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 <math> <mrow> <mi>ω</mi> <mtext> </mtext> <mo>.</mo> </mrow> </math> The coefficient of static friction between the bottom of the beaker and the surface of the disc is <math> <mrow> <mi>ω</mi> <mtext> </mtext> <mo>.</mo> </mrow> </math> The beaker will revolve with the disc if :

F. B. D of Beaker By NLM2 f = m a = m ω 2 R ⇒ m ω 2 R ≤ μ N R ≤ μ g / ω 2

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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 $ω .$ The beaker will revolve with the disc if :

Option 1 -

$R \leq \frac{μ g}{2 ω^{2}}$

Option 2 -

$R \leq \frac{μ g}{ω^{2}}$

Option 3 -

$R \geq \frac{μ g}{2 ω^{2}}$

Option 4 -

$R \geq \frac{μ g}{ω^{2}}$

2 Views | Posted 2 months ago

Answered by

Raj Pandey | Contributor-Level 9

2 months ago

Correct Option - 2

Detailed Solution:

F. B. D of Beaker

By NLM2

$\begin{array}{l} f = m a = m ω^{2} R \\ \Rightarrow m ω^{2} R \leq μ N \\ R \leq μ g / ω^{2} \end{array}$