are finding it very hard to visualise that their flying off seemingly straight out from the centre is actually them flying off at a tangent. I Hope this helps... if not then just ignore it. ;-) x 1
Robittybob1 Posted May 7, 2015 Posted May 7, 2015 (edited) Keeping up with the discussion is such a burden… You spun it. How did it tip radially outward, in the frame of the table on which you placed the device? Radially outward means no change in angle, i.e. the motion is purely radial. But if you are spinning it, there a change in angle. "Increasing in radius" ≠ "radially outward motion" Screen Shot 2015-05-07 at 7.16.35 AM.png The object increases in its distance from center because of its linear motion, as I showed in my previous drawing. Your diagram shows motion radially out for two situations one with no rotation and the other in a rotating system (with some restraint (like Mike's radial tubes)). A third situation would be like a rocket might be thought as being launched at a radial to the Earth (from the perspective of those standing under it) but from a distance there is also the motion reflecting the Earth's rotation. Edited May 7, 2015 by Robittybob1
swansont Posted May 7, 2015 Posted May 7, 2015 Your diagram shows motion radially out for two situations one with no rotation and the other in a rotating system (with some restraint (like Mike's radial tubes)). A third situation would be like a rocket might be thought as being launched at a radial to the Earth (from the perspective of those standing under it) but from a distance there is also the motion reflecting the Earth's rotation. This is a tad frustrating. What is meant by "radially out" is shown in the diagram and explained in the post. Is there something unclear about that? Do you require clarification?
Robittybob1 Posted May 8, 2015 Posted May 8, 2015 This is a tad frustrating. What is meant by "radially out" is shown in the diagram and explained in the post. Is there something unclear about that? Do you require clarification? It is a bit like what you get if you follow the point of a wheel rolling along the ground. Even though it is going around the point follows a series of humps in the horizontal direction. As traced by the blue curve in the attached YT video.
Mike Smith Cosmos Posted May 8, 2015 Author Posted May 8, 2015 (edited) If not a force radially out , a pull from an effectively flying object. Flying because it is trying to go on its way , linearly due to inertia, ( newtons laws) , it's preferred direction . This would be tangentially ( only if there was zero restraining centripetal force , but being pulled up to a higher orbit if there was some centripetal force ( eg gravity or a spring , or a gripping force by a person ) . Having reached the higher orbit, or circular trajectory , This will surely either stabilise, or continue a similar process , if an angular speed is increased , again repeating the process described in the previous sentence . This is what I was attempting to show in my rotating tubes diagram . The centripetal force here can be anything ( eg gravity or a spring , or a gripping force by a person ) . Note here, that at no time does the ascending object reach a conventional tangential straight line grazing a circle . See following model :- First ( near centre ). Second ( crossing border, whatever that is ) . Third (way , away , leaving the locality) ....... There is a danger here that we are arguing about semantics ( word ways of describing something ) rather than whether there is some " Whatever ? , Mechanism " , which can and does ( pull, push , whatever a mass upward into increasing hight / distant orbit!) which is the brief of the original question OP From what has been discussed mostly here, it is the INERTIA which appears to be responsible for the ' Lift '.? Is that not so? Mike Edited May 8, 2015 by Mike Smith Cosmos
swansont Posted May 8, 2015 Posted May 8, 2015 It is a bit like what you get if you follow the point of a wheel rolling along the ground. Even though it is going around the point follows a series of humps in the horizontal direction. As traced by the blue curve in the attached YT video. I hope you realize that this does not answer my question, and is a non-sequitur.
Robittybob1 Posted May 8, 2015 Posted May 8, 2015 I hope you realize that this does not answer my question, and is a non-sequitur. I was hoping to find a video of a radially outward motion from a rotating disc, but I didn't find it, sorry.
swansont Posted May 8, 2015 Posted May 8, 2015 I was hoping to find a video of a radially outward motion from a rotating disc, but I didn't find it, sorry. Because the motion is not radially outward. But you still haven't actually answered the question.
Robittybob1 Posted May 8, 2015 Posted May 8, 2015 Because the motion is not radially outward. But you still haven't actually answered the question. You said something like ""Increasing in radius" ≠ "radially outward motion"". There was "no change in angle" so are you saying you can't get radial outward motion from a rotating platform? I thought you could as long as the original tangential motion and outward motion simultaneously satisfied both situations where the mass has tangential velocity and the radius always aligns to the object.
swansont Posted May 8, 2015 Posted May 8, 2015 You said something like ""Increasing in radius" ≠ "radially outward motion"". There was "no change in angle" so are you saying you can't get radial outward motion from a rotating platform? No from a free release, which is what we're discussing. If there is a tangential velocity, then the motion is not purely radial.
Robittybob1 Posted May 8, 2015 Posted May 8, 2015 No from a free release, which is what we're discussing. If there is a tangential velocity, then the motion is not purely radial. Right. If it it isn't purely radial or tangential can it be a mix? Like simultaneously satisfying radial and tangential at the same time. In fact no YT physics lecture discusses "radial motion" otherwise I would have a better idea of the boundaries radial motion could have.
swansont Posted May 8, 2015 Posted May 8, 2015 Right. If it it isn't purely radial or tangential can it be a mix? Like simultaneously satisfying radial and tangential at the same time. In fact no YT physics lecture discusses "radial motion" otherwise I would have a better idea of the boundaries radial motion could have. Yes it can be a mix. (That's an underlying principle of vectors) But since the motion is no longer circular, the best coordinate system to describe the motion would be a Cartesian system. Relying on youtube is going to be a limiting factor in one's understanding. There's no substitute for actually working through a bunch of problems.
Robittybob1 Posted May 9, 2015 Posted May 9, 2015 (edited) Keeping up with the discussion is such a burden… I thought I had the solution to this problem as I drove home, so what was it again? The OP title seems to ask a slightly different question as does the post, but from recollection the reason why something weighs less at the equator than it does at the poles was part of the problem. Someone said that is because the device used to weigh the object "is falling away from under it". I didn't like that answer to begin with but it could be a useful way of looking at it. When an object is orbiting another they say the gravitational force is the centripetal force, and we have agreed that the centripetal force is equal and opposite to the centrifugal force. So what is the equal and opposite to the gravitational force? Is it still the centrifugal force? Gravity works both ways to begin with. There is equal and opposite attraction between the objects. So as we have agreed the centripetal force acts on the object and the centrifugal force acts on the center, so the centripetal force acts on the orbiting object and the centrifugal force acts on the center (the primary mass). But this next bit seems the most controversial bit. Since the centripetal force is calculated using V^2 and I think that velocity is not a relative angular velocity but an absolute one, for you can't take the rotation rate of the primary into account. Did I get that right? PS: Absolute in the sense of the velocity being the distance travelled (2*pi()*r) divided by the period. Edited May 9, 2015 by Robittybob1
tar Posted May 9, 2015 Posted May 9, 2015 SwansonT, I think in my muses I am not considering the circulating mass as a point mass. There is of course, in the real world, no such thing as a point mass, since a point has no dimension a point mass would have zero volume in which to put any mass and any such zero volume would have a mass of exactly zero grams. So a point mass might help do the calculations, but they would not reflect reality. My clay on a toothpick has a certain momentum, and as was described, its outward facing half is moving faster than its inward facing half thus imparting torque on the toothpick itself. Its mass is not confined to a point, but instead the different portions of the mass are undergoing varied motions and torques and stresses and the inertia of the large clay ball can be felt as it resists motion at first and resists stopping the next moment, and the repeated pushes to keep the disc spinning are applied. I could feel the mass and inertia of the large ball, differently from the mass and inertia of the medium ball, and that differently from the mass and inertia of the small ball, as I attempted to spin the combine. And the large ball did tip radially outward, where the medium and small ball remained upright. In another trial, I had all three similarly sized balls and none tipped outward. In another trial I did not lodge the toothpicks so deeply in the clay and they all three tipped outward. One radially, on a little ahead of radially and the third a little behind radially. You assure me though that there is no predicted...no, you assure me there is NO radially outward force allowed, and no radially outward motion possible, because there is no force acting in that direction. But now you admit that this is based on a point mass model. Perhaps you can allow some wiggle room and some possible ways for forces that start out tangential to be transferred a little inward or a little outward based on other than point masses spinning around various axis at the same time. Also the starting and stopping of my attempted spin tips the toothpick backward along the tangent as the mass tries not to move, and then forward as the mass refuses to stop as the disc does. The tangential movement backward takes the mass away from the circle, then the tangential movement forward takes it away from the circle the other way. Repeated pushes leave the mass a larger distance from the center, than where it started. The mass has moved radially outward, because of inertia, and tangential motion. This might be workable even if you assumed the mass, was a point mass. A tangential movement forward in the direction of motion of the circular rotation, added to a tangential movement backward from the direction of circular motion, adds together to be a radially outward movement. And I only say this to attempt to explain the slot made by the toothpick in the clay. Something real happened, that does not fit the mathematical model, that says it can not happen. Regards, TAR
Robittybob1 Posted May 9, 2015 Posted May 9, 2015 (edited) TAR - That must be the result of the toothpicks not being imbedded enough to stop the balls going off on their tangents. The picks embedded properly will provide sufficient centripetal force until the angular velocity rises even higher. It seems that at the limit where the mass moves along the tangent and separates from the circle there is an infinitesimally small radially outward motion. It is the summation of these movements that you see in the clay. Edited May 9, 2015 by Robittybob1
Mordred Posted May 9, 2015 Posted May 9, 2015 (edited) While your swinging the clay ball your applying a force . The string itself is limitting how far the clay ball could move. So you feel a greater force pulling on your arm. This is the action reaction forces Newtons third law. However there is less resistance on the movement on the toothpicks. Pushing the toothpicks allow greater resistance. replace the string with an elastic, the radius of the swinging ball will increase with the more force applied via angular momentum. The faster you spin the ball will cause the radius to increase, BUT remember both the string and elastic exert forces to oppose being changed from its rest configuration. The string may stretch less but with enough force can stretch a bit before breaking. Your balls mounted on clay isn't showing anything that cannot be explained as per the replies in this post. Your missing key details such as mechanical or material resistance to change. The clay is more ductile than the string The toothpick balls have less resistance than the string itself. Look up the definition on the unit of force Newton. "The SI derived unit used to measure force. One newton is equal to the force needed to accelerate a mass of one kilogram one meter per second per second. " The strings material strength due to its strong force resists the outward aceleration. The clay has less resistance. Experiment.. Take your same ball and clay design. Simply drop the clay, then note that the balls on the tooth picks will also travel further than the clay itself. How much further will depend on how deep you in bed the toothpicks. Edited May 9, 2015 by Mordred
swansont Posted May 9, 2015 Posted May 9, 2015 When an object is orbiting another they say the gravitational force is the centripetal force, and we have agreed that the centripetal force is equal and opposite to the centrifugal force. So what is the equal and opposite to the gravitational force? Is it still the centrifugal force? The reaction force to gravity is gravity. Action/reaction force pairs must be the same kind of force. Centripetal and centrifugal are not kinds of forces, they are merely labels. They must always be comprised of some actual force, or sum of forces. But this next bit seems the most controversial bit. Since the centripetal force is calculated using V^2 and I think that velocity is not a relative angular velocity but an absolute one, for you can't take the rotation rate of the primary into account. Did I get that right? PS: Absolute in the sense of the velocity being the distance travelled (2*pi()*r) divided by the period. It's from an inertial coordinate system, so the rotation of the primary would not matter; it's the speed with respect to the origin of the coordinate system. You assure me though that there is no predicted...no, you assure me there is NO radially outward force allowed, and no radially outward motion possible, because there is no force acting in that direction. But now you admit that this is based on a point mass model. Perhaps you can allow some wiggle room and some possible ways for forces that start out tangential to be transferred a little inward or a little outward based on other than point masses spinning around various axis at the same time. No. Contemplating a more complicated system will not make you right. Kinematics discussion the motion of the center of mass, which is a point. If other forces were to be present you would no longer have uniform circular motion, so the discussion becomes moot. You could have the little ball be an explosive, or add springs and pulleys, and then quite possibly have radial forces and motion, but again, this would be due to the additional complications that were added to the problem, and not due to the uniform circular motion. And we're discussing uniform circular motion.
pzkpfw Posted May 9, 2015 Posted May 9, 2015 ... My clay on a toothpick has a certain momentum, and as was described, its outward facing half is moving faster than its inward facing half thus imparting torque on the toothpick itself. ... No. Look at your base disk of clay side-on. Note that the centripetal acceleration keeping the blobs going in the circle is applied one way at the bottom of the toothpick, the momentum of the ball is applied in the other direction, at the top of the toothpick. That's what produces the rotation that makes the toothpick tip over. ... and no radially outward motion possible, because there is no force acting in that direction. You are misrepresenting the case. There is no force outward, but that doesn't mean there won't be apparent (in reference to the circle) motion outwards. Strings and toothpicks will constrain that motion. As shown in the earlier video, if a string is cut, the blob will move off at a tangent to the circle, not directly (radially) outwards. If you spun your disk fast enough that the toothpicks came right out of the base, your blobs too would fly off (more or less) at a tangent. All that point mass stuff is, well, pointless.
tar Posted May 9, 2015 Posted May 9, 2015 Thead, Well wait. If the boy on the roundabout is feeling an outward force and we are trying to figure how this is possible, and I build a clay model that is sort of representive of the situation (the boy's arm the toothpick, his grip the depth into the clay the pick is driven, the mass at the end his head...the position of the fluid in his inner ear similar to the direction indicated by the slot,) then the problem is more complex than a uniform circular motion of a point mass. Me introducing the complexities, and getting the reasoning of the different torques and directions and forces and movements at the base of the toothpick, and at the top of the toothpick is germaine to the discussion. It is not to refute what is true of uniform circular motion and center of mass figuring, it is using what is true about the laws of motion, and putting them together to explain why the boy feels like he is being pulled radially outward. The thread question is "how is this possible?" The answer "it is not possible" does not cut it. Regards, TAR -2
swansont Posted May 9, 2015 Posted May 9, 2015 Well wait. If the boy on the roundabout is feeling an outward force He isn't. The thread question is "how is this possible?" The answer "it is not possible" does not cut it. Logical fallacy aside, "How is it possible?" has been answered: your senses have been fooled, because there is no such force. 2
Robittybob1 Posted May 9, 2015 Posted May 9, 2015 (edited) The reaction force to gravity is gravity. Action/reaction force pairs must be the same kind of force. Centripetal and centrifugal are not kinds of forces, they are merely labels. They must always be comprised of some actual force, or sum of forces. It's from an inertial coordinate system, so the rotation of the primary would not matter; it's the speed with respect to the origin of the coordinate system. ..... I have heard people say the force of gravity is the centripetal force, haven't you? So by some sort of logic since the centrifugal force is the equal and opposite to the centripetal force (which is the force gravity in these cases), logically the centrifugal force is equal and opposite to the force of gravity (and is the "reaction force to gravity" but gravity acting on the other body). I seem to have repeated myself but it follows a logical path. Like when the centripetal force is provided by the tension in a string I'm not saying that is gravity. I do appreciate what you say about "kinds of forces" (equal and opposite pairs are the same kind of forces). And thanks for the right terminology; "origin of the coordinate system". So does this mean an object traveling on Earth 1012 mph to the West at the equator will weigh the same as it did at poles? For at this speed it will be stationary wrt the origin of the coordinate system. Edited May 9, 2015 by Robittybob1
swansont Posted May 9, 2015 Posted May 9, 2015 I have heard people say the force of gravity is the centripetal force, haven't you? Yes. I have said it myself, regarding objects in orbit. So by some sort of logic since the centrifugal force is the equal and opposite to the centripetal force (which is the force gravity in these cases), logically the centrifugal force is equal and opposite to the force of gravity (and is the "reaction force to gravity" but gravity acting on the other body). I seem to have repeated myself but it follows a logical path. Yes. And just like the last time this came up, it's irrelevant, since reaction forces do not dictate the motion of the object. So does this mean an object traveling on Earth 1012 mph to the West at the equator will weigh the same as it did at poles? For at this speed it will be stationary wrt the origin of the coordinate system. The answer hasn't changed. They would, if r is the same. But r isn't the same, so it still weighs less at the equator.
Robittybob1 Posted May 9, 2015 Posted May 9, 2015 (edited) Yes. I have said it myself, regarding objects in orbit. Yes. And just like the last time this came up, it's irrelevant, since reaction forces do not dictate the motion of the object. The answer hasn't changed. The would, if r is the same. But r isn't the same, so it still weighs less at the equator. I wasn't concerned about the motion of the object. I am trying to find out my own logical answer to why at suborbital speeds something weighs less. In the link you gave us your version of the answer but since you don't use the words "centrifugal force" I don't like your answer. It seems to be lacking something. .... All of the gravitational force is being "used" as the centripetal force. There's nothing "left over" so there is no normal force needed .... Doesn't seem right. Since the gravitational force and the centripetal forces are in the same direction how can they use each other? Unless you are talking of the G force that the object exerts on the primary mass, the one I'm calling the centrifugal force. Edited May 9, 2015 by Robittybob1
pzkpfw Posted May 9, 2015 Posted May 9, 2015 (edited) tar, try this: Think about artificial gravity in a space station. Like in the Movie 2001: A Space Odyssey. The space station is circular (a ring in the movie, but you could do the same with a cylinder), and is spinning. The crew can walk on the inside, of the outside wall, and seem to feel a kind of "gravity". Where does this force come from? If it's centrifugal, that implies something is actually pushing outwards (away from the centre of the circle) on them. What is that? Is it the air in the space station? No. As noted umpteen times herein, the truth is that the force applied on them is inwards. It's centripetal. Their bodies, due to their momentum (mass x velocity, remembering velocity is speed with direction), naturally want to continue in a straight line. The floor of the space station, turning in it's circle, is applying a force on them that keeps them turning in that same circle. That force is felt as the artificial gravity. Yes ... that person in the artificial gravity does seem to feel a force outwards. (The direction of the "gravity" they feel). Just like your boy on the roundabout and the finely tuned fluid in his inner ear. But ... that's not a "real force", nothing is pushing outwards on the space station crew member. What they feel is the reaction to the inwards force that's keeping them going in that circle. (And just like the ball on the string in the video a few forum pages ago - if a large section of the space station outer wall suddenly broke away, it and the crew member standing on it would not fly off into space directly away from the centre of rotation; they'd go in a direction tangential to the circle at that point. Things want to go in straight lines; at the moment the centripetal acceleration is removed, what's left is their momentum - and the tangent to the circle is the direction of that momentum.) Edited May 9, 2015 by pzkpfw
swansont Posted May 10, 2015 Posted May 10, 2015 I wasn't concerned about the motion of the object. I am trying to find out my own logical answer to why at suborbital speeds something weighs less. In the link you gave us your version of the answer but since you don't use the words "centrifugal force" I don't like your answer. It seems to be lacking something. It's lacking an incorrect analysis, which would apply to anything that cited a centrifugal force acting on the object. Doesn't seem right. Since the gravitational force and the centripetal forces are in the same direction how can they use each other? If an object is in orbit, the centripetal force is the gravity. There is no other force. If the object is not in orbit, the centripetal force is the difference between gravity and the normal force (which are in opposite directions) The centripetal force is less than the gravitational force. "Use" was in quotes. Don't take analogies literally. They are analogies. Unless you are talking of the G force that the object exerts on the primary mass, the one I'm calling the centrifugal force. Since I'm only discussing forces acting on the object, I am not talking about that. It is irrelevant.
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