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Question
A small disc is set rolling with a speed \[\nu\] on the horizontal part of the track of the previous problem from right to left. To what height will it climb up the curved part?
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Solution
Let the radius of the disc be R.
Let the angular velocity of the disc ω.
Let the height attained by the disc be h.
On applying the law of conservation of energy, we get
\[\frac{1}{2}m v^2 + \frac{1}{2}I \omega^2 = mgh\]
\[ \Rightarrow \frac{1}{2}m v^2 + \frac{1}{2} \times \frac{1}{2}m R^2 \times \left( \frac{v}{R} \right)^2 = mgh\]
\[ \Rightarrow \frac{1}{2} v^2 + \frac{1}{4} v^2 = gh\]
\[ \Rightarrow \frac{3}{4} v^2 = gh\]
\[ \Rightarrow h = \frac{3 v^2}{4g}\]
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