A particle describes a horizontal circle on the smooth inner surface of a conical 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.

Answer

`0.98 m//s`

Explanation

The forces acting on the particle are reaction R and weight mg as shown in Fig. 7.119 . So , for verticle equilibrium of the particle in horizontal plane
R cos `theta = mg " " ….(i)`
And for circular motion of the particle in horizontal plane
R sin `theta = (mv^(2)//r) ..... (ii)`
Dividing Eqn. (ii) by (i)
tan `theta = (v^(2))/("rg")`
i.e., `v = sqrt("rg tan" theta)`
or ` v = sqrt(rg xx (h //r)) = sqrt(gh)` [ as from figure tan `theta = h//r`]
`therefore " " v = sqrt(9.8 xx 9.8 xx 10^(-2)) = 0.98 m//s.`