An inverted cone is rotating about its vertical axis. A particle is kept on the inner surface of cone and it is at rest relative to the cone at a height of 0.4 m above its vertex. The coefficient of friction between the surface of cone and the particle is 0.6 and the apex angle of cone is 90°. The maximum angular velocity of revolution of the cone can be: (take g = 10/s2)

Option 1 - <p>12.5rad/s<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 2 - <p>10rad/s<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 3 - <p>7.5rad/s<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 4 - <p>5rad/s</p>
28 Views|Posted 5 months ago
Asked by Shiksha User
1 Answer
V
5 months ago
Correct Option - 2
Detailed Solution:

N = (mg/√2) + (mω²h/√2)
(mω²h/√2) = (mg/√2) + f?
(mω²h/√2) = (mg/√2) + μ (mg/√2) + (mω²h/√2)
ω = √ (g/h) (1+μ)/ (1-μ) = 10rad/s

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