Here is a concise solution to the physics problem, focusing on key concepts and essential derivation steps.

[SOLUTION]

[Q1] Critical Angle for Direction of Motion

The direction of rolling is determined by the net torque about the instantaneous point of contact, P. The spool will not roll if the net torque about P is zero. The only external force that can produce a torque about P is the tension $T$ (since gravity, normal force, and friction all act through P).

The torque is zero when the line of action of the tension force $T$ passes through the point of contact P. The force $T$ is applied tangentially to the inner hub of radius $r$. For the line of action to pass through P (a point at a vertical distance $R$ from the center C), a right triangle is formed by the center C, the point of contact P, and the point on the line of action horizontally aligned with C.

From the geometry of the diagram, this occurs when:

$$ R\cos\theta_{crit} = r $$ $$ \theta_{crit} = \arccos\left(\frac{r}{R}\right) $$

If $\theta < \theta_{crit}$, the line of action passes above P, creating a torque that rolls the spool to the right. If $\theta > \theta_{crit}$, the line of action passes below P, rolling the spool to the left.

[Q2] Horizontal Pull ($\theta = 0^\circ$)

We use the rotational equation of motion about the point of contact P, which serves as the instantaneous axis of rotation. The net external torque about P equals the moment of inertia about P times the angular acceleration, $\tau_P = I_P \alpha$. By the parallel axis theorem, $I_P = I + MR^2$.

The torque from tension $T$ about P is $\tau_T = T(R-r)$, acting clockwise (CW). Assuming the spool rolls right, the rotation is CW. Let CW be the positive direction for rotation.

$$ \tau_P = T(R-r) = (I + MR^2)\alpha $$

The angular acceleration is:

$$ \alpha = \frac{T(R-r)}{I + MR^2} $$

The linear acceleration of the center of mass, $a$, is related to $\alpha$ by the no-slip condition $a = \alpha R$.

$$ a = \frac{TR(R-r)}{I + MR^2} $$

[Q3] Pull at an Angle $\theta$

We again calculate the torque about the point of contact P. The lever arm of the tension force $T$ about P is the perpendicular distance from P to the line of action of $T$. This distance is $d = |R\cos\theta - r|$.

For forward motion ($\theta < \theta_{crit}$), the torque is CW, $\tau_P = T(R\cos\theta - r)$.

$$ \tau_P = T(R\cos\theta - r) = (I+MR^2)\alpha $$

The angular acceleration (CW positive) is:

$$ \alpha = \frac{T(R\cos\theta - r)}{I + MR^2} $$

The linear acceleration is $a = \alpha R$:

$$ a = \frac{TR(R\cos\theta - r)}{I + MR^2} $$

[Q4] Work and Energy ($\theta = 0^\circ$)

We apply the Work-Energy theorem. The net work done on the spool equals its change in kinetic energy. Static friction does no work. The work is done by the tension force $T$.

1. Work Done By You: The work you do, $W_{you}$, is the tension force multiplied by the total length of string pulled, $S_{string}$. When the CM moves a distance $L$, the spool rotates by an angle $\phi=L/R$. The length of string unwound is $L_{unwound} = r\phi = rL/R$. The point where the string is attached moves a velocity $v_Q = v_{CM}(1-r/R)$, so the string displacement is $S_{string} = L(1-r/R)$.

   $$ W_{you} = T \cdot S_{string} = T L \left(1-\frac{r}{R}\right) = \frac{TL(R-r)}{R} $$

2. Kinetic Energy: The final kinetic energy $K_f$ is the sum of translational and rotational kinetic energy. $K_f = K_{trans} + K_{rot}$.

   $$ K_f = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2 $$

   Using $v=\omega R$ and $v^2=2aL$:

   $$ K_f = \frac{1}{2}v^2\left(M + \frac{I}{R^2}\right) = aL\left(M + \frac{I}{R^2}\right) $$

   Substituting the acceleration $a$ from Q2:

   $$ K_f = \left(\frac{TR(R-r)}{MR^2+I}\right) L \left(\frac{MR^2+I}{R^2}\right) = \frac{TL(R-r)}{R} $$

   As expected, $W_{you} = K_f$.

3. Final Energies:

   • Translational: $K_{trans} = \frac{1}{2}Mv^2 = MaL = \frac{MTRL(R-r)}{MR^2+I}$
   • Rotational: $K_{rot} = \frac{1}{2}I\omega^2 = \frac{IaL}{R^2} = \frac{ITL(R-r)}{R(MR^2+I)}$

[Q5] Constant Speed Pull

Let the speed of the spool's CM be $v_s$ and the speed at which the string is pulled be $v$. The speed $v$ is the component of the velocity of the string's departure point (Q) along the direction of the string.

The velocity of point Q is $\vec{v}_Q = \vec{v}_{CM} + \vec{\omega} \times \vec{r}_Q$. With $\vec{v}_{CM}=v_s\hat{i}$ and $\vec{\omega}=-(v_s/R)\hat{k}$ (for forward motion), we find:

$$ \vec{v}_Q = v_s(1 - \frac{r}{R}\cos\theta)\hat{i} - v_s(\frac{r}{R}\sin\theta)\hat{j} $$

The direction of the string is $\hat{u}_T = \cos\theta\hat{i} + \sin\theta\hat{j}$. The speed of the string is $v = \vec{v}_Q \cdot \hat{u}_T$.

$$ v = v_s \left[ (1 - \frac{r}{R}\cos\theta)\cos\theta - (\frac{r}{R}\sin\theta)\sin\theta \right] $$ $$ v = v_s \left[ \cos\theta - \frac{r}{R}(\cos^2\theta + \sin^2\theta) \right] = v_s \left(\cos\theta - \frac{r}{R}\right) $$

Solving for the speed of the spool, $v_s$:

$$ v_s = \frac{v}{\cos\theta - r/R} = \frac{vR}{R\cos\theta - r} $$

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