At $$t=0$$, the initial angular velocity is $$\omega_0 = 120$$ rad/s. The angular acceleration is constant: $$\alpha = (\omega_f - \omega_0) / t = (0 - 120 \text{ rad/s}) / 8.0 \text{ s} = -15 \text{ rad/s}^2$$. The tangential acceleration is $$a_T = R|\alpha| = (0.25 \text{ m})(15 \text{ rad/s}^2) = 3.75 \text{ m/s}^2$$. The centripetal acceleration at $$t=0$$ is $$a_c = R\omega_0^2 = (0.25 \text{ m})(120 \text{ rad/s})^2 = 3600 \text{ m/s}^2$$. The total acceleration is $$a = \sqrt{a_T^2 + a_c^2} = \sqrt{(3.75)^2 + (3600)^2} \approx \sqrt{14 + 12960000} \approx 3600 \text{ m/s}^2$$. The centripetal component is much larger than the tangential component, so the total acceleration is very close to the centripetal acceleration.

For any point at radius $$r$$, the tangential acceleration is $$a_T = r\alpha$$ and the centripetal acceleration is $$a_c = r\omega^2$$. After time $$t$$, the angular velocity is $$\omega = \alpha t$$. The magnitude of the total acceleration is $$a = \sqrt{a_T^2 + a_c^2} = \sqrt{(r\alpha)^2 + (r\omega^2)^2} = \sqrt{r^2\alpha^2 + r^2\alpha^2t^4} = r\alpha\sqrt{1 + \alpha^2t^4}$$. Since $$\alpha$$ and $$t$$ are the same for both points, the total acceleration is directly proportional to the radius $$r$$. Therefore, the ratio $$a_B / a_A = (2R) / R = 2$$.

The total linear acceleration is the vector sum of the tangential acceleration ($$a_T$$) and the centripetal acceleration ($$a_c$$). Tangential acceleration is given by $$a_T = r\alpha$$, where $$\alpha$$ is the angular acceleration. Since the angular velocity $$\omega$$ is constant, the angular acceleration $$\alpha = d\omega/dt$$ is zero, so $$a_T = 0$$. The centripetal acceleration is given by $$a_c = r\omega^2$$. Since this is the only non-zero component, the magnitude of the total linear acceleration is $$r\omega^2$$.

The condition that the string unwinds without slipping means that the tangential acceleration of a point on the rim of the cylinder must be equal to the linear acceleration of the string. Since the string's linear motion corresponds to the cylinder's translational motion, the tangential acceleration of the rim, $$a_T$$, must equal the acceleration of the center of mass, $$a$$. The relationship between tangential and angular acceleration is $$a_T = R\alpha$$. Therefore, $$a = R\alpha$$, which can be rearranged to find the angular acceleration: $$\alpha = a/R$$.

The condition of rolling without slipping means that the point of the rolling object in contact with the surface is momentarily at rest with respect to that surface. If it were not, the object would be sliding. This applies to both velocity and acceleration. The acceleration of the contact point is the vector sum of the center of mass acceleration ($$a_{cm}$$ forward) and the tangential acceleration due to rotation ($$a_T$$ backward). For no slipping, $$a_{cm} = R\alpha$$ and $$a_T = R\alpha$$, so the magnitudes are equal. Their vector sum at the contact point is zero.

Since the belt does not slip, the linear speed of any point on the belt must be constant. This linear speed is equal to the tangential speed of the rim of each pulley. Thus, $$v_A = v_B$$. Using the relation $$v = R\omega$$, we have $$R_A \omega_A = R_B \omega_B$$. Solving for $$\omega_B$$ gives $$\omega_B = \omega_A (R_A / R_B)$$. Given that $$R_A = 2R_B$$, we get $$\omega_B = \omega_A (2R_B / R_B) = 2\omega_A$$.

The tangential acceleration $$a_T$$ is related to the angular acceleration $$\alpha$$ by $$a_T = R\alpha$$. The angular acceleration is the second time derivative of the angular position. First, find the angular velocity: $$\omega(t) = d\theta/dt = 3At^2 - B$$. Then, find the angular acceleration: $$\alpha(t) = d\omega/dt = 6At$$. Therefore, the tangential acceleration is $$a_T(t) = R\alpha(t) = R(6At) = 6ARt$$.

The angular acceleration is $$\alpha = d\omega/dt = k$$. The tangential acceleration is $$a_T = R\alpha = Rk$$. The centripetal acceleration is $$a_c = R\omega^2 = R(kt)^2 = Rk^2t^2$$. The total linear acceleration is the magnitude of the vector sum of these perpendicular components: $$a = \sqrt{a_T^2 + a_c^2} = \sqrt{(Rk)^2 + (Rk^2t^2)^2} = \sqrt{R^2k^2 + R^2k^4t^4} = Rk\sqrt{1 + k^2t^4}$$.

After one revolution, the angular displacement is $$\theta = 2\pi$$. Using rotational kinematics, $$\omega^2 = \omega_0^2 + 2\alpha\theta$$. Since it starts from rest, $$\omega_0=0$$, so $$\omega^2 = 2\alpha(2\pi) = 4\pi\alpha$$. The tangential acceleration is constant, $$a_T = R\alpha$$. The centripetal acceleration is $$a_c = R\omega^2 = R(4\pi\alpha) = 4\pi R\alpha$$. The total acceleration is the vector sum of these perpendicular components: $$a = \sqrt{a_T^2 + a_c^2} = \sqrt{(R\alpha)^2 + (4\pi R\alpha)^2} = R\alpha\sqrt{1 + 16\pi^2}$$.

All points on a rigid rotating disk have the same angular velocity, $$\omega$$. The tangential speed $$v$$ of a point at a distance $$r$$ from the axis of rotation is given by $$v = r\omega$$. Therefore, $$v_P = R\omega$$ and $$v_Q = (R/2)\omega$$. The ratio of the speeds is $$v_P / v_Q = (R\omega) / ((R/2)\omega) = 2$$, which means $$v_P = 2v_Q$$.