What is the speed of the cube on its trajectory?

**correction: friction

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Dec 25, 2017

At the point of contact of the cube with the cone, in the absence of friction two forces act on it.

1. Weight $= m g$ acting downwards.
2. Normal reaction $R$ at the point of contact perpendicular to the surface of cone.

Let $\theta$ be equal to half the angle of the vertex angle of the cone.

For circular motion required centripetal force $= \frac{m {v}^{2}}{r}$
where $m$ is mass of cube, $v$ its velocity in a horizontal circle of radius $r$.

This is provided by the horizontal component of normal reaction. We have the equation

$R \cos \theta = \frac{m {v}^{2}}{r}$ .....(1)

In the state of equilibrium vertical components of all the forces must be zero. Therefore, we get

$R \sin \theta - m g = 0$
$\implies R \sin \theta = m g$ .....(2)

Dividing (2) by (1) we get

$\tan \theta = \frac{r g}{v} ^ 2$

Solving for $v$ we get

$v = \sqrt{\frac{r g}{\tan} \theta}$

Inserting given values we get

$v = \sqrt{\frac{0.2 \times 9.81}{\tan} {55}^{\circ}}$
$v = 1.17 m {s}^{-} 1$, rounded to one decimal place.

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