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 -

12.5rad/s

Option 2 -

10rad/s

Option 3 -

7.5rad/s

Option 4 -

5rad/s

16 Views | Posted 3 weeks ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

3 weeks 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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