The only significant external force on the diver during flight is gravity, which acts on the center of mass. Therefore, there is no net external torque about the diver's center of mass. This means their angular momentum $$L$$ is conserved and remains approximately constant. When the diver tucks, they bring their mass closer to the axis of rotation, decreasing their rotational inertia $$I$$. When they extend their body, they move mass away from the axis, increasing $$I$$.

Since the putty is dropped vertically, it carries no initial horizontal momentum and thus no initial angular momentum about the disk's axis. The forces during the collision are internal to the disk-putty system, so the total angular momentum is conserved. The initial angular momentum is $$L_i = I_0 \omega_0$$. The final rotational inertia of the system is $$I_f = I_0 + I_{putty} = I_0 + mr^2$$. The final angular momentum is $$L_f = I_f \omega_f = (I_0 + mr^2)\omega_f$$. Setting $$L_i = L_f$$ gives $$I_0 \omega_0 = (I_0 + mr^2)\omega_f$$. Solving for $$\omega_f$$ yields the correct answer.

The key to conservation of angular momentum is the absence of a net external torque. The frictional forces between the two disks create torques, but these torques are internal to the system consisting of both disks. Disk 1 exerts a torque on Disk 2, and Disk 2 exerts an equal and opposite torque on Disk 1. These internal torques cancel out when considering the system as a whole, so the system's total angular momentum is conserved. The frictionless axle ensures there is no external torque from the support.

The system starts from rest, so the total initial angular momentum is zero. As there are no external torques, the total final angular momentum must also be zero. Let $$\omega_p$$ be the platform's angular velocity relative to the ground. The person's velocity relative to the ground is $$v_g = v_{rel} - v_{plat} = v - R\omega_p$$. The total final angular momentum is $$L_f = I_p\omega_p + I_{person}\omega_{person} = 0$$. The person's angular velocity is $$v_g/R$$. So, $$(\frac{1}{2}MR^2)\omega_p + (mR^2)(\frac{v - R\omega_p}{R}) = 0$$. Simplifying: $$\frac{1}{2}MR^2\omega_p + mR(v-R\omega_p) = 0 \Rightarrow \frac{1}{2}MR^2\omega_p + mRv - mR^2\omega_p = 0 \Rightarrow \omega_p(\frac{1}{2}MR^2 + mR^2) = mRv$$. Solving for $$\omega_p$$ gives the result.

The collapse is due to internal gravitational forces, so there is no net external torque. Thus, the star's angular momentum is conserved. Let $$L_i = L_f$$. The initial angular momentum is $$L_i = I_i \omega_i = (\frac{2}{5}MR^2)\omega_i$$. The final angular momentum is $$L_f = I_f \omega_f = (\frac{2}{5}M(R/2)^2)\omega_f = (\frac{2}{5}M\frac{R^2}{4})\omega_f$$. Setting them equal: $$(\frac{2}{5}MR^2)\omega_i = (\frac{1}{4})(\frac{2}{5}MR^2)\omega_f$$. This simplifies to $$\omega_i = \frac{1}{4}\omega_f$$, or $$\omega_f = 4\omega_i$$.

The conservation of angular momentum for a system depends on the absence of a net external torque. When the child walks toward the center, the forces they exert on the merry-go-round (and the forces the merry-go-round exerts on them) are internal to the defined system. Assuming a frictionless axle, there is no significant external torque, so the total angular momentum of the system is conserved.

Angular momentum is conserved about the pivot point during the collision, as the collision forces are internal and the gravitational torque is negligible over the short impact time. The initial angular momentum is solely from the ball: $$L_i = |\vec{r} \times \vec{p}| = L(mv)$$. The final rotational inertia of the system is the sum of the rod's inertia and the ball's inertia (treated as a point mass): $$I_f = I_{rod} + I_{ball} = \frac{1}{3}ML^2 + mL^2$$. The final angular momentum is $$L_f = I_f \omega = (\frac{1}{3}ML^2 + mL^2)\omega$$. Equating $$L_i = L_f$$ and solving for $$\omega$$ gives the result.

The system consists of the person, the weights, and the turntable. It starts from rest, so its initial total angular momentum is zero. All forces and torques involved in moving the arms and weights are internal to this system. With no net external torque, the total angular momentum must remain zero. As the person gives the weights a clockwise angular momentum, the person's body and the turntable must acquire an equal and opposite (counter-clockwise) angular momentum to keep the total angular momentum of the system at zero.

The rotational form of Newton's second law states that the average net external torque is equal to the rate of change of angular momentum: $$\tau_{avg} = \frac{\Delta L}{\Delta t}$$. In this case, $$\Delta L = L_f - L_i = 10.0 - 4.0 = 6.0 , \text{kg} \cdot \text{m}^2/\text{s}$$, and $$\Delta t = 2.0 , \text{s}$$. Therefore, $$\tau_{avg} = \frac{6.0}{2.0} = 3.0 , \text{N} \cdot \text{m}$$. This question highlights that a non-zero torque causes a change in angular momentum, which is the corollary to the conservation principle.

Because there is no net external torque on the skater, her angular momentum $$L = I\omega$$ is conserved. When she pulls her arms in, her moment of inertia $$I$$ decreases. To keep $$L$$ constant, her angular velocity $$\omega$$ must increase. Her rotational kinetic energy is given by $$K = \frac{1}{2}I\omega^2 = \frac{L^2}{2I}$$. Since $$L$$ is constant and $$I$$ decreases, her kinetic energy $$K$$ must increase. The increase in kinetic energy comes from the work she does to pull her arms inward against the centrifugal effects.