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प्रश्न
A particle slides on the surface of a fixed smooth sphere starting from the topmost point. Find the angle rotated by the radius through the particle, when it leaves contact with the sphere.
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उत्तर
Let the velocity be \[\nu\] when the body leaves the surface.
From the free-body diagram,
\[\frac{\text{m}\nu^2}{\text{R}} = \text{mg} \cos \theta\]
\[ [\text{ normal reaction }]\]
\[ \nu^2 = \text{Rg} \cos \theta . . . (\text{i})\]
Again, from the work-energy principle,
Change in K.E. = Work done
\[\Rightarrow \frac{1}{2}\text{m}\nu^2 - 0 = \text{mg} \left( \text{R - R} \cos \theta \right)\]
\[ \Rightarrow \nu^2 = 2\text{ gR} \left( 1 - \cos \theta \right) . . . . (ii)\]
From (i) and (ii),
\[\cos \theta = \frac{2}{3}\]
\[\theta = \cos^{- 1} \left( \frac{2}{3} \right)\]
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