michel123456 Posted November 12, 2011 Share Posted November 12, 2011 A ball A is placed on a ramp of 48 degrees. What max. height F and distance C will reach the ball, excluding any friction? The ball is 3 kg weight, 20 cm in diameter. Note: this is not homework. 1 Link to comment Share on other sites More sharing options...
John Cuthber Posted November 12, 2011 Share Posted November 12, 2011 If you round off the "corner" at the bottom where the rams meet then F is 3 metres and C is 3/tan (42) I think i.e. about 3.33 m. If you don't round it off then there might be a significant energy loss there. Why do you want to know? Link to comment Share on other sites More sharing options...
michel123456 Posted November 12, 2011 Author Share Posted November 12, 2011 (edited) Wrong answer. Hint: we are in the brain teasers & puzzles section. Edited November 12, 2011 by michel123456 1 Link to comment Share on other sites More sharing options...
TonyMcC Posted November 12, 2011 Share Posted November 12, 2011 (edited) I think you may have deliberately not drawn the angles accurately to give a false impression. Would that be right? Edited November 12, 2011 by TonyMcC Link to comment Share on other sites More sharing options...
michel123456 Posted November 12, 2011 Author Share Posted November 12, 2011 I think you may have deliberately not drawn the angles accurately to give a false impression. Would that be right? That is right. Angle. I mean right angle. The correct drawing is here below The ball will never go up, it will collide against the 90 degrees angle. That is because the sum of the angles 48 + 42 = 90, then the resulting angle between the ramps is also 90. All this to show that a good sketch is indispensable. Resolving mathematically the problem induces an error, especially when based upon a wrong sketch. Link to comment Share on other sites More sharing options...
John Cuthber Posted November 12, 2011 Share Posted November 12, 2011 Wrong answer. Hint: we are in the brain teasers & puzzles section. Actually, "If you don't round it off then there might be a significant energy loss there." is the right answer. Link to comment Share on other sites More sharing options...
michel123456 Posted November 12, 2011 Author Share Posted November 12, 2011 (edited) Actually, "If you don't round it off then there might be a significant energy loss there." is the right answer. Yes. Anyway the error is mine. The first diagram is intentionally wrong. I simply wanted to show the importance of drawing an accurate diagram. And also to relativate all blind mathematical conclusions. Edited November 12, 2011 by michel123456 Link to comment Share on other sites More sharing options...
swansont Posted November 12, 2011 Share Posted November 12, 2011 Wrong answer. Hint: we are in the brain teasers & puzzles section. F will not be 3 meters? Why? Link to comment Share on other sites More sharing options...
TonyMcC Posted November 12, 2011 Share Posted November 12, 2011 (edited) F will not be 3 meters? Why? And over horizontal distance B? (assuming no energy loss on the rebound) Edited November 12, 2011 by TonyMcC Link to comment Share on other sites More sharing options...
michel123456 Posted November 13, 2011 Author Share Posted November 13, 2011 (edited) F will not be 3 meters? Why? Because the direction of the ball is perpendicular to the 2nd ramp. The rebound will be in the reverse direction. Edited November 13, 2011 by michel123456 Link to comment Share on other sites More sharing options...
TonyMcC Posted November 13, 2011 Share Posted November 13, 2011 Because the direction of the ball is perpendicular to the 2nd ramp. The rebound will be in the reverse direction. If there is no friction or other losses the ball will bounce back to its original position. i.e. to a height of 3m. In the absence of friction and other losses I would expect the ball to bounce between these two positions indefinitely. (It is a theoretical/hypothetical situation). Link to comment Share on other sites More sharing options...
michel123456 Posted November 13, 2011 Author Share Posted November 13, 2011 If there is no friction or other losses the ball will bounce back to its original position. i.e. to a height of 3m. In the absence of friction and other losses I would expect the ball to bounce between these two positions indefinitely. (It is a theoretical/hypothetical situation). Yes. That is why I put distance C in the question. If one answers "distance C= blahblah" it means he was driven into the wrong reasoning by the wrong diagram. Link to comment Share on other sites More sharing options...
John Cuthber Posted November 13, 2011 Share Posted November 13, 2011 Yes. That is why I put distance C in the question. If one answers "distance C= blahblah" it means he was driven into the wrong reasoning by the wrong diagram. Unless one wasn't. One might spot the joke and play along with it. Link to comment Share on other sites More sharing options...
swansont Posted November 13, 2011 Share Posted November 13, 2011 Yes. That is why I put distance C in the question. If one answers "distance C= blahblah" it means he was driven into the wrong reasoning by the wrong diagram. You also asked for the height, and that part was correct. Link to comment Share on other sites More sharing options...
michel123456 Posted November 13, 2011 Author Share Posted November 13, 2011 (edited) I am really sorry. I didn't want to hurt anyone's ego. You also asked for the height, and that part was correct. Yes. O.K. Shame on me. John was correct. (...) and C is 3/tan (42) I think i.e. about 3.33 m. (...) When I am wrong, I say "sorry, I was wrong". And that's it. Everybody has the right to error, the error is human, only those who make nothing never make any error, etc. that is my position. John C. sinked in the abyss of my unconsideration for his posts #6 & #14. Not for his sympathetic post #2 that helped me in my argument. No other comment. Edited November 13, 2011 by michel123456 1 Link to comment Share on other sites More sharing options...
md65536 Posted November 27, 2011 Share Posted November 27, 2011 My take on this: F: To say that the height the ball will reach after reaching its minimum is 3m assumes that the ball is perfectly elastic and there is no loss of energy anywhere, which I don't think falls under the umbrella of "excluding any friction". However the elasticity etc is not specified, and the "maximum" height is asked for, so I think it's fair to assume best-case conditions of no loss for "maximum possible height". Technically, the question doesn't even specify that the "max height" is only after reaching its minimum, and the ball is already at 3m at the start so its max height cannot be any less than that. C: Typically a ball will roll down the incline and acquire angular momentum, which might allow it to roll up the incline after hitting the "wall". To see this effect, roll a rubber ball along the floor against a flat wall, and the ball can bounce with some vertical component to its velocity. Without knowing the properties of the ball etc it would be impossible to calculate C. However, this doesn't apply to the puzzle because it relies on friction in order to convert the ball's angular momentum into linear momentum. Excluding all friction, the ball would slide down the incline, but it would also roll (I think) because the downward gravitational force is the same throughout the volume of the ball but the upward force of the plane is off-center, allowing the ball to constantly keep tipping faster off balance as it slides down. This is beside the point of the puzzle though. Link to comment Share on other sites More sharing options...
michel123456 Posted November 28, 2011 Author Share Posted November 28, 2011 To be correct, without friction the ball will not roll, it will slide. I could have put a square in place of the ball. Link to comment Share on other sites More sharing options...
md65536 Posted November 28, 2011 Share Posted November 28, 2011 (edited) To be correct, without friction the ball will not roll, it will slide. I could have put a square in place of the ball. On an incline steeper than 45 degrees, a square shape (as in a side view like the diagram's) will always be off-balance, and I believe should always tip over even while it's sliding, so it should develop a rolling angular momentum even though it is sliding frictionlessly. For steeper inclines: The rate of acceleration of tipping or roll would approach 0 as the acceleration approached g, which would (only?) happen if the angle of the ramp approached 90 degrees. Anyway the rolling momentum would be very little on a short ramp as in the example, especially with a square shape starting flat against the ramp. On any incline a ball should be always off-balance. If there's no friction then the ball's roll should have no effect on its linear movement (the movement could be entirely described as sliding even if it is also rolling). If we replace the ball with a square, and then replace the ramp with stairs, we should be able to describe setups where a block still slides down the stairs, or where it only tips and rolls down the stairs (and slides too?, maybe necessarily), or where it doesn't move at all. Sorry, this is getting further off topic... I think it's interesting but it doesn't make a difference to the puzzle. Edited November 28, 2011 by md65536 Link to comment Share on other sites More sharing options...
md65536 Posted November 30, 2011 Share Posted November 30, 2011 (edited) To be correct, without friction the ball will not roll, it will slide. I could have put a square in place of the ball. Sorry, I was wrong. You are correct. http://www.newton.de...05/phy05139.htm My reasoning is bad because the ramp doesn't just provide an upward force that counteracts gravity (preventing the ball from accelerating at free-fall speeds), it also provides a horizontal force that causes the ball to accelerate horizontally too. It seems that the composition of these two force components will act normal to the ramp?, which for a ball will make a line between the point of contact and the ball's center of mass. This removes any "off-balance" force that could cause the ball to tip. However I don't think the same reasoning applies to a square block in general, if the block could be positioned with a corner (single point of contact) against the ramp, which could be done so that the normal force doesn't intersect the center of the block, then the normal force could provide some rolling torque to the block. In the typical starting case (probably also a stable state if the block was allowed to slide indefinitely), with the edge of the block flat against the ramp, any amount roll would cause one of the corners to become a "pivot point" which would tip the block back toward "flat against the ramp", so I guess even if it started off-balance on a corner then a rolling angular velocity couldn't accumulate for long or something, and would instead oscillate while approaching 0? Just like a block that is off-balance on a horizontal surface will roll, but only so much that it will rock back and forth and settle into a state with 0 roll. A block that is flat against the ramp will not be off-balance, no matter what the angle of the ramp is, because the force acting on the block and ramp is normal to the ramp. Edited November 30, 2011 by md65536 Link to comment Share on other sites More sharing options...
J.C.MacSwell Posted January 2, 2012 Share Posted January 2, 2012 A ball A is placed on a ramp of 48 degrees. What max. height F and distance C will reach the ball, excluding any friction? The ball is 3 kg weight, 20 cm in diameter. Note: this is not homework. The ball will reach 3.1 metres Link to comment Share on other sites More sharing options...
michel123456 Posted January 2, 2012 Author Share Posted January 2, 2012 The ball will reach 3.1 metres You didn't notice post #5. Link to comment Share on other sites More sharing options...
J.C.MacSwell Posted January 3, 2012 Share Posted January 3, 2012 (edited) You didn't notice post #5. I did. I know it bounces back, but it has potential to "reach" 3.1 when it returns (c.g.) to 3.0. since it has a diameter of 20 cm. It is, after all, touching 3.1 at the start. Post #5 really only pertains to the horizontal displacement... ...and we are in the brain teasers & puzzles section. Edited January 3, 2012 by J.C.MacSwell Link to comment Share on other sites More sharing options...
TheOnlyMaster Posted January 10, 2012 Share Posted January 10, 2012 I'm amazed how many people didn't realize the angles were wrongly displayed on the diagram, it was the first thing I noticed. The correct answer is that the ball would just bounce back to it's original position. Link to comment Share on other sites More sharing options...
michel123456 Posted January 10, 2012 Author Share Posted January 10, 2012 I'm amazed how many people didn't realize the angles were wrongly displayed on the diagram, it was the first thing I noticed. (...) That is because you are The-Only-Master. Link to comment Share on other sites More sharing options...
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