AravindG
  • AravindG
This is how i recount the question: A ball slides of from top of a hemispherical bowl..the surface is frictionless....At what height does the ball loose contact with the bowl??take radius =h..
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katieb
  • katieb
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AravindG
  • AravindG
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anonymous
  • anonymous
The surface of the hemisphere doesn't even have friction, no friction \(\to\) no rolling or rotation. So, you only need to consider the motion of the sphere's center of mass. Hint. Normal reaction from the surface on the ball would approach Zero, when the ball loses contact with the surface.
kunal
  • kunal
can anyone answer it easily......it seems to me a good question..

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anonymous
  • anonymous
Let \(\theta\) be the angle from the vertical axis at which the ball finally loses the contact. \(h\) be the radius of the hemispherical bowl. |dw:1332584128519:dw| Using the Energy theorem, \[mgh - mgh \cos \theta = \frac{1}{2}mv^2\tag 1\]Note. \(v\) is the velocity of the ball at the angle \(\theta\). At the angle \(\theta\) forces acting on the ball must be, |dw:1332584594058:dw|Normal from the surface \((N)\), gravitational pull \((mg)\) and the centripetal force \((F_{c})\). \[mg \cos \theta = N + F_{c}\]The Normal force must approach Zero for the ball to leave the hemispherical bowl, \(N \to 0\). \[\implies mg \ cos \theta = \frac{mv^2}{h}\]\[\implies v^2 = gh \cos \theta\tag 2\] Use equation 1 and 2 to get the value of \(\theta \). Note. \(F_c = \frac{mv^2}{h}\)
kunal
  • kunal
good question + good answer = brilliant

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