This question evaluates whether angular momentum is conserved when internal changes occur with negligible external torque. Angular momentum remains L₀ because external torque is negligible, unaffected by the student's movement increasing I from I to 2I. The internal repositioning changes I but not the total L of the isolated system. The new angular momentum is still L₀, as conservation holds. Choice A (2L₀) could result from mistakenly thinking L increases with I, ignoring that ω decreases. A transferable strategy is to recognize that internal forces cannot change total L; always confirm external torque is zero before applying conservation.

This problem demonstrates conservation of angular momentum in everyday situations. When external torque is negligible, angular momentum L = Iω remains constant for the person-stool system. Moving dumbbells farther from the rotation axis increases the system's rotational inertia I because more mass is distributed at larger distances from the axis. Since L must remain constant and I increases, the angular speed ω must decrease proportionally. Choice B incorrectly claims angular speed stays constant, confusing it with angular momentum. The strategy is to recognize that extending mass outward always increases rotational inertia, requiring decreased angular speed to conserve angular momentum.

This problem tests conservation of angular momentum with changing radial positions. As masses slide inward, the moment of inertia becomes 1/4 of initial (since I proportional to r², and r halves), and no external torque conserves L. Initial L = I₀ω₀ = final (I₀/4)ω, so ω = 4ω₀, speeding up. Reduced I demands higher ω for constant L. Choice D errs by claiming constant ω from negligible torque, but torque absence conserves L, not ω. A transferable tip: For variable radius systems, recall I ∝ r² and use L conservation to predict ω adjustments.

This problem tests conservation of angular momentum when moment of inertia increases. With negligible external torque about the rotation axis, angular momentum L = Iω must remain constant as the student extends arms. Initially: L = I₀ω₀. After extending arms: L = (4I₀)ω_final. Setting these equal: I₀ω₀ = (4I₀)ω_final, yielding ω_final = ω₀/4. The quadrupling of moment of inertia causes angular speed to decrease by a factor of four. Choice D incorrectly claims extending arms adds external torque—arm extension involves only internal forces that cannot change total angular momentum. The strategy is recognizing that I and ω change inversely to maintain constant L.

This problem demonstrates conservation of angular momentum in a system with changing angular momentum directions. Initially, the system (student + turntable + wheel) has angular momentum equal to the wheel's spin angular momentum pointing upward. When the student flips the wheel, its angular momentum reverses to point downward. Since external torque is negligible, total system angular momentum must remain constant (pointing upward). To compensate for the wheel's downward angular momentum, the student-turntable must acquire upward angular momentum by rotating opposite to the wheel's original spin direction. Choice C incorrectly claims that internal torques violate conservation—internal forces cannot change total angular momentum. The strategy is to track angular momentum as a vector quantity and ensure the total remains constant.

This problem examines conservation of angular momentum during aerial motion. With negligible air resistance and external torque about the center of mass, the gymnast's angular momentum remains constant throughout the somersault. When tucking reduces the moment of inertia I about the rotation axis, the angular speed ω must increase to maintain constant angular momentum L = Iω. This allows gymnasts to control rotation rate without external forces. Choice B incorrectly claims internal muscle torques affect total angular momentum—internal forces cannot change a system's total angular momentum. The key insight is that athletes can redistribute their mass to change I and thereby control ω while conserving L.

This question assesses the conservation of angular momentum in a system with negligible external torque. Angular momentum is conserved because no external torque acts on the system about the vertical axis, so the initial angular momentum I₀ω₀ equals the final angular momentum (I₀/4)ω_f. When the student pulls the dumbbells inward, the moment of inertia decreases to one-fourth, causing the angular speed to increase to maintain the same angular momentum. Therefore, solving I₀ω₀ = (I₀/4)ω_f gives ω_f = 4ω₀. A common distractor like choice A (ω₀/4) might result from mistakenly thinking angular speed is directly proportional to moment of inertia instead of inversely. To solve similar problems, remember that when external torque is negligible, angular momentum L = Iω is conserved, and changes in I lead to inverse changes in ω.

This problem explores angular momentum conservation about a specific axis during reconfiguration. Pulling the wheel closer reduces the moment of inertia to 2/3 I₀, and with negligible external torque about the vertical axis, L is conserved. Initial L = I₀ω₀ = final (2/3 I₀)ω, so ω = (3/2)ω₀, increasing speed. The decreased I requires faster rotation to maintain L. Choice D is wrong as it assumes kinetic energy conservation, but energy isn't conserved here due to internal work. When analyzing axial momentum, ensure torque is negligible and apply Iω constant for varying I.

This question tests conservation of angular momentum when the moment of inertia changes in a rotating system with negligible external torque. Angular momentum is conserved since external torque is negligible, so initial $I \omega_0$ equals final $\frac{I}{3} \omega_f$. The rider's position change reduces the moment of inertia to one-third, increasing the angular speed to maintain L. Thus, $\omega_f = 3 \omega_0$ as the system spins faster. Choice A ($\omega_0 / 3$) might arise from inverting the relationship and thinking speed decreases with smaller I. In general, verify no external torque, then use the inverse proportionality of $\omega$ and I to predict rotational changes.

This question explores conservation of angular momentum during a star's contraction with negligible external torque. Angular momentum is conserved as the star contracts uniformly, so initial I_i ω_i equals final I_f ω_f. For a uniform sphere, I ∝ R², so halving the radius quarters I, requiring angular speed to quadruple to keep L constant. Thus, ω_f = 4 ω_i, meaning it quadruples. Choice D (stays constant) might be chosen by confusing negligible torque with constant speed instead of constant L. Remember to calculate I changes based on geometry and use ω_f / ω_i = I_i / I_f for conserved angular momentum problems.