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Q. A particle describes a horizontal circle in a conical funnel whose inner surface is smooth with speed of $0.5 \,m/s$. What is the height of the plane of circle from vertex of the funnel?

Laws of Motion

Solution:

image
The particle is moving in circular path
From the figure, $mg = R\, sin\, \theta \, \dots(i)$
$\frac{mv^{2}}{r} =R \,cos\,\theta \, \dots (ii)$
From equations (i) and (ii) we get
$tan\,\theta =\frac{rg}{v^{2}}$ but $tan \,\theta =\frac{r}{h}$
$\therefore h=\frac{v^{2}}{g} = \frac{(0.5)^{2}}{10}$
$=0.025\,m = 2.5\,cm$