Qn on ratational motion.

  • Thread starter Wen
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  • #1
Wen
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Could someone help me with these problems?

A uniform solid sphere of radius a is placed on top of a stationary spherical dome of radius R, rolls down the dome from rest without slipping. At what height, measured from the bottom, in term of a and R will the solid sphere leave the surface of the dome?

I can only think of resolving mg into mgsinQ and mgcosQ.
Can anyone give me some clues and hints?
 

Answers and Replies

  • #2
HallsofIvy
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That's a good start. At what angle will the velocity vector be above the tangent vector to the sphere?
 
  • #3
mukundpa
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1. As the sphere rolls down the domb its center of mass moves on a circular path, requirs centripetal force, How it gets centripetal force?
2. At the point of leaving, the normal reaction between the two surfaces just becoms zoro.
 
  • #4
Wen
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sorry but i still couldn't get it.
 
  • #5
mukundpa
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ok
consider the sphere at a position where the radius from the center of the dome to the center of the sphere makes an angle q(theeta) with the vertical, and then draw the forces acting on the sphere.
The forces are weight of the sphere mg, normal reaction of the dome N along the radius (upward)and friction which is tangential. Now write the equation for forces along radial direction.
also find the speed of CM of the sphere using energy conservation...
 

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