Question

A horizontal smooth rod $AB$ rotates with a constant angular velocity $\omega =2.00rad/s$ about a vertical axis passing through its end $A$. A freely sliding sleeve of mass $m=0.50kg$ moves along the rod from the point $A$ with the initial velocity $v_0=1.00m/s$. Find the Coriolis force acting on the sleeve (in the reference frame fixed to the rotating rod) at the moment when the sleeve is located at the distance $r=50cm$ from the rotation axis in Newton. (Round off to the nearest integer.)

Answer 1

The sleeve is free to slide along the rod $AB$. Thus only the centrifugal force acts on it. The equation is,
$m\dot{v}=m{\omega }^{2}r$ where $v=\frac{dr}{dt}$.
But $\dot{v}=v\frac{dv}{dr}=\frac{d}{dr}\left(\frac{1}{2}{v}^2\right)$
so, $\frac{1}{2}{v}^2=\frac{1}{2}{\omega }^2{r}^2+constant$
or, $v^2=v_0^2+{\omega }^2{r}^2$
$v_0$ being the initial velocity when $r=0$. The Corialis force is then,
$2m\omega \sqrt{v_0^2+{\omega }^2{r}^2}=2m{\omega }^{2}r\sqrt{1+{\left(\frac{v_0}{\omega r}\right)}^2}$
$=2.83N$ on putting the values.