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

[TSny]

I need the component of $\vec{a}_{c/p}$ that's vertical to the rope, that's in the direction of $\vec{e_\theta}$
View attachment 89816 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 $r\dot\theta^2$.

I'm not following what you did there. The center, c, of the wheel is always at a fixed distance, r, from the moving point p. So the only motion that c can have relative to p is circular motion about p. Thus, the acceleration of c relative to p can be thought of as having a centripetal component (towards p) and a tangential component (tangent to the circular path of c relative to p). Which of these components is perpendicular to the string?

 

[Karol]

The centripetal component is perpendicular to the string: $a_{\bot}=r\dot\theta^2$

 

[TSny]

The centripetal component is perpendicular to the string: $a_{\bot}=r\dot\theta^2$

Yes, good.

 

[Karol]

The second independent equation:
$$mg\sin\theta=-m(r\dot\theta^2+s\ddot\theta+2\dot s\dot\theta)$$
The first independent equation was:
$$\frac{I_c+mr^2}{r}\ddot s+2ms\dot s\dot\theta+[I_c+m(s^2+r^2)]\ddot\theta=mg(r\cos\theta-s\cdot \sin\theta)$$

So you get two coupled second-order differential equations for ss and θ\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)=s0s(0) = s_0, s˙(0)=θ(0)=θ˙(0)=0\dot{s}(0) = \theta(0) = \dot{\theta}(0) = 0. The solution keeps θ=0\theta = 0 as the wheel descends.

How do i do that?

 

[TSny]

Your two equations look correct to me.

How do i do that?

I believe you can show that the string always remains vertical in this case by just using the one equation $$mg\sin\theta=-m(r\dot\theta^2+s\ddot\theta+2\dot s\dot\theta)$$
Solve it for $\ddot{\theta}$ to get an equation of the form $\ddot{\theta} = f(s, \dot{s}, \theta, \dot{\theta})$.

Use this equation to evaluate $\ddot{\theta}$ at time $t = 0$.

Next, take the time derivative of $\ddot{\theta} = f(s, \dot{s}, \theta, \dot{\theta})$ to get an expression for $\frac{d^3 \theta}{dt^3}$. Use this to evaluate $\frac{d^3 \theta}{dt^3}$ at $t = 0$.

Continue this until you can see what all of the time derivatives of $\theta$ will be at $t = 0$.

 

[Karol]

$$\ddot\theta=-\frac{r\dot\theta^2+2\dot s\dot\theta+g\sin\theta}{s}$$
Since i know that at the start the wheel is left from stand still $\ddot\theta(t=0)=0$
Why do i need the third derivative $\frac{d^3 \theta}{dt^3}$? it doesn't have any physical meaning, the acceleration is the last

Continue this until you can see what all of the time derivatives of θ\theta will be at t=0t = 0.

I know the velocity, $\dot\theta(t=0)=0$, and the acceleration $\ddot\theta(t=0)=0$, what does it help if later it swings?

 

[TSny]

If we assume that the function $\theta(t)$ has a Taylor expansion of the form $\theta(t) = \theta(0) + \dot{\theta}(0)t + \frac{1}{2!} \ddot{\theta}(0) t^2+ \ldots$
what would $\theta(t)$ be if $\theta$ and all time derivatives of $\theta$ are zero at $t = 0$?

 

[TSny]

If the previous argument is not satisfactory, then you can appeal to uniqueness theorems of systems of differential equations as mentioned in post #11. You now have the differential equations that the system must satisfy and you have the initial conditions $s(0) = s_0$, $\dot{s}(0) = \theta(0) = \dot{\theta}(0) = 0$. If you find one solution to the differential equations that satisfies these initial conditions, then it is the only solution. You can easily find a solution where $\theta(t) = 0$ for all $t$.

 

[Karol]

$$\frac{d^3 \theta}{dt^3}=-\frac{s(2r\dot\theta\ddot\theta+2\dot\theta\ddot s+2\dot s\ddot\theta+g\dot\theta\cos\theta)-(r\dot\theta^2+2\dot s\dot\theta+g\sin\theta)\dot s}{s^2}$$
$$\frac{d^3 \theta}{dt^3}(t=0)=0$$
Since all members in the numerator include θ or $\dot\theta$ all the higher derivatives will include at least $\dot\theta$ which, at t=0 are 0, so all higher derivatives will be 0 at t=0.

 

[TSny]

Yes. This strongly suggests that $\theta$ will remain zero. However, there are examples of functions where all derivatives are zero at a point yet the function is not constant. See https://en.wikipedia.org/wiki/Flat_function

 

[Karol]

You can easily find a solution where $\theta(t) = 0$ for all t.

I don't know how to solve differential equations, not even one alone, so how will i find a solution for two?

 

[TSny]

OK. By the uniqueness theorem, we just need to find any solution. It will then be the only solution. So, try to find a solution where $\theta$ is always zero. Can you satisfy the two differential equations with this assumption?

 

[Karol]

The first equation:
$$mg\sin\theta=-m(r\dot\theta^2+s\ddot\theta+2\dot s\dot\theta)$$
The second:
$$\frac{I_c+mr^2}{r}\ddot s+2ms\dot s\dot\theta+[I_c+m(s^2+r^2)]\ddot\theta=mg(r\cos\theta-s\cdot \sin\theta)$$
If i take $\theta(t)=0$ the first equation is satisfied but the second becomes:
$$\frac{I_c+mr^2}{r}\ddot s=mgr$$

 

[TSny]

OK. The second equation gives you information about the acceleration of the disk as it moves straight down. You can then get the velocity and position of the disk at any time.

 

[Karol]

Yes, correct:
$$M=I_A\alpha,\quad \alpha r=\ddot s$$
$$\frac{I_c+mr^2}{r}\ddot s=mgr\quad \rightarrow\quad I_A\ddot s=mgr^2$$
$$I_A\alpha r=mgr\cdot r$$

 

[TSny]

Looks good.

 

[Karol]

I have no more questions about this problem, i thank you TSny and Haruspex very much!