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=0.50g` 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 rotating rod) at the moment when the sleeve is located at the distance `r=50cm` from the rotation axis.

Answer 1

The sleeve is free to slide along the rod AB. Thus only the centrifugal force acts on it.
The equation is,
`moverset(.)v=momega^2r` where `v=(dr)/(dt)`.
But `overset(.)v=v(dv)/(dr)=(d)/(dr)(1/2v^2)`
so, `1/2v^2=1/2omega^2r^2+const ant`
or, `v^2=v_0^2+omega^2r^2`
`v_0` being the initial velocity when `r=0`. The Coriolis force is then,
`2momegasqrt(v_0^2+omega^2r^2)=2momega^2rsqrt(1+v_0^2//omega^2r^2)`
`=2*83N` on putting the values.s