Shoot I took freshman physics and then a bunch after that, but it was a long time ago. Assuming that there is no slipping (which is why maybe d is right), the yo-yo will roll left unwinding further, right? Won't it create a moment around the center of mass (assuming equal mass distribution) creating counter clockwise motion. But if the force was so great so as to create horizontal motion to the right un-counter-acted by the rolling to the left (we need to know coefficient of friction) then it will be spinning as if to unwind itself but still the center of mass will be moving to the right.
But I guess we might need to know the mass of the yo-yo, the force of the pull, and the coefficients of friction between the string and the yo-yo and the yo-yo and the surface. So my answer is d...final answer.
I'll never make it to grad school. :-(
I forgot to mention that I won't comment on any answers until tomorrow.
I think the angle has something to do with it too. No matter what angle you pull at, the moment that creates the spin (causing the net force to be left-ward) will be the same. A perfectly horizontal pull of the string will have a great tendency to have the yo-yo move right, while a perfectly verticle (angle =90 degrees) pull will have the yo-yo move left (zero force in the rightward direction). So there has to be some point between 0 and 90 degrees at which the yo-yo will go in different directions after the initial force is applied and that angle is determined by those things mentioned in answer d.
So my final answer is d.
I hope I pass.
I vote d also. In the limit of zero friction, the center of mass moves right while the unbalanced torque causes the yoyo to rotate CCW. In the limit of infinite coefficient of friction, the yo-yo must roll without slipping, which means we can use the point touching the table as the center of moments. In that case, the torque from the pull leads to CW rotations. Either way, the yo-yo moves right, but it may either unwind or wind up.
Strangely, my intuition tells me that the yo-yo simply refuses to move.
d.
I'll have to see if I can draw up the final question from my Statics final. It's essentially the same problem, but it looks very different. Two students got it right. I was the "other one."
Didn't anyone see the Matrix? There is no yoyo.
Heh...Assuming the string is not completely wound out already...
(disclaimer..it is way too long since I did physics at uni - a decade or more)
I think (d) can be excluded as the mass and friction is irrelevant. (As friction has to be overcome in whatever direction it goes)
The real question is whether the forces exerted by pulling the string has more of a left or a right component. The higher the angle, the more to the left. (So I believe (c) can be excluded as an answer too) At the angle shown, I would guess it was going to head left, but I haven't put any formula into it.
So (b) for me David.
A gentle pull equates to no slipping between the yoyo and the surface. If there is no slipping the yoyo will move in the direction of the pull. Since the string is "attatched" to the lower part of the yoyo the string will wind up. This is regardless of the angle of the string as long as the pull is to the right and it is gentle enough.
I vote a)
I am a homeschooled fourth grader. My Dad showed this to me. I vote b since the picture looks like the yo-yo has enough string to unwind just a little. It will move to the left, then it will stop and answer a will happen if you would have pulled harder.
I believe it will "roll" left until your force vector is aligned through the center of the mass along the radius, at which point you will be pulling the yo-yo.
Until that time you are essentially 'pushing' the right side "up".
I think it is a trick question. Answers c) and d) both say “Cannot say WITH…” where one would expect them to say “Cannot say WITHOUT…” So my choice, based on the answers as written, is ‘d) Cannot say WITH the mass and the coefficient of friction.’ The reason is that the mass and coefficient of friction have to be known, but that is still not enough information to solve the problem. The mass and the coefficient of friction information have to be specified in order to know if the yo-yo will initially roll left or right or will simply be dragged to the right without rolling at all. However, a “gentle pull” is not specific enough to know if the pulling force will be sufficient enough to overcome the combination of mass and friction. So it then becomes apparent that specifying the magnitude of the pulling force is also a necessary component in order to know if anything will move at all.
OK, so I suppose I missed the boat altogether and it really has something to do with the radii. However, I am absolutely certain that the correct answer depends on whether the experiment takes place in the northern or southern hemispheres and what is the phase of the moon. 
It's been a while since I've done any physics, so I'll admit I had to dust off my old textbook to figure this out, but I think I have the solution.
Four different forces are at work here: the normal force of the surface pushing up (N), the weight of the yoyo pushing down (should be equal to N), the friction between the yoyo and the surface (mu * N), and the tension on the string (T). I'll call the angle alpha, the inner radius r0, and the outer radius r1. Two of the forces produce torque about the center of the yoyo (the tension on the string and the friction). The torques can be equated: mu * N * r1 = T * r0. Two of the forces have horizontal components (again, the friction and the tension on the string). The friction is all horizontal, and the horizontal component of the string tension is T * cos(alpha), so the horizontal forces can be equated: mu * N = T * cos(alpha).
A funny thing happens when we divide these two equations: the normal force (N) and the coefficient of friction (mu) cancel out, giving us r0 / r1 = cos(alpha). This means the direction the yoyo rolls has nothing to do with mass or friction. It depends solely on the ratio of the two radii and the angle of the string.
Assuming the drawing is correct, I'd say the left side of this equation is 1/2. This means if the angle is less than 45 degrees the yoyo will move right, and if it's more than 45 degrees the yoyo will move left. The angle looks like less than 45 degrees to me, so my answer is (a) the yoyo will roll to the right.
sal was right that C and D should have said "without" -- but it was not a trick question--I hate trick question problems.
i'm confident in saying d) , because without knowing how much string there is, also effects the weight, and the heavier it is the more likely it is to roll towards the person and roll up, if it's too light becasue there's very little string on it it'll continue to unravel
I would answer (a).
In the frictionless case, there is no torque. Why? Because all the forces lie on two intersecting lines. Thus the yo-yo doesn't roll at all; it just slides right.
Friction will create a torque, but that torque will make the yo-yo roll to the right, possibly with some amount of slipping, depending on the coefficient of friction.
There are also some other cases not really suggested by the diagram, such as when the inner and outer radii are equal. In this case, it is possible that the tension of the string will not be enough to overcome friction, so the yo-yo will not move at all.
Anyway, that's what I think.
If it were to unroll to the left, then the net force on the wheel would have to be to the left. There are three forces acting: gravity, tension in the string, and friction. Because tension is to the right, the frictional force will have to be equal to the horizontal component of the tension. If the frictional force is equal to the tension, then the frictional force will exert a lot more torque because it is acting at a great radius. Frictional torque acts to spin the yoyo so that it rolls to the right. So there is a contradiction. If the inner radius is small enough, the yoyo has to roll to the right.
However, notice that the tension in the string has a vertical component. Thus if the ratio of the inner and outer radii are close, the tension exerts more torque than the frictional force because the frictional force only equals the horizontal component of the tension.
For small ratios the yoyo will slide forward. (Consider the case where the inner radius is zero!)
So that gives me c!
The yo-yo will roll right (a)
The yo-yo cannot roll left, unless the anchor of the string (your hand) moves left also. One revolution to the left will move the entire yo-yo a distance equal to the circumferance of the outer circle, but only unwind a length of string equal to the circumference of the inner circle (which is less). Since you are pulling on the string in the right direction, the yo-yo would need some outside force to cause it to spin fast enough (and slip on the surface) to unwind string faster than your hand is pulling.
Conversely, if the yo-yo were to roll to the right, it would 'wind up' the string, but each revolution would see the yo-yo moving a greater distance than the length of string that it winds. Thus the anchor at the end of the string would need to move right to maintain tension. This is consistent with a gently pull.
ratio of the inner and outer circles are unimportant, except to clarify that the inner is in fact smaller (but this is already implied by the diagram and the name).
mass and coefficient of friction are unimportant, unless friction was so low that the yo-yo could travel to the right while unwinding (spinning towards the left). Even in this situation, I would deem (A) to be the best answer.
WELL I TRIED IT AND IT ROLLED TO THE RIGHT
WHY IS THERE A PICTURE OF A CAR NEXT TO MY NAME?
ugliest car in history too...
It can't roll right because it would have to advance along the floor a greater distance than the length of string it winds up. The string would instantly go slack if it rolled right. You can see that by just inspecting the angle and the relative sizes of the circles. It can't roll right because the string can't pay out fast enough to keep up with the yo yo. You don't need to calculate torque or friction coefficients or any of that. Its completely intuitive if you just consider the consequences of the yo yo rolling. So the answer is that its neither a nor b and in the real world, you'll drag the stupid thing across the floor.
Sorry, I got a left mixed up with a righ in this. The correct version is this.
It can't roll right because it would have to advance along the floor a greater distance than the length of string it winds up. The string would instantly go slack if it rolled right. You can see that by just inspecting the angle and the relative sizes of the circles. It can't roll left because the string can't pay out fast enough to keep up with the yo yo. You don't need to calculate torque or friction coefficients or any of that. Its completely intuitive if you just consider the consequences of the yo yo rolling. So the answer is that its neither a nor b and in the real world, you'll drag the stupid thing across the floor.
I just realized this question is suckier than I first thought. If you had to answer it, then you have to assume by gentle tug, you meant you just deliver an initial impulse to the yo yo. In that case, it will roll right and the string will go slack and then it will stop rolling. So you're kind of screwing us over with the question, in typical pedagog style.
b.. duhhh
what an ugly car
Mark Catan: Have you ever tried this? Pull slowly and gently on the string and the yo-yo will wind up to the right. You do not have to assume an initial impulse force, and the string remains taut throughout. The question is a good one: (a).
D'oh! Its a. Sorry for my "pedagog" comment. Fair question. Think of the yo yo as a tower on an infinitessimal base. When a pull is exerted, it would tip the tower to the right. As you keep pulling (gently) you keep taking up slack and tipping agin. Even with a low friction coefficient, if the pull is gentle, it will roll to the right.
Andy: Yes you're right. It stayed on my mind while I was trying to sleep with a cold and I realized that, in principle, there's no resistance to the rightward roll. I regret saying it was a bad problem. I was just all wet. Its quite fine. In fact, I believe it will roll to the right even if you pull almost straight up and only angled slightly to the right, as long as there's a component in the rightward direction. You just have to pull very gently and the yo yo and the floor have to be very smooth.
I'm currently in my lat year of school so i may be wrong but from what i can work out is that:
it can't be a) for the very reason that the force is applied to the bottom of the inner circle which could only cause it to rotate in an anti-clockwise direction.
I don't think it can be b) either because say the limiting friction on the yo-yo(which would be acting to the left) was much larger than the force on the string multiplied by the cosine of the angle it is being pulled at, the yo-yo just simply wouldn't slip.
it can't be c) because it is irrelevant to this scenario because suppose the ratio of the inner radius to the outer radius was something in the order of 1:1000000 the yo-yo would still move in the same direction but would require greater force for the same acceleration.
Therefore in my opinion i think the answer is option d).
If any one knows why I'm wrong can you please e-mail me, it would be off great help to me.
Thanks.
Ed
Ed Abound,
The answer is A, there solution is here.
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