encipher Posted December 18, 2006 Posted December 18, 2006 Hi, I am working with a couple of highschool students on a physics project their class is doing. They have to construct a yoyo that will fall a distance of 2 meters in the GREATEST amount of time. Mechanical friction cannot be used as a means of slowing the yoyo down. The maximum allowed mass is 1000grams and the maximum diameter / length is 20cm. We came up with a square shape and a thin rod in between. We acheived a time of about 3 minutes. Now today one of the students talked to me about the 'competition'. Apparently a rivaling group claimed to have gotten the yoyo slowed down to 45 minutes over the 2 meter drop. I have thought about how this is possible without friction being a factor and within the guidlines of the rules, but have yet to come up with a workable idea. Any thoughts on the setup they might have used? Also,it cannot be lighter than air. Thank you
swansont Posted December 18, 2006 Posted December 18, 2006 Hi, I am working with a couple of highschool students on a physics project their class is doing. They have to construct a yoyo that will fall a distance of 2 meters in the GREATEST amount of time. Mechanical friction cannot be used as a means of slowing the yoyo down. The maximum allowed mass is 1000grams and the maximum diameter / length is 20cm. We came up with a square shape and a thin rod in between. We acheived a time of about 3 minutes. Now today one of the students talked to me about the 'competition'. Apparently a rivaling group claimed to have gotten the yoyo slowed down to 45 minutes over the 2 meter drop. I have thought about how this is possible without friction being a factor and within the guidlines of the rules, but have yet to come up with a workable idea. Any thoughts on the setup they might have used? Also,it cannot be lighter than air. Thank you Ah, but can it be as light as air? Is air resistance counted as mechanical friction? If not, cross-vanes (making it look like a water-wheel) will help slow it down and dissipate energy. You want to maximize the moment of inertia, so that it takes up the maximum energy with minimal rotation speed. You also want to choose your axle to maximize rotation at the expense of linear motion.
encipher Posted December 18, 2006 Author Posted December 18, 2006 Hi, Well, their paper says it must be heavier than air, and no, air resistance cannot be a factor. What we went with was a Square, with the insides cut out, except for rods to connect the axle to (which is very thin) that way we have a large moment of inertia. Now with that design, I mathematically calculated it to be something around 147 seconds (using conservation of energy) and the test was pretty close. Now I don't know what could possibly give 45 minutes.
EvoN1020v Posted December 18, 2006 Posted December 18, 2006 Maybe you could input gum at the ends of the axle to slow down the rotation speed?
encipher Posted December 18, 2006 Author Posted December 18, 2006 As I mentioned earlier, one of their requirements is that no mechanical friction can slow the yoyo down, that includes peices of gum stuck on the axel any other thoughts?
swansont Posted December 18, 2006 Posted December 18, 2006 Why a square and not a circle? The only thing I can think of to extend the time is to come up with some kind of winding scheme to make the yo-yo partly rewind itself before the two meters have been reached.
ecoli Posted December 18, 2006 Posted December 18, 2006 this is strange, neglecting friction, all objects fall at the same rate. How is exercise possible without using friction?
insane_alien Posted December 18, 2006 Posted December 18, 2006 ecoli, remember its a torque that gets applied, there is tension in the string.
Jacques Posted December 18, 2006 Posted December 18, 2006 Use a round geometry, concentrate the mass on the edge and minimize the axel diameter, maximize the yoyo diameter. Use 2 large rings, a few spokes to connect the rings to the axel.
encipher Posted December 18, 2006 Author Posted December 18, 2006 Why a square and not a circle? The only thing I can think of to extend the time is to come up with some kind of winding scheme to make the yo-yo partly rewind itself before the two meters have been reached. Because the moment of inertia of a square is greater than that of a circle (within the given dimentions) I calculated the moment of inertia of the square to be [math]\frac{1}{3}MA^2[/math] Where A is the length of one side. As opposed to that of a hoop which would be: [math]MR^2[/math] where R is the radius of the hoop. Since the maximum length of the side or radius can be 20cm.. and the mass doesnt matter (sine according to conservation of energy it cancels out).. We figured the square shape would be the shape with the largest moment of inertia.
Rocket Man Posted December 19, 2006 Posted December 19, 2006 does the yoyo need to come back up? i think you're getting stuck on linear inertia, radial intertia is more along the lines of levers. if the mass is concentrated around the rim and the spindle is 1 cm radius, the mass will need to travel 20x faster than the centre of the yoyo. so you then divide G by 20 to get your final acceleration, roughly 0.5 m/s. if the mass were closer in, the ratio is proportionally less so the mass needn't accelerate as much so the yoyo accelerates faster. mechanical dampenning is out... how about electrical dampenning... use an inductive load on a counter-rotating wheel. kinetic energy is turned into heat via ohmic heating so it's not "mechanical friction"
encipher Posted December 19, 2006 Author Posted December 19, 2006 no, the yoyo does not need to go back up. Yeah, concentrating mass on the edge is key, but really making the yoyo heavier will not matter, since mass cancels out according to the following equation: [math]mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2[/math] therefore what really matters is what the radius of the object is (for its moment of inertia). Or maybe I am wrong, any thoughts?
swansont Posted December 19, 2006 Posted December 19, 2006 Because the moment of inertia of a square is greater than that of a circle (within the given dimentions) I calculated the moment of inertia of the square to be [math]\frac{1}{3}MA^2[/math] Where A is the length of one side. As opposed to that of a hoop which would be: [math]MR^2[/math] where R is the radius of the hoop. Since the maximum length of the side or radius can be 20cm.. and the mass doesnt matter (sine according to conservation of energy it cancels out).. We figured the square shape would be the shape with the largest moment of inertia. Ok. I had interpreted maximum dimension to be the diagonal, in which case the circle wins, but if it's the side length then you are indeed correct. The mass does matter in the limit of low density, where you can't ignore the bouyant force of the air.
encipher Posted December 19, 2006 Author Posted December 19, 2006 yeah, of course the density will play a part in the challenge, but nothing as significant as making the yoyo take 45 minutes. I have tried all possible solutions, but nothing comes to mind.
Jacques Posted December 19, 2006 Posted December 19, 2006 I calculated the moment of inertia of the square to be [math]\frac{1}{3}MA^2[/math] Where A is the length of one side. How did you found that ? I google "moment of inertia and I found [math]\frac{1}{6}MA^2[/math] on that site http://www.physics.uoguelph.ca/tutorials/torque/Q.torque.inertia.html If you bring A to R you have [math]\frac{2}{3}MR^2[/math] I am not very good in math but I would like to know theorically how long it will take the yoyo if we use a 2 mm radius axel with the cylinder geometry [math]MR^2[/math] Thanks
Jacques Posted December 19, 2006 Posted December 19, 2006 Ok I will give it a try using g=10m/s^2 the force is mg = 1kg*10m/s^2 = 10 Newton the torque t on the axel of 1mm radius = F * r = 10 N * .001 m = .01 N.m the moment of inertia for a 10 cm radius cylinder is MR^2 = 1kg * (.1m)^2 = .01 angular acceleration = torque / moment of inertia = .01/.01= 1 rad/sec^2 on the axel 1mm = 1 radian then 2m = 2000 radian 2000 radian = 1/2 angular acceleration * time^2 sqrt 4000 = time = 62sec It look like a short time... What do you think of it ?
encipher Posted December 19, 2006 Author Posted December 19, 2006 Well, its kinda messy, but it looks ok. Yes the moment of inertia of a square is 1/6 MA^2. But I mentioned earlier that I had cut out the inside of the square, making it like a hoop. But square. and according to the general formula: [math]I = \int r^2 dm[/math] I derived that moment of inertia (if you would like to see me work i would be glad to post it. ) Now it would actually be that + [math] 4 (\frac{1}{3}ML^2)[/math] for the 4 spokes that connect it to the axle. I talked to a few of the students, and they mentioned that their friends said something about a counterweight being used to achieve the slow speed. Anyone have any thoughts on that?
Rocket Man Posted December 20, 2006 Posted December 20, 2006 sorry, i thought you meant 20cm max radius, a square has a max width of ~28cm, so the optimum would be to concentrate the mass at the corners of a square of side length 20cm. try to get as close as possible to the 1kg, the more the mass is concentrated, the less the mass of the spindle effects the descent. if you had a winding arrangement with a pin such that it moves down a distance, then back up over and over again, unavoidable natural losses can be utilised. a counter weight can be used to lock a gearing system to vertical, the spindle turns relative to the counter weight, transferring the torque through the gears to spin the yoyo X times faster than normal dividing the descent rate by the gearing ratio.
Glider Posted December 20, 2006 Posted December 20, 2006 Why not use a counterweight? As the yoyo moves down. spinning out on its string, have another string attached to the axle pulling a counterweight up. The balance between the weights will determine how fast the yoyo goes. If they are nearly equivalent, the yoyo will move very slowly. You might need to use a long weight on two strings that attach each side of the main yoyo string to keep it balanced, but that's easy enough. As long as the yoyo is slightly heavier than the counterweight, it will move down and wind the counterweight up. You'd just have to play with the different weights a bit. Just a thought.
Jacques Posted December 20, 2006 Posted December 20, 2006 So if my calculations are good, 62 sec is far less than what encipher did and by a factor that is more than any improvement only by work on the geometry. If you do the calculation with the square geometry, you will still have less than the 3 minutes you achive and by the others result. Some other factor made the difference between the theory and the experiment. I came up with the concept of a yoyo that will take more than an infinite time to travel the 2 meter But it is agains the defenition of a yoyo. A yoyo must rotate. I thought that an inbalance in the real yoyo made the difference. A small imbalance will displace the center of mass from the geometric center. The torque will vary like an offset sine. Chirstmass diner is waiting will continue later....
encipher Posted December 20, 2006 Author Posted December 20, 2006 Actually, my calculation came up to be pretty close to what I got in real life. i started off with the folowing: [math]mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2[/math] by subsituting in [math]\frac{v}{r}[/math] for omega, where V is the linear velocity of the yoyo and r is the radius of the AXLE. and subsituting in [math]\frac{1}{3}MA^2[/math] for the moment of inertia (I) the Mass cancels out, and A is the length of one side of the square. Re-arranging the equation for the velocity will give you the following: [math]\large v^2 = \frac{gh}{\frac{1}{2} + \frac{A^2}{6r^2}}[/math] then using kinematics, the general formula [math] v_f^2 = v_i^2 + 2ax[/math] We know that the initial velocity is zero so that cancels out Solving for a and plugging in the above equation gives: [math]\large{a = \frac{gh}{2h(\frac{1}{2} + \frac{A^2}{6r^2})}}[/math] Then Plugging in all the numeical values: g = 9.81 meters/s/s h = 2 meters A = 20 centimeters = .2 meters r = .5 millimeters = .0005 meters The acceleration is: .000184 meters/s/s (Yes, I hate sig figs) Then, using the equation [math] x = v_it + \frac{1}{2}at^2[/math] subtituting in the above value for a, and 2 for x, and solving for t I get the following: 147.47 seconds.. divide by 60 find remainder, i get: 2 minutes and 27 seconds. Very close the the experimental value acheived. I will be attempting to setup a counterweight yoyo and will post my results.
Jacques Posted December 20, 2006 Posted December 20, 2006 Ok one big difference is that you use a 1mm diameter axel and in my setup I used a 2mm axel. My idea of moving the center of mass out of the axel will create a two time cycle: The first half of the cycle begin when the center of mass is just pass over the axel, the torque is in the direction to unwind the yoyo. The rotation speed increase and also the linear speed. When the center of mass is at the lowess point in the rotation the torque is zero and the rotation bring the center of mass on the other side of the axel. The torque is in the opposite direction slowing the rotation until the center of mass pass over the axel again. Hope you understand the idea The first half cycle convert potential energy to angular momentum and the second half of the cycle convert back some angular momentum back to potential energy. I think that by fine tunning the unbalance you can achive any speed. If to much unbalance the yoyo won't go down so there is a limit to the unbalance. I would use a perfectly balance yoyo an used some putty to unbalance it and do some experiment to see how much putty to add and where to add it. Important to keep it balanced horizontally
swansont Posted December 20, 2006 Posted December 20, 2006 The first half cycle convert potential energy to angular momentum and the second half of the cycle convert back some angular momentum back to potential energy. Careful, you're mixing terms here. You can convert potential energy into rotational and translational kinetic energy in this system, but you only change angular momentum by exerting a torque.
Jacques Posted January 3, 2007 Posted January 3, 2007 You are right Swansont Encipher any new developpement in your slow yoyo competition ? Thanks
encipher Posted January 7, 2007 Author Posted January 7, 2007 Yes actually, I spoke to them after I came back from my vacation. The design they were talking about was an axle, and two cylinders on each end. Inside the cylinders on each end was water, and a series of segments. (imagine a waterwheel, but closed and there is water in one of the sections.) as the yoyo rotates, there are tiny holes in each flaps that allows water to flow slowly from one section to the other. That is how it took a very long time to reach its destination.
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