TimeSpaceLightForce Posted March 27, 2015 Report Share Posted March 27, 2015 J -- -- -- --o | | | | | | o Attached at ends of a one unit long and weightless cord are two identical balls with a unit mass. If the cord slides on hook J w/out friction.. How far can the higher ball can be pulled horizontally away from J ,so that when released it can hit the lower ball? 1 Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted March 29, 2015 Report Share Posted March 29, 2015 I assumed a constant tension on the string. Looking at conservation of energy is simpler, but naively applying it gives an initial sideways distance of 1/2. That would mean the lower weight would fall, at first, then rise back to its initial height. Is that the right answer? Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted March 28, 2015 Report Share Posted March 28, 2015 If I modeled it correctly, The higher ball will have fallen 84.44% of its horizontal distance when its string reaches vertical. So if it is pulled a distance of .5921 from the hook it will fall a distance of .5 when it reaches vertical. The balls will then collide. Nice puzzle. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted March 28, 2015 Report Share Posted March 28, 2015 A little more fun: Let the lower mass be a fraction f of the upper mass. What are the values of f for which the upper ball should be pulled distances of [a] 0.5 and 1.0 from the hook in order for the balls to collide? 1 Quote Link to comment Share on other sites More sharing options...
0 TimeSpaceLightForce Posted March 28, 2015 Author Report Share Posted March 28, 2015 If I modeled it correctly,The higher ball will have fallen 84.44% of its horizontal distance when its string reaches vertical. So if it is pulled a distance of .5921 from the hook it will fall a distance of .5 when it reaches vertical. The balls will then collide.Nice puzzle. Is the lower ball at rest when they collide? Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted March 28, 2015 Report Share Posted March 28, 2015 Is the lower ball at rest when they collide? No. Since the upper ball ends up closer to the hook than when it started, the lower ball is moving downward. Quote Link to comment Share on other sites More sharing options...
0 TimeSpaceLightForce Posted March 29, 2015 Author Report Share Posted March 29, 2015 Is the lower ball at rest when they collide? No. Since the upper ball ends up closer to the hook than when it started, the lower ball is moving downward. Yes, the lower ball moves downward..but it decelerate to v=0 as the opposing up pull cause by the swinging ball becomes equal to its weight upon impact (both vertical)..So the raised lower ball start from v=0 to v=max to v=0. Unless with experiments, I'm unsure about my model.. Since potential energy = kinetic energy of the system J | | | | | | | | | | | | | | | | | | | | oo 1/2L initial energy =0 J --- --- --- --- --- --- --- --- --- ---o Ep=mg(1/2) | | | | | | | o Ep=mg(h) h -- raised energies: mg(1/2+h) = 1/2m V^2 (swinging ball) ...a perpetual motion Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted March 29, 2015 Report Share Posted March 29, 2015 I will recheck my model. Momentum comes into play. If we assume the string is not sliding in the hook at impact, then the lower ball is stationary. But we can't just assume it, we somehow have to derive it. Quote Link to comment Share on other sites More sharing options...
0 TimeSpaceLightForce Posted March 30, 2015 Author Report Share Posted March 30, 2015 I assumed a constant tension on the string. Looking at conservation of energy is simpler, but naively applying it gives an initial sideways distance of 1/2. That would mean the lower weight would fall, at first, then rise back to its initial height. Is that the right answer? Maybe you are right. And I don't rely much on the ff: a) the swinging ball is 1/2+h away from when released. b) its distance from the hook always > 1/2 before impact. c) it has no vertical velocity on impact d) it follows a curve non radial or radial but on different center e) the sliding ball move downward only f) the tension on string is lowest at the instant of release g)the given values on the problem are complete I post the same problem on a science forum (classic physics section) just to comfirm a thing. A member claimed to have done an experiment where the lower ball remain at rest until collision and that is to prove the earlier post of a moderator. But I doubt if there had been an experiment, and still hoping for the calculated values. Yes , a nice puzzle bonanova..thanks Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted March 31, 2015 Report Share Posted March 31, 2015 A very, very cool puzzle, indeed. This arrangement of weights and pulleys is known as the Atwood Machine. When one of the weights is allowed to swing in pendulum fashion it's called the Swinging Atwood Machine. Its analysis is difficult, and yields best to Lagrangian methods. The motion of the swinging weight is governed by the tension in the string, gravity, and a centrifugal force of varying magnitude. The path can be chaotic or periodic, and its shape changes qualitatively depending on the ratio of the hanging and swinging weights. The swinging weight acquires an initial downward momentum during the part of its path when its fall is unopposed by the force of the hanging weight (when the cord is horizontal) When the mass ratio is unity, the force of the hanging weight never overcomes this downward momentum, with the result that the swinging weight continues to descend until the hanging weight rises, and eventually hits the pulley. For the equal-weight case, the length of the cord of the swinging weigh decreases, initially, owing to the initial tension in the cord. Ir reaches its minimum length of 79.76% at an angle of 54 degrees. As it acquires velocity, however its centrifugal force overcomes the tension, and its cord length increases. At vertical, it has lengthened to 87% of its original length. Because of its continuing downward momentum, its path is only approximately horizontal at this point: it has a small downward component. For a unit-length string, the weights will collide if the swinging weight has initial cord length of .5/.87 = 0.574. As it continues swinging, its cord length continually increases, until the hanging weight eventually rises at hits its pulley, after which the swinging weight becomes a simple pendulum. If the hanging weight is only slightly heavier, however, it will eventually overcome the downward momentum of the swinging weight, which, while still swinging, will begin to rise, eventually to attain its original height. If this occurs at an extremity of its swinging motion, it will retrace its path, and the motion will be periodic. In general this does not happen and the motion is aperiodic. I used my simulation model to verify the case for equal weights that the swinging weight has a slight downward component at vertical. Then I did some searching on the web and eventually found a delightful java program that plots things for various input parameters. That model uses two pulleys, just so the hanging weight does not interfere with the swinging weight. Here is the start: the ratio of the hanging weight to the swinging weight is 1.0 When the weight reaches vertical the motion is nearly horizontal, but with a slight downward component. The cord length has increase to 87% of initial length. Later, it's still falling. Eventually the hanging weight hits the hook. Now, slightly increase the ratio of the hanging to swinging weight to 1.119. After 3 1/2 swings, it hits its lowest point. It then rises, and comes back to horizontal, but on the opposite side of the hook It then falls back, retraces its path, and continues in periodic motion. Now make the weight ratio = 1.2 The weight does not fall as far before its downward momentum is reversed. And for this ratio, the motion is aperiodic. This is after about 14 oscillations. Now make the weight ratio = 2.4 To expand the figure, I made the swinging cord length twice as long. The heavier hanging weight creates enough centrifugal force to launch the swinging weight over the top, creating a really weird, but very pleasant, smiling face. And, for this ratio, the motion is periodic. Quote Link to comment Share on other sites More sharing options...
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