For angular momentum to be conserved, torque $\vec{\tau} = \vec{r} \times \vec{F}$ must be zero, meaning $\vec{r}$ and $\vec{F}$ are collinear. Taking the ratio of their components: $\frac{2}{\alpha} = \frac{-6}{3} = \frac{-12}{9}$, which simplifies to $\frac{2}{\alpha} = -2$, giving $\alpha = -1$.

Using $K = \frac{1}{2} I \omega^2$, we substitute the values to get $1500 = \frac{1}{2}(1.2) \omega^2$, giving $\omega = 50 \text{ rad/s}$. Applying kinematics equation $\omega = \omega_0 + \alpha t$, we find $50 = 0 + 25t$, yielding $t = 2 \text{ s}$.

Given $I_{\text{diameter}} = \frac{MR^2}{4} = I$, which means $MR^2 = 4I$. Using the parallel axis theorem, the moment of inertia about a perpendicular axis on the rim is $I_{\text{rim}} = \frac{MR^2}{2} + MR^2 = \frac{3}{2} MR^2 = \frac{3}{2} (4I) = 6I$.

The formula for rotational kinetic energy is $K = \frac{1}{2} I \omega^2$. Substituting the given values, $360 = \frac{1}{2} I (30)^2 = 450 I$. Solving for $I$ gives $I = \frac{360}{450} = 0.8 \text{ kgm}^2$.

Moment of inertia $I = \frac{1}{2}MR^{2} = 6.25\text{ kg m}^{2}$. Angular acceleration $\alpha = 2\text{ rev/s}^{2} = 4\pi\text{ rad/s}^{2}$. Torque $\tau = I\alpha = 25\pi = 78.5\text{ N m}$. Tension $T = \frac{\tau}{R} = \frac{78.5}{0.5} = 157\text{ N}$.

Angular velocity of man $\omega_{m} = \frac{v}{R} = 0.5\text{ rad/s}$. By conservation of angular momentum, $I_{p}\omega_{p} + I_{m}\omega_{m} = 0$, giving $\omega_{p} = -0.5\text{ rad/s}$. Relative angular velocity $\omega_{rel} = \omega_{m} - \omega_{p} = 1\text{ rad/s}$. Time $T = \frac{2\pi}{\omega_{rel}} = 2\pi\text{ s}$.

Angular momentum is defined as $\vec{L} = \vec{r} \times \vec{p}$. According to the properties of the cross product, the vector $\vec{L}$ is directed perpendicular to the plane containing the position vector $\vec{r}$ and momentum vector $\vec{p}$.

Using the parallel axis theorem, $I = I_{cm} + Md^{2}$. Here, the center of mass moment of inertia is $I_{cm} = I_{0}$ and the distance to the parallel axis is $d = \frac{L}{2}$. Thus, $I = I_{0} + M(\frac{L}{2})^{2} = I_{0} + \frac{ML^{2}}{4}$.

Angular velocity $\omega = \frac{d\theta}{dt} = 6t^{2} - 12t$. Angular acceleration $\alpha = \frac{d\omega}{dt} = 12t - 12$. Torque is zero when $\alpha = 0$, which implies $12t - 12 = 0$, or $t = 1\text{ s}$.

Initial moment of inertia $I_{original} = \frac{1}{2}(9M)R^{2} = \frac{9}{2}MR^{2}$. Moment of inertia of the removed part is $I_{removed} = \frac{1}{2}M(\frac{R}{3})^{2} = \frac{1}{18}MR^{2}$. The remaining moment of inertia is $I = I_{original} - I_{removed} = \frac{9}{2}MR^{2} - \frac{1}{18}MR^{2} = \frac{40}{9}MR^{2}$.