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A small particle of mass $m$ is given an initial horizontal velocity $ν_{0}$ tangent to the rim of a smooth cone at a radius $r_{0}$ from the vertical centerline as shown at point A. As the particle slides to point B, a vertical distance $h$ below A and a distance $r$ from the vertical centerline, its velocity $ν$ makes an angle $θ$ with the horizontal tangent to the cone at B, then

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the speed of particle at point B is $\sqrt{ν_{0}^{2} + 2 g h}$

the speed of particle at point B is $\sqrt{ν_{0}^{2} + g h}$

The value of $ν_{0}$ for which particle will be moving in a horizontal circle of radius $r_{0}$ is $\sqrt{\frac{g r_{0}}{\tan α}}$

The value of $ν_{0}$ for which particle will be moving in a horizontal circle of radius $r_{0}$ is $\sqrt{\frac{2 g r_{0}}{\tan α}}$

answer is A, C.

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Detailed Solution

Use energy conservation and conservation of angular momentum about axis of cone.

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