Question

A smooth horizontal disc rotates with a constant angular velocity $\omega$ about a stationary vertical axis passing through its centre, the point $O$. At a moment $t=0$ a disc is set in motion from that point with velocity $v_0$. Find the angular momentum $M\left(t\right)$ of the disc relative to the point $O$ in the reference frame fixed to the disc. Make sure that this angular momentum is caused by the Coriolis force.

Answer 1

The Coriolis force is $(2m{\overrightarrow{v}}^{\prime }\times \overrightarrow{\omega })$.
Here $\overrightarrow{\omega }$ is along the z-axis (vertical). The moving disc is moving with velocity $v_0$ which is constant. The motion is along the x-axis say. Then the Coriolis force is along y-axis and has the magnitude $2m{v}_0\omega$. At time $t$, the distance of the centre of moving disc from $O$ is $v_{0}t$ (along x-axis). Thus the torque $N$ due to the coriolis force is $N=2m{v}_0\omega \cdot v_{0}t$ along the z-axis.
Hence equating this to $\frac{dM}{dt}$
$\frac{dM}{dt}=2m{v}_0^2\omega t$ or $M=m{v}_0^2\omega t^2+$ constant.
The constant is irrelevant and may be put equal to zero if the disc is originally set in motion from the point $O$.
This discussion is approximate. The Coriolis force will cause the disc to swerve from straight line motion and thus cause deviation from the above formula which will be substantial for large $t$.