Question 154 of 210

A disc with a flat small bottom beaker placed on it at a distance $R$ from its center is revolving about an axis passing through the center and perpendicular to its plane with an angular velocity $\omega$. The coefficient of static friction between the bottom of the beaker and the surface of the disc is $\mu$. The beaker will revolve with the disc if :

A$R\leq \dfrac{\mu g}{2{\omega }^{2}}$B$R\leq \dfrac{\mu g}{{\omega }^{2}}$C$R\geq \dfrac{\mu g}{2{\omega }^{2}}$D$R\geq \dfrac{\mu g}{{\omega }^{2}}$

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Answer: (B) $R\leq \dfrac{\mu g}{{\omega }^{2}}$

As the disc is revolving in circular motion about an axis passing through centre and perpendicular to its plane, thus centrifugal force must act on it. The forces acting on disc is shown below.
From FBD

For circular motion, frictional force must balance the required centripetal force.
So, $f=m{\omega }^{2}R$
$\Rightarrow \mu N\geq m{\omega }^{2}R$
$\Rightarrow \mu mg\geq m{\omega }^{2}R$
$\Rightarrow R\leq \dfrac{\mu g}{{\omega }^{2}}$