Subjective Type

A horizontal smooth rod $$AB$$ rotates with a constant angular velocity $$\omega=2.00\:rad/s$$ about a vertical axis passing through its end $$A$$. A freely sliding sleeve of mass $$m=0.50\;kg$$ moves along the rod from the point $$A$$ with the initial velocity $$v_0=1.00\:m/s$$. 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 $$r=50\:cm$$ from the rotation axis in Newton. (Round off to the nearest integer.)

Solution

The sleeve is free to slide along the rod $$AB$$. Thus only the centrifugal force acts on it. The equation is,

$$m\dot{v}=m\omega^2r$$ where $$\displaystyle v=\frac{dr}{dt}$$.

But $$\displaystyle\dot{v}=v\frac{dv}{dr}=\frac{d}{dr}\left(\frac{1}{2}v^2\right)$$

so, $$\displaystyle\frac{1}{2}v^2=\frac{1}{2}\omega^2r^2+constant$$

or, $$v^2=v_0^2+\omega^2r^2$$
$$v_0$$ being the initial velocity when $$r=0$$. The Corialis force is then,

$$\displaystyle2m\omega\sqrt{v_0^2+\omega^2r^2}=2m\omega^2r\sqrt{1+{\left(\frac{v_0}{\omega r}\right)}^2}$$

$$=2.83\:N$$ on putting the values.

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