Question

A solid body starts rotating about a stationary axis with an angular acceleration $\beta ={\beta }_0\cos \phi$, where ${\beta }_0$ is a constant vector and $\phi$ is an angle of rotation from the initial position. If the angular velocity of the body as a function of the angle $\phi$ is ${\omega }_z=\pm \sqrt{x{\beta }_0\sin \phi }$. Find $x$.

Answer 1

Let us choose the positive direction of z-axis (stationary rotation axis) along the vector ${\beta }_0$. In accordance with the equation
$\frac{d{\omega }_z}{dt}={\beta }_z$ or ${\omega }_z\frac{d{\omega }_z}{d\phi }={\beta }_z$
or, ${\omega }_{z}d{\omega }_z={\beta }_{z}d\phi =\beta \cos \phi d\phi$,
Integrating this equation within its limit for ${\omega }_z\left(\phi \right)$
or, ${\int }_0^{{\omega }_z}d{\omega }_z={\beta }_0{\int }_0^{\phi }\cos \phi d\phi$
or, $\frac{{\omega }_z^2}{2}={\beta }_0\sin \phi$
Hence ${\omega }_z=\pm \sqrt{2{\beta }_0\sin \phi }$
The plot ${\omega }_z\left(\phi \right)$ is shown in figure below. It can be seen that as the angle $\phi$ grows, the vector $\overrightarrow{\omega }$ first increases, coinciding with the direction of the vector $\overrightarrow{{\beta }_0}({\omega }_z>0)$, reaches the maximum at $\phi =\frac{\phi }{2}$, then starts decreasing and finally turns into zero at $\phi =\pi$. After that the body starts rotating in the opposite direction in a similar fashion $({\omega }_z<0)$. As a result, the body will oscillate about the position $\phi =\frac{\phi }{2}$ with an amplitude equal to $\frac{\pi }{2}$.