Passage: A uniform circular disk of radius 0.2 m and mass 1 kg is...

Moment of inertia of the disk about the pivot $C$ is $I_C = \dfrac{1}{2}MR^2 + MR^2 = \dfrac{3}{2}MR^2$.

Substituting $M = 1$ kg and $R = 0.2$ m, we get $I_C = \dfrac{3}{2}(1)(0.2)^2 = 0.06$ kg m$^2$.

The initial angular momentum of the particle about $C$ is $L_i = m u y_{\perp}$, where $y_{\perp}$ is the perpendicular distance from $C$ to the initial line of motion.

$L_i = m u (R + R\cos 45^\circ) = (0.02)(100)(0.2)\left(1 + \dfrac{1}{\sqrt{2}}\right) = 0.4 + 0.2\sqrt{2}$ kg m$^2$/s.

This initial angular momentum is in the clockwise direction.

The final angular momentum of the particle about $C$ is $L_f = m v x_{\perp}$, where $x_{\perp}$ is the perpendicular distance from $C$ to the final line of motion.

$L_f = m v (R\sin 45^\circ) = (0.02)(90)(0.2)\left(\dfrac{1}{\sqrt{2}}\right) = 0.18\sqrt{2}$ kg m$^2$/s.

This final angular momentum is also in the clockwise direction.

By conservation of angular momentum about $C$ during the collision, $L_i = L_f + I_C \omega$, where $\omega$ is the angular velocity of the disk just after the collision.

$0.4 + 0.2\sqrt{2} = 0.18\sqrt{2} + 0.06 \omega$

$0.06 \omega = 0.4 + 0.02\sqrt{2}$

$\omega = \dfrac{40 + 2\sqrt{2}}{6} = \dfrac{20 + \sqrt{2}}{3}$ rad/s.

Applying conservation of mechanical energy for the disk to find the maximum change in height $h$ of its center $O$:

$\dfrac{1}{2}I_C \omega^2 = M g h$

$h = \dfrac{I_C \omega^2}{2 M g} = \dfrac{0.06}{2(1)(10)} \left(\dfrac{20 + \sqrt{2}}{3}\right)^2$

$h = 0.003 \left(\dfrac{400 + 40\sqrt{2} + 2}{9}\right)$

$h = \dfrac{3}{1000} \left(\dfrac{402 + 40\sqrt{2}}{9}\right)$

$h = \dfrac{402 + 40\sqrt{2}}{3000} = \dfrac{201 + 20\sqrt{2}}{1500}$ = 0.15