vcsharp2003 said:
Why is my reasoning not correct?
My attempt is as in screenshot below.
View attachment 281416
It is important that you understand the reason that makes your reasoning not correct.
Point P can't have any of those U velocity vectors.
The points of the strings next to point P don't have velocity U, they will move more or less as P will.
Point P is limited to move vertically only, because the symmetrical nature of the forces and geometry in this case.
The only velocity vector for point P, respect to frame of reference formed by the axis of the pulleys, has value V and its direction is aligned with the vertical.
We could also simplify the problem to analyze the left side only if we can imagine that the point P is forced to slide within a vertical slot that is fixed respect to the axes of the pulleys, and that the radii of the pulleys is very small.
Imagine that the section of string that runs between the left small pulley and point P has 10 sections of length=1 unit each.
Section #10 is the closest to the pulley and section #1 is the closest to point P.
The velocity vector for section #10 has value U and its direction is aligned with the string.
The velocity vector for section #1 has value V and its direction is aligned with the vertical (the slot that guides P).
That means that while section #9 takes the place of section #10, the string shortens and rotates counter-clockwise some simultaneously.
That rotation responds to the different directions of vectors U (for #9) and V (for #1).
After resolving the problem graphically, the following is my understanding regarding the relation between U and V based on the geometry of the problem, in case it can help you:
1) As the left side string gets shorter at a constant rate because U of the vertical string beyond the left pulley, the distance that P moves up within our vertical slot for each of our ten sections of the string is bigger than the length of each section.
2) The vertical distances increase respect to each section as P moves up.
3) As the vertical distance that P moves is greater than the length that the string shortens in the same time, the average value of V is greater than U.
4) As the individual values of V increase as the height reached by P increases, there is a positive vertical acceleration until the moment the string reaches a horizontal position.
5) The angular velocity of the string increases as the height reached by P increases; therefore, the string has a positive angular acceleration.
6) Both accelerations rapidly increase as the string gets closer to the horizontal position.
Considering
H the hypotenuse of the triangles successively formed,
h the distances between point P and the top of the slot (90° angle of the triangles) and
k the horizontal distance between the left pulley and the slot, which is constant and has the length towards which H tends,
we have:
##h=\sqrt {H^2-k^2}##
Considering k=4 units:
For t=0, H=10, then h=9.16 units
For t=1, H=9, then h=8.06 units and Δh=1.10
For t=2, H=8, then h=6.92 units and Δh=1.13
For t=3, H=7, then h=5.74 units and Δh=1.17
For t=4, H=6, then h=4.47 units and Δh=1.27
For t=5, H=5, then h=3.00 units and Δh=1.47
For t=6, H=4 (string is horizontal), then h=0 units and Δh=3.00 (note the sudden increase).
