Kinetic energy of a disk that is rolling and not slipping

[Mohamed BOUCHAKOUR]

Homework Statement

Given a disk of a radius $r$ rolling slip-free on a circular surface of radius $R$ (as shown in the figure below). We need to determine the period of the oscillations. To solve this problem, we use the Lagrange equation, Calculating the potential energy of the system was no problem, but when trying to find the kinetic energy, I found a problem. I find different results using different methods, and I'm not able to find an error with either one of them.
(Sorry for the bad diagrams).

Relevant Equations

$$L=T-U$$
$$\vec{V(B)}=\vec{V(A)}+\vec{\omega} \wedge \vec{AB}$$
$$T= \frac 1 2 mV^2 +\frac 1 2 I\omega ^2$$

Let $\Theta$ be the angle, following the movement of the center of the disk.
In order to find the kinetic energy, we brake the movement of the disk into 2: The translation of the center of mass, and the rotation of the disk around it.
So, the kinetic energy will be given by:
$$T= \frac 1 2 mV^2 +\frac 1 2 I\omega ^2$$
Where $\omega$ is the angular velocity of the disk around it's center, and $V$ is the velocity of the center.
Finding $V$ is quite simple, $V=(R-r)\dot{\Theta}$.
The problem is in finding the angular velocity of the disk, I used two methods, and each one ended up with a different result, and I can't find the error in either one of them.-1st method:

If the center of the disk moves by and angle $\Theta$, the displacement of the disk will be $d$, and we can clearly say the $d=\Theta R$.
And given the slipping-free condition, we can also say that :$d=\psi r$. ($\frac{d\psi}{dt}=\omega$).
So we'll find: $$\omega =\frac{\dot{\Theta}R}{r}$$-2nd method:
If we consider the disk at it's lowest point, name $A$ the point of contact of the disk with the surface, and $C$ it's center.
So, we'll have: $\vec{V(A)}=\vec{V(C)}+\vec{\omega} \wedge \vec{CA}$...(1)

.The no slipping condition will give us: $\vec{V(A)}=\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$.

.And we know that: $\vec{V(C)}=\begin{pmatrix} (R-r)\dot{\Theta} \\ 0 \\ 0 \end{pmatrix}$, and: $\vec{CA}=\begin{pmatrix} 0 \\ r \\ 0 \end{pmatrix}$

.And clearly, $\vec{\omega}$ only has a component following the $z$ axis. ($\vec{\omega}=\begin{pmatrix} 0 \\ 0 \\ \omega \end{pmatrix}$

By replacing all these in equation (1), and solving for $\omega$ we'll find:
$$\omega =\frac{\dot{\Theta}(R-r)}{r}$$

 

[TSny]

The 1st method has the mistake. If $\psi$ is the angle that the disk rotates relative to the "lab", then it is not true that $d = r \psi$. A picture might help to see why.

First, imagine the disk rotates clockwise, without slipping, along a horizontal line a distance $d$ as shown on the left. The red line rotates through some angle $\beta$ which in this figure is quite a bit greater than 90 degrees. Now imagine bending the horizontal segment into a circular arc of arclength $d$ and which subtends an angle $\theta$, as shown on the right. During this bending process, the disk rises upward and rotates counterclockwise. This counterclockwise rotation subtracts from the initial clockwise rotation $\beta$. Relative to the lab, the net rotation of the disk is some angle $\psi$ (which is about 90 degrees in the figure).

 

[Mohamed BOUCHAKOUR]

The 1st method has the mistake. If $\psi$ is the angle that the disk rotates relative to the "lab", then it is not true that $d = r \psi$. A picture might help to see why.

View attachment 258820

First, imagine the disk rotates clockwise, without slipping, along a horizontal line a distance $d$ as shown on the left. The red line rotates through some angle $\beta$ which in this figure is quite a bit greater than 90 degrees. Now imagine bending the horizontal segment into a circular arc of arclength $d$ and which subtends an angle $\theta$, as shown on the right. During this bending process, the disk rises upward and rotates counterclockwise. This counterclockwise rotation subtracts from the initial clockwise rotation $\beta$. Relative to the lab, the net rotation of the disk is some angle $\psi$ (which is about 90 degrees in the figure).

In my reasoning, the angle $\psi$ was not taken relative to a constant vertical line, but rather relative to the line that goes from the point of contact to the center of the bigger circle, so in this case, the mouvement that I will find, will be the relative mouvement of the disk (relative to a moving reference), and not it's absolute mouvement relative to a fixed reference, that's it?

 

[TSny]

In my reasoning, the angle $\psi$ was not taken relative to a constant vertical line, but rather relative to the line that goes from the point of contact to the center of the bigger circle, so in this case, the mouvement that I will find, will be the relative mouvement of the disk (relative to a moving reference), and not it's absolute mouvement relative to a fixed reference, that's it?

Yes, sounds good.

 

[Mohamed BOUCHAKOUR]

Yes, sounds good.

Okay great! I think I got it... Thanks for the help!