Question

A particle describe a horizontal circle on the smooth inner surface of a conical funnel. The height of the circle above the vertex is $9.8$cm. Find the speed of the particle (Take $g=9.8m{s}^2$)

Answer 1

Since mg is in the downward direction so the gravitational force and $nsin\theta$ is the balance that force

$Nsin\theta =mg........1$

$Ncos\theta =\frac{m{v}^2}{R}...................2$ R is the radius of the circle.

Dividing equation 1 and 2 then we get,

$\frac{Nsin\theta }{Ncos\theta }=tan\theta =\frac{gE}{v^2}$

$tan\theta =\frac{R}{h}=\frac{gR}{v^2}$ Where h is the height of the plane from the vertex,

$v^2=gh\Rightarrow v=\sqrt{gh}=\sqrt{0.098\times 9.8}=0.98m/s$