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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 $ω$ . The coefficient of static friction between the bottom of the beaker and the surface of the disc is $μ$ . The beaker will revolve with the disc if :

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$R \leq \frac{μg}{2 ω^{2}}$

$R \geq \frac{μg}{2 ω^{2}}$

$R \geq \frac{μg}{ω^{2}}$

$R \leq \frac{μg}{ω^{2}}$

answer is B.

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Detailed Solution

For beaker to move with disc

$f_{s} = {mω}^{2} R$

$We know that f_{s} \leq f_{smax}$

${mω}^{2} R \leq μmg$

$R \leq \frac{μg}{ω^{2}}$

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