Conservation of Angular Momentum - AP Physics C Mechanics

A playground merry-go-round (uniform disk) of mass $M = 200 \text{ kg}$ and radius $R = 2.0 \text{ m}$ rotates at $\omega = 0.5 \text{ rad/s}$ . A child of mass $m = 30 \text{ kg}$ jumps onto the edge moving tangentially at $v = 3.0 \text{ m/s}$ in the same direction as the rim's motion. The rim's speed before the child jumps on is $v_{rim} = \omega R = 1.0 \text{ m/s}$ . What is the final angular velocity?
Answer: $0.65 \text{ rad/s}$ Apply conservation of angular momentum about the center. Initially: $L_i = I_{disk}\omega + mvR = \frac{1}{2}MR^2\omega + mvR = \frac{1}{2}(200)(4)(0.5) + (30)(3)(2) = 200 + 180 = 380 \text{ kg⋅m}^2\text{/s}$ . Finally: $L_f = (I_{disk} + mR^2)\omega_f = [\frac{1}{2}(200)(4) + (30)(4)]\omega_f = (400 + 120)\omega_f = 520\omega_f$ . Therefore: $\omega_f = \frac{380}{520} = 0.65 \text{ rad/s}$ . Choice A incorrectly uses the rim speed instead of the child's actual speed. Choice C forgets to include the child's initial angular momentum. Choice D incorrectly calculates the final moment of inertia.
Two disks on a common frictionless axle are brought into contact. Disk 1 has rotational inertia $I_1$ and is initially spinning. Disk 2 has rotational inertia $I_2$ and is initially at rest. They eventually rotate together at a common final angular velocity. Why is the total angular momentum of the two-disk system conserved during this process?
Answer: Because the frictional torques that bring them to a common speed are internal to the two-disk system. 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.
A comet in an elliptical orbit around the Sun has a speed $v_A$ at its farthest point from the Sun (aphelion), a distance $r_A$ . What is its speed, $v_P$ , at its closest point (perihelion), a distance $r_P$ ?
Answer: $v_P = v_A \frac{r_A}{r_P}$ The torque on the comet about the Sun is zero, so its angular momentum is conserved. At aphelion and perihelion, the velocity vector is perpendicular to the radius vector. The angular momentum magnitude is $L = rmv$ . Conservation of angular momentum implies $L_A = L_P$ , so $m r_A v_A = m r_P v_P$ . Solving for $v_P$ gives $v_P = v_A (r_A / r_P)$ .
An ice skater is spinning on a frictionless surface with her arms extended. She then pulls her arms in close to her body. Which of the following correctly describes the change in her angular velocity and her rotational kinetic energy?
Answer: Her angular velocity increases, and her rotational kinetic energy increases because she does positive work to pull her arms inward. 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.
A horizontal disk with rotational inertia $I_0$ is rotating freely with angular velocity $\omega_0$ about a vertical axis. A small piece of putty of mass $m$ is dropped vertically and sticks to the disk at a distance $r$ from the axis. What is the final angular velocity of the disk-putty system?
Answer: $\frac{I_0 \omega_0}{I_0 + mr^2}$ 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.
A circular platform of mass $M$ and radius $R$ is free to rotate friction-free about its center. A person of mass $m$ stands at the edge. The system is initially at rest. The person begins to walk along the edge with a speed $v$ relative to the platform. What is the magnitude of the angular velocity of the platform relative to the ground? The rotational inertia of the platform is $I_p = \frac{1}{2}MR^2$ .
Answer: $\frac{mv}{R(\frac{1}{2}M + m)}$ 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.
A rotating, uniform spherical star of mass $M$ and radius $R$ has an initial angular velocity $\omega_i$ . It undergoes a gravitational collapse, shrinking to a final radius of $R/2$ while retaining all its mass. Assuming the star's density remains uniform, what is its final angular velocity $\omega_f$ ? The rotational inertia of a uniform sphere is $I = \frac{2}{5}MR^2$ .
Answer: $4\omega_i$ 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 $\omega_f = 4\omega_i$ .
A spinning top with a large angular momentum $\vec{L}$ pointing nearly vertically is subject to a small torque $\vec{\tau}$ due to gravity, which points horizontally. This torque causes the top to precess. The change in the angular momentum vector, $\Delta\vec{L}$ , over a short time interval $\Delta t$ is best described as:
Answer: a vector parallel to $\vec{\tau}$ , causing the direction of $\vec{L}$ to change. The relationship between torque and angular momentum is $\vec{\tau} = d\vec{L}/dt$ , which can be approximated as $\Delta\vec{L} \approx \vec{\tau} \Delta t$ . This means the change in the angular momentum vector, $\Delta\vec{L}$ , is in the same direction as the torque vector $\vec{\tau}$ . Since the torque is horizontal, the change in angular momentum is horizontal. Adding this small horizontal change to the large vertical original momentum causes the momentum vector's direction to shift, which is the phenomenon of precession.
A child is sitting on the edge of a merry-go-round that is rotating freely. The child then slowly walks toward the center of the merry-go-round. The angular momentum of the child-merry-go-round system is conserved because:
Answer: the forces between the child and the merry-go-round are internal to the system, resulting in zero net external torque. 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.
A sticky lump of clay and a high-bouncing rubber ball have identical masses and are thrown with identical horizontal velocities toward the edge of a hinged door that is initially at rest. The clay sticks to the door. The ball bounces straight back with nearly its original speed. Which object causes the door to acquire a greater angular velocity?
Answer: The rubber ball, because it has a larger change in angular momentum, imparting a larger angular impulse. The change in the door's angular momentum (and thus its final angular velocity) is equal to the angular impulse imparted to it. This angular impulse is equal and opposite to the change in the projectile's angular momentum. The clay's angular momentum changes from $L_i$ to 0, a change of magnitude $|L_i|$ . The rubber ball's angular momentum changes from $L_i$ to approximately $-L_i$ , a change of magnitude $|-L_i - L_i| = 2|L_i|$ . Since the ball experiences a larger change in angular momentum, it imparts a larger angular impulse to the door, causing a greater final angular velocity.
A thin uniform rod of mass $M$ and length $L$ is pivoted at one end and hangs vertically at rest. A small ball of mass $m$ traveling horizontally with speed $v$ strikes the rod at its bottom end and embeds itself. What is the angular velocity of the rod-ball system immediately after the collision? The rotational inertia of the rod about the pivot is $I_{rod} = \frac{1}{3}ML^2$ .
Answer: $\frac{mvL}{\frac{1}{3}ML^2 + mL^2}$ 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.
A person stands at the center of a frictionless turntable that is initially at rest. They hold two heavy weights. They extend their arms horizontally and swing them in a clockwise circle. What happens to the turntable?
Answer: It rotates counter-clockwise to keep the total angular momentum of the system zero. 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.
A uniform rod of mass $M$ and length $L$ is pivoted at one end and hangs vertically at rest. A small ball of mass $m$ moving horizontally with speed $v$ strikes the rod at distance $\frac{3L}{4}$ from the pivot and sticks to it. What is the angular velocity of the rod-ball system immediately after the collision?
Answer: $\frac{3mvL}{4(\frac{1}{3}ML^2 + \frac{9}{16}mL^2)}$ Apply conservation of angular momentum about the pivot. Initially, only the ball has angular momentum: $L_i = mv \cdot \frac{3L}{4}$ . After collision, the rod rotates about its end with $I_{rod} = \frac{1}{3}ML^2$ , and the ball rotates at distance $\frac{3L}{4}$ with $I_{ball} = m(\frac{3L}{4})^2 = \frac{9}{16}mL^2$ . So $L_f = (\frac{1}{3}ML^2 + \frac{9}{16}mL^2)\omega$ . Setting $L_i = L_f$ : $\frac{3mvL}{4} = (\frac{1}{3}ML^2 + \frac{9}{16}mL^2)\omega$ , giving $\omega = \frac{3mvL}{4(\frac{1}{3}ML^2 + \frac{9}{16}mL^2)}$ . Choice A uses the wrong moment of inertia for the rod. Choice B incorrectly calculates the final moment of inertia. Choice C has an error in the numerator coefficient.
A student sits at rest on a frictionless rotating stool, holding a bicycle wheel that is spinning with its axis vertical, creating an upward angular momentum $\vec{L}_w$ . The student flips the wheel over so its axis is vertical but pointing downward. What is the final angular velocity of the student-stool system?
Answer: A non-zero value, causing rotation in the same direction as the wheel's initial spin. The total angular momentum of the student-stool-wheel system is conserved. Initially, $\vec{L}_{total} = \vec{L}_{student} + \vec{L}_{wheel} = 0 + \vec{L}_w = \vec{L}_w$ . After the flip, the wheel's angular momentum is $-\vec{L}_w$ . To conserve total angular momentum, the student and stool must acquire an angular momentum $\vec{L}_{student}'$ such that $\vec{L}_{total} = \vec{L}_{student}' - \vec{L}_w$ . For this to equal the initial momentum $\vec{L}_w$ , we must have $\vec{L}_{student}' = 2\vec{L}_w$ . Thus, the student and stool rotate in the same direction as the wheel's original spin.
A satellite in space has initial rotational inertia $I_i=120\,\text{kg}\cdot\text{m}^2$ and spins at $\omega_i=1.5\,\text{rad/s}$ . It deploys solar panels, increasing inertia to $I_f=300\,\text{kg}\cdot\text{m}^2$ with negligible external torque. Angular momentum is conserved during deployment. Considering the given conditions, calculate the new speed of rotation when the configuration changes.
Answer: 0.60 rad/s This question tests AP Physics C: Mechanics skills, specifically the conservation of angular momentum. Angular momentum is conserved in a closed system without external torques; it is calculated as the product of rotational inertia and angular velocity (L = Iω). In this scenario, a satellite deploys solar panels, increasing its moment of inertia from 120 to 300 kg·m², which causes the angular velocity to decrease proportionally. Choice A is correct because L_initial = $I_i$ ·ω_i = 120·1.5 = 180 kg·m²/s, and ω_final = L_initial/I_final = 180/300 = 0.60 rad/s. Choice B is incorrect as it represents the inverse calculation (300/120)·1.5, a common error when students confuse the relationship between I and ω. To help students: Use the ice skater analogy - extending arms increases I and decreases ω. Practice problems with changing configurations, emphasizing that L = Iω remains constant while I and ω change inversely.
Two uniform disks, each of mass $M$ and radius $R$ , are mounted on a common axle. Initially, the upper disk rotates at $\omega_1 = 10 \text{ rad/s}$ and the lower disk rotates at $\omega_2 = -6 \text{ rad/s}$ (opposite direction). A clutch mechanism suddenly connects the disks so they rotate together. What is their common angular velocity after engagement?
Answer: $2 \text{ rad/s}$ Apply conservation of angular momentum. Before engagement: $L_i = I_1\omega_1 + I_2\omega_2 = \frac{1}{2}MR^2(10) + \frac{1}{2}MR^2(-6) = \frac{1}{2}MR^2(10 - 6) = 2MR^2$ . After engagement: $L_f = (I_1 + I_2)\omega_f = (\frac{1}{2}MR^2 + \frac{1}{2}MR^2)\omega_f = MR^2\omega_f$ . Setting $L_i = L_f$ : $2MR^2 = MR^2\omega_f$ , so $\omega_f = 2 \text{ rad/s}$ . Choice B incorrectly adds the magnitudes instead of considering the opposite directions. Choice C forgets that angular momentum is conserved, not angular velocity. Choice D incorrectly multiplies instead of adding angular momenta.
A diver performs a somersault after jumping from a diving board. They initially leave the board with their body extended, then tuck into a compact shape, and finally extend their body again before entering the water. Neglecting air resistance, which of the following correctly describes the changes in their rotational inertia $I$ and angular momentum $L$ about their center of mass during this process?
Answer: $I$ decreases, then increases; $L$ remains approximately constant. 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 $I$ .
A uniform disk with rotational inertia $I$ is spinning freely with angular velocity $\omega$ . A uniform ring with the same mass and radius as the disk, having a rotational inertia of $2I$ , is dropped coaxially onto the spinning disk. If there are no external torques, what is the final common angular velocity of the combined system?
Answer: $\omega/3$ The total angular momentum of the system is conserved because there are no external torques. The initial angular momentum is $L_i = I_{disk}\omega = I\omega$ . The final rotational inertia of the combined system is $I_f = I_{disk} + I_{ring} = I + 2I = 3I$ . The final angular momentum is $L_f = I_f \omega_f = (3I)\omega_f$ . Setting $L_i = L_f$ gives $I\omega = (3I)\omega_f$ . Solving for the final angular velocity yields $\omega_f = \omega/3$ .
A horizontal turntable of moment of inertia $I_t = 0.5 \text{ kg⋅m}^2$ rotates freely at $\omega_0 = 2.0 \text{ rad/s}$ . A block of mass $m = 0.2 \text{ kg}$ slides radially outward on the turntable from radius $r_1 = 0.3 \text{ m}$ to radius $r_2 = 0.8 \text{ m}$ while the turntable continues to rotate. Assuming the block maintains the same angular velocity as the turntable throughout its motion, what is the turntable's angular velocity when the block reaches $r_2$ ?
Answer: $1.7 \text{ rad/s}$ Angular momentum is conserved about the turntable's axis. Initially: $L_i = I_t\omega_0 + mr_1^2\omega_0 = (0.5 + 0.2 \times 0.09) \times 2.0 = (0.5 + 0.018) \times 2.0 = 1.036 \text{ kg⋅m}^2\text{/s}$ . Finally: $L_f = (I_t + mr_2^2)\omega_f = (0.5 + 0.2 \times 0.64)\omega_f = (0.5 + 0.128)\omega_f = 0.628\omega_f$ . Therefore: $\omega_f = \frac{1.036}{0.628} = 1.65 \text{ rad/s} \approx 1.7 \text{ rad/s}$ .
The total angular momentum of a system of particles is conserved under which of the following conditions?
Answer: The net external torque on the system about a particular axis is zero. The condition for the conservation of angular momentum is that the net external torque acting on the system is zero. This is the rotational analogue of Newton's second law, $\vec{\tau}_{net, ext} = d\vec{L}/dt$ . If $\vec{\tau}_{net, ext} = 0$ , then $d\vec{L}/dt = 0$ , which means $\vec{L}$ is constant (conserved). Zero net external force conserves linear momentum, not necessarily angular momentum. Constant kinetic energy is not a prerequisite, and the moment of inertia can change while angular momentum is conserved.

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