College Physics Quiz: Angular Momentum And Angular Impulse
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Angular Momentum And Angular ImpulseQuestion 1 of 20

A horizontal rod of length 2L2L and mass MM can rotate about a vertical axis through its center. A ball of mass mm moving horizontally with speed vv 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?

mvL112ML2+mL2\frac{mvL}{\frac{1}{12}ML^2 + mL^2}
3mvLML2+3mL2\frac{3mvL}{ML^2 + 3mL^2}
mvL13ML2+mL2\frac{mvL}{\frac{1}{3}ML^2 + mL^2}
6mvLML2+12mL2\frac{6mvL}{ML^2 + 12mL^2}
12mvLML2+12mL2\frac{12mvL}{ML^2 + 12mL^2}
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College Physics Quiz

College Physics Quiz: Angular Momentum And Angular Impulse

Practice Angular Momentum And Angular Impulse in College Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Angular Momentum And Angular Impulse, giving you a quick way to practice the rules, question types, and explanations that matter most for College Physics.

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Question 1

A horizontal rod of length 2L2L and mass MM can rotate about a vertical axis through its center. A ball of mass mm moving horizontally with speed vv 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?

  1. mvL112ML2+mL2\frac{mvL}{\frac{1}{12}ML^2 + mL^2}
  2. 3mvLML2+3mL2\frac{3mvL}{ML^2 + 3mL^2}
  3. mvL13ML2+mL2\frac{mvL}{\frac{1}{3}ML^2 + mL^2} (correct answer)
  4. 6mvLML2+12mL2\frac{6mvL}{ML^2 + 12mL^2}
  5. 12mvLML2+12mL2\frac{12mvL}{ML^2 + 12mL^2}
Explanation: 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: Linitial=LfinalL_{initial} = L_{final}. Initially, only the ball has momentum, so its angular momentum about the rod's center is mvLmvL (since it's moving at distance LL from the rotation axis). After collision, both the rod and ball rotate together with angular velocity ω\omega. The final angular momentum is ItotalωI_{total}\omega, where ItotalI_{total} is the combined moment of inertia. For a rod of length 2L2L rotating about its center, Irod=13M(L)2=13ML2I_{rod} = \frac{1}{3}M(L)^2 = \frac{1}{3}ML^2 (using half-length LL). The ball, now at distance LL from the axis, contributes Iball=mL2I_{ball} = mL^2. Setting up conservation: mvL=(13ML2+mL2)ωmvL = (\frac{1}{3}ML^2 + mL^2)\omega Solving for ω\omega: ω=mvL13ML2+mL2\omega = \frac{mvL}{\frac{1}{3}ML^2 + mL^2} Answer A incorrectly uses 112ML2\frac{1}{12}ML^2, which would be the moment of inertia for a rod of length 2L2L about its center, but this problem uses the rod's half-length. Answer B has the wrong coefficient (3 instead of 13\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.

Question 2

A solid disk of mass MM and radius RR is rotating about its center with angular velocity ω\omega. A small mass mm moving with velocity vv tangentially strikes and sticks to the rim of the disk. What is the angular momentum of the system immediately after the collision?

  1. 12MR2ω+mvR+12mv2\frac{1}{2}MR^2\omega + mvR + \frac{1}{2}mv^2
  2. 12MR2ω+mvR\frac{1}{2}MR^2\omega + mvR (correct answer)
  3. (12MR2+mR2)ω(\frac{1}{2}MR^2 + mR^2)\omega
  4. 12MR2ω+mR2ω\frac{1}{2}MR^2\omega + mR^2\omega
  5. MR2ω+mvRMR^2\omega + mvR
Explanation: 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 Idiskω=12MR2ωI_{disk}\omega = \frac{1}{2}MR^2\omega, where 12MR2\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 mvRmvR. This comes from L=r×pL = r \times p, where the perpendicular distance is RR and the momentum is mvmv. Since the mass moves tangentially, the full momentum contributes to angular momentum. Therefore, the total angular momentum immediately after collision is 12MR2ω+mvR\frac{1}{2}MR^2\omega + mvR, which is answer B. Answer A incorrectly adds 12mv2\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 ω\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 mR2ωmR^2\omega term, incorrectly treating the small mass as if it were already rotating at ω\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.

Question 3

Two identical solid cylinders, each with moment of inertia II and angular velocity ω\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?

  1. ω\omega
  2. ω2\frac{\omega}{2}
  3. 00 (correct answer)
  4. −ω2-\frac{\omega}{2}
  5. 2ω2\omega
Explanation: 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 +ω+\omega and −ω-\omega. Since they're identical with moment of inertia II, their angular momenta are IωI\omega and −Iω-I\omega. The total angular momentum of the system is Iω+(−Iω)=0I\omega + (-I\omega) = 0. After friction causes them to rotate together, both cylinders have the same final angular velocity ωf\omega_f. The total moment of inertia is now 2I2I, so the final angular momentum is 2Iωf2I\omega_f. By conservation of angular momentum: 0=2Iωf0 = 2I\omega_f, which gives ωf=0\omega_f = 0. Answer A (ω\omega) incorrectly assumes one cylinder's motion dominates completely. Answer B (ω2\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 (−ω2-\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.

Question 4

A wheel with moment of inertia II is spinning with angular velocity ω0\omega_0. A constant torque τ\tau is applied in the opposite direction to the rotation. After time tt, the wheel has angular velocity ω0/3\omega_0/3 in the original direction. What was the angular impulse applied to the wheel?

  1. 2Iω03\frac{2I\omega_0}{3}
  2. −2Iω03-\frac{2I\omega_0}{3} (correct answer)
  3. Iω03\frac{I\omega_0}{3}
  4. −Iω03-\frac{I\omega_0}{3}
  5. Iω0I\omega_0
Explanation: 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 Li=Iω0L_i = I\omega_0 and ends with Lf=I(ω0/3)=Iω03L_f = I(\omega_0/3) = \frac{I\omega_0}{3}. The change in angular momentum is: ΔL=Lf−Li=Iω03−Iω0=Iω0−3Iω03=−2Iω03\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 −2Iω03-\frac{2I\omega_0}{3}. The negative sign correctly indicates that the impulse opposes the original rotation direction. Answer A gives 2Iω03\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 Iω03\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 −Iω03-\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: −2Iω03-\frac{2I\omega_0}{3}. Remember: Angular impulse always equals the change in angular momentum (ΔL\Delta L), not the initial or final values themselves. Pay careful attention to signs - they indicate direction.

Question 5

A massless rod of length LL has two point masses: mass 2m2m at one end and mass mm at the other end. The rod rotates about an axis through its center. A torque τ\tau is applied for time tt. What is the magnitude of angular momentum gained by the system?

  1. τt\tau t (correct answer)
  2. τtL2\frac{\tau t L}{2}
  3. τtL26\frac{\tau t L^2}{6}
  4. τtI\frac{\tau t}{I} where I=3mL24I = \frac{3mL^2}{4}
  5. τt3mL24\tau t \sqrt{\frac{3mL^2}{4}}
Explanation: 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: τ=dLdt\tau = \frac{dL}{dt}. When a constant torque τ\tau acts for time tt, the change in angular momentum is simply ΔL=τ⋅t\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⋅t=ΔpF \cdot t = \Delta p in linear motion regardless of the object's mass, τ⋅t=ΔL\tau \cdot t = \Delta L in rotational motion regardless of the moment of inertia. Answer A (τt\tau t) is correct because it directly applies this fundamental relationship. Answer B (τtL2\frac{\tau t L}{2}) incorrectly introduces the length dimension, suggesting confusion between torque and force. Answer C (τtL26\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 (τtI\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.

Question 6

A uniform solid sphere of mass MM and radius RR is rolling without slipping with velocity vv. What is the magnitude of its angular momentum about the contact point with the ground?

  1. 25MvR\frac{2}{5}MvR
  2. 75MvR\frac{7}{5}MvR (correct answer)
  3. MvRMvR
  4. 35MvR\frac{3}{5}MvR
  5. 710MvR\frac{7}{10}MvR
Explanation: 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=Lcm+LorbitalL = L_{cm} + L_{orbital}, where LcmL_{cm} is the angular momentum about the center of mass, and LorbitalL_{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=25MR2I = \frac{2}{5}MR^2. Since v=ωRv = \omega R for rolling without slipping, we have ω=vR\omega = \frac{v}{R}. The angular momentum about the center of mass is Lcm=Iω=25MR2⋅vR=25MvRL_{cm} = I\omega = \frac{2}{5}MR^2 \cdot \frac{v}{R} = \frac{2}{5}MvR. The orbital angular momentum is Lorbital=MvRL_{orbital} = MvR (treating the center of mass as a point mass at distance RR from the contact point). Therefore, the total angular momentum about the contact point is L=25MvR+MvR=75MvRL = \frac{2}{5}MvR + MvR = \frac{7}{5}MvR. Answer A (25MvR\frac{2}{5}MvR) only accounts for rotation about the center of mass. Answer C (MvRMvR) only considers the orbital motion. Answer D (35MvR\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.

Question 7

A uniform rod of mass MM and length LL is initially at rest on a frictionless horizontal surface. A bullet of mass mm moving horizontally with velocity vv strikes the rod at distance dd 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?

  1. mvdmvd
  2. MmvdM+m\frac{Mmvd}{M + m}
  3. mv⋅MdM+mmv \cdot \frac{Md}{M + m} (correct answer)
  4. mvdM+m\frac{mvd}{M + m}
  5. mvd⋅MM+mmvd \cdot \frac{M}{M + m}
Explanation: 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 mvmv and strikes the rod at distance dd 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 mdM+m\frac{md}{M+m} from the rod's original center. This means the collision point is now at distance d−mdM+m=MdM+md - \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⋅MdM+mL = mv \cdot \frac{Md}{M+m}. Answer A (mvdmvd) incorrectly uses the original distance dd without accounting for the center of mass shift. Answer B (MmvdM+m\frac{Mmvd}{M+m}) has the masses multiplied incorrectly and doesn't represent any meaningful physical quantity. Answer D (mvdM+m\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.

Question 8

A thin rod of mass mm and length LL 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 θ\theta from the horizontal, what is its angular momentum magnitude about the pivot?

  1. 13mL2⋅mgL(1−cos⁡θ)\sqrt{\frac{1}{3}mL^2 \cdot mgL(1 - \cos\theta)}
  2. 23mL2⋅mgL(1−cos⁡θ)\sqrt{\frac{2}{3}mL^2 \cdot mgL(1 - \cos\theta)}
  3. mL2⋅mgL(1−cos⁡θ)\sqrt{mL^2 \cdot mgL(1 - \cos\theta)}
  4. 13mL23g(1−cos⁡θ)L\frac{1}{3}mL^2\sqrt{\frac{3g(1 - \cos\theta)}{L}} (correct answer)
  5. mLgL(1−cos⁡θ)mL\sqrt{gL(1 - \cos\theta)}
Explanation: 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 L2(1−cos⁡θ)\frac{L}{2}(1 - \cos\theta). This gives a potential energy change of ΔPE=mg⋅L2(1−cos⁡θ)\Delta PE = mg \cdot \frac{L}{2}(1 - \cos\theta). By conservation of energy, this potential energy converts to rotational kinetic energy: 12Iω2=mgL(1−cos⁡θ)2\frac{1}{2}I\omega^2 = \frac{mgL(1 - \cos\theta)}{2}. For a rod rotating about one end, I=13mL2I = \frac{1}{3}mL^2, so: 12⋅13mL2⋅ω2=mgL(1−cos⁡θ)2\frac{1}{2} \cdot \frac{1}{3}mL^2 \cdot \omega^2 = \frac{mgL(1 - \cos\theta)}{2} Solving for ω: ω2=3g(1−cos⁡θ)L\omega^2 = \frac{3g(1 - \cos\theta)}{L}, so ω=3g(1−cos⁡θ)L\omega = \sqrt{\frac{3g(1 - \cos\theta)}{L}} Angular momentum is L=Iω=13mL23g(1−cos⁡θ)LL = 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 mL2mL^2 and mgLmgL under the square root, which doesn't match the physics. These forms suggest incorrectly applying L=I⋅somethingL = \sqrt{I \cdot \text{something}} rather than the proper L=IωL = I\omega relationship. Remember: angular momentum equals moment of inertia times angular velocity (L=IωL = I\omega), not the square root of their product. Always derive ω from energy conservation first, then multiply by I.

Question 9

A torque of 8.0 N⋅m8.0 \text{ N⋅m} is applied to a wheel for 3.0 s3.0 \text{ s}. If the wheel's moment of inertia is 2.0 kg⋅m22.0 \text{ kg⋅m}^2 and it starts from rest, what is its angular momentum after the torque is removed?

  1. 12.0 kg⋅m2/s12.0 \text{ kg⋅m}^2\text{/s}
  2. 24.0 kg⋅m2/s24.0 \text{ kg⋅m}^2\text{/s} (correct answer)
  3. 36.0 kg⋅m2/s36.0 \text{ kg⋅m}^2\text{/s}
  4. 48.0 kg⋅m2/s48.0 \text{ kg⋅m}^2\text{/s}
  5. 16.0 kg⋅m2/s16.0 \text{ kg⋅m}^2\text{/s}
Explanation: 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: τΔt=ΔL\tau \Delta t = \Delta L, where τ\tau is torque, Δt\Delta t is time, and ΔL\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: τΔt=8.0 N⋅m×3.0 s=24.0 kg⋅m2/s\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 II 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.

Question 10

A solid cylinder of mass MM and radius RR rolls without slipping down an inclined plane. At the bottom, it has translational kinetic energy KtK_t and rotational kinetic energy KrK_r. What is the ratio Kr/KtK_r/K_t?

  1. 13\frac{1}{3}
  2. 12\frac{1}{2} (correct answer)
  3. 23\frac{2}{3}
  4. 11
  5. 32\frac{3}{2}
Explanation: 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=ωRv = \omega R, where vv is the translational velocity and ω\omega is the angular velocity. The translational kinetic energy is Kt=12Mv2K_t = \frac{1}{2}Mv^2, and the rotational kinetic energy is Kr=12Iω2K_r = \frac{1}{2}I\omega^2. For a solid cylinder, the moment of inertia is I=12MR2I = \frac{1}{2}MR^2. Substituting the rolling constraint into the rotational energy expression: Kr=12⋅12MR2⋅(vR)2=14Mv2K_r = \frac{1}{2} \cdot \frac{1}{2}MR^2 \cdot \left(\frac{v}{R}\right)^2 = \frac{1}{4}Mv^2 Therefore: KrKt=14Mv212Mv2=12\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 (13\frac{1}{3}) would apply to a solid sphere, which has I=25MR2I = \frac{2}{5}MR^2. Answer C (23\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 (11) 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: 12MR2\frac{1}{2}MR^2, solid sphere: 25MR2\frac{2}{5}MR^2, hollow cylinder: MR2MR^2). Rolling problems always use the constraint v=ωRv = \omega R, and the energy ratio depends entirely on the object's shape through its moment of inertia.

Question 11

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 rad/s2.0 \text{ rad/s} to 6.0 rad/s6.0 \text{ rad/s}. If her moment of inertia with arms extended is 2.4 kg⋅m22.4 \text{ kg⋅m}^2, what is her moment of inertia with arms pulled in?

  1. 0.8 kg⋅m20.8 \text{ kg⋅m}^2 (correct answer)
  2. 1.2 kg⋅m21.2 \text{ kg⋅m}^2
  3. 7.2 kg⋅m27.2 \text{ kg⋅m}^2
  4. 14.4 kg⋅m214.4 \text{ kg⋅m}^2
  5. 3.6 kg⋅m23.6 \text{ kg⋅m}^2
Explanation: 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=IωL = I\omega remains constant throughout the motion. Setting up the conservation equation: I1ω1=I2ω2I_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 kg⋅m2×2.0 rad/s=I2×6.0 rad/s2.4 \text{ kg⋅m}^2 \times 2.0 \text{ rad/s} = I_2 \times 6.0 \text{ rad/s}. This gives us 4.8=I2×6.04.8 = I_2 \times 6.0, so I2=0.8 kg⋅m2I_2 = 0.8 \text{ kg⋅m}^2. Looking at the wrong answers: Choice B (1.2 kg⋅m21.2 \text{ kg⋅m}^2) might come from incorrectly using the ratio of angular velocities as ω2/ω1=3\omega_2/\omega_1 = 3, then dividing the initial moment of inertia by 2 instead of 3. Choice C (7.2 kg⋅m27.2 \text{ kg⋅m}^2) results from multiplying instead of dividing, getting 2.4×3=7.22.4 \times 3 = 7.2. Choice D (14.4 kg⋅m214.4 \text{ kg⋅m}^2) comes from incorrectly multiplying the initial angular momentum by the final angular velocity: 4.8×3=14.44.8 \times 3 = 14.4. The correct answer is A (0.8 kg⋅m20.8 \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 II and ω\omega are inversely related. When one increases, the other must decrease proportionally to keep IωI\omega constant.

Question 12

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?

  1. Angular momentum increases, kinetic energy decreases
  2. Angular momentum decreases, kinetic energy increases
  3. Angular momentum is conserved, kinetic energy decreases (correct answer)
  4. Angular momentum is conserved, kinetic energy is conserved
  5. Angular momentum decreases, kinetic energy decreases
Explanation: 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 L0=Iω0L_0 = I\omega_0. After they stick together, both disks rotate at the same final angular velocity ωf\omega_f. Since total angular momentum is conserved: Iω0=2IωfI\omega_0 = 2I\omega_f, which gives ωf=ω0/2\omega_f = \omega_0/2. However, kinetic energy is not conserved during this inelastic collision. Initially: KEi=12Iω02KE_i = \frac{1}{2}I\omega_0^2. Finally: KEf=12(2I)ωf2=12(2I)(ω0/2)2=14Iω02KE_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.

Question 13

A spinning ice skater has angular momentum L0L_0 and rotational kinetic energy K0K_0. She then pulls in her arms, reducing her moment of inertia by a factor of 4. What is her new rotational kinetic energy?

  1. K0K_0
  2. 2K02K_0
  3. 4K04K_0 (correct answer)
  4. K02\frac{K_0}{2}
  5. K04\frac{K_0}{4}
Explanation: 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=IωL = I\omega and K=12Iω2K = \frac{1}{2}I\omega^2. Since angular momentum is conserved, L0=LfinalL_0 = L_{final}, which means I0ω0=IfinalωfinalI_0\omega_0 = I_{final}\omega_{final}. Given that the moment of inertia reduces by a factor of 4, we have Ifinal=I04I_{final} = \frac{I_0}{4}. From conservation: I0ω0=I04ωfinalI_0\omega_0 = \frac{I_0}{4}\omega_{final}, so ωfinal=4ω0\omega_{final} = 4\omega_0. The skater spins four times faster. Now for the kinetic energy: Kfinal=12Ifinalωfinal2=12⋅I04⋅(4ω0)2=12⋅I04⋅16ω02=4⋅12I0ω02=4K0K_{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) K0K_0 incorrectly assumes kinetic energy is conserved like angular momentum. B) 2K02K_0 might come from incorrectly thinking energy doubles when angular velocity doubles. D) K02\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.

Question 14

Two identical uniform rods, each of mass MM and length LL, 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 τ\tau is applied for 3.0 s3.0 \text{ s}, starting from rest, what is the final angular momentum?

  1. 3.0τ3.0\tau (correct answer)
  2. 3.0τ16ML2\frac{3.0\tau}{\frac{1}{6}ML^2}
  3. 3.0τ13ML2\frac{3.0\tau}{\frac{1}{3}ML^2}
  4. 3.0τ⋅16ML23.0\tau \cdot \frac{1}{6}ML^2
  5. 3.0τ⋅ML26\frac{3.0\tau \cdot ML^2}{6}
Explanation: 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: ΔL=τ⋅Δt\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 τ\tau applied for 3.0 s3.0 \text{ s}, the final angular momentum is: Lf=τ⋅3.0=3.0τL_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, 3.0τ16ML2\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=IωL = I\omega) and the direct impulse-momentum relationship. Choice C, 3.0τ13ML2\frac{3.0\tau}{\frac{1}{3}ML^2}, makes the same conceptual error but with an incorrect moment of inertia value. Choice D, 3.0τ⋅16ML23.0\tau \cdot \frac{1}{6}ML^2, multiplies angular impulse by moment of inertia, which would give units of kg⋅m4⋅s−1\text{kg⋅m}^4\text{⋅s}^{-1} rather than the correct units for angular momentum. Remember: when given torque and time directly, use ΔL=τΔt\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.

Question 15

A thin rod of length LL and mass MM is free to rotate about one end. It is initially at rest when struck by a ball of mass mm moving horizontally with speed vv at the other end of the rod. The collision is perfectly inelastic. What is the angular velocity of the rod immediately after the collision?

  1. ω=mvML\omega = \frac{mv}{ML}, applying conservation of linear momentum at the collision point
  2. ω=mvL13ML2+mL2\omega = \frac{mvL}{\frac{1}{3}ML^2 + mL^2}, applying conservation of angular momentum about the pivot (correct answer)
  3. ω=3mv(M+m)L\omega = \frac{3mv}{(M + m)L}, using the center of mass approach for the collision
  4. ω=mvL12ML2+mL2\omega = \frac{mvL}{\frac{1}{2}ML^2 + mL^2}, applying conservation of angular momentum with incorrect rod inertia
Explanation: This is an angular impulse-momentum problem best solved using conservation of angular momentum about the pivot point. Initially, the ball has angular momentum Li=mvLL_i = mvL about the pivot (treating the ball as a point mass at distance LL). 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=Irod+Iball=13ML2+mL2I = I_{rod} + I_{ball} = \frac{1}{3}ML^2 + mL^2. By conservation of angular momentum: mvL=Iω=(13ML2+mL2)ωmvL = I\omega = (\frac{1}{3}ML^2 + mL^2)\omega, giving ω=mvL13ML2+mL2\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 13ML2\frac{1}{3}ML^2, not 12ML2\frac{1}{2}ML^2).

Question 16

A horizontal platform rotates about a vertical axis with angular velocity ω0\omega_0. A person of mass mm walks radially outward from the center to the edge at radius RR with constant speed vrelv_{rel} relative to the platform. If the platform's moment of inertia (without the person) is I0I_0, what is the angular velocity when the person reaches the edge?

  1. ω=ω0(1−mR2I0)\omega = \omega_0\left(1 - \frac{mR^2}{I_0}\right), using the change in moment of inertia
  2. ω=ω0−mvrelRI0+mR2\omega = \omega_0 - \frac{mv_{rel}R}{I_0 + mR^2}, accounting for the person's radial motion
  3. ω=I0ω0I0+mR2\omega = \frac{I_0\omega_0}{I_0 + mR^2}, applying conservation of angular momentum (correct answer)
  4. ω=I0ω0−mvrelRI0+mR2\omega = \frac{I_0\omega_0 - mv_{rel}R}{I_0 + mR^2}, including both angular momentum conservation and radial motion effects
Explanation: 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 Li=I0ω0L_i = I_0\omega_0 (just the platform rotating). When the person reaches the edge, the total moment of inertia becomes If=I0+mR2I_f = I_0 + mR^2 (platform plus person at radius RR). Since angular momentum is conserved: I0ω0=(I0+mR2)ωI_0\omega_0 = (I_0 + mR^2)\omega. Solving for the final angular velocity gives ω=I0ω0I0+mR2\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 mvrelRmv_{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 Li=LfL_i = L_f. Don't overthink by adding velocity terms that are already accounted for in the conservation law.

Question 17

A uniform rod of mass MM and length LL is pivoted at its center and initially at rest. Two equal forces FF are applied simultaneously: one upward at the left end and one downward at the right end, both perpendicular to the rod. After time tt, what is the rod's angular momentum about the pivot?

  1. L=Ft⋅L2L = Ft \cdot \frac{L}{2}, using impulse from one force only
  2. L=0L = 0, because the forces are equal and opposite, canceling each other
  3. L=FtLL = FtL, but the angular momentum direction depends on which force is stronger
  4. L=2Ft⋅L2=FtLL = 2Ft \cdot \frac{L}{2} = FtL, summing impulses from both forces (correct answer)
Explanation: 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 FF creates a torque about the pivot. The upward force at the left end creates torque τ1=F⋅L2\tau_1 = F \cdot \frac{L}{2} (clockwise), while the downward force at the right end creates torque τ2=F⋅L2\tau_2 = F \cdot \frac{L}{2} (also clockwise). Both torques work in the same direction, so the total torque is τtotal=F⋅L2+F⋅L2=FL\tau_{total} = F \cdot \frac{L}{2} + F \cdot \frac{L}{2} = FL. The angular momentum after time tt equals the angular impulse: L=τtotal⋅t=FLtL = \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 FtLFtL 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⋅L2=FtLL = 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.

Question 18

A merry-go-round (modeled as a uniform disk) has mass MM and radius RR. It rotates freely about its center with angular velocity ω0\omega_0. A child of mass mm runs tangentially and jumps onto the edge of the merry-go-round with speed vv 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?

  1. ωf=12MRω0+mv12MR+mR\omega_f = \frac{\frac{1}{2}MR\omega_0 + mv}{\frac{1}{2}MR + mR}, using incorrect units for moment of inertia
  2. ωf=Mω0+mv/RM+m\omega_f = \frac{M\omega_0 + mv/R}{M + m}, treating this as a conservation of angular velocity problem
  3. ωf=12MR2ω0+mvR12MR2+mR2\omega_f = \frac{\frac{1}{2}MR^2\omega_0 + mvR}{\frac{1}{2}MR^2 + mR^2}, conserving angular momentum about the center (correct answer)
  4. ωf=ω0+mvMR\omega_f = \omega_0 + \frac{mv}{MR}, adding the child's contribution to the original rotation
Explanation: 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: Linitial=LfinalL_{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): Ldisk=Idiskω0=12MR2ω0L_{disk} = I_{disk}\omega_0 = \frac{1}{2}MR^2\omega_0 For the child running tangentially: Lchild=mvRL_{child} = mvR (since angular momentum equals mvrmvr for linear motion at distance rr from the axis) After the collision, the total moment of inertia becomes Itotal=12MR2+mR2I_{total} = \frac{1}{2}MR^2 + mR^2 (disk plus point mass at radius RR). Setting up conservation: 12MR2ω0+mvR=(12MR2+mR2)ωf\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 MRMR instead of MR2MR^2 for moment of inertia—a units error. Choice B incorrectly treats this as conservation of angular velocity rather than angular momentum, missing the RR 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.

Question 19

A wheel of radius RR and moment of inertia II is initially spinning with angular velocity ω0\omega_0. A constant friction torque τ\tau acts on the wheel for time tt, during which the wheel's angular velocity decreases to ωf\omega_f. Which expression correctly relates the angular impulse to the change in angular momentum?

  1. τ⋅t=I(ωf−ω0)\tau \cdot t = I(\omega_f - \omega_0), where τ\tau is the magnitude of the friction torque
  2. τ⋅t=I(ωf2−ω02)2\tau \cdot t = \frac{I(\omega_f^2 - \omega_0^2)}{2}, using the work-energy theorem instead of impulse-momentum
  3. τ⋅t=Iωf−Iω0+12Iω02\tau \cdot t = I\omega_f - I\omega_0 + \frac{1}{2}I\omega_0^2, including rotational kinetic energy effects
  4. −τ⋅t=I(ωf−ω0)-\tau \cdot t = I(\omega_f - \omega_0), where τ\tau is the magnitude of the friction torque (correct answer)
Explanation: 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: ∫τ dt=ΔL=Iωf−Iω0\int \tau \, dt = \Delta L = I\omega_f - I\omega_0. For constant torque, this becomes τnet⋅t=I(ωf−ω0)\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 (ωf<ω0\omega_f < \omega_0), the friction torque acts opposite to the initial rotation direction. If we define τ\tau as the magnitude of the friction torque, then the actual torque acting on the wheel is −τ-\tau. Therefore: −τ⋅t=I(ωf−ω0)-\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 ω2\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.

Question 20

Two identical disks, each with moment of inertia II about their centers, are initially rotating in opposite directions with angular velocities +ω+\omega and −ω-\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?

  1. ωf=0\omega_f = 0, because the initial angular momenta cancel exactly (correct answer)
  2. ωf=ω2\omega_f = \frac{\omega}{2}, because angular momentum is conserved and distributed equally
  3. ωf=ω\omega_f = \omega, because the faster disk maintains its motion
  4. ωf=−ω2\omega_f = -\frac{\omega}{2}, depending on which disk dominates the interaction
Explanation: This is a conservation of angular momentum problem. Initially, the total angular momentum is Li=I(+ω)+I(−ω)=Iω−Iω=0L_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: Lf=Iωf+Iωf=2Iωf=0L_f = I\omega_f + I\omega_f = 2I\omega_f = 0, which gives ωf=0\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.