Question

A wire ring of radius R rotates with a constant angular velocity
w about a vertical axis 00' as shown. A small sleeve can slide
along the ring without friction. The angle e corresponding to
the equilibrium position of the sleeve is
sleeve​

Answer 1

Given:

A wire ring of radius R rotates with a constant angular velocity $\omega$ about a vertical axis 00' as shown. A small sleeve can slide

along the ring without friction.

To find:

Angle corresponding to equilibrium position of sleeve.

Calculation:

First of all , the sleeve will experience centrifugal force equivalent to :

$\boxed{F_{c} = \dfrac{m {v}^{2} }{r} = m { \omega}^{2} r}$

Lets consider that the angle of equilibrium of sleeve with vertical be $\theta$.

Now , refer to the diagram for all the components of forces acting on the sleeve.

• The sleeve will be in equilibrium when the tangential component of forces cancel out each other.

$\therefore \: mg \sin( \theta) = m { \omega}^{2} r \cos( \theta)$

$\implies \: mg \sin( \theta) = m { \omega}^{2} \{R \sin( \theta) \} \cos( \theta)$

$\implies \: g \sin( \theta) = { \omega}^{2} \{R \sin( \theta) \} \cos( \theta)$

$\implies \: g = { \omega}^{2} R \cos( \theta)$

$\implies \: \cos( \theta) = \dfrac{g}{{ \omega}^{2}R }$

$\implies \: \theta= { \cos}^{ - 1} \bigg( \dfrac{g}{{ \omega}^{2}R } \bigg)$

So, final answer is:

$\boxed{ \bold{\: \theta= { \cos}^{ - 1} \bigg( \dfrac{g}{{ \omega}^{2}R } \bigg)}}$