- #1

Mohamed BOUCHAKOUR

- 15

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- 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:

$$\omega =\frac{\dot{\Theta}(R-r)}{r}$$

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