A disc of radius $$R$$ and mass $$M$$ is rolling horizontally without slipping with a speed $$v$$. It then moves up an inclined smooth surface as shown in the figure. The maximum height that the disc can go up the incline is

Initial Total Energy:

On the horizontal track, the disc has both translational and rotational kinetic energy:

$$E_i = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2$$

Final Total Energy:

At the maximum height $$h$$, the translational velocity is zero, but the rotational kinetic energy remains unchanged due to the lack of friction:

$$E_f = Mgh + \frac{1}{2}I\omega^2$$

Conservation of Mechanical Energy (since there is no friction):

Equating the initial and final energy states ($$E_i = E_f$$):

$$\frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2 = Mgh + \frac{1}{2}I\omega^2$$

$$\frac{1}{2}Mv^2 = Mgh$$

$$h = \frac{v^2}{2g}$$