What is the velocity of the ice cube when it loses contact with the sphere?

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SUMMARY

The discussion centers on calculating the velocity of an ice cube as it loses contact with a half-spherical surface. The ice cube starts from rest at the top of the sphere with no friction. The centripetal force equation, F=(mv^2)/R, is utilized to derive the conditions under which the ice cube will lose contact. The participant seeks clarification on expressing the forces and acceleration in a cylindrical coordinate system and how to relate the angle of the ice cube to time and gravitational force.

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  • Understanding of centripetal force and its application in mechanics
  • Familiarity with cylindrical coordinate systems
  • Basic knowledge of kinematics and dynamics
  • Concept of gravitational acceleration and its effects on motion
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faidros
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Hello,

I've encountered some problems while doing a mechanics project, and was hoping for some help here.
The problem is what follows:

An icecube starts to move at the top of a half spherical building (start velocity 0, no friction, and can be regarded as a particle). What's the icecube's velocity at the time it looses contact with the sphere and begins to fall?

Any ideas?
 
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The centripetal force for the particle is F=(mv^2)/R
Could this force be expressed as F = ma, where a is the acceleration in a cylindrical coordinate system?
i.e:
r=R(cosa, sina)
v=Ra'(-sina, cosa)
a=Ra''(-cosa,-sina) <----

Am I way off here?
Does anybody have an approach on this problem? Any ideas how to express the angle between the vertical and the particle in terms of time and gravity?

thanks for any help
 

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