Use the rotational kinematic equation $$\omega^2 = \omega_0^2 + 2\alpha\Delta\theta$$. The final angular velocity is $$\omega = 0$$, the initial angular velocity is $$\omega_0 = 10 , \text{rad/s}$$, and the angular acceleration is $$\alpha = -2.0 , \text{rad/s}^2$$ (negative because it's slowing down). Solving for $$\Delta\theta$$: $$0^2 = (10)^2 + 2(-2.0)\Delta\theta$$. This gives $$0 = 100 - 4.0\Delta\theta$$, so $$4.0\Delta\theta = 100$$, and $$\Delta\theta = 25 , \text{rad}$$.

The wheel reverses its direction of rotation when its angular velocity $$\omega$$ changes sign, which occurs when $$\omega(t)=0$$. To find $$\omega(t)$$, we differentiate $$\theta(t)$$ with respect to time: $$\omega(t) = \frac{d\theta}{dt} = t^2 - 5t + 6$$. Setting $$\omega(t) = 0$$ gives $$t^2 - 5t + 6 = 0$$, which factors to $$(t-2)(t-3) = 0$$. The solutions are $$t=2 , \text{s}$$ and $$t=3 , \text{s}$$. The first time this occurs for $$t>0$$ is at $$t=2.0 , \text{s}$$.

Angular position $$\theta$$ is the integral of angular velocity $$\omega$$ with respect to time. $$\theta(t) = \int_0^t \omega(t') dt' + \theta(0)$$. Given $$\theta(0)=0$$, we have $$\theta(t) = \int_0^t \omega_0 e^{-kt'} dt' = \omega_0 \left[ -\frac{1}{k}e^{-kt'} \right]_0^t = -\frac{\omega_0}{k}(e^{-kt} - e^0) = -\frac{\omega_0}{k}(e^{-kt} - 1) = \frac{\omega_0}{k}(1-e^{-kt})$$.

To find when the direction reverses, we must find when the angular velocity $$\omega$$ is zero (and changing sign). First, find $$\omega(t)$$ by integrating $$\alpha(t)$$: $$\omega(t) = \int(6t-12) dt = 3t^2 - 12t + C$$. Since the wheel starts from rest, $$\omega(0)=0$$, which implies $$C=0$$. So, $$\omega(t) = 3t^2 - 12t = 3t(t-4)$$. The angular velocity is zero at $$t=0$$ and $$t=4 , \text{s}$$. For $$0 < t < 4$$, $$\omega(t)$$ is negative. For $$t > 4$$, $$\omega(t)$$ is positive. Therefore, the direction of rotation reverses at $$t=4 , \text{s}$$.

By definition, $$\alpha = \frac{d\omega}{dt}$$. Rearranging gives $$d\omega = \alpha dt$$. To find the total change in angular velocity, $$\Delta\omega$$, we integrate this expression over the time interval: $$\Delta\omega = \int_{0}^{t_f} \alpha(t) dt$$. The definite integral of a function represents the area under the curve of that function. Thus, the change in angular velocity is the area under the $$\alpha$$ versus $$t$$ graph.

To find the angular displacement, we must integrate the angular acceleration twice. First, find the angular velocity $$\omega(t)$$ by integrating $$\alpha(t)$$: $$\omega(t) = \int \alpha(t) dt = \int 6t dt = 3t^2 + C$$. Since the object starts from rest, $$\omega(0) = 0$$, so $$C=0$$. Thus, $$\omega(t) = 3t^2$$. Next, find the angular displacement $$\Delta\theta$$ by integrating $$\omega(t)$$ from $$t=0$$ to $$t=2.0$$ s: $$\Delta\theta = \int_0^2 3t^2 dt = \left[ t^3 \right]_0^2 = 2^3 - 0^3 = 8.0 , \text{rad}$$.

This question tests AP Physics C: Mechanics concepts on rotational kinematics and dynamics, specifically calculating required torque for a desired angular acceleration. The problem requires working backwards from kinematic information to find the necessary torque. For a wheel with I = 1.2 kg·m² accelerating from ω0 = 5.0 rad/s to ωf = 17 rad/s in t = 4.0 s, we first find the angular acceleration. Choice B is correct because α = (ωf - ω0)/t = (17 - 5.0)/4.0 = 12/4.0 = 3.0 rad/s², and then τ = Iα = 1.2 × 3.0 = 3.6 N·m. Choice C at 14 N·m might result from calculation errors or using wrong values. To help students: emphasize the two-step process of finding acceleration first, then torque; practice problems that work backwards from desired motion to required forces/torques; and always verify that calculated values make physical sense.

This question tests AP Physics C: Mechanics concepts on rotational kinematics and dynamics, specifically calculating angular displacement under constant angular acceleration. This parallels the translational kinematic equation for displacement under constant acceleration. For a disk with I = 0.30 kg·m² experiencing torque τ = 1.5 N·m for t = 2.0 s starting from rest, we need to find total angular displacement. Choice A is correct because first we find α = τ/I = 1.5/0.30 = 5.0 rad/s², then use θ = ½αt² = ½(5.0)(2.0)² = ½(5.0)(4.0) = 10 rad. Choice C at 20 rad might result from forgetting the factor of ½ in the kinematic equation. To help students: draw parallels between rotational and translational kinematic equations; emphasize that the ½ factor comes from integration just as in linear motion; and practice problems starting from rest to simplify calculations.

This question tests AP Physics C: Mechanics concepts on rotational kinematics and dynamics, specifically gyroscopic precession. Precession occurs when a torque acts perpendicular to a spinning object's angular momentum, causing the spin axis to rotate. For a gyroscope with angular momentum L = 0.40 kg·m²/s experiencing torque τ = 0.080 N·m, the precession rate is found using the gyroscopic equation. Choice A is correct because Ω = τ/L = 0.080/0.40 = 0.20 rad/s. Choice B at 5.0 rad/s would result from inverting the fraction or other calculation errors. To help students: explain that precession is different from ordinary rotation; emphasize that the precession equation Ω = τ/L applies when spin angular momentum is much larger than precession angular momentum; and use demonstrations or videos to visualize gyroscopic motion.

Angular velocity $$\omega$$ is the time derivative of angular position $$\theta$$. Taking the derivative of the given function $$\theta(t) = At^3 - Bt$$ with respect to time $$t$$ gives $$\omega(t) = \frac{d\theta}{dt} = \frac{d}{dt}(At^3 - Bt) = 3At^2 - B$$.