# What is the speed of the cube on its trajectory?

##

**correction: friction

**correction: friction

##### 1 Answer

#### Answer

#### Answer:

#### Explanation

#### Explanation:

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

- Weight
#=mg# acting downwards. - Normal reaction
#R# at the point of contact perpendicular to the surface of cone.

Let

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=sqrt((rg)/tantheta)#

Inserting given values we get

#v=sqrt((0.2xx9.81)/tan55^@)#

#v=1.17ms^-1# , rounded to one decimal place.

Describe your changes (optional) 200