swansont Posted May 4, 2015 Posted May 4, 2015 If the centripetal force and the centrifugal are equal and opposite, how can one be real but not the other? The centrifugal force is acting on the center. It is acting on the center, not on itself. It is a force pulling the center to itself. The centrifugal force is not a real force acting on you. This whole issue is that people thing there's a centrifugal force acting on them when they move in a circle. It's fictitious. How can the scales be dropping away when they are firmly on the surface? The surface is also dropping away. It's moving in a circle.
pzkpfw Posted May 4, 2015 Posted May 4, 2015 (edited) How can the scales be dropping away when they are firmly on the surface? Because you're standing on a circle (slice of the sphere), that's turning. The surface is dropping away. The natural inclination of your body is to keep moving in a straight line. The reason you don't, as the surface drops away, is gravity. So while you'd "weigh less" at the equator than at one of the poles *, it's not because there's some force pulling you up (as hinted at in post #168 by Tar), it's because the surface, relative to you, is moving down. (* actually it's a lot more compicated than that, as Earth isn't a perfect spehere of even density, but that's beyond this thread. (And me.)) Edit (not directly about the post above but very much about this thread): Watch this video: https://www.youtube.com/watch?v=O6yJG_fnJRc Then think about why, when the string is cut, the ball goes up (in the direction of the tangent to the wheel), not horizontal, in the direction of the centre to the ball (i.e. the string). Edited May 4, 2015 by pzkpfw
tar Posted May 4, 2015 Posted May 4, 2015 Hum, I suppose the defintion of force on an object, as opposed to inertia within an object is at the root of the problem here. If a force causes an object to move, and an object is moving in a straight line, the force has not disappeared, it has been transferred to the momentum or inertia within the object. That force, is the "pull" you feel on the rope, with the rock at the end, that you are swinging around your head, in a circle. While the direction that the rock goes when it slips out of your knot is a straight line tangent to the circle you got it going in, you had to get it going in the circle by getting the rope to go taught. And you have to get the rope to go taught by lifting the rock off the ground against the force of gravity. Then you coax the rock in a small circle around your feet by inscribing a circle with your hand around your head, with the other end of the rope, in it. Letting the rock "fall" in only one direction tangent to the circle you are inscribing around your head as you move the fulcrum point of this one way pendulum in the circle around your head. You keep leading the rock, faster and faster, in this manner, pulling it in a tangent to the circle direction, by utilizing a combination of its weight (the force of gravity) and its momentum (desire to continue in a straight line). The rock's momentum, when you get it going fast enough to "lift" above your waist, "feels" like an outward pull. When it slips your knot or you release it like a rock from a slingshot, it goes in a straight line, tangent to the arc at the moment of release, but it travels away from you, effectively "radially outward" from the circumference of the circle you were describing. Regards, TAR
swansont Posted May 4, 2015 Posted May 4, 2015 Hum, I suppose the defintion of force on an object, as opposed to inertia within an object is at the root of the problem here. If a force causes an object to move, and an object is moving in a straight line, the force has not disappeared, it has been transferred to the momentum or inertia within the object. Forces cause changes in motion (speed and/or direction). A force is not required to maintain a constant velocity. That force, is the "pull" you feel on the rope, with the rock at the end, that you are swinging around your head, in a circle. While the direction that the rock goes when it slips out of your knot is a straight line tangent to the circle you got it going in, you had to get it going in the circle by getting the rope to go taught. Yes. You exert a force on the rock, inward. Toward the center. It also exerts a force on you, but who cares, since it is only the forces on the rock that dictate its motion. And you have to get the rope to go taught by lifting the rock off the ground against the force of gravity. Then you coax the rock in a small circle around your feet by inscribing a circle with your hand around your head, with the other end of the rope, in it. Letting the rock "fall" in only one direction tangent to the circle you are inscribing around your head as you move the fulcrum point of this one way pendulum in the circle around your head. You keep leading the rock, faster and faster, in this manner, pulling it in a tangent to the circle direction, by utilizing a combination of its weight (the force of gravity) and its momentum (desire to continue in a straight line). The rock's momentum, when you get it going fast enough to "lift" above your waist, "feels" like an outward pull. When it slips your knot or you release it like a rock from a slingshot, it goes in a straight line, tangent to the arc at the moment of release, but it travels away from you, effectively "radially outward" from the circumference of the circle you were describing. Again, who cares about the force on you, since in this example you are not the one moving in a circle. The rock is. And you are continually pulling it inward.
Robittybob1 Posted May 4, 2015 Posted May 4, 2015 Because you're standing on a circle (slice of the sphere), that's turning. The surface is dropping away. .... Then think about why, when the string is cut, the ball goes up (in the direction of the tangent to the wheel), not horizontal, in the direction of the centre to the ball (i.e. the string). I am doing some math that shows the ball goes off in a direction that the distance from the center keeps increasing. The distance from the center increases rapidly to begin with, then slowly up to the value of its velocity away from the center. (I would like to introduce gravitational attraction to make the path curved rather than straight.)
swansont Posted May 4, 2015 Posted May 4, 2015 I am doing some math that shows the ball goes off in a direction that the distance from the center keeps increasing. That's the nature of having a velocity: distance keeps increasing. It's not really tied in with centripetal motion.
pzkpfw Posted May 4, 2015 Posted May 4, 2015 (edited) ... While the direction that the rock goes when it slips out of your knot is a straight line tangent to the circle you got it going in, you had to get it going in the circle by getting the rope to go taught. ... The rest of your post goes a bit off the deep end, I think, but here you do have the crux of it. When you say "by getting the rope to go taught" you are showing the force needed to make the object go in a circle is towards the centre. At any moment, the direction of momentum of the object is at a tangent to the circle. It's the force towards the centre, whether by string or gravity or the thing is on the inside of a rim (e.g. space station "artificial gravity", see "2001: A Space Odyssey" for visuals) that keeps the thing going in a circle. There is no real force outwards, that's just the "opposite" of the force needed to make the thing keep going inwards. Edited May 4, 2015 by pzkpfw
tar Posted May 5, 2015 Posted May 5, 2015 pzkpfw, Well OK, if the force ON the object is what we are counting, and the momentum does not count as a force, because an object in constant linear motion requires no further force to maintain such, then the "pull" on my arms as I hold on to the carosel is my arms pulling me inward, and no force is pulling me outward...except outward is where I wind up should I let go. Robittybob1 seems to be stuck on the same idea I am, in terms of the tangent motion being "away" from the circle. In the video the cut string allows the ball to go up tangent to the circle, but if you imagine an elastic thread still attached to the ball and the center of the circle, one can imagine the thread on a line radially outward from the center which seems to maintain for close to a quarter turn of the wheel. For instance, let's say a lasar pointer is attached to the center of the wheel, which keeps turning even after the string is cut. If the laser was pointed at the ball, would it continue to shine on the ball, after the string is cut and the ball flies off on the tangent? At least for almost a quarter turn. Regards, TAR
pzkpfw Posted May 5, 2015 Posted May 5, 2015 (edited) ... except outward is where I wind up should I let go. ... No. You're the same as the ball in that video. ... For instance, let's say a lasar pointer is attached to the center of the wheel, which keeps turning even after the string is cut. If the laser was pointed at the ball, would it continue to shine on the ball, after the string is cut and the ball flies off on the tangent? At least for almost a quarter turn. Regards, TAR Not sure. I'd have to do some math. It's not an "exact alignment", as the ball moves off at a tangent, and the wheel rotates. Curves are different to straight lines. The width of the ball will determine how long the laser hits it. Certainly, the ball moves away from the wheel, but a tangent is different to a radial motion, and the main thing to get over in thinking about all this is that there is no real force outwards radially in this situation - otherwise the ball in that video would not have gone up when that string was cut. Edited May 5, 2015 by pzkpfw
Robittybob1 Posted May 5, 2015 Posted May 5, 2015 (edited) Not sure. I'd have to do some math. It's not an "exact alignment", as the ball moves off at a tangent, and the wheel rotates. Curves are different to straight lines. The width of the ball will determine how long the laser hits it...... Certainly, the ball moves away from the wheel, but a tangent is different to a radial motion,..... That is the basis of the math I'm trying out for as the travelling mass follows the tangent the rate of rotation of the center will decrease to keep the angular momentum of the travelling mass the same (well that is the concept.) It looked as if the position of the travelling mass, the point it departed from the perimeter and the center point always remains in a straight line. That was if the rotating center had no mass but just a rotational rate (that rapidly decreases to close to zero.) So in effect it looks as if it is moving radially from the departure point. It the central object has mass it will not slow down simply because something has separated from it. pzkpfw, Well OK, if the force ON the object is what we are counting, and the momentum does not count as a force, because an object in constant linear motion requires no further force to maintain such, then the "pull" on my arms as I hold on to the carosel is my arms pulling me inward, and no force is pulling me outward...except outward is where I wind up should I let go. Robittybob1 seems to be stuck on the same idea I am, in terms of the tangent motion being "away" from the circle. In the video the cut string allows the ball to go up tangent to the circle, but if you imagine an elastic thread still attached to the ball and the center of the circle, one can imagine the thread on a line radially outward from the center which seems to maintain for close to a quarter turn of the wheel. For instance, let's say a lasar pointer is attached to the center of the wheel, which keeps turning even after the string is cut. If the laser was pointed at the ball, would it continue to shine on the ball, after the string is cut and the ball flies off on the tangent? At least for almost a quarter turn. Regards, TAR Yes it will as long as the rotation rate of the core drops to virtually zero over the first 1/4 of the rotation. So the attachment point and the center stays inline with the mass. But there are virtually no physical situations where the center system will have no mass. When you say "by getting the rope to go taught" you are showing the force needed to make the object go in a circle is towards the centre. .... To get the object moving the forces have to be more than just centripetal forces. You won't get tangential velocity just from centripetal force. That's the nature of having a velocity: distance keeps increasing. It's not really tied in with centripetal motion. The distance keeps on increasing and also the rate it moves away increases too. How it ties in with centripetal motion is that I think it will show that the tangential vector component of the motion will be the velocity needed to keep the angular momentum constant. I'm not that good at doing the math but I'm still working on it. Edited May 5, 2015 by Robittybob1
pzkpfw Posted May 5, 2015 Posted May 5, 2015 (edited) That is the basis of the math I'm trying out for as the travelling mass follows the tangent the rate of rotation of the center will decrease to keep the angular momentum of the travelling mass the same (well that is the concept.) It looked as if the position of the travelling mass, the point it departed from the perimeter and the center point always remains in a straight line. That was if the rotating center had no mass but just a rotational rate (that rapidly decreases to close to zero.) So in effect it looks as if it is moving radially from the departure point. It the central object has mass it will not slow down simply because something has separated from it. This is where you seem so keen to find something "clever" you descend into gibberish. You're way over-complicating this. Yes it will as long as the rotation rate of the core drops to virtually zero over the first 1/4 of the rotation. So the attachment point and the center stays inline with the mass. If you need to specify exotic conditions that were not stated in the question, why then even bother saying "yes"? Edit: and on second thought, "yes" can't be right anyway, as the object is moving off at a tangent to the circle, and that laser is shining out from the centre of the circle. It can't work regardless of where the wheel stops. The ball will either be moving parallel to the laser, or the path of the ball and the laser will cross at exactly one point. But there are virtually no physical situations where the center system will have no mass. Again, over thinking the whole thing. To get the object moving the forces have to be more than just centripetal forces. You won't get tangential velocity just from centripetal force. Of course yes, and no (in that order). That's all pretty obvious. But none of that was really the original point of this thread. The distance keeps on increasing and also the rate it moves away increases too. Why does the rate increase? How it ties in with centripetal motion is that I think it will show that the tangential vector component of the motion will be the velocity needed to keep the angular momentum constant. I'm not that good at doing the math but I'm still working on it. Taking that video from a few posts ago as an example, the speed at which the ball leaves the wheel (tangentally) will be the speed of the circumference of the wheel at the time the ball was let go. Edited May 5, 2015 by pzkpfw
swansont Posted May 5, 2015 Posted May 5, 2015 pzkpfw, Well OK, if the force ON the object is what we are counting, and the momentum does not count as a force, because an object in constant linear motion requires no further force to maintain such, then the "pull" on my arms as I hold on to the carosel is my arms pulling me inward, and no force is pulling me outward...except outward is where I wind up should I let go. Except that it isn't. That's a misconception you have. You would travel tangent to the circle, not radially outward. Robittybob1 seems to be stuck on the same idea I am, in terms of the tangent motion being "away" from the circle. In the video the cut string allows the ball to go up tangent to the circle, but if you imagine an elastic thread still attached to the ball and the center of the circle, one can imagine the thread on a line radially outward from the center which seems to maintain for close to a quarter turn of the wheel. What you can imagine is not constrained by reality. The reality is that the ball leaves at a tangent and moves straight up. It does not move left or right (depending on which view you have). There is NO MOTION in the radial direction at the point of the cut. For instance, let's say a lasar pointer is attached to the center of the wheel, which keeps turning even after the string is cut. If the laser was pointed at the ball, would it continue to shine on the ball, after the string is cut and the ball flies off on the tangent? At least for almost a quarter turn. Regards, TAR So? That's not due to a force in the radial direction. The distance keeps on increasing and also the rate it moves away increases too. How it ties in with centripetal motion is that I think it will show that the tangential vector component of the motion will be the velocity needed to keep the angular momentum constant. I'm not that good at doing the math but I'm still working on it. If the speed is increasing then there must be a force. What is that force? That also doesn't jibe with "The distance from the center increases rapidly to begin with, then slowly up to the value of its velocity away from the center" which sort of implies it is slowing (which it is in the video, owing to gravity) That is the basis of the math I'm trying out for as the travelling mass follows the tangent the rate of rotation of the center will decrease to keep the angular momentum of the travelling mass the same (well that is the concept.) Angular momentum will be conserved if there is no torque, which is not true if a motor is driving the wheel. If it's a free-spinning wheel then yes, you can apply conservation of angular momentum to a wheel+mass system, but I think you'll find no effect on the rotation rate by releasing a mass from it. The angular momentum of the mass doesn't change in the instant after it's released, so neither does the angular momentum of the wheel.
tar Posted May 5, 2015 Posted May 5, 2015 SwansonT, "Except that it isn't. That's a misconception you have. You would travel tangent to the circle, not radially outward." On the playground carosel, if I lost my grip for a moment (clinging to a point near the center of the merry-go-round) I would slide outward on the deck, increasing the radius of the circle I was traveling and getting further from the center. I would have to fight inward again, pulling toward the center. What Robittybob1 and I are stuck on, is that during the release, when you go off in a tangent direction, you are still attached to the deck by a little bit of friction, and take on some of the tangential directional attributes of the deck as you slide out through the concentric, larger and larger circles. The motion outward is effectively an outward trek. As you attach yourself to a bar circulating at a larger radius, it gets harder to hold on and your adreneline shoots up and you beg the pushers to slow the thing as you do not wish to continue the outward trek, past the outer circumference of the ride. If, when determining whether your direction of motion is tangent or radial you were to be on a rotating disc 100 meters in radius, and you were to draw a line on the deck from your position near the center where you let go (you are wearing teflon clothing and the deck is oiled and there are no holding points past a meter out,) to the point where you are on the 50 meter circle, you would find that your direction of motion through the 50 meter radius circle line, would be roughly normal to the tangent line drawn at that point. So, when determining the direction a fella goes in when he falls off the carosel, one can figure a direction tangent to the outer circumference of the ride. But what direction is he going when he is still on the ride, but being pulled toward the outer circumference? Regards, TAR -2
swansont Posted May 5, 2015 Posted May 5, 2015 SwansonT, "Except that it isn't. That's a misconception you have. You would travel tangent to the circle, not radially outward." On the playground carosel, if I lost my grip for a moment (clinging to a point near the center of the merry-go-round) I would slide outward on the deck, increasing the radius of the circle I was traveling and getting further from the center. I would have to fight inward again, pulling toward the center. Which is not radially outward in an inertial coordinate system. It's a spiral, which is what you get when you increase r while having a tangential velocity. And motion along the initial tangent increases r What Robittybob1 and I are stuck on, is that during the release, when you go off in a tangent direction, you are still attached to the deck by a little bit of friction, and take on some of the tangential directional attributes of the deck as you slide out through the concentric, larger and larger circles. The motion outward is effectively an outward trek. As you attach yourself to a bar circulating at a larger radius, it gets harder to hold on and your adreneline shoots up and you beg the pushers to slow the thing as you do not wish to continue the outward trek, past the outer circumference of the ride. Once you are released, you are no longer attached. You can't have it both ways. And, again, your motion is not radially outward. "Effectively outward" is not radial. If, when determining whether your direction of motion is tangent or radial you were to be on a rotating disc 100 meters in radius, and you were to draw a line on the deck from your position near the center where you let go (you are wearing teflon clothing and the deck is oiled and there are no holding points past a meter out,) to the point where you are on the 50 meter circle, you would find that your direction of motion through the 50 meter radius circle line, would be roughly normal to the tangent line drawn at that point. No, it absolutely would not. Just stating that does not make it true. First of all, if there was no friction. You wouldn't move away from the center. At the outset there is no force, so there is no change from rest. If you were not at the center, you would not start moving, because you have no friction to start you moving. You can't analyze this from a dynamic point of view. The conditions conflict with each other — you can't require friction but have a frictionless surface. You have to start this from a point of already moving at some v and some r, such that you are not moving relative to the circle that is rotating. Then you can "turn friction off" and analyze the motion — that's why cutting the string is a good example. You will not move radially outward, as measured by an inertial frame of reference. You will move in a straight line, tangent to the circle. So, when determining the direction a fella goes in when he falls off the carosel, one can figure a direction tangent to the outer circumference of the ride. But what direction is he going when he is still on the ride, but being pulled toward the outer circumference? Regards, TAR He is not pulled toward the outer circumference. There is no force in that direction. He moves in that direction because he is moving in a straight line, tangent to the release point, which has acquires a radial component once the linear motion starts, because that's how linear motion works in a circular coordinate system. Constant v linear motion not only does not require a force, it precludes it. (Newton's first law) ———— A snapshot of the release point in that video. I've added a larger circle to accommodate the scenario of doing this in the interior. Motion is tangent to the circle. Notice how the radius would increase as it follows this path. Also notice how nothing is happening in the radial direction. Distance from the center of the circle increases without an outward push. 3
Robittybob1 Posted May 5, 2015 Posted May 5, 2015 ...... If the speed is increasing then there must be a force. What is that force? That also doesn't jibe with "The distance from the center increases rapidly to begin with, then slowly up to the value of its velocity away from the center" which sort of implies it is slowing (which it is in the video, owing to gravity) Angular momentum will be conserved if there is no torque, which is not true if a motor is driving the wheel. If it's a free-spinning wheel then yes, you can apply conservation of angular momentum to a wheel+mass system, but I think you'll find no effect on the rotation rate by releasing a mass from it. The angular momentum of the mass doesn't change in the instant after it's released, so neither does the angular momentum of the wheel. The Tangential velocity stays constant but I am measuring the distance from the center point, which is the length of the Hypotenuse, calculated by the Pythagoras Formula. The length of this hypotenuse starts off as "r" which is one side of the right angled triangle and velocity * time is the other (along the tangent). This distance increases at a non-linear rate and distance from the center over time has a speeding up curvature. This is always radial (i.e. joins the center point to the object) At all times the mass travelling along the tangent line at the time of "release" will also be on a line that is along the same radius. The way this can happen is the rotation rate declines (but this is the rotation rate required if the object was reattached at that point. I am trying to show that relationship with maths. I have seen similar math used to calculate angular momentum of unattached objects.) (This could be wrong as the tangential velocity will be added to the orbital velocity (will it or won't it?), that will give me something to work on.) This works if there is a mass less center wheel (assume string has no mass and the turntable has no mass either). If it has mass its rotational inertia would resist the slowing that is possible with the mass less system i.e one without angular momentum.
imatfaal Posted May 5, 2015 Posted May 5, 2015 Rob - please leave your angular momentum exposition to another thread. This thread is confused enough as it is. Additionally -- In this case the scalar magnitude of the angular momentum is the mass times speed times the perpendicular distance - this does not change when the string is cut. There is no need for any angular momentum calculations as |L|= m*v*r stays the same all the way through. If you wish to investigate this - do it in a new thread
swansont Posted May 5, 2015 Posted May 5, 2015 The Tangential velocity stays constant but I am measuring the distance from the center point, which is the length of the Hypotenuse, calculated by the Pythagoras Formula. The length of this hypotenuse starts off as "r" which is one side of the right angled triangle and velocity * time is the other (along the tangent). This distance increases at a non-linear rate and distance from the center over time has a speeding up curvature. This is always radial (i.e. joins the center point to the object) Once the object is released from circular motion, "tangential" and "radial" cease to be the proper terminology. The circle is described by the motion of the object, and with a straight line there is no circle.
Mike Smith Cosmos Posted May 5, 2015 Author Posted May 5, 2015 (edited) Once the object is released from circular motion, "tangential" and "radial" cease to be the proper terminology. The circle is described by the motion of the object, and with a straight line there is no circle.Yes. But surely there are conditions where ( a) there is a transition time when total restrained moves smoothly across through partial restrain to unrestrained . Where the circulating mass has different trajectories other than circular or tangential . Also where a force is inward , toward the Centre ( centrifugal ) And another force acting in the opposite direction is not equal (magnitude wise) , whereby the inward force is either reduced or indeed overcome . This gives rise to an outward circulating force. The title of which seems elusive ? ................. . ........... dare I say it centrifugal in some wise .........? Mike Edited May 5, 2015 by Mike Smith Cosmos
swansont Posted May 5, 2015 Posted May 5, 2015 Yes. But surely there are conditions where ( a) there is a transition time when total restrained moves smoothly across through partial restrain to unrestrained . Where the circulating mass has different trajectories other than circular or tangential . Yes, and those are just complications from the simplified case where the concepts are more clearly demonstrated. Given the obvious difficulty in getting this concept straight, going the complicated route doesn't seem to be advisable. Also where a force is inward , toward the Centre ( centrifugal ) And another force acting in the opposite direction is not equal (magnitude wise) , whereby the inward force is either reduced or indeed overcome . This gives rise to an outward circulating force. The title of which seems elusive ? ................. . ........... dare I say it centrifugal in some wise .........? Centrifugal means "center--fleeing", so a centrifugal force is never directed toward the center. If an object is traveling in a circle, there is a net force directed toward the center. The net force is the total. i.e. what you have after adding everything up. There will NEVER be a resulting outward force. No amount of rigging an example will change that.
tar Posted May 6, 2015 Posted May 6, 2015 SwansonT, OK, no net radial outward force, if an object is moving in a circle. How about if an object is moving in a spiral? With an increasing distance from the center of the spiral? I am not sure if you can have a tangent to a spiral, but it seems if you do, such a tangent is not normal to a line drawn from the origin of the spiral, but it has an outward component. When dealing with the playground merry-go-round, it is harder to cause your body to follow an inwardly directed spiral (pulling yourself closer to the center) than to relax your arms and slide or lean outward. When holding fast to a circular path, your hair and loose clothing and blood seems to be drawn outward. There is this situational result of everything wanting to maintain its inertia that makes the outward spiral more likely to occur than the inward one. It takes effort or force to maintain the circular path, more to pull your body closer to the center. The net force might be directed toward the center, and there is no force directed radially outward, but if there is motion along the path of a larger and larger spiral, then some component of the net force on the object must be directed outward. Well, maybe not must, but the straight line inertial momentum causes the spiral to widen. Regards, TAR
Mike Smith Cosmos Posted May 6, 2015 Author Posted May 6, 2015 Yes, and those are just complications from the simplified case where the concepts are more clearly demonstrated. Given the obvious difficulty in getting this concept straight, going the complicated route doesn't seem to be advisable. Mike Smith Cosmos, on 05 May 2015 - 10:11 PM, said: Also where a force is inward , toward the Centre ( centrifugal ) And another force acting in the opposite direction is not equal (magnitude wise) , whereby the inward force is either reduced or indeed overcome . This gives rise to an outward circulating force. The title of which seems elusive ? ................. . ........... dare I say it centrifugal in some wise .........? Centrifugal means "center--fleeing", so a centrifugal force is never directed toward the center. If an object is traveling in a circle, there is a net force directed toward the center. The net force is the total. i.e. what you have after adding everything up. There will NEVER be a resulting outward force. No amount of rigging an example will change that. Yes . That was clearly a typing error .. I meant to say .. Mike Smith Cosmos, on 05 May 2015 - 10:11 PM, said: Also where a force is inward , toward the Centre. ... ( centripetal ) ....And another force acting in the opposite direction is not equal (magnitude wise) , whereby the inward force is either reduced or indeed overcome . This gives rise to an outward circulating force. The title of which seems elusive ? ................. . ........... dare I say it centrifugal in some wise .........?
pzkpfw Posted May 6, 2015 Posted May 6, 2015 No. There is no outward force, there's only the inclination of an object to move in a straight line.
Robittybob1 Posted May 6, 2015 Posted May 6, 2015 Once the object is released from circular motion, "tangential" and "radial" cease to be the proper terminology. The circle is described by the motion of the object, and with a straight line there is no circle. Why would tangential ever stop being tangential? I know radial is harder to define. Rob - please leave your angular momentum exposition to another thread. This thread is confused enough as it is. Additionally -- In this case the scalar magnitude of the angular momentum is the mass times speed times the perpendicular distance - this does not change when the string is cut. There is no need for any angular momentum calculations as |L|= m*v*r stays the same all the way through. If you wish to investigate this - do it in a new thread Thanks I'll do that, for I think I've come up with something strange.
Mike Smith Cosmos Posted May 6, 2015 Author Posted May 6, 2015 (edited) No. There is no outward force, there's only the inclination of an object to move in a straight line. Yes, I agree with that , it is the straight line inertia which indirectly gives this mass an outward going trajectory ( or pull from the centre, if it is being restrained in any way, say by a spring ) if you look at it in the context of the circle it was being restrained to travel . Mike Edited May 6, 2015 by Mike Smith Cosmos
Robittybob1 Posted May 6, 2015 Posted May 6, 2015 Yes, I agree with that , it is the straight line inertia which indirectly gives this mass an outward going trajectory ( or pull from the centre, if it is being restrained in any way, say by a spring ) if you look at it in the context of the circle it was being restrained to travel . image.jpg Mike Good one Mike. But remember if you had that situation as the mass stretches the spring the rate of rotation will slow.
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