What is the speed of the cube on its trajectory?

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**correction: friction

enter image source here

**correction: friction

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A08 Share
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 #=mg# 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 #=(mv^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

#Rcostheta=(mv^2)/r# .....(1)

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

#Rsintheta-mg=0#
#=>Rsintheta=mg# .....(2)

Dividing (2) by (1) we get

#tantheta=(rg)/v^2#

Solving for #v# we get

#v=sqrt((rg)/tantheta)#

Inserting given values we get

#v=sqrt((0.2xx9.81)/tan55^@)#
#v=1.17ms^-1#, rounded to one decimal place.

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