InterviewSolution
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InterviewSolution
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A particle `P` is initially at rest on the top `pf`a smooth hemispherical surface which is fixed on a horizontal plane. The particle is given a velocity `u` horizontally. Radius of spherical surface is `a`. .A. If the particle leaves the sphere, when it has fallen vertically by a distance of `(a)/(4)m u = (sqrt(ga))/(2)`.B. If the particle leaves the sphere at angle `theta` (fig) where `cos theta = (sqrt(3))/(2)`, then `u = (sqrt(ag))/(3)`C. If `u=0` and the particle just slides down the hemispherical surface, it will leave the surface when `cos theta = (2)/(3)`.D. The minimum value of `u`, for the object to leave the sphere without sliding over the surface is `sqrt(ag)`. |
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Answer» Correct Answer - A::C::D Applying conservation of total energy `(1)/(2) m u^(2) + mga (1 -c os theta) = (1)/(2) mv^(2)` `mg cos theta -N =(mv^(2))/(a)` for particle to lose contact `N = 0` `v^(2) =ag cos theta , u^(2)+ga (2-3 cos theta) = 0`. |
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