Question:
A small disc is set rolling with a speed $v$ 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?
Solution:
By energy conservation,
$\frac{1}{2} I \omega^{2}+\frac{1}{2} m v^{2}=m g h$
$\frac{1}{2} m v^{2}+\frac{1}{2}\left(\frac{m R^{2}}{2}\right)\left(\frac{v^{2}}{R^{2}}\right)=m g h$
$h=\frac{3 v^{2}}{4 g}$