forcesnewtonian-mechanicsrotational-dynamics
This question has been around the net for a while, and I haven't seen a good explanation for it:
A yo-yo is initially at rest on a horizontal surface. A string is pulled in the direction shown in the figure. In what direction will the yo-yo rotate and move?
I think the way to look at the problem is this: The tricky thing about this problem is that the force $F$ is doing two things. First, it is pulling the entire bicycle backwards and, secondly, it is exerting a torque on the pedals. So let's try to separate those two effects. Consider, then, a person just sitting normally on the bicycle with his feet on the pedals and exerting the same force $F$ on the pedals (or, equivalently, a torque $rF$ on the pedal gear where r is the effective radius at which the pedals are located with respect to the center of the pedal gear). Also, assume that there is another man pulling backwards on the bicycle with a rope attached not to a pedal but to the frame of the bicycle. Now we have a situation which is equivalent to the problem originally posed. We have (1) the same torque as originally stated acting on the pedal gear and (2) we have the same force as originally stated acting to pull the bicycle backwards. Any problems with this picture thus far?
OK, so with this new but equivalent situation, what is the answer? Imagine that you are on the bicycling trying to pedal forward while exerting the torque stated above on the pedals while at the same time someone is trying to pull you backwards by pulling with a force $F$ on a rope attached to the frame of the bicycle. Remember that he is required to pull with exactly the same force $F$ that you are exerting on the pedals. No more and no less. Do you go forward or does he pull you backwards?
I think that the answer is clearly that it all depends on what bicycle gear you are in. If you are in low gear, you will be able to move forward. If you are in high gear, he will pull you backwards. In physics terms, it depends on whether the (clockwise) torque that you are exerting on the rear drive wheel by means of your feet on the pedals is greater than or less than the (counter-clockwise) torque that the other man is able to effectively exert on the rear drive wheel by pulling the bicycle in the backwards direction with a rope.
P.S.: With the single-speed bicycle shown in the picture, the gearing is such that the bicycle would most likely move backwards.
EDIT - SOLUTION FOUND
Found a discussion and solution for this puzzle on the Scientific American web site. It turns out that the bicycle can go either backwards or forwards, depending on the gearing ratio (which is the same result that I arrived at above). For normal gearing the bicycle will most probably go backwards, but for very low gearing the bicycle can also go forward. The argument they use is based on noting whether the point where the rope is attached to the pedal goes forward or backwards with respect to the ground. Here's the web link with an explanation video: Scientific American: Bicycle Puzzle
We will start with much more general case of your question. Where there is a spring block system as shown in the figure $m_{2} > m_{1}$, initially both the blocks are at rest and the spring is not stretched (at its natural length). Now we will try to figure out the motion of both the blocks when they are released.
System in initial state
The blue block is heavier.
NOTE
The following arrangements shown in the diagram below are equivalent ( I have drawn block around Springs just for the sake of my convenience in making diagrams so please ignore the blocks at the ends of the spring ).
Let us now analyze the given situation and try to figure out the motions of the blocks.
When released the blocks will try to move downwards due to gravity but when they start moving the string becomes taut and tension force comes into the play. Since in the string is inextensible so when the block tries to moves downward the string will also try to move but since it is attached to the spring so it will start to stretch the spring. Let us analyze the situation when a small instant of time ( say $dt$ ) is passed.
The spring will stretch more at that end where it is connected to the heavier block as shown in figure below, note that the spring have stretched by half square more on the side where it is connected to the heavier block.
Let's say that the spring is stretched by a small length $dx$ so,
$$
\vec{F}(x)=\vec{T}=k d x
$$
where $k$ is the spring constant. Note that tension in string $\vec{T} \neq m_{1} \vec{g}\neq m_{2} \vec{g}$ because if $\vec{T} = m_{1} \vec{g}$ then block $m_{1}$ would not move at all which is obviously incorrect.
At this moment both the blocks are moving downward because the forces acting on them are unbalanced. As both the block move down further the elongation in the spring will increase. It will stretch more at end attached to the heavier block and will stretch less at the end attached to the lighter block. Over time following changes will occur in the system:
Eventually a time will come when $\vec{T} = kx = m_{1} \vec{g} $ $$\Rightarrow \vec{F_{net}}(m_{1})=m_{1} \vec{g} - \vec{T} = 0$$ $$\Rightarrow \vec{F_{net}}(m_{2})=m_{2} \vec{g} - \vec{T} = (m_{2}-m_{1}) \vec{g} $$
Now, since both the blocks still have their velocity downwards so the spring will stretch further, therefor increasing tension force which will further decrease the acceleration of block $m_{2}$ (still the velocity of $m_{2}$ is increasing) but decelerating block $m_{1}$ (as now the net force on block $m_{1}$ is upwards ).
Eventually a time will come when the velocity of block $m_{1}$ will becomes zero, at this point there are several possibilities for the acceleration of block $m_{2}$ depending on the values of $m_{1}$ & $m_{2}$.
The velocity of block $m_{2}$ will downwards ( for any value of $m_{1}$ & $m_{2}$ since $ \vec{V_{2}} > \vec{V_{1}}$ and block $m_{1}$ starts to decelerate earlier than $m_{2}$ ) but the net force on the block $m_{2}$ maybe upwards or downwards.
CASE 1 ( when net force on $m_{2}$ is upwards)
In this case, block $m_{2}$ will decelerate and block $m_{1}$ accelerate, the spring will first stretch ( when $ \vec{V_{2}} > \vec{V_{1}}$) then elongation of spring stop ( when $ \vec{V_{2}} = \vec{V_{1}}$) and then the spring will shrink ( when $ \vec{V_{2}} < \vec{V_{1}}$ as the block $m_{1}$ accelerates and the block $m_{2}$ decelerates ) and so on ... the motion will continue. I hope you will be able to figure out the rest of the motion in this case.
Here I have created a simulation for your help. Heavier block is on the right side.
Increasing the length of rope of lighter block.
Note: The spring block system appears to move right, it is because stretching of spring at the right side was more when blocks were moving downwards due to higher acceleration of block $m_{2}$ whereas while coming upwards the compression for spring is more at the left end due to higher velocity and acceleration of block $m_{1}$.
CASE 2 ( when net force on $m_{2}$ is downwards)
In this case, block $m_{2}$ will accelerate downwards and block $m_{1}$ accelerate upwards, the spring will further stretch as $ \vec{V_{2}} > \vec{V_{1}}$ eventually $ \vec{V_{2}} = 0 $ then $ \vec{V_{1}}$ will be upwards so the spring start to shrink and so on ... the motion will continue. I hope you will be able to figure out the rest of the motion in this case also.
Here I have created a simulation for your help. Heavier block is on the right side.
This time also the spring block system seem to move right and you can figure out yourselves Why is it so.
After such long discussion you will be able to answer your questions yourselves but I am writing them here in brief.
Question 1
Since in this scenario the extra weight of heavier block is balanced by some external force and whole system is in a state of rest initially so it means that the net force on the lighter block is zero which implies that the tension force in the string is equals to weight of lighter block. When we release the heavier block its speed will increase downward and the speed of lighter block will increase upward since acceleration of heavier block is higher so it will gain speed faster stretching the spring eventually the heavier block will start to decelerate and all this will lead to a motion very similar to the one shown in case one.
Question 2 (I am bit skeptic about this one but here is what I think will happen in this case. Any suggestions would be appreciated.)
In this scenario also since the system is at rest initially it means that the net force on the the blocks is zero which implies that the tension in the string is equals to their weight and the spring is already stretched now as we start adding small weights on one of the blocks this block will try to move down and stretches the spring for the more just a bit more which makes the force on the other block and balance accelerating it up since the spring was stretched only by a very small length so it reaches its natural length very quickly giving both the blocks very small velocities which results in the very small amplitude SHM (simple harmonic motion) of these blocks almost like small vibration and as we keep putting weights this velocity is increased so is the vibrations.
Best Answer
Imagine the string runs over a pulley and is connected to a weight, like this:
The weight must fall, which means the red dot must move to the right. To make that happen, the yo-yo must roll to the right.