A particle describes a horizontal circule on the smooth inner surface of a coincal funnel as shown in Fig. If the height of the plane of the circle above the vertex is `9.8 cm`, find the speed of the particle.

Explanation

As is clear from Fig. forces acting on the particle are : weight `(mg)`of particle acting vertically downwards and normal reaction `R` of funnel acting perpendicular to wall of funnel.
`R cos theta = mg and R sin theta = (m v^(2))/(r )`, the centripetal force required

`:. (R sin theta)/(R cos theta) = (m v^(2))/(r mg) or tan theta = (v^(2))/(r g)`
`v = sqrt(r g tan theta)`
but `tan theta = (h)/(r )`
`:. v = sqrt(rg((h)/(r ))) = sqrt(g h)`
`= sqrt(9.8 xx (9.8 xx 10^(-2))) = 0.98 m//s`