Conservation of Angular Momentum - College Physics

1. A potter's wheel with moment of inertia $0.15 \text{ kg⋅m}^2$ spins at $120 \text{ rpm}$. The potter places a $0.50 \text{ kg}$ lump of clay at the center and then moves it outward to $0.20 \text{ m}$ from the center. What is the wheel's angular velocity when the clay is at $r = 0.20 \text{ m}$?
   Answer: $106 \text{ rpm}$ When you encounter rotating objects where mass is redistributed, you're dealing with conservation of angular momentum. This fundamental principle states that if no external torques act on a system, the total angular momentum remains constant. Initially, the wheel spins alone with angular momentum $L_i = I_{wheel} \omega_i$. Converting the initial speed: $\omega_i = 120 \text{ rpm} \times \frac{2\pi}{60} = 4\pi \text{ rad/s}$. So $L_i = 0.15 \times 4\pi = 0.6\pi \text{ kg⋅m}^2\text{/s}$. When the clay moves to $r = 0.20 \text{ m}$, the system's moment of inertia increases. The clay contributes $I_{clay} = mr^2 = 0.50 \times (0.20)^2 = 0.02 \text{ kg⋅m}^2$. The total final moment of inertia becomes $I_f = 0.15 + 0.02 = 0.17 \text{ kg⋅m}^2$. Using conservation of angular momentum: $L_i = L_f$, so $0.6\pi = 0.17\omega_f$. Solving: $\omega_f = \frac{0.6\pi}{0.17} = 3.53\pi \text{ rad/s}$. Converting back to rpm: $\omega_f = 3.53\pi \times \frac{60}{2\pi} = 106 \text{ rpm}$, confirming answer B. Choice A (100 rpm) underestimates the final speed. Choice C (112 rpm) overestimates it, possibly from calculation errors. Choice D (120 rpm) incorrectly assumes angular velocity stays constant, ignoring that adding mass farther from the rotation axis increases the moment of inertia and must slow the rotation. Remember: when mass moves outward in a rotating system, the increased moment of inertia always decreases angular velocity to conserve angular momentum.
2. A horizontal disk rotating at $10 \text{ rad/s}$ has moment of inertia $0.20 \text{ kg⋅m}^2$. A second identical disk, initially at rest, is dropped onto it and they stick together. What is the final angular velocity of the combined system?
   Answer: $5.0 \text{ rad/s}$ When you encounter a problem involving objects sticking together during rotation, you're dealing with conservation of angular momentum. Just like linear momentum is conserved in collisions, angular momentum remains constant when no external torques act on the system. Initially, the first disk has angular momentum $L_1 = I_1\omega_1 = (0.20 \text{ kg⋅m}^2)(10 \text{ rad/s}) = 2.0 \text{ kg⋅m}^2\text{/s}$. The second disk contributes zero angular momentum since it's at rest. The total initial angular momentum is $2.0 \text{ kg⋅m}^2\text{/s}$. After collision, both disks rotate together with the same final angular velocity $\omega_f$. The combined moment of inertia is $I_{total} = 0.20 + 0.20 = 0.40 \text{ kg⋅m}^2$. Using conservation of angular momentum: $L_{initial} = L_{final}$, so $2.0 = (0.40)\omega_f$. Solving gives $\omega_f = 5.0 \text{ rad/s}$, which is answer B. Answer A (2.5 rad/s) results from incorrectly dividing the initial angular velocity by 4 instead of 2. Answer C (7.5 rad/s) might come from averaging the two initial angular velocities (5 rad/s each), which ignores the physics entirely. Answer D (10 rad/s) incorrectly assumes angular momentum isn't conserved and the system maintains the original disk's speed. Remember: in rotational "collision" problems, always apply conservation of angular momentum. The key insight is that when objects stick together, their moments of inertia add, but the total angular momentum stays constant.
3. A student sits on a rotating stool holding two $2.0 \text{ kg}$ masses, each $0.50 \text{ m}$ from the rotation axis. The system rotates at $1.5 \text{ rad/s}$. If the student pulls the masses to $0.25 \text{ m}$ from the axis, and the moment of inertia of the student and stool remains $0.50 \text{ kg⋅m}^2$, what is the new angular velocity?
   Answer: $3.0 \text{ rad/s}$ When you see a rotating system where masses move closer to or farther from the rotation axis, think conservation of angular momentum. Since no external torques act on this system, angular momentum must remain constant: $L_i = L_f$. Angular momentum equals moment of inertia times angular velocity ($L = I\omega$). Initially, the total moment of inertia includes the student/stool ($0.50 \text{ kg⋅m}^2$) plus the two masses. Each mass contributes $I = mr^2 = 2.0 \times (0.50)^2 = 0.50 \text{ kg⋅m}^2$, so the two masses together contribute $1.0 \text{ kg⋅m}^2$. The initial total moment of inertia is $I_i = 0.50 + 1.0 = 1.5 \text{ kg⋅m}^2$. After pulling the masses inward, each mass contributes $I = 2.0 \times (0.25)^2 = 0.125 \text{ kg⋅m}^2$, so both masses contribute $0.25 \text{ kg⋅m}^2$. The final moment of inertia is $I_f = 0.50 + 0.25 = 0.75 \text{ kg⋅m}^2$. Using conservation: $I_i\omega_i = I_f\omega_f$, so $1.5 \times 1.5 = 0.75 \times \omega_f$. Solving: $\omega_f = \frac{2.25}{0.75} = 3.0 \text{ rad/s}$. Answer B ($3.0 \text{ rad/s}$) is correct. Answer A ($2.4 \text{ rad/s}$) likely comes from calculation errors. Answer C ($4.5 \text{ rad/s}$) might result from forgetting to include the student/stool's moment of inertia. Answer D ($6.0 \text{ rad/s}$) suggests using only the masses' contribution without the constant $0.50 \text{ kg⋅m}^2$ term. Remember: in rotational problems, always account for ALL rotating components when calculating total moment of inertia.
4. A space station in the form of a ring with radius $100 \text{ m}$ rotates to provide artificial gravity. If the rotation rate decreases by $20\%$ due to mechanical failure, by what percentage does the angular momentum decrease?
   Answer: $20\%$ This problem tests your understanding of angular momentum and how it relates to rotational motion. When analyzing rotating systems like space stations, you need to distinguish between quantities that depend linearly on rotation rate versus those with higher-order dependencies. Angular momentum is defined as $L = I\omega$, where $I$ is the moment of inertia and $\omega$ is the angular velocity. For a space station with fixed structure, the moment of inertia remains constant during the rotation rate change. This means angular momentum depends linearly on the rotation rate. If the rotation rate decreases by 20%, the new angular velocity becomes $\omega_{new} = 0.8\omega_{original}$. Since $L = I\omega$ and $I$ stays constant, the new angular momentum is $L_{new} = I(0.8\omega) = 0.8L_{original}$. This represents a 20% decrease in angular momentum. Answer A (16%) likely comes from incorrectly calculating $(0.8)^2 - 1 = -0.36$, then somehow getting 16%. Answer C (25%) might result from the misconception that if something decreases to 80% of its original value, it decreased by 25% (confusing the relationship between 4/5 and 1/4). Answer D (36%) comes from incorrectly applying the square relationship: $(0.8)^2 = 0.64$, giving a 36% decrease, which would apply to kinetic energy, not angular momentum. Remember: Angular momentum has a linear relationship with rotation rate, while rotational kinetic energy has a quadratic relationship. Always identify which quantity the problem is asking for before applying mathematical relationships.
5. A figure skater spinning with her arms extended has an angular velocity of $2.0 \text{ rad/s}$. When she pulls her arms in, her moment of inertia decreases from $3.0 \text{ kg⋅m}^2$ to $1.5 \text{ kg⋅m}^2$. What is her final angular velocity?
   Answer: $4.0 \text{ rad/s}$ When you encounter a spinning object that changes shape while no external torques act on it, you're dealing with conservation of angular momentum. This fundamental principle states that $L = I\omega$ remains constant when the system is isolated. Since no external forces act on the skater, her initial angular momentum equals her final angular momentum: $I_1\omega_1 = I_2\omega_2$. Substituting the given values: $(3.0 \text{ kg⋅m}^2)(2.0 \text{ rad/s}) = (1.5 \text{ kg⋅m}^2)\omega_2$. This gives us $6.0 = 1.5\omega_2$, so $\omega_2 = 4.0 \text{ rad/s}$. Choice A ($1.0 \text{ rad/s}$) represents a common error where students might think the angular velocity decreases proportionally with the moment of inertia, perhaps dividing $2.0$ by $2$ since the moment of inertia halved. Choice B ($2.0 \text{ rad/s}$) suggests the misconception that angular velocity remains constant when moment of inertia changes. This ignores the conservation principle entirely. Choice C ($3.0 \text{ rad/s}$) might result from incorrectly applying ratios or misunderstanding the relationship between the variables. Choice D ($4.0 \text{ rad/s}$) is correct because when moment of inertia decreases by a factor of 2, angular velocity must increase by the same factor to conserve angular momentum. Remember this key pattern: in conservation of angular momentum problems, the quantities $I$ and $\omega$ are inversely related. When one decreases, the other increases proportionally to keep their product constant.
6. A bicycle wheel with moment of inertia $I$ spins freely with angular velocity $\omega_0$. A brake pad is pressed against the rim with constant friction force $f$ for time $t$. If the wheel's radius is $R$, what is the wheel's angular momentum after braking?
   Answer: $I\omega_0 - fRt$ When you encounter rotational motion problems involving torque and time, you're dealing with the rotational analog of impulse-momentum theory. Just as linear impulse equals change in linear momentum, angular impulse equals change in angular momentum. The brake pad applies friction force $f$ at the wheel's rim, creating a torque $\tau = fR$ that opposes the rotation. This torque acts for time $t$, producing an angular impulse of $fRt$. Since angular impulse equals the change in angular momentum, the wheel's final angular momentum is its initial value minus the braking impulse: $L_f = I\omega_0 - fRt$. Looking at the wrong answers: Choice B ($I\omega_0 - ft$) incorrectly omits the radius $R$ from the torque calculation. Remember that torque equals force times the perpendicular distance from the rotation axis, so you need $fR$, not just $f$. Choice C ($I\omega_0 - \frac{fRt}{I}$) confuses angular momentum with angular velocity—this expression has units of angular velocity, not angular momentum. Choice D ($I(\omega_0 - \frac{fRt}{I})$) is mathematically equivalent to choice A when expanded, but it's written in terms of final angular velocity times moment of inertia rather than directly as final angular momentum. The key insight is recognizing that this is an angular impulse-momentum problem. When you see constant torque applied over time in rotational motion, immediately think: angular impulse = change in angular momentum = $\tau \cdot t$. The correct answer is A.
7. A gymnast performs a somersault by tucking into a ball, reducing her moment of inertia from $15 \text{ kg⋅m}^2$ to $3.0 \text{ kg⋅m}^2$. If she starts with angular velocity $2.0 \text{ rad/s}$, what is her angular velocity when tucked?
   Answer: $10 \text{ rad/s}$ When you encounter a problem involving a gymnast or figure skater changing body position mid-rotation, you're dealing with conservation of angular momentum. Since no external torques act on the gymnast during her somersault, her angular momentum $L = I\omega$ must remain constant throughout the motion. Setting up the conservation equation: $I_1\omega_1 = I_2\omega_2$, where the initial state has $I_1 = 15 \text{ kg⋅m}^2$ and $\omega_1 = 2.0 \text{ rad/s}$, and the tucked state has $I_2 = 3.0 \text{ kg⋅m}^2$. Solving for the final angular velocity: $\omega_2 = \frac{I_1\omega_1}{I_2} = \frac{(15)(2.0)}{3.0} = 10 \text{ rad/s}$ Looking at the wrong answers: Choice A ($0.40 \text{ rad/s}$) comes from incorrectly multiplying the ratios instead of dividing: $\frac{I_2}{I_1} \times \omega_1$. Choice B ($2.0 \text{ rad/s}$) assumes angular velocity stays constant, ignoring the moment of inertia change entirely. Choice C ($6.0 \text{ rad/s}$) results from using the wrong ratio of moments of inertia: $\frac{3.0}{15} \times 2.0 \times 15 = 6.0$. The correct answer is D ($10 \text{ rad/s}$). Remember: when moment of inertia decreases, angular velocity must increase proportionally to conserve angular momentum. Figure skaters spin faster when they pull their arms in for exactly this reason. Always check that your final angular velocity is higher when the object becomes more compact.
8. A uniform rod of mass $M$ and length $L$ initially at rest can rotate about its center. Two identical balls of mass $m$ each, moving with speed $v$ in opposite directions perpendicular to the rod, simultaneously hit the rod at distances $\frac{L}{4}$ from the center and stick. What is the angular velocity of the system after the collisions?
   Answer: $0$ This problem tests conservation of angular momentum in rotational collisions. When objects collide and stick in rotational motion, the total angular momentum before collision equals the total angular momentum after collision. Initially, the rod is at rest, so its angular momentum is zero. Each ball has angular momentum $L = r \times p = r \times mv$, where $r = \frac{L}{4}$ is the distance from the rotation axis. Since the balls move in opposite directions but both create clockwise or counterclockwise rotation about the center, their angular momenta add: $L_{initial} = 2 \times \frac{L}{4} \times mv = \frac{mvL}{2}$. After collision, you need the moment of inertia of the entire system. For a uniform rod rotating about its center: $I_{rod} = \frac{ML^2}{12}$. Each ball, now at distance $\frac{L}{4}$ from the axis, contributes: $I_{ball} = m(\frac{L}{4})^2 = \frac{mL^2}{16}$. The total moment of inertia is: $I_{total} = \frac{ML^2}{12} + 2 \times \frac{mL^2}{16} = \frac{ML^2}{12} + \frac{mL^2}{8}$. Using conservation of angular momentum: $\frac{mvL}{2} = I_{total} \times \omega$, so $\omega = \frac{mvL/2}{ML^2/12 + mL^2/8}$. The wrong answers likely come from: (A) ignoring one ball or the rod's inertia, (B) miscounting the balls or doubling something incorrectly, (C) and (D) making algebraic errors in the moment of inertia calculation or forgetting to account for both balls properly. Always remember: in rotational collisions, carefully account for every object's moment of inertia and ensure angular momentum directions are consistent.
9. Two ice skaters, initially at rest, push off each other. Skater A ($60 \text{ kg}$) moves with speed $4.0 \text{ m/s}$ at radius $3.0 \text{ m}$ from their common center of mass. What is the magnitude of the total angular momentum of the system about their center of mass?
   Answer: $0 \text{ kg⋅m}^2\text{/s}$ When you encounter problems involving objects moving around a center of mass after they interact, think about conservation of momentum and angular momentum. This question tests your understanding of how angular momentum behaves in an isolated system. Since both skaters start at rest and no external torques act on the system, the total angular momentum must be conserved. Initially, both skaters are at rest, so the initial angular momentum is zero. By conservation of angular momentum, the final total angular momentum must also be zero. You can verify this mathematically. Angular momentum is $L = mvr$, where $m$ is mass, $v$ is tangential speed, and $r$ is distance from the rotation axis. For skater A: $L_A = (60 \text{ kg})(4.0 \text{ m/s})(3.0 \text{ m}) = 720 \text{ kg⋅m}^2\text{/s}$. Since the total must be zero, skater B must have angular momentum of $-720 \text{ kg⋅m}^2\text{/s}$ (opposite direction). The vector sum gives zero total angular momentum, confirming answer A. Answer B ($240 \text{ kg⋅m}^2\text{/s}$) might result from calculation errors or misunderstanding the problem setup. Answer C ($480 \text{ kg⋅m}^2\text{/s}$) could come from incorrectly adding partial contributions. Answer D ($720 \text{ kg⋅m}^2\text{/s}$) is the magnitude of each skater's individual angular momentum, but ignores that they rotate in opposite directions. Remember: in isolated systems with no external torques, angular momentum is conserved. If the system starts with zero angular momentum, it must end with zero total angular momentum, regardless of how the individual parts move.
10. A horizontal turntable with moment of inertia $0.80 \text{ kg⋅m}^2$ rotates at $2.5 \text{ rad/s}$. A $0.50 \text{ kg}$ block slides radially outward on the turntable from $r = 0.20 \text{ m}$ to $r = 0.60 \text{ m}$. What is the turntable's angular velocity when the block reaches $r = 0.60 \text{ m}$?
    Answer: $2.2 \text{ rad/s}$ When you see a rotating system where mass moves radially while the system spins, you're dealing with conservation of angular momentum. The key insight is that as the block slides outward, the system's moment of inertia changes, which affects the angular velocity. The total angular momentum must remain constant since no external torques act on the system. Initially, you have the turntable plus the block at $r = 0.20 \text{ m}$: $L_i = I_{table}\omega_i + m r_i^2 \omega_i = (0.80 + 0.50 \times 0.20^2) \times 2.5 = (0.80 + 0.02) \times 2.5 = 2.05 \text{ kg⋅m}^2\text{⋅rad/s}$ When the block reaches $r = 0.60 \text{ m}$, the moment of inertia increases: $L_f = I_{table}\omega_f + m r_f^2 \omega_f = (0.80 + 0.50 \times 0.60^2) \omega_f = (0.80 + 0.18) \omega_f = 0.98 \omega_f$ Setting $L_i = L_f$: $2.05 = 0.98 \omega_f$, so $\omega_f = 2.09 \approx 2.2 \text{ rad/s}$. This confirms answer C. Answer A (1.5 rad/s) represents too large a decrease, possibly from incorrectly treating this as energy conservation. Answer B (2.0 rad/s) might come from rounding errors or approximating the block's contribution incorrectly. Answer D (2.5 rad/s) assumes no change in angular velocity, ignoring that the block's increased moment of inertia must slow the system down. Remember: in rotational problems involving changing mass distribution, always check whether angular momentum (external torques absent) or energy (friction/collision present) is conserved.
11. A door of width $w$ and moment of inertia $I$ about its hinges is initially at rest. A person applies a constant force $F$ perpendicular to the door at distance $\frac{3w}{4}$ from the hinges for time $t$. What is the angular momentum of the door after time $t$?
    Answer: $\frac{3Fwt}{4}$ This problem tests rotational dynamics, specifically the relationship between torque, angular impulse, and angular momentum. When you see a rotating object with an applied force, think about how torque creates angular acceleration and changes angular momentum. To find the angular momentum after time $t$, you need to use the angular impulse-momentum theorem: the change in angular momentum equals the angular impulse (torque multiplied by time). Since the door starts at rest, its final angular momentum equals the angular impulse. First, calculate the torque. Torque equals force times the perpendicular distance from the axis of rotation (the hinges). Here, $\tau = F \times \frac{3w}{4} = \frac{3Fw}{4}$. The angular impulse is this torque multiplied by time: $\frac{3Fw}{4} \times t = \frac{3Fwt}{4}$. Therefore, the angular momentum after time $t$ is $\frac{3Fwt}{4}$. Answer A ($\frac{Fwt}{4}$) incorrectly uses $\frac{w}{4}$ as the distance, perhaps confusing the given distance $\frac{3w}{4}$ with its complement. Answer B ($\frac{Fwt}{2}$) might result from incorrectly using $\frac{w}{2}$ (the door's center) as the distance. Answer D ($Fwt$) uses the full width $w$ as the distance, which would only be correct if the force were applied at the door's edge opposite the hinges. Remember: torque depends on the actual distance from the rotation axis to where the force is applied, not the total dimension of the object. Always identify this distance carefully from the problem statement.
12. A satellite in orbit has angular momentum $L$ about Earth's center. Due to atmospheric drag, its orbital radius decreases by a factor of 4. Assuming the orbit remains circular and using conservation of angular momentum, by what factor does the orbital speed change?
    Answer: $4$ This question tests your understanding of orbital mechanics and angular momentum conservation when atmospheric forces are present. The key insight is recognizing that while drag removes energy from the system, angular momentum is still approximately conserved during the gradual orbital decay process. Start with the angular momentum formula: $L = mvr$, where $m$ is the satellite's mass, $v$ is orbital speed, and $r$ is orbital radius. Since angular momentum is conserved and mass remains constant, we have $mvr = \text{constant}$, which means $vr = \text{constant}$. If the orbital radius decreases by a factor of 4, then $r_{\text{final}} = \frac{r_{\text{initial}}}{4}$. Using the conservation relationship: $v_{\text{initial}} \cdot r_{\text{initial}} = v_{\text{final}} \cdot r_{\text{final}}$ Substituting: $v_{\text{initial}} \cdot r_{\text{initial}} = v_{\text{final}} \cdot \frac{r_{\text{initial}}}{4}$ Solving for the final velocity: $v_{\text{final}} = 4v_{\text{initial}}$ Therefore, the orbital speed increases by a factor of 4, making D correct. A ($\frac{1}{4}$) incorrectly assumes speed decreases proportionally with radius. B ($\frac{1}{2}$) might come from confusing this with energy relationships or taking a square root incorrectly. C ($2$) could result from using $r^{1/2}$ scaling instead of the direct inverse relationship from angular momentum conservation. Study tip: Remember that for orbital problems with atmospheric drag, angular momentum conservation gives you $vr = \text{constant}$, so velocity and radius are inversely proportional. Smaller orbits mean faster satellites!
13. A playground merry-go-round with moment of inertia $150 \text{ kg⋅m}^2$ rotates at $0.80 \text{ rad/s}$. A $25 \text{ kg}$ child jumps on at the edge, $\sqrt{2} \text{ m}$ from the center, with velocity perpendicular to the radius. What is the angular velocity after the child lands?
    Answer: $0.60 \text{ rad/s}$ When you see a rotational collision problem like this, you're dealing with conservation of angular momentum. The key insight is that no external torques act on the system, so the total angular momentum before and after the child jumps on must be equal. Initially, only the merry-go-round is rotating with angular momentum $L_i = I_{mgr}\omega_i = 150 \times 0.80 = 120 \text{ kg⋅m}^2\text{/s}$. The child has zero angular momentum since they're not yet part of the rotating system. After the child lands, both the merry-go-round and child rotate together. The child's moment of inertia about the center is $I_{child} = mr^2 = 25 \times (\sqrt{2})^2 = 50 \text{ kg⋅m}^2$. The total moment of inertia becomes $I_{total} = 150 + 50 = 200 \text{ kg⋅m}^2$. Using conservation of angular momentum: $L_i = L_f$, so $120 = 200\omega_f$. Solving gives $\omega_f = 0.60 \text{ rad/s}$, which is answer B. Answer A (0.50 rad/s) might result from incorrectly calculating the child's moment of inertia or making arithmetic errors. Answer C (0.70 rad/s) could come from neglecting the child's contribution to the total moment of inertia. Answer D (0.80 rad/s) incorrectly assumes angular velocity stays constant, ignoring that adding mass increases the system's rotational inertia. Remember: in rotational collisions, angular momentum is conserved, but kinetic energy typically decreases. Always identify all rotating masses and calculate their moments of inertia about the rotation axis.
14. A rod of length $2.0 \text{ m}$ and mass $3.0 \text{ kg}$ can rotate about one end. A $1.0 \text{ kg}$ ball moving at $6.0 \text{ m/s}$ perpendicular to the rod hits the free end and sticks. What is the angular velocity immediately after the collision?
    Answer: $1.5 \text{ rad/s}$ When you encounter a collision problem involving rotation, think conservation of angular momentum. Since no external torques act on the system during the collision, angular momentum before equals angular momentum after. Before collision, only the ball has angular momentum about the pivot point. The ball's angular momentum is $L = mvr = (1.0 \text{ kg})(6.0 \text{ m/s})(2.0 \text{ m}) = 12.0 \text{ kg⋅m²/s}$. After collision, both the rod and stuck ball rotate together with the same angular velocity $\omega$. You need the total moment of inertia. For a rod rotating about one end: $I_{rod} = \frac{1}{3}ML^2 = \frac{1}{3}(3.0)(2.0)^2 = 4.0 \text{ kg⋅m²}$. For the ball (now a point mass at the end): $I_{ball} = mr^2 = (1.0)(2.0)^2 = 4.0 \text{ kg⋅m²}$. Total: $I_{total} = 8.0 \text{ kg⋅m²}$. Setting initial and final angular momentum equal: $12.0 = 8.0\omega$, so $\omega = 1.5 \text{ rad/s}$. Choice A ($1.0 \text{ rad/s}$) likely comes from incorrectly using $I = \frac{1}{2}ML^2$ for the rod instead of $\frac{1}{3}ML^2$. Choice C ($2.0 \text{ rad/s}$) might result from forgetting to include the ball's moment of inertia after collision. Choice D ($2.4 \text{ rad/s}$) could come from using only the rod's inertia in the denominator. Remember: in rotational collisions, carefully identify all rotating objects afterward and use the correct moment of inertia formula for each geometry. The rod-about-end formula ($\frac{1}{3}ML^2$) is frequently tested and often confused with the rod-about-center formula.
15. Two identical solid cylinders, each with moment of inertia $I$ about their central axes, are initially at rest. One cylinder is given angular velocity $\omega_0$ and then placed against the other so they roll without slipping in contact. What is the final angular velocity of each cylinder?
    Answer: $\frac{\omega_0}{2}$ When two rotating objects come into contact and reach equilibrium, you need to apply conservation of angular momentum while considering the constraint that they roll without slipping in contact. Initially, only one cylinder rotates with angular velocity $\omega_0$, so the total angular momentum is $L_i = I\omega_0 + I(0) = I\omega_0$. After contact, both cylinders rotate with the same final angular velocity $\omega_f$ (since they're identical and in rolling contact), giving final angular momentum $L_f = I\omega_f + I\omega_f = 2I\omega_f$. By conservation of angular momentum: $I\omega_0 = 2I\omega_f$, which simplifies to $\omega_f = \frac{\omega_0}{2}$. Looking at the wrong answers: Choice A ($\frac{\omega_0}{4}$) would result if you incorrectly assumed the angular momentum gets divided by 4, perhaps by confusing this with energy considerations. Choice B ($\frac{\omega_0}{3}$) might come from incorrectly applying conservation of energy instead of angular momentum, or from a calculation error involving the moment of inertia. Choice D ($\frac{2\omega_0}{3}$) could result from incorrectly setting up the conservation equation, perhaps by not properly accounting for both cylinders in the final state. The correct answer is C: $\frac{\omega_0}{2}$. Study tip: In rotational collision problems, always start with conservation of angular momentum and carefully identify what's rotating before and after the interaction. Remember that when identical objects reach the same final state, the initial angular momentum gets shared equally between them.
16. A uniform disk of mass $M$ and radius $R$ spins with angular velocity $\omega$ about its center. A small mass $m$ is dropped onto the disk at distance $\frac{R}{2}$ from the center and sticks. What is the final angular velocity?
    Answer: $\frac{2M\omega}{2M + m}$ When you encounter problems involving rotating objects and collisions, you're dealing with conservation of angular momentum. Since no external torques act on the disk-mass system, angular momentum before and after the collision must be equal. Initially, only the disk rotates with angular momentum $L_i = I_{disk}\omega = \frac{1}{2}MR^2\omega$. After the collision, both the disk and the point mass (now at distance $\frac{R}{2}$ from the center) rotate together with final angular velocity $\omega_f$. The final moment of inertia is $I_f = I_{disk} + I_{mass} = \frac{1}{2}MR^2 + m(\frac{R}{2})^2 = \frac{1}{2}MR^2 + \frac{mR^2}{4}$. Factoring out $\frac{R^2}{4}$: $I_f = \frac{R^2}{4}(2M + m)$. Applying conservation of angular momentum: $\frac{1}{2}MR^2\omega = \frac{R^2}{4}(2M + m)\omega_f$ Solving for $\omega_f$: $\omega_f = \frac{2M\omega}{2M + m}$ Choice A incorrectly treats the added mass as if it were at the center (zero moment of inertia contribution). Choice B uses $\frac{m}{4}$ instead of the correct $\frac{mR^2}{4}$ contribution to moment of inertia, confusing mass with rotational inertia. Choice D doubles the mass term incorrectly, possibly from algebraic errors in the setup. Study tip: For rotating collision problems, always write out both the initial and final moments of inertia explicitly, remembering that $I = mr^2$ for point masses. The distance from the rotation axis is crucial—it's squared in the moment of inertia formula.

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