Angular Momentum and Angular Impulse - College Physics

A uniform solid sphere of mass $M$ and radius $R$ is rolling without slipping with velocity $v$ . What is the magnitude of its angular momentum about the contact point with the ground?
Answer: $\frac{7}{5}MvR$ When analyzing rolling motion problems involving angular momentum, you need to consider that angular momentum depends on your choice of reference point. This question asks about angular momentum relative to the contact point, not the center of mass. Angular momentum about any point equals $L = L_{cm} + L_{orbital}$ , where $L_{cm}$ is the angular momentum about the center of mass, and $L_{orbital}$ is the angular momentum due to the motion of the center of mass about the reference point. For a solid sphere rolling without slipping, the moment of inertia about its center is $I = \frac{2}{5}MR^2$ . Since $v = \omega R$ for rolling without slipping, we have $\omega = \frac{v}{R}$ . The angular momentum about the center of mass is $L_{cm} = I\omega = \frac{2}{5}MR^2 \cdot \frac{v}{R} = \frac{2}{5}MvR$ . The orbital angular momentum is $L_{orbital} = MvR$ (treating the center of mass as a point mass at distance $R$ from the contact point). Therefore, the total angular momentum about the contact point is $L = \frac{2}{5}MvR + MvR = \frac{7}{5}MvR$ . Answer A ( $\frac{2}{5}MvR$ ) only accounts for rotation about the center of mass. Answer C ( $MvR$ ) only considers the orbital motion. Answer D ( $\frac{3}{5}MvR$ ) might result from incorrectly combining terms or using wrong moment of inertia values. Study tip: Always identify your reference point for angular momentum calculations. When the reference point isn't the center of mass, remember to add both rotational and orbital contributions.
A torque of $8.0 \text{ N⋅m}$ is applied to a wheel for $3.0 \text{ s}$ . If the wheel's moment of inertia is $2.0 \text{ kg⋅m}^2$ and it starts from rest, what is its angular momentum after the torque is removed?
Answer: $24.0 \text{ kg⋅m}^2\text{/s}$ When you encounter rotational motion problems involving torque and time, you're dealing with angular impulse and momentum. The key relationship here is that torque applied over time changes angular momentum, just like force applied over time changes linear momentum. The fundamental equation you need is the angular impulse-momentum theorem: $\tau \Delta t = \Delta L$ , where $\tau$ is torque, $\Delta t$ is time, and $\Delta L$ is the change in angular momentum. Since the wheel starts from rest, its initial angular momentum is zero, so the final angular momentum equals the change in angular momentum. Calculating the angular impulse: $\tau \Delta t = 8.0 \text{ N⋅m} \times 3.0 \text{ s} = 24.0 \text{ kg⋅m}^2\text{/s}$ . This equals the final angular momentum, confirming answer B. Answer A (12.0) represents a common error where students might divide instead of multiply, or use only half the time period. Answer C (36.0) could result from incorrectly including the moment of inertia in the calculation—remember, you don't need $I$ when using the impulse-momentum approach directly. Answer D (48.0) might come from somehow doubling the correct answer or confusing this with a different rotational quantity. The key insight is recognizing that you can solve this problem directly using angular impulse without needing to find the angular acceleration first. When you see torque applied over time, think "angular impulse equals change in angular momentum" as your fastest path to the solution.
A thin rod of mass $m$ and length $L$ can rotate freely about a pivot at one end. The rod is released from rest in a horizontal position. When it has rotated through an angle $\theta$ from the horizontal, what is its angular momentum magnitude about the pivot?
Answer: $\frac{1}{3}mL^2\sqrt{\frac{3g(1 - \cos\theta)}{L}}$ When tackling rotational motion problems involving energy and angular momentum, you need to connect the rod's changing gravitational potential energy to its rotational kinetic energy, then use that to find angular momentum. As the rod rotates from horizontal through angle θ, its center of mass (located at L/2 from the pivot) drops by a height of $\frac{L}{2}(1 - \cos\theta)$ . This gives a potential energy change of $\Delta PE = mg \cdot \frac{L}{2}(1 - \cos\theta)$ . By conservation of energy, this potential energy converts to rotational kinetic energy: $\frac{1}{2}I\omega^2 = \frac{mgL(1 - \cos\theta)}{2}$ . For a rod rotating about one end, $I = \frac{1}{3}mL^2$ , so: $\frac{1}{2} \cdot \frac{1}{3}mL^2 \cdot \omega^2 = \frac{mgL(1 - \cos\theta)}{2}$ Solving for ω: $\omega^2 = \frac{3g(1 - \cos\theta)}{L}$ , so $\omega = \sqrt{\frac{3g(1 - \cos\theta)}{L}}$ Angular momentum is $L = I\omega = \frac{1}{3}mL^2 \sqrt{\frac{3g(1 - \cos\theta)}{L}}$ , which matches answer D. Options A, B, and C all have the wrong structure—they factor out terms like $mL^2$ and $mgL$ under the square root, which doesn't match the physics. These forms suggest incorrectly applying $L = \sqrt{I \cdot \text{something}}$ rather than the proper $L = I\omega$ relationship. Remember: angular momentum equals moment of inertia times angular velocity ( $L = I\omega$ ), not the square root of their product. Always derive ω from energy conservation first, then multiply by I.
A solid disk of mass $M$ and radius $R$ is rotating about its center with angular velocity $\omega$ . A small mass $m$ moving with velocity $v$ tangentially strikes and sticks to the rim of the disk. What is the angular momentum of the system immediately after the collision?
Answer: $\frac{1}{2}MR^2\omega + mvR$ When you encounter collision problems involving rotating objects, you need to apply conservation of angular momentum. The key insight is that angular momentum is conserved during the collision, and you must calculate the total angular momentum of the system immediately after impact. The system after collision consists of two parts: the original rotating disk and the small mass now stuck to the rim. The disk's angular momentum is $I_{disk}\omega = \frac{1}{2}MR^2\omega$ , where $\frac{1}{2}MR^2$ is the moment of inertia for a solid disk about its center. The small mass contributes angular momentum equal to $mvR$ . This comes from $L = r \times p$ , where the perpendicular distance is $R$ and the momentum is $mv$ . Since the mass moves tangentially, the full momentum contributes to angular momentum. Therefore, the total angular momentum immediately after collision is $\frac{1}{2}MR^2\omega + mvR$ , which is answer B. Answer A incorrectly adds $\frac{1}{2}mv^2$ , which is kinetic energy, not angular momentum. This represents confusion between energy and momentum concepts. Answer C treats the entire system as rotating with the same angular velocity $\omega$ , but this assumes the final state rather than the immediate post-collision moment when the small mass still has its original tangential velocity. Answer D makes the same error as C but adds an extra $mR^2\omega$ term, incorrectly treating the small mass as if it were already rotating at $\omega$ . Remember: in collision problems, distinguish between the immediate moment after impact (when conservation laws apply) and the final equilibrium state after internal forces act.
A merry-go-round (modeled as a uniform disk) has mass $M$ and radius $R$ . It rotates freely about its center with angular velocity $\omega_0$ . A child of mass $m$ runs tangentially and jumps onto the edge of the merry-go-round with speed $v$ relative to the ground. Immediately after landing, the child moves with the merry-go-round (no slipping). What is the final angular velocity of the system?
Answer: $\omega_f = \frac{\frac{1}{2}MR^2\omega_0 + mvR}{\frac{1}{2}MR^2 + mR^2}$ , conserving angular momentum about the center When you encounter problems involving rotating objects and collisions, immediately think about conservation of angular momentum. This principle applies when no external torques act on the system, which is the case here since the merry-go-round rotates freely. To solve this correctly, you need to apply conservation of angular momentum: $L_{initial} = L_{final}$ . The initial angular momentum has two components: the merry-go-round's rotational momentum and the child's linear momentum converted to angular momentum about the center. For the merry-go-round (uniform disk): $L_{disk} = I_{disk}\omega_0 = \frac{1}{2}MR^2\omega_0$ For the child running tangentially: $L_{child} = mvR$ (since angular momentum equals $mvr$ for linear motion at distance $r$ from the axis) After the collision, the total moment of inertia becomes $I_{total} = \frac{1}{2}MR^2 + mR^2$ (disk plus point mass at radius $R$ ). Setting up conservation: $\frac{1}{2}MR^2\omega_0 + mvR = (\frac{1}{2}MR^2 + mR^2)\omega_f$ This gives choice C as correct. Choice A uses $MR$ instead of $MR^2$ for moment of inertia—a units error. Choice B incorrectly treats this as conservation of angular velocity rather than angular momentum, missing the $R$ factors entirely. Choice D simply adds terms without proper consideration of the physics, ignoring the need to account for the changed total moment of inertia. Study tip: Always identify what's conserved (energy, momentum, angular momentum) first, then carefully write the correct expressions for each object's contribution before solving.
Two identical disks, each with moment of inertia $I$ about their centers, are initially rotating in opposite directions with angular velocities $+\omega$ and $-\omega$ respectively. They are brought into contact along their edges and allowed to slip against each other until they reach the same angular velocity. What is the final angular velocity of each disk?
Answer: $\omega_f = 0$ , because the initial angular momenta cancel exactly This is a conservation of angular momentum problem. Initially, the total angular momentum is $L_i = I(+\omega) + I(-\omega) = I\omega - I\omega = 0$ . Since no external torques act on the system (the contact forces are internal), angular momentum is conserved. Therefore, the final total angular momentum must also be zero: $L_f = I\omega_f + I\omega_f = 2I\omega_f = 0$ , which gives $\omega_f = 0$ . Both disks come to rest. Choice B incorrectly assumes the angular velocities simply average out without considering the vector nature of angular momentum. Choice C ignores conservation principles. Choice D incorrectly suggests the direction matters for the final result when the initial angular momenta are exactly equal and opposite.
A spinning ice skater has angular momentum $L_0$ and rotational kinetic energy $K_0$ . She then pulls in her arms, reducing her moment of inertia by a factor of 4. What is her new rotational kinetic energy?
Answer: $4K_0$ This problem tests conservation of angular momentum during rotational motion, a fundamental principle when no external torques act on a system. When the skater pulls in her arms, she's changing her moment of inertia but no external forces are acting, so her angular momentum must remain constant. Start with the key relationships: $L = I\omega$ and $K = \frac{1}{2}I\omega^2$ . Since angular momentum is conserved, $L_0 = L_{final}$ , which means $I_0\omega_0 = I_{final}\omega_{final}$ . Given that the moment of inertia reduces by a factor of 4, we have $I_{final} = \frac{I_0}{4}$ . From conservation: $I_0\omega_0 = \frac{I_0}{4}\omega_{final}$ , so $\omega_{final} = 4\omega_0$ . The skater spins four times faster. Now for the kinetic energy: $K_{final} = \frac{1}{2}I_{final}\omega_{final}^2 = \frac{1}{2} \cdot \frac{I_0}{4} \cdot (4\omega_0)^2 = \frac{1}{2} \cdot \frac{I_0}{4} \cdot 16\omega_0^2 = 4 \cdot \frac{1}{2}I_0\omega_0^2 = 4K_0$ . The answer is C. Looking at the wrong answers: A) $K_0$ incorrectly assumes kinetic energy is conserved like angular momentum. B) $2K_0$ might come from incorrectly thinking energy doubles when angular velocity doubles. D) $\frac{K_0}{2}$ could result from confusing the direction of the moment of inertia change. Remember: Angular momentum is always conserved in isolated rotating systems, but rotational kinetic energy can change. When moment of inertia decreases, the system spins faster and gains kinetic energy.
A horizontal rod of length $2L$ and mass $M$ can rotate about a vertical axis through its center. A ball of mass $m$ moving horizontally with speed $v$ strikes and sticks to one end of the rod. Before the collision, the rod was at rest. What is the angular velocity of the rod-ball system immediately after the collision?
Answer: $\frac{mvL}{\frac{1}{3}ML^2 + mL^2}$ This problem tests conservation of angular momentum during an inelastic collision, where rotational motion is involved. When you see a collision problem with rotation, immediately think about angular momentum being conserved while kinetic energy is not. To solve this, you need to apply conservation of angular momentum: $L_{initial} = L_{final}$ . Initially, only the ball has momentum, so its angular momentum about the rod's center is $mvL$ (since it's moving at distance $L$ from the rotation axis). After collision, both the rod and ball rotate together with angular velocity $\omega$ . The final angular momentum is $I_{total}\omega$ , where $I_{total}$ is the combined moment of inertia. For a rod of length $2L$ rotating about its center, $I_{rod} = \frac{1}{3}M(L)^2 = \frac{1}{3}ML^2$ (using half-length $L$ ). The ball, now at distance $L$ from the axis, contributes $I_{ball} = mL^2$ . Setting up conservation: $mvL = (\frac{1}{3}ML^2 + mL^2)\omega$ Solving for $\omega$ : $\omega = \frac{mvL}{\frac{1}{3}ML^2 + mL^2}$ Answer A incorrectly uses $\frac{1}{12}ML^2$ , which would be the moment of inertia for a rod of length $2L$ about its center, but this problem uses the rod's half-length. Answer B has the wrong coefficient (3 instead of $\frac{1}{3}$ ) in the numerator and denominator. Answer D uses 6 and 12 as coefficients, suggesting confusion about the rod's moment of inertia formula. Remember: for rod rotation problems, carefully identify the rotation axis and use the appropriate moment of inertia formula – it changes dramatically based on the axis location.
A figure skater is spinning with her arms extended. When she pulls her arms close to her body, her angular velocity increases from $2.0 \text{ rad/s}$ to $6.0 \text{ rad/s}$ . If her moment of inertia with arms extended is $2.4 \text{ kg⋅m}^2$ , what is her moment of inertia with arms pulled in?
Answer: $0.8 \text{ kg⋅m}^2$ When you see a spinning figure skater changing position, you're dealing with conservation of angular momentum. Since no external torques act on the skater, her angular momentum $L = I\omega$ remains constant throughout the motion. Setting up the conservation equation: $I_1\omega_1 = I_2\omega_2$ , where the subscripts 1 and 2 represent the initial (arms extended) and final (arms pulled in) states respectively. Substituting the given values: $2.4 \text{ kg⋅m}^2 \times 2.0 \text{ rad/s} = I_2 \times 6.0 \text{ rad/s}$ . This gives us $4.8 = I_2 \times 6.0$ , so $I_2 = 0.8 \text{ kg⋅m}^2$ . Looking at the wrong answers: Choice B ( $1.2 \text{ kg⋅m}^2$ ) might come from incorrectly using the ratio of angular velocities as $\omega_2/\omega_1 = 3$ , then dividing the initial moment of inertia by 2 instead of 3. Choice C ( $7.2 \text{ kg⋅m}^2$ ) results from multiplying instead of dividing, getting $2.4 \times 3 = 7.2$ . Choice D ( $14.4 \text{ kg⋅m}^2$ ) comes from incorrectly multiplying the initial angular momentum by the final angular velocity: $4.8 \times 3 = 14.4$ . The correct answer is A ( $0.8 \text{ kg⋅m}^2$ ), showing that the moment of inertia decreases as mass moves closer to the rotation axis. Study tip: In angular momentum conservation problems, always remember that $I$ and $\omega$ are inversely related. When one increases, the other must decrease proportionally to keep $I\omega$ constant.
A uniform rod of mass $M$ and length $L$ is initially at rest on a frictionless horizontal surface. A bullet of mass $m$ moving horizontally with velocity $v$ strikes the rod at distance $d$ from the center and embeds in it. What is the angular momentum of the rod-bullet system about the center of mass of the system immediately after the collision?
Answer: $mv \cdot \frac{Md}{M + m}$ When you encounter collision problems involving rotation, the key is recognizing that angular momentum is conserved about any point, but choosing the right reference point simplifies the calculation significantly. Initially, the bullet has linear momentum $mv$ and strikes the rod at distance $d$ from the rod's center. After collision, both objects move together, but you need to find the angular momentum about the system's center of mass, not the rod's center. First, find where the center of mass is located. Using the rod's center as reference, the center of mass shifts toward the bullet by $\frac{md}{M+m}$ from the rod's original center. This means the collision point is now at distance $d - \frac{md}{M+m} = \frac{Md}{M+m}$ from the system's center of mass. The initial angular momentum about the system's center of mass equals the bullet's linear momentum times this perpendicular distance: $L = mv \cdot \frac{Md}{M+m}$ . Answer A ( $mvd$ ) incorrectly uses the original distance $d$ without accounting for the center of mass shift. Answer B ( $\frac{Mmvd}{M+m}$ ) has the masses multiplied incorrectly and doesn't represent any meaningful physical quantity. Answer D ( $\frac{mvd}{M+m}$ ) incorrectly divides the entire expression by the total mass, which has no physical justification. Study tip: In collision problems involving rotation, always identify the center of mass first. Angular momentum conservation is much cleaner when calculated about the center of mass, and remember that the collision point's effective distance changes when you shift reference points.
A wheel with moment of inertia $I$ is spinning with angular velocity $\omega_0$ . A constant torque $\tau$ is applied in the opposite direction to the rotation. After time $t$ , the wheel has angular velocity $\omega_0/3$ in the original direction. What was the angular impulse applied to the wheel?
Answer: $-\frac{2I\omega_0}{3}$ When you encounter rotational motion problems involving changing angular velocity, think about angular impulse and momentum. Angular impulse equals the change in angular momentum, just like linear impulse equals change in linear momentum. The wheel starts with angular momentum $L_i = I\omega_0$ and ends with $L_f = I(\omega_0/3) = \frac{I\omega_0}{3}$ . The change in angular momentum is: $\Delta L = L_f - L_i = \frac{I\omega_0}{3} - I\omega_0 = \frac{I\omega_0 - 3I\omega_0}{3} = -\frac{2I\omega_0}{3}$ Since angular impulse equals change in angular momentum, the angular impulse is $-\frac{2I\omega_0}{3}$ . The negative sign correctly indicates that the impulse opposes the original rotation direction. Answer A gives $\frac{2I\omega_0}{3}$ - this has the correct magnitude but wrong sign. This would represent an impulse in the same direction as the original rotation, which contradicts the problem statement that torque is applied opposite to rotation. Answer C gives $\frac{I\omega_0}{3}$ - this is actually the final angular momentum, not the change in angular momentum. This confuses the final state with the impulse applied. Answer D gives $-\frac{I\omega_0}{3}$ - this also represents the final angular momentum but with an incorrect negative sign. The final angular velocity is still in the original direction, so final angular momentum should be positive. Answer B is correct: $-\frac{2I\omega_0}{3}$ . Remember: Angular impulse always equals the change in angular momentum ( $\Delta L$ ), not the initial or final values themselves. Pay careful attention to signs - they indicate direction.
Two identical uniform rods, each of mass $M$ and length $L$ , are joined at their centers to form a cross. The cross rotates about an axis through the center, perpendicular to both rods. If a constant torque $\tau$ is applied for $3.0 \text{ s}$ , starting from rest, what is the final angular momentum?
Answer: $3.0\tau$ When you encounter rotational motion problems involving torque and time, think about the relationship between torque, angular impulse, and angular momentum. This is analogous to how force and time create linear momentum changes. The key insight here is recognizing that angular impulse equals the change in angular momentum: $\Delta L = \tau \cdot \Delta t$ . Since the cross starts from rest, the initial angular momentum is zero, so the final angular momentum simply equals the angular impulse applied. Given a constant torque $\tau$ applied for $3.0 \text{ s}$ , the final angular momentum is: $L_f = \tau \cdot 3.0 = 3.0\tau$ This makes choice A correct. Now let's see where the other options go wrong. Choice B, $\frac{3.0\tau}{\frac{1}{6}ML^2}$ , incorrectly divides the angular impulse by a moment of inertia. This suggests confusion between angular momentum ( $L = I\omega$ ) and the direct impulse-momentum relationship. Choice C, $\frac{3.0\tau}{\frac{1}{3}ML^2}$ , makes the same conceptual error but with an incorrect moment of inertia value. Choice D, $3.0\tau \cdot \frac{1}{6}ML^2$ , multiplies angular impulse by moment of inertia, which would give units of $\text{kg⋅m}^4\text{⋅s}^{-1}$ rather than the correct units for angular momentum. Remember: when given torque and time directly, use $\Delta L = \tau \Delta t$ . You only need to calculate moment of inertia when you're asked to find angular velocity or when torque isn't given directly.
A thin rod of length $L$ and mass $M$ is free to rotate about one end. It is initially at rest when struck by a ball of mass $m$ moving horizontally with speed $v$ at the other end of the rod. The collision is perfectly inelastic. What is the angular velocity of the rod immediately after the collision?
Answer: $\omega = \frac{mvL}{\frac{1}{3}ML^2 + mL^2}$ , applying conservation of angular momentum about the pivot This is an angular impulse-momentum problem best solved using conservation of angular momentum about the pivot point. Initially, the ball has angular momentum $L_i = mvL$ about the pivot (treating the ball as a point mass at distance $L$ ). The rod is initially at rest. After the inelastic collision, both the rod and ball move together. The moment of inertia of the system is $I = I_{rod} + I_{ball} = \frac{1}{3}ML^2 + mL^2$ . By conservation of angular momentum: $mvL = I\omega = (\frac{1}{3}ML^2 + mL^2)\omega$ , giving $\omega = \frac{mvL}{\frac{1}{3}ML^2 + mL^2}$ . Choice A incorrectly uses linear momentum concepts. Choice C uses an incorrect mass combination. Choice D uses the wrong moment of inertia for a rod about its end (should be $\frac{1}{3}ML^2$ , not $\frac{1}{2}ML^2$ ).
A solid cylinder of mass $M$ and radius $R$ rolls without slipping down an inclined plane. At the bottom, it has translational kinetic energy $K_t$ and rotational kinetic energy $K_r$ . What is the ratio $K_r/K_t$ ?
Answer: $\frac{1}{2}$ When you encounter rolling motion problems, remember that objects rolling without slipping have both translational and rotational kinetic energy, and these are related through the rolling constraint. For a rolling cylinder, the key relationship is $v = \omega R$ , where $v$ is the translational velocity and $\omega$ is the angular velocity. The translational kinetic energy is $K_t = \frac{1}{2}Mv^2$ , and the rotational kinetic energy is $K_r = \frac{1}{2}I\omega^2$ . For a solid cylinder, the moment of inertia is $I = \frac{1}{2}MR^2$ . Substituting the rolling constraint into the rotational energy expression: $K_r = \frac{1}{2} \cdot \frac{1}{2}MR^2 \cdot \left(\frac{v}{R}\right)^2 = \frac{1}{4}Mv^2$ Therefore: $\frac{K_r}{K_t} = \frac{\frac{1}{4}Mv^2}{\frac{1}{2}Mv^2} = \frac{1}{2}$ This confirms answer B is correct. Answer A ( $\frac{1}{3}$ ) would apply to a solid sphere, which has $I = \frac{2}{5}MR^2$ . Answer C ( $\frac{2}{3}$ ) might result from incorrectly using the sphere's moment of inertia in the numerator while keeping the cylinder's in the denominator. Answer D ( $1$ ) would incorrectly assume equal translational and rotational energies, ignoring the specific moment of inertia. Study tip: Memorize the moments of inertia for common shapes (solid cylinder: $\frac{1}{2}MR^2$ , solid sphere: $\frac{2}{5}MR^2$ , hollow cylinder: $MR^2$ ). Rolling problems always use the constraint $v = \omega R$ , and the energy ratio depends entirely on the object's shape through its moment of inertia.
A wheel of radius $R$ and moment of inertia $I$ is initially spinning with angular velocity $\omega_0$ . A constant friction torque $\tau$ acts on the wheel for time $t$ , during which the wheel's angular velocity decreases to $\omega_f$ . Which expression correctly relates the angular impulse to the change in angular momentum?
Answer: $-\tau \cdot t = I(\omega_f - \omega_0)$ , where $\tau$ is the magnitude of the friction torque When you encounter rotational dynamics problems involving torque and time, think about the rotational analog of Newton's second law and the impulse-momentum theorem. Just as linear impulse equals change in linear momentum, angular impulse equals change in angular momentum. The angular impulse-momentum theorem states that the net angular impulse equals the change in angular momentum: $\int \tau \, dt = \Delta L = I\omega_f - I\omega_0$ . For constant torque, this becomes $\tau_{net} \cdot t = I(\omega_f - \omega_0)$ . The key insight is recognizing the sign convention. Since friction opposes motion and the wheel is slowing down ( $\omega_f < \omega_0$ ), the friction torque acts opposite to the initial rotation direction. If we define $\tau$ as the magnitude of the friction torque, then the actual torque acting on the wheel is $-\tau$ . Therefore: $-\tau \cdot t = I(\omega_f - \omega_0)$ , which matches answer D. Answer A incorrectly treats the friction torque as positive, ignoring that it opposes rotation. Answer B confuses the impulse-momentum theorem with the work-energy theorem—angular impulse relates to momentum change, not energy change (which would involve $\omega^2$ terms). Answer C incorrectly mixes momentum and energy concepts, adding a kinetic energy term that doesn't belong in the impulse-momentum relationship. Remember: always pay careful attention to signs in rotational problems. Friction torques oppose motion, so if you're given the magnitude of a friction torque, you must include the negative sign to represent its opposing direction.
Two identical solid cylinders, each with moment of inertia $I$ and angular velocity $\omega$ , are rotating in opposite directions about parallel axes. They are brought into contact so their surfaces touch and friction causes them to eventually rotate together. What is the final angular velocity of each cylinder?
Answer: $0$ This problem tests conservation of angular momentum in rotational collisions. When two rotating objects interact through friction until they move together, you need to consider the system's total angular momentum before and after contact. Initially, the cylinders rotate in opposite directions with angular velocities $+\omega$ and $-\omega$ . Since they're identical with moment of inertia $I$ , their angular momenta are $I\omega$ and $-I\omega$ . The total angular momentum of the system is $I\omega + (-I\omega) = 0$ . After friction causes them to rotate together, both cylinders have the same final angular velocity $\omega_f$ . The total moment of inertia is now $2I$ , so the final angular momentum is $2I\omega_f$ . By conservation of angular momentum: $0 = 2I\omega_f$ , which gives $\omega_f = 0$ . Answer A ( $\omega$ ) incorrectly assumes one cylinder's motion dominates completely. Answer B ( $\frac{\omega}{2}$ ) might tempt you if you mistakenly averaged the speeds rather than properly applying momentum conservation, or if you forgot they rotate in opposite directions. Answer D ( $-\frac{\omega}{2}$ ) makes a similar error but assumes the negative direction dominates. The key insight is that equal and opposite angular momenta cancel out completely, just like equal masses moving in opposite directions with equal speeds would come to rest after a perfectly inelastic collision. Remember: when applying conservation laws, always account for the vector nature of the quantities involved—direction matters as much as magnitude.
A uniform rod of mass $M$ and length $L$ is pivoted at its center and initially at rest. Two equal forces $F$ are applied simultaneously: one upward at the left end and one downward at the right end, both perpendicular to the rod. After time $t$ , what is the rod's angular momentum about the pivot?
Answer: $L = 2Ft \cdot \frac{L}{2} = FtL$ , summing impulses from both forces When you encounter rotational motion problems involving multiple forces, think about how each force contributes to the total angular momentum through the concept of angular impulse. Angular impulse equals the torque multiplied by time, and it changes the angular momentum of an object. Here, each force $F$ creates a torque about the pivot. The upward force at the left end creates torque $\tau_1 = F \cdot \frac{L}{2}$ (clockwise), while the downward force at the right end creates torque $\tau_2 = F \cdot \frac{L}{2}$ (also clockwise). Both torques work in the same direction, so the total torque is $\tau_{total} = F \cdot \frac{L}{2} + F \cdot \frac{L}{2} = FL$ . The angular momentum after time $t$ equals the angular impulse: $L = \tau_{total} \cdot t = FLt$ . Option A incorrectly considers only one force, missing that both forces create torque in the same rotational direction. Option B falls into the trap of thinking equal and opposite forces always cancel—while the net translational force is zero, the torques add because they act at different distances from the pivot. Option C correctly calculates $FtL$ but incorrectly suggests the direction depends on which force is stronger, when both forces are equal and both contribute to the same rotational direction. Option D correctly recognizes that both forces create angular impulse in the same direction, giving $L = 2Ft \cdot \frac{L}{2} = FtL$ . Study tip: In rotational problems, equal and opposite forces don't necessarily cancel if they're applied at different positions—always check whether the torques add or subtract based on their rotational directions.
A horizontal platform rotates about a vertical axis with angular velocity $\omega_0$ . A person of mass $m$ walks radially outward from the center to the edge at radius $R$ with constant speed $v_{rel}$ relative to the platform. If the platform's moment of inertia (without the person) is $I_0$ , what is the angular velocity when the person reaches the edge?
Answer: $\omega = \frac{I_0\omega_0}{I_0 + mR^2}$ , applying conservation of angular momentum When you encounter rotating systems with changing mass distributions, immediately think about conservation of angular momentum. This fundamental principle states that angular momentum remains constant when no external torques act on the system. Initially, the system's angular momentum is $L_i = I_0\omega_0$ (just the platform rotating). When the person reaches the edge, the total moment of inertia becomes $I_f = I_0 + mR^2$ (platform plus person at radius $R$ ). Since angular momentum is conserved: $I_0\omega_0 = (I_0 + mR^2)\omega$ . Solving for the final angular velocity gives $\omega = \frac{I_0\omega_0}{I_0 + mR^2}$ , which is answer C. Answer A incorrectly subtracts the person's contribution to moment of inertia, when it should be added. The person doesn't reduce the system's rotational inertia—they increase it by moving farther from the axis. Answer B attempts to account for radial motion but misapplies it. The $mv_{rel}R$ term represents angular momentum in the lab frame, but conservation of angular momentum already accounts for all motion effects. Adding this term double-counts the physics. Answer D makes the same error as B, incorrectly including the radial velocity term. While the person does have radial motion, this doesn't create additional angular momentum changes beyond what's already captured by the changing moment of inertia. Study tip: For rotating systems with changing mass distributions, stick to the basics—identify initial and final moments of inertia, then apply $L_i = L_f$ . Don't overthink by adding velocity terms that are already accounted for in the conservation law.
A uniform disk is rotating about its center when a second identical disk (initially not rotating) is gently placed on top of it. Friction between the disks causes them to eventually rotate together. During this process, what happens to the total angular momentum and total rotational kinetic energy of the two-disk system?
Answer: Angular momentum is conserved, kinetic energy decreases When you encounter problems involving collisions or objects coming together, you need to identify which quantities are conserved and which are not. The key is recognizing whether external torques act on the system. In this problem, the two-disk system experiences no external torques about the rotation axis. Friction acts only between the disks (internal forces), so angular momentum must be conserved. Initially, only the bottom disk rotates with angular momentum $L_0 = I\omega_0$ . After they stick together, both disks rotate at the same final angular velocity $\omega_f$ . Since total angular momentum is conserved: $I\omega_0 = 2I\omega_f$ , which gives $\omega_f = \omega_0/2$ . However, kinetic energy is not conserved during this inelastic collision. Initially: $KE_i = \frac{1}{2}I\omega_0^2$ . Finally: $KE_f = \frac{1}{2}(2I)\omega_f^2 = \frac{1}{2}(2I)(\omega_0/2)^2 = \frac{1}{4}I\omega_0^2$ . The final kinetic energy is half the initial value, so energy decreases due to friction converting mechanical energy to heat. Choice A incorrectly suggests angular momentum increases, but conservation laws prevent this without external torques. Choice B wrongly claims angular momentum decreases while kinetic energy increases, violating both conservation principles and energy considerations. Choice D incorrectly assumes this is an elastic collision where kinetic energy is conserved, but friction always dissipates energy. Remember: Angular momentum is conserved when no external torques act, but kinetic energy is only conserved in perfectly elastic collisions—friction always indicates energy loss.
A massless rod of length $L$ has two point masses: mass $2m$ at one end and mass $m$ at the other end. The rod rotates about an axis through its center. A torque $\tau$ is applied for time $t$ . What is the magnitude of angular momentum gained by the system?
Answer: $\tau t$ When you encounter rotational dynamics problems involving torque and angular momentum, the fundamental relationship to remember is the rotational analog of Newton's second law: the impulse-momentum theorem for rotation. The key insight here is that torque is the rate of change of angular momentum: $\tau = \frac{dL}{dt}$ . When a constant torque $\tau$ acts for time $t$ , the change in angular momentum is simply $\Delta L = \tau \cdot t$ . This is the rotational version of the impulse-momentum theorem, where torque impulse equals the change in angular momentum. Notice that this relationship is completely independent of the moment of inertia or the specific mass distribution. Just as $F \cdot t = \Delta p$ in linear motion regardless of the object's mass, $\tau \cdot t = \Delta L$ in rotational motion regardless of the moment of inertia. Answer A ( $\tau t$ ) is correct because it directly applies this fundamental relationship. Answer B ( $\frac{\tau t L}{2}$ ) incorrectly introduces the length dimension, suggesting confusion between torque and force. Answer C ( $\frac{\tau t L^2}{6}$ ) appears to mix the torque-impulse relationship with moment of inertia calculations, which aren't needed here. Answer D ( $\frac{\tau t}{I}$ ) represents a common misconception—dividing by moment of inertia as if finding angular acceleration, then forgetting to multiply by time again. Remember: torque impulse directly equals change in angular momentum, just like force impulse equals change in linear momentum. The moment of inertia only matters when you need to convert between angular momentum and angular velocity.

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