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 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 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.

This problem tests understanding of which quantities are conserved during rotational motion. With negligible external torque in deep space, angular momentum about the spin axis must remain constant as the masses move inward. While angular momentum L = Iω is conserved, the angular speed ω increases as moment of inertia I decreases. Rotational kinetic energy K = ½Iω² actually increases because work is done pulling the masses inward against centrifugal effects. Choice D incorrectly suggests torque must increase—no external torque is needed to maintain rotation in the absence of friction. The fundamental principle is that angular momentum is the conserved quantity when external torque is negligible, not angular speed or energy.

This question assesses conservation of angular momentum in a sticking collision between wheels with negligible external torque. Angular momentum remains constant because external torque is negligible, with initial L = I ω₀ + 0 equaling final 2I ω_f for identical wheels. The second wheel adding to the system doubles the moment of inertia, halving the angular speed. Solving yields ω_f = ω₀ / 2 as they rotate together slower. Choice D (zero) is a distractor assuming sticking violates conservation, but momentum is conserved in inelastic rotational collisions. A key strategy is to treat the final system as combined and set L_i = L_f, solving for ω_f using the total I.