What is the Kinetic and Linear Dynamics of a Falling Wheel with a Rope?

[Karol]

Homework Statement

A wheel is made of two hoops, one of radius R and mass $\frac{2M}{3}$ and the inner is of radius R/2 and mass M/3.
1. the wheel rotates around it's axis with angular velocity ω but remains in place. what is it's kinetic energy
2. the wheel rotates without friction at velocity V, what's it's kinetic energy
3. a rope is wound around the inner hoop. the rope's end is attached to the ceiling and the wheel falls. what is it's angular velocity if the velocity of the center is u.
4. what is the linear velocity of the center if the wheel descended the height h
5. what is the linear acceleration of the center

Homework Equations

Moment of inertia of a thin hoop: I=mr²
Torque and angular acceleration of a rigid body: T=Iα

The Attempt at a Solution

The moment of inertia of the entire wheel:
$$I=\left( \frac{2M}{3}+\frac{M}{3} \right)R^2=\frac{3}{4}MR^2$$
Total mass=M
1. When the wheel rotates at ω:
$$E=I\omega^2=\frac{3}{4}MR^2\omega^2$$
2. $$V=\omega R,\quad E=MV^2+I\omega^2$$
$$E=MV^2+\frac{3}{4}MV^2=\frac{7}{4}MV^2$$
3.
$$u=\omega \frac{R}{2}$$
4. The energy calculated in paragraph 2 equals the loss in potential energy:
$$\frac{7}{4}MV^2=Mgh\quad\rightarrow\quad V=\frac{4}{9}gh$$
5. I take the moment of inertia round the small radius.
$$Mg\frac{R}{2}=\left( \frac{3}{4}MR^2+M\frac{R^2}{4} \right)\alpha \quad\rightarrow\quad a=\frac{g}{2}$$

 

[haruspex]

Where you obtain the moment of inertia of the system, you get the right answer but there appears to be a typo along the way. I assume this is an error introduced in making the post.

In the energy formulae, you've left out a constant factor.

In the statement of q2, I am guessing it should say the wheel descends at speed V. But the rope is not mentioned until q3, so maybe that's wrong. Perhaps it is supposed to be rotating about a fixed centre here, with V being the tangential velocity.

There is a subtlety in the last two parts. Strictly speaking, the string would not remain vertical. To some extent, the whole system would behave like a pendulum. However, I'm sure you are not expected to get into that complication, so I agree with your answer to q5.

 

[Karol]

The moment of inertia of the entire wheel about the center:
$$I=\frac{2M}{3}R^2+\frac{M}{3}\frac{R^2}{4}=\frac{3}{4}MR^2$$
When the wheel rotates at ω:
$$E=\frac{1}{2}I\omega^2=\frac{3}{8}MR^2\omega^2$$
$$V=\omega R,\quad E=\frac{1}{2}MV^2+\frac{1}{2}I\omega^2$$
$$E=\frac{1}{2}MV^2+\frac{1}{2}\frac{3}{4}MV^2=\frac{7}{8}MV^2$$
The energy calculated in paragraph 2 equals the loss in potential energy:
$$\frac{7}{8}MV^2=Mgh\quad\rightarrow\quad V=\frac{2}{9}gh$$

There is a subtlety in the last two parts. Strictly speaking, the string would not remain vertical. To some extent, the whole system would behave like a pendulum. However, I'm sure you are not expected to get into that complication, so I agree with your answer to q5.

Moments around A: the weight Mg creates torque T:
$$T=-I\alpha\quad\rightarrow\quad Mgx=I\alpha$$
The rope unwinds and the distance y increases, so the moment of inertia increases. $I_c$ is the moment of inertia round the center.
$$Mgx=-(I_c+My^2)\alpha$$
Trigonometry:
$$y\alpha=\ddot x\quad\rightarrow\quad \alpha=\frac{\ddot x}{y}$$
$$\rightarrow Mgx=-(I_c+My^2)=\frac{\ddot x}{y}$$
Is this the equation for the oscillation?

 

[TSny]

I agree with your answers to parts 1 and 3.

I don't understand part 2. Could it be asking for the KE when the wheel is rolling without slipping along a surface?

I don't agree with your answers to parts 4 and 5. Did you use the result of part 3?

I also don't see why there would be any pendulum motion. What force has a horizontal component that would move the center of mass of the wheel horizontally?

 

[Karol]

I don't understand part 2. Could it be asking for the KE when the wheel is rolling without slipping along a surface?

Yes

I don't agree with your answers to parts 4 and 5. Did you use the result of part 3?

When the wheel rotates at ω:
$$E=\frac{1}{2}I\omega^2=\frac{3}{8}MR^2\omega^2$$
$$E=\frac{1}{2}MV^2+\frac{1}{2}I\omega^2=\frac{1}{2}MV^2+\frac{3}{8}MR^2\omega^2$$
When the velocity of the center is V the angular velocity is $V=\frac{R}{2}\omega\rightarrow \omega=\frac{2V}{R}$
$$E=\frac{1}{2}MV^2+\frac{3}{8}MR^2\frac{4V^2}{R^2}=2MV^2$$

I also don't see why there would be any pendulum motion. What force has a horizontal component that would move the center of mass of the wheel horizontally?

When a weight is hang not along the center of mass it rotates:

You mean the wheel will descend right down, and the rope will change gradually it's angle, like in the drawing:

 

[haruspex]

I also don't see why there would be any pendulum motion. What force has a horizontal component that would move the center of mass of the wheel horizontally?

Yes, I had second thoughts about my comment there. It seemed intuitively true, but I cannot justify it.

 

[TSny]

When the wheel rotates at ω:
$$E=\frac{1}{2}I\omega^2=\frac{3}{8}MR^2\omega^2$$
$$E=\frac{1}{2}MV^2+\frac{1}{2}I\omega^2=\frac{1}{2}MV^2+\frac{3}{8}MR^2\omega^2$$
When the velocity of the center is V the angular velocity is $V=\frac{R}{2}\omega\rightarrow \omega=\frac{2V}{R}$
$$E=\frac{1}{2}MV^2+\frac{3}{8}MR^2\frac{4V^2}{R^2}=2MV^2$$

That looks good.

When a weight is hang not along the center of mass it rotates:View attachment 89188

Yes, if you release the block when it is in the position shown on the right in your figure, then as it "falls" the block will rotate about the point of attachment with the string and the string will not stay vertical. The string will initially swing to the left a little. At least that's what I think will happen.

You mean the wheel will descend right down, and the rope will change gradually it's angle, like in the drawing:View attachment 89189

In this case, I think the string will stay vertical as the wheel descends.

 

[Karol]

In this case, I think the string will stay vertical as the wheel descends.

Interesting why do you think it's so, can you explain your intuition? i will also think about that

 

[TSny]

In the case of the block, the string has a fixed length. As the center of mass of the block falls the block must rotate. The point of attachment of the string to the block must then move to the left. The string is thus pulled away from vertical to the left. The resultant horizontal component of the tension in the string will then cause the center of mass of the block to start swinging to the right. So, you get some swinging motion in the string as the block pivots about the point of attachment.

In the case of the wheel, the string unwraps and increases in length. As the center of mass of the wheel falls the wheel must rotate. But now the point where the string unwraps from the wheel always stays at the same point relative to the center of mass of the wheel. The string is not pulled to the left as in the case of the block. My main reason for believing that the string remains vertical for the wheel is that you get a solution that is consistent with Newton's laws when you assume the string remains vertical.

 

[Karol]

My main reason for believing that the string remains vertical for the wheel is that you get a solution that is consistent with Newton's laws when you assume the string remains vertical.

I understand the power of such a reasoning but i don't see which laws, specifically, are satisfied and how could any other solution not be based on those laws.
I know the equation $T=I\alpha$ that we used is parallel to Newton's law F=ma, but how does it assure you that the rope must stay vertical?

 

[TSny]

I understand the power of such a reasoning but i don't see which laws, specifically, are satisfied and how could any other solution not be based on those laws.
I know the equation $T=I\alpha$ that we used is parallel to Newton's law F=ma, but how does it assure you that the rope must stay vertical?

The mathematical proof that the string must stay vertical for this problem would come from invoking a uniqueness theorem for a system of ordinary differential equations. The degrees of freedom for the system can be taken to be the angle θ that the string makes to the vertical and the length s of string that is unwrapped from the wheel. You can derive the coupled differential equations for θ and s as functions of time from the Lagrangian of the system. The differential equations will satisfy the conditions necessary for the uniqueness theorem to hold. So, if you find any solution that satisfies the differential equations for initial conditions s(0) = s_o, s'(0) = 0, θ(0) = 0, and θ'(0) = 0, then the theorem guarantees that the solution is the only solution. [Edited to replace α by θ.]

 

[Karol]

So, if you find any solution that satisfies the differential equations for initial conditions s(0) = so, s'(0) = 0, θ(0) = 0, and θ'(0) = 0, then the theorem guarantees that the solution is the only solution.

I don't know about lagrangian and the rest about differential equations and the theorem, and i won't question you about it, but if the initial conditions s'(0) = 0, θ(0) = 0 are satisfied, does it say θ won't obtain other values? the rope also starts from rest and s'(0) = 0, θ(0) = 0 are fulfilled.
I also don't see the differential equation that includes θ, besides the one i tried to derive, is it true? in the analysis we made so far there was no differential equation involved

 

[TSny]

I don't know about lagrangian and the rest about differential equations and the theorem, and i won't question you about it, but if the initial conditions s'(0) = 0, θ(0) = 0 are satisfied, does it say θ won't obtain other values? the rope also starts from rest and s'(0) = 0, θ(0) = 0 are fulfilled.
I also don't see the differential equation that includes θ, besides the one i tried to derive, is it true? in the analysis we made so far there was no differential equation involved

You don't need to use the Lagrangian approach. You can just apply Newton's laws for the general configuration shown below. However, the torque equation should be written $\tau = \frac{dL}{dt}$ where $L$ is the total angular momentum of the system. The angular momentum can be broken up into two parts: angular momentum due to motion of CM and angular momentum due to rotation about the CM.

The torque equation will give a differential equation involving $s, \dot{s}, \ddot{s}, \theta, \dot{\theta}$ and $\ddot{\theta}$

A second differential equation can be obtained by applying $F= ma$ in the direction perpendicular to the string. (Choose this direction to avoid bringing in the tension in the string.) This differential equation will involve $s, \dot{s}, \theta, \dot{\theta}$ and $\ddot{\theta}$.

So you get two coupled second-order differential equations for $s$ and $\theta$. The equations are too complicated to solve in closed form for general initial conditions. But, you can solve them for the special initial conditions $s(0) = s_0$, $\dot{s}(0) = \theta(0) = \dot{\theta}(0) = 0$. The solution keeps $\theta = 0$ as the wheel descends.

 

[Karol]

I change the markings for clarity: m is the total mass, r is the radius of the small rim, I_C is the moment of inertia round the center and I is the moment of inertia at the unwrapping point.
The component of the wight parallel to the rope:
$$L=I\omega=(I_c+mr^2)\omega=\frac{I_c+mr^2}{r}\dot s$$
$$L'=\frac{I_c+mr^2}{r}\ddot s=mg\cdot\cos\theta$$
This equation doesn't contain $\dot{s}, \dot{\theta}, \ddot{\theta}$
The component of the gravity vertical to the rope:
$$F=ma:\quad mg\sin\theta=m\dot v=m(\theta\sqrt{s^2-r^2})'$$
The derivative is complicated and it won't contain $\ddot s, \dot{\theta}, \ddot{\theta}$

 

[TSny]

View attachment 89520 I change the markings for clarity: m is the total mass, r is the radius of the small rim, I_C is the moment of inertia round the center and I is the moment of inertia at the unwrapping point.
The component of the wight parallel to the rope:
$$L=I\omega=(I_c+mr^2)\omega=\frac{I_c+mr^2}{r}\dot s$$

This expression for $L$ is incomplete. Note that if you imagine $s$ as held constant as $\theta$ changes, you would have nonzero angular momentum about the point where the string attaches to the support. But your expression gives $L = 0$ if $s$ is held constant. You can fix your expression by adding appropriate terms that include the effect of changing $\theta$. These terms will involve $s$ and $\dot{\theta}$. So, when you take the time derivative of $L$ you will get something involving $\dot{s}, \dot{\theta}$, and $\ddot{\theta}$.

The component of the gravity vertical to the rope:
$$F=ma:\quad mg\sin\theta=m\dot v=m(\theta\sqrt{s^2-r^2})'$$
The derivative is complicated and it won't contain $\ddot s, \dot{\theta}, \ddot{\theta}$

OK, $F_{\bot} = mg\sin\theta$. But I don't understand your expression for $a_{\bot}$. I don't think it's right.

 

[Karol]

$$L=I_1\omega+I_2\dot\theta=(I_c+mr^2)\omega+[I_c+m(s^2-r^2)]\dot\theta=\frac{I_c+mr^2}{r}\dot s+[I_c+m(s^2-r^2)]\dot\theta$$
$$L'=\frac{I_c+mr^2}{r}\ddot s=mg\cdot\cos\theta+2ms\dot s\dot\theta[I_c+m(s^2-r^2)]\ddot\theta$$
But which force and torque arms to take? for I₁ i take the tangential component and for I₂-the vertical. can i take mg as a whole? to where to stretch the arm?
for $a_{\bot}:\quad v=\dot \theta\sqrt{s^2-r^2}$
$$mg\cdot\sin\theta=m\left[ \dot \theta\sqrt{s^2-r^2} \right]'$$

 

[TSny]

Your final expression for $L$ almost agrees with what I got using a different approach. But shouldn't your $(s^2 - r^2)$ be replaced by $(s^2+r^2)$?
$s$ and $r$ are perpendicular.

It looks like your expression for $\dot{L}$ has some typographical errors and I'm not sure how you got the $mg\cos \theta$ term.

Your method of getting the component of acceleration of the CM that's perpendicular to the string does not look correct.

-----------------------------------------

For me, the least confusing way to approach the problem was to express the location of the center of mass of the wheel in Cartesian components where I chose the origin of Cartesian axes at the point where the string attaches to the support. I chose the positive $x$ direction to be to the right and the positive $y$ direction to be upward.

Let $x_c$ and $y_c$ denote the Cartesian coordinates of the center of mass of the wheel. These coordinates can be expressed in terms of $s$, $r$, and $\theta$. The part of the angular momentum that is due to motion of the center of mass can be expressed as $L_c = M(x_c \dot{y}_c - y_c \dot{x}_c)$. It is then straightforward but a little tedious to crank out $\dot{L}_c$.

The only torque about the origin is due to the force of gravity and you can see that the lever arm is just $x_c$.

When applying F = ma perpendicular to the string, you can first find the acceleration vector from $x_c$ and and $y_c$ and then project the acceleration along the direction perpendicular to the string.

 

[Karol]

$$L=I_1\omega+I_2\dot\theta=(I_c+mr^2)\omega+[I_c+m(s^2+r^2)]\dot\theta=\frac{I_c+mr^2}{r}\dot s+[I_c+m(s^2+r^2)]\dot\theta$$
$$L'=\frac{I_c+mr^2}{r}\ddot s+2ms\dot s\dot\theta+[I_c+m(s^2-r^2)]\ddot\theta=mg\sqrt{s^2+r^2}\sin\theta$$
For $a_{\bot}$ i tend to say $v=\dot \theta s\rightarrow a=\ddot\theta s+\dot\theta \dot s$ but it's not true since s changes and i have to get an expression with $\ddot s$
I will try to work on the solution with Cartesian coordinates as soon as i finish this one.

 

[TSny]

$$L=I_1\omega+I_2\dot\theta=(I_c+mr^2)\omega+[I_c+m(s^2+r^2)]\dot\theta=\frac{I_c+mr^2}{r}\dot s+[I_c+m(s^2+r^2)]\dot\theta$$

OK

$$L'=\frac{I_c+mr^2}{r}\ddot s+2ms\dot s\dot\theta+[I_c+m(s^2-r^2)]\ddot\theta=mg\sqrt{s^2+r^2}\sin\theta$$

The factor in front of $\ddot{\theta}$ should have $s^2 + r^2$ instead of $s^2 - r^2$. The expression for the torque on the right is not quite right. The force $mg$ does not make an angle $\theta$ to the line connecting the point of support to the center of the wheel.

 

[Karol]

$$L'=\frac{I_c+mr^2}{r}\ddot s+2ms\dot s\dot\theta+[I_c+m(s^2+r^2)]\ddot\theta=mg[s\cdot \sin\theta-r\sin(90-\theta)]$$
$$L'=\frac{I_c+mr^2}{r}\ddot s+2ms\dot s\dot\theta+[I_c+m(s^2+r^2)]\ddot\theta=mg(s\cdot \sin\theta-r\cos\theta)$$
$$F=ma:\quad mg\sin\theta=m\dot v=m(\dot\theta s)'=m(\dot\theta^2\dot s+s\ddot\theta)$$
Do i need the F=ma equation? the information about the CM motion is already included in the member $I_2\dot\theta$ of $L=I_1\omega+I_2\dot\theta$
I understand i need 2 equations for s and θ but doesn't the information in F=ma repeat itself?
For the approach $L_c = M(x_c \dot{y}_c - y_c \dot{x}_c)$ i take y directed downward:
$$x_c=s\cdot \sin\theta-r\cos\theta$$
$$y_c=s\cdot \cos\theta+r\sin\theta$$
$$\dot{x_c}=(s\cos\theta+r\sin\theta)\dot\theta$$
$$\dot{y_c}=(r\cos\theta-s\cdot \sin\theta)\dot\theta$$
$$L_c = M(x_c \dot{y}_c - y_c \dot{x}_c)=m\dot\theta[(s\cdot \sin\theta-r\cos\theta)(r\cos\theta-s\cdot \sin\theta)+(s\cdot \cos\theta+r\sin\theta)(s\cos\theta+r\sin\theta)]$$
$$L_c = M(x_c \dot{y}_c - y_c \dot{x}_c)=m\dot\theta[2sr\sin\theta\cos\theta+(s^2-r^2)(\cos^2\theta-\sin^2\theta)]$$
$$L=I_1\omega+m\dot\theta[2sr\sin\theta\cos\theta+(s^2-r^2)(\cos^2\theta-\sin^2\theta)]$$
$$L=\frac{I_c+mr^2}{r}\dot s+m\dot\theta[2sr\sin\theta\cos\theta+(s^2-r^2)(\cos^2\theta-\sin^2\theta)]$$

 

[TSny]

$$L'=\frac{I_c+mr^2}{r}\ddot s+2ms\dot s\dot\theta+[I_c+m(s^2+r^2)]\ddot\theta=mg[s\cdot \sin\theta-r\sin(90-\theta)]=mg(s\cdot \sin\theta-r\cos\theta)$$

Check the overall sign of the right side is wrong. What is the direction of the torque due to mg? Otherwise, it looks good to me.

$$F=ma:\quad mg\sin\theta=m\dot v=m(\dot\theta s)'=m(\dot\theta^2\dot s+s\ddot\theta)$$

I don't think your expression is correct for the component of acceleration of the CM that's perpendicular to the string.

Do i need the F=ma equation? the information about the CM motion is already included in the member $I_2\dot\theta$ of $L=I_1\omega+I_2\dot\theta$
I understand i need 2 equations for s and θ but doesn't the information in F=ma repeat itself?

I believe the F = ma equation will give you an independent equation.

 

[Karol]

The torque is a restoring force:
$$L'=\frac{I_c+mr^2}{r}\ddot s+2ms\dot s\dot\theta+[I_c+m(s^2+r^2)]\ddot\theta=-mg(s\cdot \sin\theta-r\cos\theta)$$
$$L'=\frac{I_c+mr^2}{r}\ddot s+2ms\dot s\dot\theta+[I_c+m(s^2+r^2)]\ddot\theta=mg(r\cos\theta-s\cdot \sin\theta)$$
$$F=ma:\quad mg\sin\theta=m\dot v=m(\theta s)''=m(\dot\theta s+\dot s \theta)'=mg(s\ddot\theta+\dot\theta\dot s+\theta\ddot s+\dot s\ddot\theta)$$
That is the magnitude but what is the direction? and how do i project it on r?
If i try cylindrical coordinates:
$$a=(\theta \hat{\theta}+s \hat{r})''=(\dot \theta \hat{\theta}+\dot s \hat{r})'=\ddot \theta \hat{\theta}+\ddot s \hat{r}$$
The projection is a dot product:
$$(\ddot \theta \hat{\theta}+\ddot s \hat{r})\centerdot(\hat{\theta})=\ddot{\theta}$$
And it's not true

 

[TSny]

Using polar (cylindrical) coordinates is a little tricky because $\theta$ does not represent the angle of the line from the origin to the CM.
Your expression for the acceleration in cylindrical coordinates is not correct. You did not take into account the time derivatives of the unit vectors $\hat{r}$ and $\hat{\theta}$. See http://ocw.mit.edu/courses/aeronaut...fall-2009/lecture-notes/MIT16_07F09_Lec05.pdf

A sure way to get the correct expression for the acceleration vector of the CM is to work in Cartesian coordinates : $\vec{a}_{cm} = \ddot{X}_{cm} \hat{i} + \ddot{Y}_{cm} \hat{j}$. Then you can project that along the direction perpendicular to the string.

 

[Karol]

$$\dot{x_c}=(s\cos\theta+r\sin\theta)\dot\theta$$
$$\dot{y_c}=(r\cos\theta-s\cdot \sin\theta)\dot\theta$$
$$\ddot{x_c}=(s\cdot\cos\theta+r\sin\theta)\ddot\theta+(\dot s\cos\theta-s\dot\theta\sin\theta+r\dot\theta\cos\theta)\dot\theta$$
$$\ddot{y_c}=(r\cos\theta-s\cdot\sin\theta)\ddot\theta+(-r\dot\theta\sin\theta-\dot s\sin\theta-s\dot\theta\cos\theta)\dot\theta$$
The unit vector of CM (the unwrapping point, but for the acceleration it's the same as for CM) is $(\sin\theta,\cos\theta)$
The perpendicular to a line $y=mx$ is $y=-\frac{1}{m}x$ so the unit vector of the perpendicular to the rope is $(\cos\theta,\sin\theta)$
$$a_{\bot}=(\ddot{x_c},\ddot{y_c})\centerdot(\cos\theta,\sin\theta)$$
$$a_{\bot}=(2r\ddot\theta-2s\dot\theta^2-\dot s\dot\theta)\sin\theta\cos\theta+(s\ddot\theta+r\dot\theta^2+\dot s\dot\theta)(\cos^2\theta-\sin^2\theta)$$

 

[TSny]

$$\dot{x_c}=(s\cos\theta+r\sin\theta)\dot\theta$$
$$\dot{y_c}=(r\cos\theta-s\cdot \sin\theta)\dot\theta$$

You've left out terms involving $\dot{s}$.

The unit vector of CM (the unwrapping point, but for the acceleration it's the same as for CM) is $(\sin\theta,\cos\theta)$
The perpendicular to a line $y=mx$ is $y=-\frac{1}{m}x$ so the unit vector of the perpendicular to the rope is $(\cos\theta,\sin\theta)$

You can check that the vector $(\sin\theta, \cos\theta)$ is not perpendicular to the vector $(\cos\theta, \sin\theta)$. (Their scalar product is not zero.)

 

[Karol]

$$x_c=s\cdot \sin\theta-r\cos\theta,\quad y_c=s\cdot \cos\theta+r\sin\theta$$
$$\dot{x_c}=\dot s\sin\theta+s\dot\theta\cos\theta+r\dot\theta\sin\theta$$
$$\dot{y_c}=\dot s\cos\theta-s\dot\theta\sin\theta+r\dot\theta\cos\theta$$
The second derivative is long. the unit vector of the rope is $(\sin\theta, \cos\theta)$ and the perpendicular $(-\cos\theta, \sin\theta)$

 

[TSny]

Looks good. If you push it through, you should find that $a_{\bot}$ reduces to just three terms.

 

[Karol]

After i make $a_{\bot}=(\ddot{x_c},\ddot{y_c})\centerdot(-\cos\theta,\sin\theta)=-(r\dot\theta^2+s\ddot\theta+2\dot s\dot\theta)$
I will try the polar coordinates now. from the link you gave me:
$$a=(\ddot r-r\ddot\theta^2)\vec{e_r}+(r\ddot\theta+2\dot r\dot\theta)\vec{e_\theta}$$
But i marked θ as extruding from the y axis, so when my θ increases the θ in the formula above decreases.
$$(90^0-\theta)'=-\dot\theta,\quad (-\dot\theta)'=-\ddot\theta$$
So the above formula for my problem should be:
$$a=(\ddot r-r\ddot\theta^2)\vec{e_r}-(r\ddot\theta+2\dot r\dot\theta)\vec{e_\theta}$$
I need only the component of $a_{\bot}$ that's perpendicular to the rope, so i need only the $\vec{e_\theta}$ component: $a_{\bot}=-(r\ddot\theta+2\dot r\dot\theta)\vec{e_\theta}=-(r\ddot\theta+2\dot r\dot\theta)$
But this term has only 2 of my previously found terms: $s\ddot\theta$ and $2\dot s\dot\theta$. in the cartesian analysis there were 3 terms.

 

[TSny]

After i make $a_{\bot}=(\ddot{x_c},\ddot{y_c})\centerdot(-\cos\theta,\sin\theta)=-(r\dot\theta^2+s\ddot\theta+2\dot s\dot\theta)$

Good. That's what I got, too. This is the component of the acceleration of the center of the wheel in a direction perpendicular to the string and pointing back toward the direction of decreasing $\theta$.

I will try the polar coordinates now. from the link you gave me:
$$a=(\ddot r-r\ddot\theta^2)\vec{e_r}+(r\ddot\theta+2\dot r\dot\theta)\vec{e_\theta}$$
But i marked θ as extruding from the y axis, so when my θ increases the θ in the formula above decreases.
$$(90^0-\theta)'=-\dot\theta,\quad (-\dot\theta)'=-\ddot\theta$$
So the above formula for my problem should be:
$$a=(\ddot r-r\ddot\theta^2)\vec{e_r}-(r\ddot\theta+2\dot r\dot\theta)\vec{e_\theta}$$

There is no need to change the sign of the $\vec{e_\theta}$ component. If you do change the sign the way you did, then you would need to interpret $\vec{e_\theta}$ as a unit vector pointing toward decreasing $\theta$ rather than the usual definition of pointing toward increasing $\theta$.

I need only the component of $a_{\bot}$ that's perpendicular to the rope, so i need only the $\vec{e_\theta}$ component: $a_{\bot}=-(r\ddot\theta+2\dot r\dot\theta)\vec{e_\theta}=-(r\ddot\theta+2\dot r\dot\theta)$
But this term has only 2 of my previously found terms: $s\ddot\theta$ and $2\dot s\dot\theta$. in the cartesian analysis there were 3 terms.

The third term comes from noting that the coordinates $s$ and $\theta$ are not the polar coordinates for the center of the wheel. They are the polar coordinates for the point p shown in the figure below where p is a point that moves such that it is always coincident with the point where the string is unwrapping from the wheel. To get the acceleration of the center of the wheel, c, you can use the relative acceleration formula $\vec{a}_c = \vec{a}_p + \vec{a}_{c/p}$. Your expression gives $\vec{a}_p$. See if you can show that $\vec{a}_{c/p}$ gives the missing third term.

 

[Karol]

See if you can show that $\vec{a}_{c/p}$ gives the missing third term.

I need the component of $\vec{a}_{c/p}$ that's vertical to the rope, that's in the direction of $\vec{e_\theta}$

i copy r parallel to itself from point A to B. the CM has rotated $d\theta$ . the vertical change of CM is $r(1-\cos\theta)$ and if i take the derivative i don't get the third member $r\dot\theta^2$.