Suppose, the particle P is in motion in an inverted cone. Velocity of the particle is v and radius of its circular path is r. In equilibrium, the horizontal component R cosθ of normal reaction R, provides the require centripetal force \(\frac{mv^2}{r}\) and its vertical component Rsinθ balances the weight mg of the particle.
\(\therefore\ R\cos\theta = \frac{mv^2}{r}\) .....(1)
and R cos θ = mg .......(2)
on dividing equation (2) by (1), we get
\(\tan\theta = \frac{rg}{v^2} \ ...(2)\)
If the vertical height of plane of circular path for the vertex of the cone be h, then
\(\tan\theta = \frac{r}{h}\ ..(4)\)
Now equating equations (3) and (4), we have,
