Angular Impulse-Momentum - Statics and Dynamics

1. What is the angular velocity of gear B immediately after the impulsive torque?
   Answer: $\omega_B = 5.00 \text{ rad/s}$, found by computing the effective system moment of inertia referred to gear A's shaft as $I_{\text{eff}} = I_A + I_B(r_A/r_B)^2 = 0.5 + 2.0(0.5)^2 = 0.5 + 0.5 = 1.0 \text{ kg}\cdot\text{m}^2$, solving $\omega_A = \hat{T}_A\,\Delta t / I_{\text{eff}} = 10/1.0 = 10 \text{ rad/s}$, then applying the gear ratio $\omega_B = \omega_A(r_A/r_B) = 10(0.2/0.4) = 5.00 \text{ rad/s}$. 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 $(r_A/r_B)$, because kinetic energy must be preserved in the transformation. This gives an effective inertia of $I_{\text{eff}} = I_A + I_B\left(\frac{r_A}{r_B}\right)^2 = 0.5 + 2.0(0.5)^2 = 0.5 + 0.5 = 1.0 \text{ kg}\cdot\text{m}^2$. Applying the angular impulse-momentum theorem: $\omega_A = \hat{T}_A\,\Delta t\,/\,I_{\text{eff}} = 10/1.0 = 10 \text{ rad/s}$. Then the no-slip constraint gives $\omega_B = \omega_A(r_A/r_B) = 10(0.5) = 5.00 \text{ rad/s}$. That's answer B, the correct choice. A is wrong because it adds $I_A + I_B$ 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 $\omega_A$. D attempts to transfer the torque impulse to gear B's axis correctly (multiplying by $r_B/r_A$) but then divides by the raw sum $I_A + I_B$, 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 $(r_{\text{input}}/r_{\text{output}})^2$ — ratios square, velocities don't.
2. Which expression correctly gives the skater's new spin rate $\omega_2$ 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 $R_1$?
   Answer: $\omega_2 = \omega_1 \cdot \dfrac{\tfrac{1}{2}MR_1^2 + 2\left(\tfrac{1}{3}mL^2 + mR_1^2\right)}{\tfrac{1}{2}MR_1^2 + 2\left(\tfrac{1}{3}mL'^2 + mR_1^2\right)}$, derived by conserving angular momentum with the correct parallel-axis form for each arm about the spin axis. When a spinning skater pulls her arms in, no external torques act, so angular momentum is conserved: $I_1\omega_1 = I_2\omega_2$, giving $\omega_2 = \omega_1 \cdot I_1/I_2$. 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 $R_1$ from the spin axis. The moment of inertia of a rod about one of its own ends is $\tfrac{1}{3}mL^2$. But that end isn't on the spin axis — it's offset by $R_1$. Applying the parallel-axis theorem properly requires shifting from the rod's own end to the spin axis, adding $mR_1^2$. This gives $I_{\text{arm}} = \tfrac{1}{3}mL^2 + mR_1^2$ 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 $R_1 + L/2$ from the spin axis), which ignores the distributed rotational inertia of the rod — a conceptually simpler but less accurate model. C uses $\tfrac{1}{3}mL^2$ as if the rod's near end were on the spin axis, completely omitting the $mR_1^2$ offset. This undercounts each arm's inertia. D uses the centroidal formula $\tfrac{1}{12}mL^2$ and then shifts by only $R_1$ (the distance to the arm's inner end), forgetting that the arm's center is actually $R_1 + L/2$ 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.
3. What is the angular velocity at $t = 3 \text{ s}$?
   Answer: $\omega(3) = +17.0 \text{ rad/s}$, found by correctly computing the angular impulse $\int_0^3(6t^2-4t)\,dt = [2t^3-2t^2]_0^3 = 54-18 = 36 \text{ N}\cdot\text{m}\cdot\text{s}$, applying $\Delta\omega = 36/I = 36/2 = 18 \text{ rad/s}$, and adding to $\omega_0 = -1 \text{ rad/s}$, giving $\omega(3) = -1 + 18 = +17 \text{ rad/s}$. When a time-varying torque acts on a rotating body, the cleanest approach uses the angular impulse-momentum theorem: $\int_{t_1}^{t_2} \tau\, dt = I(\omega_f - \omega_0)$. This tells you that the change in angular momentum equals the angular impulse — so you must integrate the torque, divide by $I$, 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: $\int_0^3 (6t^2 - 4t)\,dt = \left[2t^3 - 2t^2\right]_0^3 = 2(27) - 2(9) = 54 - 18 = 36 \text{ N}\cdot\text{m}\cdot\text{s}$ Divide by $I = 2 \text{ kg}\cdot\text{m}^2$ to get $\Delta\omega = 18 \text{ rad/s}$. Adding the initial condition gives $\omega(3) = -1 + 18 = +17 \text{ rad/s}$. A arrives at the same integral value (36 N·m·s) and the same $\Delta\omega = 18$ 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 $\Delta\omega = 18 \text{ rad/s}$ but forgets to add $\omega_0 = -1 \text{ rad/s}$, 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; $\Delta\omega$ and $\omega_f$ are not the same thing.

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