kingwinner Posted April 6, 2006 Report Posted April 6, 2006 I have some questions relating to proportionality and circular motion. 1) Given a set of data points:x-values: 1.2, 2.5, 3.4, 4.2, 5.5y-values: 11.38889, 10.32, 10.17301, 10.113379, 10.06612 I know that for example if y=2(1/x^2), y is definitely proportional to 1/x^2, what if there is a y-intercept that is not zero? I found that the equation that relates the above data points is y=2(1/x^2) +10, the y-intercept is not zero, does that still mean y is proportional to 1/x^2 ? 2) Has anyone done this lab before? ;) In this UNIFORM circular motion lab, I have to measure the time taken for 20 cycles, varying the number of washers hanging at the bottom. The radius measured to be 50cm at the beginning. The mass of the rubber stopper is constant at 11.94g. I need to plot a graph of force (force of gravity of the washers) vs speed and try to find the relationship. But I don't get how this lab is a controlled experiment. How fast the rubber stopper is moving depends on how much (how violently) you whirl the string, right? What I mean is that your hand, instead of the weight of the washers, is controlling the speed of the rubber stopper. Then, what's the point of doing all this experiment? How is it possible to find the relationship between force and speed if speed depends on how violently you whirl it? Thank you for answering! ;) Quote
Qfwfq Posted April 7, 2006 Report Posted April 7, 2006 does that still mean y is proportional to 1/x^2 ?Strictly speaking, it isn't. But, if there is a known uncertainty on the data (which typically is the result of measurement), this uncertainty can be propagated to the intercept value. Therefore, if that intercept is actually 10 + or minus 15, for instance, it is compatible with zero so the law could be of proportionality. How is it possible to find the relationship between force and speed if speed depends on how violently you whirl it?Methinks the thing would settle into an orbit in which radius and velocity give a force balanced by the weight down below. Remember angular momentum conservation; if the whirl is insufficient, the decreasing radius will speed the whirl up. Quote
kingwinner Posted April 10, 2006 Author Report Posted April 10, 2006 2) This is a uniform circular motion lab, the mass and the radius are kept constant, and we time the 10 cycles as soon as the speed of whirling becomes constant. We repeat this using different number of washers. And then we are to determine the relationship between the force of gravity of the washers and the speed of the rubber stopper. "Espeically when using a low number of washers, the radius is not 50cm. but less than this. Explain why." (Note that in my lab, I were to fix the radius at 50cm.) I was given this problem and asked to explain why. Well, I really can't think of a reason why...does anyone have any idea? 3) The moon orbits the Earth in a nearly circular path without the aid of a string. What force causes the moon to orbit the Earth? If we were to remove this force, what would happen to the moon? If we somehow decreased the moon's mass, what would happen in the long term to the moon as compared to the Earth? I can only answer the first question :cup: ...it's the force of gravity between the moon and the Earth that causes the moon to orbit the Earth. But what will happen if this force is removed? Will the moon fly out of its orbit? in what direction? And what will happen if the moon's mass is decreased? Will the moon fly out, have a larger orbit radius, or the same orbit radius? How can I justify these answers better, instead of doing it intuitively? Quote
kingwinner Posted April 17, 2006 Author Report Posted April 17, 2006 2) This is a uniform circular motion lab, the mass and the radius are kept constant, and we time the 10 cycles as soon as the speed of whirling becomes constant. We repeat this using different number of washers. And then we are to determine the relationship between the force of gravity of the washers and the speed of the rubber stopper. "Espeically when using a low number of washers, the radius is not 50cm. but less than this. Explain why." (Note that in my lab, I were to fix the radius at 50cm.) I was given this problem and asked to explain why. Well, I really can't think of a reason why...does anyone have any idea? 3) The moon orbits the Earth in a nearly circular path without the aid of a string. What force causes the moon to orbit the Earth? If we were to remove this force, what would happen to the moon? If we somehow decreased the moon's mass, what would happen in the long term to the moon as compared to the Earth? I can only answer the first question 0.o ...it's the force of gravity between the moon and the Earth that causes the moon to orbit the Earth. But what will happen if this force is removed? Will the moon fly out of its orbit? in what direction? And what will happen if the moon's mass is decreased? Will the moon fly out, have a larger orbit radius, or the same orbit radius? How can I justify these answers better, instead of doing it intuitively?Can someone please help me? I have no clue... Quote
kingwinner Posted April 17, 2006 Author Report Posted April 17, 2006 2) I was thinking, the radius is measured horizontally and the string can't be exactly horizontal, so at any point the radius would be less than 50cm. But where can I go from there? Would it have a smaller (horizontal) radius if the centripetal force (which depends on number of suspended washers) is smaller? That is, the string's angle to the horizontal will be larger if the speed is slower? How can I prove this mathematically? 3) Since centripetal force=force of gravity in this case.Fc=Fgmv^2/r = GmM/r^2 where m is mass of moon and M is mass of Earthv^2=(G)(M)/r Does this suggest that NOTHING will happen (i.e. speed and radius remains the same?) if the moon's mass is decreased? I have another question...just curious :eek_big: ...what would happen if we decreased the mass of the earth? v^2=(G)(M)/r, how can I use this equation to predict the result? G is constant, and we decreased M, but then we are left with 2 variables (v and r), how can I know what would happen? I can solve 1 equation with 2 variables...... Quote
CraigD Posted April 17, 2006 Report Posted April 17, 2006 I’ll answer the easiest questions first3) The moon orbits the Earth in a nearly circular path without the aid of a string. What force causes the moon to orbit the Earth?As kingwinner already answered, the force of gravity.If we were to remove this force, what would happen to the moon?It would behave as does any object not subject to a force – it would travel in a straight line at an unchanging speed. The direction of its travel would depend on the instant the force was removed. The moon is not only subject to gravitational force from the Earth, but from the Sun, and to a lesser extent, the other planets, minor solar bodies, and distant, extra-solar bodies. If only the force of gravity due to the Earth were removed, the moon would orbit the Sun in an elliptical path determined by the instant the Earth’s force was removed. Only if all gravitational force was removed would it move in a straight line.If we somehow decreased the moon's mass, what would happen in the long term to the moon as compared to the Earth?(Almost) nothing. Since both the inertia of the moon and the force exerted on it by the earth are proportional to the moon’s mass, increasing or decreasing it would have almost no effect on its orbit. Note the “almost” in the above. If we consider the mass of the moon to be negligible compared to the mass of the Earth, we can ignore this “almost”. Otherwise, we must consider the slight change in the center of gravity of the Earth-moon system when reducing the moon’s mass, and predict a slight increase in the difference between the lengths of the major and semimajor axes of its orbit – that is, that its orbit would become a slightly less circular ellipse. To kingwinner’s original question:The first step is to more carefully define your terms – diagramming them, if that’s helpful.What I mean is that your hand, instead of the weight of the washers, is controlling the speed of the rubber stopper. Then, what's the point of doing all this experiment? How is it possible to find the relationship between force and speed if speed depends on how violently you whirl it?Imagine that, rather than whirling the stopper by holding the tube, the tube is fixed, and you carefully launch the washer by hand in as nearly a circular path as possible. Would this be any different than whirling the washer, then holding it still and measuring its period? In other words, when you’ve reached roughtly the speed you desire, stop whirling, and hold the tube still. Quote
kingwinner Posted April 18, 2006 Author Report Posted April 18, 2006 Hi CraigD,To answer this question:""Espeically when using a low number of washers, the radius is not 50cm. but less than this. Explain why." (Note that in my lab, I were to fix the radius at 50cm.)" Now I understand why the radius is less than 50cm for all of my trials (e.g. 8 washers, 12 washers, 40 washers, etc...), it's because the string is actually a bit below horizontal. But I am still not too sure why the radius would be particularly small when a low number of washers are used. For a particular number of washers, say 8 washers, the angle theta of the string below horizontal would stay constant because we whirl the stopper at a constant speed. I understand that. But in this experiment, we actually varied the number of washers, i.e. speed is not constant throughout the experiment. In the case where 40 washers are used, would angle theta of the string below horizontal be smaller than that of 8 washers? Say, for example, 8 washers --> theta=30 degrees ?40 washers --> theta=10 degrees ?(will be less than the angle for 8 washers?) Note that, we are whriling faster when using 40 washers to try and maintain the string's length (not actaully radius) to be 50cm and I sense that as we whirl the stopper faster, the stopper stays higher, and if we whirl it slower, the stopper drops to a lower point (i.e. theta is larger)......I don't know if that is ture, and how can I justify this mathematically? Quote
kingwinner Posted April 18, 2006 Author Report Posted April 18, 2006 (Almost) nothing. Since both the inertia of the moon and the force exerted on it by the earth are proportional to the moon’s mass, increasing or decreasing it would have almost no effect on its orbit. hi, CraigD, To keep this simple because this is part of a uniform circular motion lab, so I should have to assume the moon's orbit is circlar. Mathematically, is this the right way to show that if we decreased the moon's mass, nothing will happen? Since centripetal force=force of gravity in this case.Fc=Fgmv^2/r = GmM/r^2 where m is mass of moon and M is mass of Earthv^2=(G)(M)/r Quote
CraigD Posted April 19, 2006 Report Posted April 19, 2006 Mathematically, is this the right way to show that if we decreased the moon's mass, nothing will happen? Since centripetal force=force of gravity in this case.Fc=Fgmv^2/r = GmM/r^2 where m is mass of moon and M is mass of Earthv^2=(G)(M)/rYes. You could take it a step further, and rearrangemv^2/r = GmM/r^2intor = GmM/(mv^2)simplify intor = GM/v^2and clearly show that r doesn’t depend on m. Quote
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