Robittybob1 Posted May 23, 2015 Posted May 23, 2015 The ball wants to move in a straight line, which would take it further from the origin. The tension arises because the string (ideally) does not stretch, and thus exerts a force. OK as I've conceded above that there is only one force on the ball and that is the centripetal force, let's cast our minds back to the sliding issue discussed during the week, will the coin slide along a radial line, with only slight curve back from the radial line as I expect or as you said a transverse or transverse component. I had asked you about how much forward or backward from the radial line you'd expect after slipping 300 mm. If you answered that I missed it sorry, for today is good day to run the test. Would it be forward or backward of the radial line? The above statement sounds a bit like radial line movement. Do you still stick to the transverse concept? Remember in the video the slip line was drawn perpendicular and forward wrt to the radial line. I'll have a flat bottomed pint sized glass jar filled with water, and spin the merry go round by hand. I'll draw a line out from the center, place the jar on the line 300 mm from the edge and then "attempt to spin the whole system fast enough to generate enough centrifugal force to throw the jar off the edge". (I'll slow it down when it is near the edge as I don't want broken glass.) I'll then slow it to a stop and see what path the jar took. Predictions please.
DrP Posted May 23, 2015 Posted May 23, 2015 (edited) OK as I've conceded above that there is only one force on the ball and that is the centripetal force, let's cast our minds back to the sliding issue discussed during the week, will the coin slide along a radial line, with only slight curve back from the radial line as I expect or as you said a transverse or transverse component. I had asked you about how much forward or backward from the radial line you'd expect after slipping 300 mm. If you answered that I missed it sorry, for today is good day to run the test. Would it be forward or backward of the radial line? The above statement sounds a bit like radial line movement. Do you still stick to the transverse concept? Remember in the video the slip line was drawn perpendicular and forward wrt to the radial line. I'll have a flat bottomed pint sized glass jar filled with water, and spin the merry go round by hand. I'll draw a line out from the center, place the jar on the line 300 mm from the edge and then "attempt to spin the whole system fast enough to generate enough centrifugal force to throw the jar off the edge". (I'll slow it down when it is near the edge as I don't want broken glass.) I'll then slow it to a stop and see what path the jar took. Predictions please. The glass will spin with the wheel in a circle at lower speeds all the time the friction between the glass and the wheel is high enough to keep the glass moving in a circle. When a high enough speed is reached by the wheel the increased momentum of the glass (from being spun faster) will over come the friction and the glass will keep going in a straight line (instead of following the wheel in a circle), straight off the edge at a tangent, as it carries on in a straight line whilst the part of the wheel that it was sitting on continues in a circle. This might look like it has been flung straight outward, but it hasn't, it has gone off sideways. If you slow the wheel before the glass flies off but just after it slips (as you suggested above) the it would have slipped back from the radial line as the glass goes straight and the radial line continues round with the wheel. Thus the glass finishes further from the center of the wheel since it slipped. How far exactly will depend upon the weight of glass, the speed of rotation, the co-efficient of friction between the glass and the wheel, etc., so can not be answered. and then "attempt to spin the whole system fast enough to generate enough centrifugal force to throw the jar off the edge". .. so you can see that there is no centrifugal force throwing it off the edge - it just slips off when it gets fast enough. (because it keeps going in a straight line and not in a circle) Edited May 23, 2015 by DrP
Robittybob1 Posted May 23, 2015 Author Posted May 23, 2015 (edited) The glass will spin with the wheel in a circle at lower speeds all the time the friction between the glass and the wheel is high enough to keep the glass moving in a circle. When a high enough speed is reached by the wheel the increased momentum of the glass (from being spun faster) will over come the friction and the glass will keep going in a straight line (instead of following the wheel in a circle), straight off the edge at a tangent, as it carries on in a straight line whilst the part of the wheel that it was sitting on continues in a circle. This might look like it has been flung straight outward, but it hasn't, it has gone off sideways. If you slow the wheel before the glass flies off but just after it slips (as you suggested above) the it would have slipped back from the radial line as the glass goes straight and the radial line continues round with the wheel. Thus the glass finishes further from the center of the wheel since it slipped. How far exactly will depend upon the weight of glass, the speed of rotation, the co-efficient of friction between the glass and the wheel, etc., so can not be answered. ... so you can see that there is no centrifugal force throwing it off the edge - it just slips off when it gets fast enough. Some good points there. If it was extremely low friction the jar might just not move at all and it would transcribe a circle drawn in the opposing direction to the rotation. If there was extreme friction no matter how fast I turn it there will be no slippage. So definitely the coefficient of friction will be a factor. Weight of the jar??? Mass is involved with friction. Friction force is proportional to mass but also the centrifugal force will be proportional to mass as well. So mass might not be a factor as long as it didn't exceed my ability to give it angular momentum. Is the speed of rotation a factor? Keeping the other factors the same there will be a speed at which the jar will slip at a certain radius. You won't be able to change that. Will the speed that is needed to make it slip be the same speed no matter where it is put? I am assuming there will be no slipping back from the radial line as the acceleration I'll be able to give it will not be great enough. So at all times the speed the jar is going is the speed of the surface at that particular radius. If it started at 1m and progresses to a radius of 1.300m at the same angular rotation of 1 Hz the force required to hold it in circular motion is greater as the radius increases so once it starts to slip it shouldn't stop, for at all times the centrifugal force is greater than the force of friction (being constant). So the idea of slowing the merry go round down once it begins to slip is rather impossible as the slipping off will happen faster than my ability to slow it down again. I'll have to have an edge so the jar doesn't smash. Edited May 23, 2015 by Robittybob1
DrP Posted May 23, 2015 Posted May 23, 2015 (edited) Pretty much... except that it is centripetal force, not centrifugal. The friction is providing the centripetal force. When the centripetal force required to keep the glass/(any object) in a circular motion becomes greater than the friction then it slips and the glass and the wheel part company. You edited your post - so my reply might not make sense now.. anyway. I'm off to watch Mad Max in 3D... ttfn. So the idea of slowing the merry go round down once it begins to slip is rather impossible as the slipping off will happen faster than my ability to slow it down again. I'll have to have an edge so the jar doesn't smash. Or catch it... or just don't bother and take our word for it. Watch that vid again - it explains things really clearly. ;-) - bfn. Edited May 23, 2015 by DrP
Robittybob1 Posted May 23, 2015 Author Posted May 23, 2015 (edited) Pretty much... except that it is centripetal force, not centrifugal. The friction is providing the centripetal force. When the centripetal force required to keep the glass/(any object) in a circular motion becomes greater than the friction then it slips and the glass and the wheel part company. You edited your post - so my reply might not make sense now.. anyway. I'm off to watch Mad Max in 3D... ttfn. Or catch it... or just don't bother and take our word for it. Watch that vid again - it explains things really clearly. ;-) - bfn. There will be plenty examples of centrifugal forces in that movie. Cheers. Minus 2 for a bit of humour I like the spirit around here! I was trying to understand this friction thing. Once friction is overcome what direction does something move? Edited May 23, 2015 by Robittybob1 -2
swansont Posted May 23, 2015 Posted May 23, 2015 OK as I've conceded above that there is only one force on the ball and that is the centripetal force, let's cast our minds back to the sliding issue discussed during the week, will the coin slide along a radial line, with only slight curve back from the radial line as I expect or as you said a transverse or transverse component. I had asked you about how much forward or backward from the radial line you'd expect after slipping 300 mm. If you answered that I missed it sorry, for today is good day to run the test. Would it be forward or backward of the radial line? I did answer it. Part of the problem with the entire discussion has been having to repeat answers, multiple times. I'll have a flat bottomed pint sized glass jar filled with water, and spin the merry go round by hand. I'll draw a line out from the center, place the jar on the line 300 mm from the edge and then "attempt to spin the whole system fast enough to generate enough centrifugal force to throw the jar off the edge". (I'll slow it down when it is near the edge as I don't want broken glass.) I'll then slow it to a stop and see what path the jar took. I guess the acceptance of there not being a centrifugal force was short-lived.
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