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Question
A solid sphere of mass m is released from rest from the rim of a hemispherical cup so that it rolls along the surface. If the rim of the hemisphere is kept horizontal, find the normal force exerted by the cup on the ball when the ball reaches the bottom of the cup.
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Solution
The radius of the solid sphere (r) is negligible w.r.t. the radius of the hemispherical cup (R).
Let the ball reaches the bottom of the cup with a velocity v and angular velocity ω.
On applying the law of conservation of energy, we get
\[mgR = \frac{1}{2}I \omega^2 + \frac{1}{2}m v^2\]
\[\Rightarrow mgR = \frac{1}{2} \times \frac{2}{5} m \nu^2 + \frac{1}{2} m \nu^2 \]
\[ \Rightarrow \frac{7}{10} m \nu^2 = mgR\]
\[ \Rightarrow \nu^2 = \frac{10}{7}gR\]
As shown in the free body diagram, we have
Total normal force on the ball \[= mg + \frac{m \nu^2}{R}\]
\[= mg + \frac{m\left( \frac{10}{7}gR \right)}{R}\]
\[ = mg + mg \left( \frac{10}{7} \right)\]
\[ = \frac{17}{7}mg\]
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