Question

A bobbin rolls without slipping over a horizontal surface sot hat the velocity $v$ of the end of the thread (point A) is directed along the horizontal. A board hinged at B leans against the bobbin are $e$ and R respectively. Determine the angular velocity $\omega$ of the board as a function of angle $\alpha$

Answer 1

Let the board touch the bobbin at point $C$ at a certain instant of time. The velocity of point $C$ is the sum of the velocity ${\nu }_0$ of the axis $O$ of the bobbin and the velocity of point $C$ (relative to point $O$, which is tangent to the circle at point $C$ and equal in magnitude to ${\nu }_0$ (since there is no slipping). If the angular velocity of the board at this instant is $\omega$, the linear velocity of the point of the board touching the bobbin will be $\omega R{\tan }^{-1}\left(\alpha /2\right)$ Fig. Since the board remains in contact with the bobbin all the time, the velocity of point $C$ relative to the board is directed along the board, whence $\omega R{\tan }^{-1}\left(\alpha /2\right)={\nu }_0\sin \alpha$ Since there is no slipping of the bobbin over the horizontal surface, we can write
$\frac{{\nu }_0}{R}=\frac{\nu }{R+r}$.
Therefore, we obtain the following expression for the angular velocity $\omega$:
$\omega =\frac{\nu }{R+r}\sin \alpha .\tan \frac{\alpha }{2}=\frac{2\nu {\sin }^2\left(\alpha /2\right)}{(R+r)\cos \left(\alpha /2\right)}$