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Question

A particle inside the rough surface of a rotating cone about its axis is at rest, relative to it as a height of 1m above its vertex. Friction coefficient is μ=0.5. If half angle of cone is 450, the maximum angular velocity of revolution of cone can be:

A
10 rad/s
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B
30 rad/s
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C
403 rad/s
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D
50 rad/s
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Solution

The correct option is A 10 rad/s
Given :- Mass of particle be ( m )
Half angle of cone = 45°
Coefficient of friction ( μ ) = 0.5
Acceleration due to gravity (g) = 10 m/s2
Height of particle from vertex (r) = 1m

To Find :- Maximum angular velocity ( ω )

Solution :- we know , for max ω , friction (f) = 0
Now since the particle is at rest ,
mgsin45°=FCentripetalsin45°
mg=FCentripetal
mg=mrω2
ω=gr
ω=10rad/s––––––––––––––
Hence , Option A(10rad/s) is correct.–––––––––––––––––––––––––––––––––


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