Statics and Dynamics Quiz: Angular Impulse Momentum
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Angular Impulse MomentumQuestion 1 of 3

A rigid body rotates about a fixed axis. A time-varying torque τ(t)=(6t2−4t) N⋅m\tau(t) = (6t^2 - 4t) \text{ N}\cdot\text{m} is applied from t=0t = 0 to t=3 st = 3 \text{ s}. The body's moment of inertia about the rotation axis is I=2 kg⋅m2I = 2 \text{ kg}\cdot\text{m}^2. At t=0t = 0, the angular velocity is ω0=−1 rad/s\omega_0 = -1 \text{ rad/s} (clockwise, taken as negative).

What is the angular velocity at t=3 st = 3 \text{ s}?

ω(3)=+11.0 rad/s\omega(3) = +11.0 \text{ rad/s}, found by computing the angular impulse ∫03τ dt=∫03(6t2−4t) dt=[2t3−2t2]03=54−18=36 N\cdotpm\cdotps\int_0^3 \tau\,dt = \int_0^3(6t^2 - 4t)\,dt = [2t^3 - 2t^2]_0^3 = 54 - 18 = 36 \text{ N·m·s}, then applying Δω=36/2=18 rad/s\Delta\omega = 36/2 = 18 \text{ rad/s}, and adding to ω0=−1\omega_0 = -1: ω=17 rad/s\omega = 17 \text{ rad/s}. Error: arithmetic mistake in the integral.
ω(3)=+17.0 rad/s\omega(3) = +17.0 \text{ rad/s}, found by correctly computing the angular impulse ∫03(6t2−4t) dt=[2t3−2t2]03=54−18=36 N\cdotpm\cdotps\int_0^3(6t^2-4t)\,dt = [2t^3-2t^2]_0^3 = 54-18 = 36 \text{ N·m·s}, applying Δω=36/I=36/2=18 rad/s\Delta\omega = 36/I = 36/2 = 18 \text{ rad/s}, and adding to ω0=−1 rad/s\omega_0 = -1 \text{ rad/s}, giving ω(3)=−1+18=+17 rad/s\omega(3) = -1 + 18 = +17 \text{ rad/s}.
ω(3)=+18.0 rad/s\omega(3) = +18.0 \text{ rad/s}, found by computing the angular impulse as 36 N·m·s, dividing by I=2I = 2 to get Δω=18 rad/s\Delta\omega = 18 \text{ rad/s}, but neglecting the initial angular velocity ω0=−1 rad/s\omega_0 = -1 \text{ rad/s} and reporting Δω\Delta\omega alone as the final angular velocity.
ω(3)=+9.0 rad/s\omega(3) = +9.0 \text{ rad/s}, found by evaluating the torque function at the midpoint t=1.5 st = 1.5 \text{ s} to get τ(1.5)=6(2.25)−4(1.5)=13.5−6=7.5 N\cdotpm\tau(1.5) = 6(2.25) - 4(1.5) = 13.5 - 6 = 7.5 \text{ N·m}, then computing the impulse as 7.5×3=22.5 N\cdotpm\cdotps7.5 \times 3 = 22.5 \text{ N·m·s}, giving Δω=11.25\Delta\omega = 11.25 rad/s and ω(3)≈10.25\omega(3) \approx 10.25 rad/s, rounded and combined with a sign error.
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Statics and Dynamics Quiz

Statics and Dynamics Quiz: Angular Impulse Momentum

Practice Angular Impulse Momentum in Statics and Dynamics 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 Impulse Momentum, giving you a quick way to practice the rules, question types, and explanations that matter most for Statics and Dynamics.

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Question 1

A rigid body rotates about a fixed axis. A time-varying torque τ(t)=(6t2−4t) N⋅m\tau(t) = (6t^2 - 4t) \text{ N}\cdot\text{m} is applied from t=0t = 0 to t=3 st = 3 \text{ s}. The body's moment of inertia about the rotation axis is I=2 kg⋅m2I = 2 \text{ kg}\cdot\text{m}^2. At t=0t = 0, the angular velocity is ω0=−1 rad/s\omega_0 = -1 \text{ rad/s} (clockwise, taken as negative).

What is the angular velocity at t=3 st = 3 \text{ s}?

  1. ω(3)=+11.0 rad/s\omega(3) = +11.0 \text{ rad/s}, found by computing the angular impulse ∫03τ dt=∫03(6t2−4t) dt=[2t3−2t2]03=54−18=36 N\cdotpm\cdotps\int_0^3 \tau\,dt = \int_0^3(6t^2 - 4t)\,dt = [2t^3 - 2t^2]_0^3 = 54 - 18 = 36 \text{ N·m·s}, then applying Δω=36/2=18 rad/s\Delta\omega = 36/2 = 18 \text{ rad/s}, and adding to ω0=−1\omega_0 = -1: ω=17 rad/s\omega = 17 \text{ rad/s}. Error: arithmetic mistake in the integral.
  2. ω(3)=+17.0 rad/s\omega(3) = +17.0 \text{ rad/s}, found by correctly computing the angular impulse ∫03(6t2−4t) dt=[2t3−2t2]03=54−18=36 N\cdotpm\cdotps\int_0^3(6t^2-4t)\,dt = [2t^3-2t^2]_0^3 = 54-18 = 36 \text{ N·m·s}, applying Δω=36/I=36/2=18 rad/s\Delta\omega = 36/I = 36/2 = 18 \text{ rad/s}, and adding to ω0=−1 rad/s\omega_0 = -1 \text{ rad/s}, giving ω(3)=−1+18=+17 rad/s\omega(3) = -1 + 18 = +17 \text{ rad/s}. (correct answer)
  3. ω(3)=+18.0 rad/s\omega(3) = +18.0 \text{ rad/s}, found by computing the angular impulse as 36 N·m·s, dividing by I=2I = 2 to get Δω=18 rad/s\Delta\omega = 18 \text{ rad/s}, but neglecting the initial angular velocity ω0=−1 rad/s\omega_0 = -1 \text{ rad/s} and reporting Δω\Delta\omega alone as the final angular velocity.
  4. ω(3)=+9.0 rad/s\omega(3) = +9.0 \text{ rad/s}, found by evaluating the torque function at the midpoint t=1.5 st = 1.5 \text{ s} to get τ(1.5)=6(2.25)−4(1.5)=13.5−6=7.5 N\cdotpm\tau(1.5) = 6(2.25) - 4(1.5) = 13.5 - 6 = 7.5 \text{ N·m}, then computing the impulse as 7.5×3=22.5 N\cdotpm\cdotps7.5 \times 3 = 22.5 \text{ N·m·s}, giving Δω=11.25\Delta\omega = 11.25 rad/s and ω(3)≈10.25\omega(3) \approx 10.25 rad/s, rounded and combined with a sign error.
Explanation: When a time-varying torque acts on a rotating body, the cleanest approach uses the angular impulse-momentum theorem: ∫t1t2τ dt=I(ωf−ω0)\int_{t_1}^{t_2} \tau\, dt = I(\omega_f - \omega_0). This tells you that the change in angular momentum equals the angular impulse — so you must integrate the torque, divide by II, and then add the result to your initial angular velocity, not replace it. Here's the correct path (answer B): integrate the torque from 0 to 3 s: ∫03(6t2−4t) dt=[2t3−2t2]03=2(27)−2(9)=54−18=36 N\cdotpm\cdotps\int_0^3 (6t^2 - 4t)\,dt = \left[2t^3 - 2t^2\right]_0^3 = 2(27) - 2(9) = 54 - 18 = 36 \text{ N·m·s} Divide by I=2 kg\cdotpm2I = 2 \text{ kg·m}^2 to get Δω=18 rad/s\Delta\omega = 18 \text{ rad/s}. Adding the initial condition gives ω(3)=−1+18=+17 rad/s\omega(3) = -1 + 18 = +17 \text{ rad/s}. A arrives at the same integral value (36 N·m·s) and the same Δω=18\Delta\omega = 18 rad/s, then somehow reports 17 rad/s as an "arithmetic mistake" — but 17 is actually the correct answer. The label in A is internally inconsistent; it calls the right number wrong. C correctly computes Δω=18 rad/s\Delta\omega = 18 \text{ rad/s} but forgets to add ω0=−1 rad/s\omega_0 = -1 \text{ rad/s}, treating the change in angular velocity as if it were the final angular velocity. This is a classic "forgot initial conditions" error. D approximates a varying torque using a single midpoint value — valid only for linear functions, not quadratics. Here it produces a meaningless estimate. Study tip: Whenever you see a time-varying torque, integrate — never sample a single point. And always track initial conditions separately; Δω\Delta\omega and ωf\omega_f are not the same thing.

Question 2

Two gears, A and B, mesh without slipping. Gear A has moment of inertia IA=0.5 kg⋅m2I_A = 0.5 \text{ kg}\cdot\text{m}^2 and radius rA=0.2 mr_A = 0.2 \text{ m}. Gear B has moment of inertia IB=2.0 kg⋅m2I_B = 2.0 \text{ kg}\cdot\text{m}^2 and radius rB=0.4 mr_B = 0.4 \text{ m}. Both gears are initially at rest. An impulsive torque T^A Δt=10 N⋅m⋅s\hat{T}_A\,\Delta t = 10 \text{ N}\cdot\text{m}\cdot\text{s} is applied to gear A. The no-slip constraint at the mesh point means rA ωA=rB ωBr_A\,\omega_A = r_B\,\omega_B.

What is the angular velocity of gear B immediately after the impulsive torque?

  1. ωB=3.33 rad/s\omega_B = 3.33 \text{ rad/s}, found by computing an effective system moment of inertia referred to gear A's shaft as Ieff=IA+IB(rB/rA)2=0.5+2.0(2.0)2=8.5 kg\cdotpm2I_{\text{eff}} = I_A + I_B(r_B/r_A)^2 = 0.5 + 2.0(2.0)^2 = 8.5 \text{ kg·m}^2, erroneously using the inverse gear ratio (rB/rA)(r_B/r_A) instead of (rA/rB)(r_A/r_B), giving ωA=10/8.5=1.18 rad/s\omega_A = 10/8.5 = 1.18 \text{ rad/s}, then ωB=1.18(rA/rB)=0.59 rad/s\omega_B = 1.18(r_A/r_B) = 0.59 \text{ rad/s} — actually yielding a much smaller value. Alternatively, using Ieff=IA+IB=2.5 kg\cdotpm2I_{\text{eff}} = I_A + I_B = 2.5 \text{ kg·m}^2 (adding inertias without accounting for gear ratio) gives ωA=10/2.5=4 rad/s\omega_A = 10/2.5 = 4 \text{ rad/s} and ωB=4(0.5)=2.0 rad/s\omega_B = 4(0.5) = 2.0 \text{ rad/s}.
  2. ωB=5.00 rad/s\omega_B = 5.00 \text{ rad/s}, found by computing the effective system moment of inertia referred to gear A's shaft as Ieff=IA+IB(rA/rB)2=0.5+2.0(0.5)2=0.5+0.5=1.0 kg\cdotpm2I_{\text{eff}} = I_A + I_B(r_A/r_B)^2 = 0.5 + 2.0(0.5)^2 = 0.5 + 0.5 = 1.0 \text{ kg·m}^2, solving ωA=T^A Δt/Ieff=10/1.0=10 rad/s\omega_A = \hat{T}_A\,\Delta t / I_{\text{eff}} = 10/1.0 = 10 \text{ rad/s}, then applying the gear ratio ωB=ωA(rA/rB)=10(0.2/0.4)=5.00 rad/s\omega_B = \omega_A(r_A/r_B) = 10(0.2/0.4) = 5.00 \text{ rad/s}. (correct answer)
  3. ωB=10.00 rad/s\omega_B = 10.00 \text{ rad/s}, found by applying the impulsive torque only to gear A in isolation, ignoring gear B's inertia entirely: ωA=T^A Δt/IA=10/0.5=20 rad/s\omega_A = \hat{T}_A\,\Delta t / I_A = 10/0.5 = 20 \text{ rad/s}, then applying the gear ratio ωB=20(rA/rB)=20(0.5)=10.00 rad/s\omega_B = 20(r_A/r_B) = 20(0.5) = 10.00 \text{ rad/s}.
  4. ωB=2.00 rad/s\omega_B = 2.00 \text{ rad/s}, found by incorrectly transferring the torque impulse to gear B's axis by multiplying by the gear ratio to get an equivalent impulse of T^B=10(rB/rA)=20 N\cdotpm\cdotps\hat{T}_B = 10(r_B/r_A) = 20 \text{ N·m·s}, then dividing by the total system inertia IA+IB=2.5 kg\cdotpm2I_A + I_B = 2.5 \text{ kg·m}^2, giving ωB=20/2.5=8.0 rad/s\omega_B = 20/2.5 = 8.0 \text{ rad/s}... but then applying an additional ratio factor, yielding an inconsistent result of 2.00 rad/s.
Explanation: When two meshed gears are struck by an impulsive torque, the key challenge is properly accounting for both gears' inertia in a single equation of motion. The standard technique is to refer all inertias to one shaft using the gear ratio, then solve for that shaft's angular velocity before converting. To refer gear B's inertia to gear A's shaft, you use the square of the speed ratio (rA/rB)(r_A/r_B), because kinetic energy must be preserved in the transformation. This gives an effective inertia of Ieff=IA+IB(rArB)2=0.5+2.0(0.5)2=0.5+0.5=1.0 kg\cdotpm2I_{\text{eff}} = I_A + I_B\left(\frac{r_A}{r_B}\right)^2 = 0.5 + 2.0(0.5)^2 = 0.5 + 0.5 = 1.0 \text{ kg·m}^2. Applying the angular impulse-momentum theorem: ωA=T^A Δt / Ieff=10/1.0=10 rad/s\omega_A = \hat{T}_A\,\Delta t\,/\,I_{\text{eff}} = 10/1.0 = 10 \text{ rad/s}. Then the no-slip constraint gives ωB=ωA(rA/rB)=10(0.5)=5.00 rad/s\omega_B = \omega_A(r_A/r_B) = 10(0.5) = 5.00 \text{ rad/s}. That's answer B, the correct choice. A is wrong because it adds IA+IBI_A + I_B directly without the gear-ratio correction — you cannot simply add inertias on different shafts as if they spin at the same speed. C ignores gear B's inertia entirely, treating the gears as uncoupled. A meshed gear always loads the driving gear; omitting that load grossly overestimates ωA\omega_A. D attempts to transfer the torque impulse to gear B's axis correctly (multiplying by rB/rAr_B/r_A) but then divides by the raw sum IA+IBI_A + I_B, mixing up reference frames inconsistently — you must stay on one shaft throughout the calculation. Study tip: Whenever you refer inertia across a gear mesh, always use (rinput/routput)2(r_{\text{input}}/r_{\text{output}})^2 — ratios square, velocities don't.

Question 3

A figure skater is modeled as a cylinder of mass M=55 kgM = 55 \text{ kg} and radius R1=0.18 mR_1 = 0.18 \text{ m} spinning at ω1=2 rev/s\omega_1 = 2 \text{ rev/s} with arms extended. Each arm is modeled as a uniform slender rod of mass m=3.5 kgm = 3.5 \text{ kg} and length L=0.65 mL = 0.65 \text{ m}, held horizontally outward from the body surface. When the skater pulls her arms in, each arm is modeled as a rod of length L′=0.20 mL' = 0.20 \text{ m} (still extending from the body surface). Ice friction is negligible.

Which expression correctly gives the skater's new spin rate ω2\omega_2 after pulling in the arms, assuming the body cylinder's inertia is unchanged and each arm's moment of inertia is computed about the spin axis with the arm's inner end at radius R1R_1?

  1. ω2=ω1⋅12MR12+2(13mL2+mR12)12MR12+2(13mL′2+mR12)\omega_2 = \omega_1 \cdot \dfrac{\tfrac{1}{2}MR_1^2 + 2\left(\tfrac{1}{3}mL^2 + mR_1^2\right)}{\tfrac{1}{2}MR_1^2 + 2\left(\tfrac{1}{3}mL'^2 + mR_1^2\right)}, derived by conserving angular momentum with the correct parallel-axis form for each arm about the spin axis. (correct answer)
  2. ω2=ω1⋅12MR12+2m(R1+L2)212MR12+2m(R1+L′2)2\omega_2 = \omega_1 \cdot \dfrac{\tfrac{1}{2}MR_1^2 + 2m\left(R_1 + \tfrac{L}{2}\right)^2}{\tfrac{1}{2}MR_1^2 + 2m\left(R_1 + \tfrac{L'}{2}\right)^2}, derived by conserving angular momentum and treating each arm as a point mass located at its center of mass, neglecting the distributed nature of the arm's inertia.
  3. ω2=ω1⋅12MR12+2⋅13mL212MR12+2⋅13mL′2\omega_2 = \omega_1 \cdot \dfrac{\tfrac{1}{2}MR_1^2 + 2\cdot\tfrac{1}{3}mL^2}{\tfrac{1}{2}MR_1^2 + 2\cdot\tfrac{1}{3}mL'^2}, derived by conserving angular momentum using 13mL2\tfrac{1}{3}mL^2 for each arm about its own near end, but omitting the parallel-axis offset mR12mR_1^2 that accounts for the arm's attachment point being at R1R_1 rather than on the spin axis.
  4. ω2=ω1⋅12MR12+2(112mL2+mR12)12MR12+2(112mL′2+mR12)\omega_2 = \omega_1 \cdot \dfrac{\tfrac{1}{2}MR_1^2 + 2\left(\tfrac{1}{12}mL^2 + m R_1^2\right)}{\tfrac{1}{2}MR_1^2 + 2\left(\tfrac{1}{12}mL'^2 + m R_1^2\right)}, derived by conserving angular momentum using the centroidal moment of inertia 112mL2\tfrac{1}{12}mL^2 for each arm and shifting to the spin axis by the distance R1R_1 from spin axis to arm center, omitting the arm's half-length in the shift distance.
Explanation: When a spinning skater pulls her arms in, no external torques act, so angular momentum is conserved: I1ω1=I2ω2I_1\omega_1 = I_2\omega_2, giving ω2=ω1⋅I1/I2\omega_2 = \omega_1 \cdot I_1/I_2. The critical skill here is correctly computing each arm's moment of inertia about the spin axis (the skater's central axis), not about the arm's own end or center. Each arm is a uniform rod whose inner end sits at radius R1R_1 from the spin axis. The moment of inertia of a rod about one of its own ends is 13mL2\tfrac{1}{3}mL^2. But that end isn't on the spin axis — it's offset by R1R_1. Applying the parallel-axis theorem properly requires shifting from the rod's own end to the spin axis, adding mR12mR_1^2. This gives Iarm=13mL2+mR12I_{\text{arm}} = \tfrac{1}{3}mL^2 + mR_1^2 per arm. Choice A applies this correctly, making it the right answer. B treats each arm as a point mass at its center of mass (distance R1+L/2R_1 + L/2 from the spin axis), which ignores the distributed rotational inertia of the rod — a conceptually simpler but less accurate model. C uses 13mL2\tfrac{1}{3}mL^2 as if the rod's near end were on the spin axis, completely omitting the mR12mR_1^2 offset. This undercounts each arm's inertia. D uses the centroidal formula 112mL2\tfrac{1}{12}mL^2 and then shifts by only R1R_1 (the distance to the arm's inner end), forgetting that the arm's center is actually R1+L/2R_1 + L/2 from the spin axis — a partial parallel-axis shift. Study tip: Whenever you apply the parallel-axis theorem, always identify the distance from the object's own center of mass to the reference axis — not just to a convenient nearby point.