Question

A horizontal smooth rod AB rotates with a constant angular velocity w = 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 u = 1.00 ms-¹. Find the Cariolis 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 horizontal axis.


Thanks.​

Answer 1

Answer:

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

mv˙=mω2r where v=dtdr.

But v˙=vdrdv=drd(21v2)

so, 21v2=21ω2r2+constant

or, v2=v02+ω2r2

v0 being the initial velocity when r=0. The Corialis force is then,

2mωv02+ω2r2=2mω2r1+(ωrv0)2

=2.83N on putting the values.

Answer 2

Only the centripetal acceleration, as acceleration, is acting on the sleeve and is given by,

$\sf{\longrightarrow a=\omega^2r}$

or,

$\sf{\longrightarrow \dfrac{dv}{dt}=\omega^2r}$

By chain rule,

$\sf{\longrightarrow \dfrac{dv}{dr}\cdot\dfrac{dr}{dt}=\omega^2r}$

Since $\sf{\dfrac{dr}{dt}=v,}$

$\sf{\longrightarrow v\,\dfrac{dv}{dr}=\omega^2r}$

$\sf{\longrightarrow v\,dv=\omega^2r\ dr}$

Integrating,

$\displaystyle\sf{\longrightarrow\int\limits_u^vv\,dv=\omega^2\int\limits_0^rr\ dr}$

$\displaystyle\sf{\longrightarrow\dfrac{v^2-u^2}{2}=\dfrac{\omega^2r^2}{2}}$

$\displaystyle\sf{\longrightarrow v=\sqrt{u^2+\omega^2r^2}}$

Now, coriolis force is given by,

$\sf{\longrightarrow F=2m\omega v}$

$\sf{\longrightarrow F=2m\omega\sqrt{u^2+\omega^2r^2}}$

In the question,

• $\sf{m=0.5\ kg}$

• $\sf{\omega=2\ rad\,s^{-1}}$

• $\sf{u=1\ m\,s^{-1}}$

• $\sf{r=0.5\ m}$

Hence,

$\sf{\longrightarrow F=2m\omega\sqrt{u^2+\omega^2r^2}}$

$\sf{\longrightarrow F=2\times0.5\times2\sqrt{1^2+2^2(0.5)^2}}$

$\sf{\longrightarrow\underline{\underline{F=2\sqrt{2}\ N}}}$