To solve this, note the question states that the rod's length is fixed, and is not permitted to lose contact with the wall and ground. Therefore, the constraint equation is given by the Pythagorean theorem: \[\begin{aligned} x_P^2 + y_Q^2 = \ell^2 \end{aligned}\]

This becomes a related rates problem from calculus 1.

Differentiating with respect to time, keeping in mind the chain rule: \[\begin{aligned} 2x_P\dot{x}_P + 2y_Q\dot{y}_Q = 0 \end{aligned}\]

The quantity we're interested in is $\dot{y}_Q$, or $v_Q$, and $y_Q = \sqrt{\ell^2 - x_P^2}$.

Solving yields $\vec{v}_Q = -8\hat{\jmath} \rm\ m/s$.

Another way to solve this is using the rigid body relations, knowing that the distance and angles from the top to the bottom of the rod remains constant, which is a rigid body by definition. Using rigid body relations: \[\begin{aligned} \vec{v}_Q &= \vec{v}_P + \vec\omega \times \vec{r}_{PQ} \\ &= \vec{v}_P + \omega_z \vec{r}_{PQ}^{\perp} \end{aligned}\] where we have two unknowns: $\vec{v}_Q$, and $\vec\omega$, and $\vec{r}_{PQ} = -\ell \cos\theta \hat{\imath} + \ell \sin\theta \hat{\jmath} \rm\ m$, where $\theta = \cos^{-1}\left(\frac{x_P}{\ell}\right)$

Therefore, $\vec{r}_{PQ}^{\perp} = -\ell \sin\theta \hat{\imath} - \ell \cos\theta \hat{\jmath} \rm\ m$

Expanding the vector equation into its two scalar equations yields: \[\begin{aligned} v_{Qx} &= v_{Px} - \omega_z \ell \sin\theta \\ v_{Qy} &= v_{Py} - \omega_z \ell \cos\theta \end{aligned}\]

Since the top of the rod can only move vertically, $v_{Qx} = 0$, and $\omega_z = \frac{v_{Px}}{\ell \sin\theta} \rm\ rad/s$

Since the bottom of the rod can only move horizontally, $v_{Py} = 0$, and $v_{Qy} = -\omega_z \ell \cos\theta \rm\ m/s$

Plugging in $\omega_z$ expression: \[ v_{Qy} = -\frac{v_{Px}}{\ell \sin\theta} \ell \cos\theta = -v_{Px}\cot\theta \rm\ m/s \]

Plugging in the numbers yields the same result of $\vec{v}_{Q} = -8\hat{\jmath} \rm\ m/s$.

Note in the rigid body relations method, we implicitly used the constraint equation for a point. Namely, we know that the top of the rod can only move vertically, so $\vec{v}_Q \cdot \hat{n}_Q = 0$, where $\hat{n}_Q = \hat{\imath}$ and $\vec{v}_P \cdot \hat{n}_P = 0$, where $\hat{n}_P = \hat{\jmath}$.

We will begin by underlining important information given in the question. Namely:

1. The rope is inextensible, meaning the total length of the rope remains constant. This is our constraint.
2. The pulleys are massless, meaning we do not need to consider any rotational motion they may exhibit.
3. The pulleys radii' are negligbile, meaning we do not need to consider their radii as a part of the total length of the rope.

Using the solution procedure highlighted in #rkc-cp:

1. Write down constraint equation in terms of distances/positions. If the rope's length remains constant, we need to choose a reference line to relate the distances of the masses to that reference line and set it equal to the rope's length. The following figure shows the reference line, as well as the masses' distances from that line.

Note that $y_1$ is only extended to pulley $A$. Since the pulleys are massless, they will not rotate, and therefore $m_1$ and pulley $A$ will translate vertically together at the same rate.

Now, we can write the total length of the rope in terms of those two distances. The total length of the rope, $L$ will be: \[ 2y_1 + y_2 = L \]

2. Differentiate once or twice with respect to time to obtain $\vec{v}$ and $\vec{a}$ relationships. Now, we can take two derivatives with respect to time to obtain the relationship between the speeds $v_1, v_2$, and the accelerations $a_1, a_2$. \[\begin{aligned} 2v_1 + v_2 &= 0 \\ 2a_1 + a_2 &= 0 \end{aligned}\]

Notice that the length of the rope does not matter ultimately. This makes sense, as the length of the rope is constant, and therefore does not change with time.

3. Solve for the desired quantity. We are interested in both the velocity and acceleration of $m_1$.
Solving for $\vec{v}_1 = \frac{-\vec{v}_2}{2} = 3\hat{\jmath}\rm\ m/s$ and $\vec{a}_1 = \frac{-\vec{a}_2}{2} = \hat{\jmath}\rm\ m/s^2$

Reflecting on the result obtained, it makes sense. The distance $m_1$ has to travel up the rope is two times more than the length $m_2$ has to travel. We can also draw a free-body diagram to understand this result further. We will return to this example further down when constraint forces are covered.

We will begin by drawing a free body diagram on the rod.

Note that due to the constraint placed on the rod, point $P$ can only slide along the ground and therefore by #rkc-pc, $\vec{F}_P$ acts vertically. Similarly for point $Q$, $\vec{F}_Q$ acts horizontally.

Applying Newton's second law: \[ \sum \vec{F}_{ext} = m\vec{a}_C \]

Breaking it up into the scalar equations. Note we will use $\ddot{x}_C, \ddot{y}_C$ to denote the acceleration of the center of mass $C$ in each direction. \[\begin{aligned} F_Q &= m\ddot{x}_C \\ F_P - mg &= m\ddot{y}_C \end{aligned}\]

Now we can apply the solution procedure detailed in #rkc-cp to find the acceleration of point $C$. Due to our constraint, the following applies at all times: \[ x_C = \frac{\ell}{2}\sin\phi \qquad \qquad y_C = \frac{\ell}{2}\cos\phi \]

Differentiating twice, keeping in mind the chain rule: \[\begin{aligned} \ddot{x}_C &= -\frac{\ell\dot\theta^2}{2}\sin\phi + \frac{\ell\ddot\theta}{2}\cos\phi \\ \ddot{y}_C &= -\frac{\ell\dot\theta^2}{2}\cos\phi - \frac{\ell\ddot\theta}{2}\sin\phi \end{aligned}\]

Plugging into Newton's second law equations and solving for each constraint force yields: \[\begin{aligned} F_Q &= -\frac{m\ell\dot\theta^2}{2}\sin\phi + \frac{m\ell\ddot\theta}{2}\cos\phi \\ F_P &= -\frac{m\ell\dot\theta^2}{2}\cos\phi - \frac{m\ell\ddot\theta}{2}\sin\phi + mg \end{aligned}\]

While there is contact with the wall, our constraint still holds. This means our constraint forces expressions from #rkc-cf still hold. We are only interested in the constraint force from the wall on the rod, $\vec{F}_Q$ at $\theta$: \[ F_Q = -\frac{m\ell\dot\theta^2}{2}\sin\theta + \frac{m\ell\ddot\theta}{2}\cos\theta \]

If the rod loses contact with the wall, then $F_Q = 0$. We have two unknowns: the angular velocity $\dot\theta$ and the angular acceleration $\ddot\theta$. However, we know that $\ddot\theta = \frac{d\dot\theta}{dt}$, so if we obtain an expression for $\dot\theta$, then finding $\ddot\theta$ is possible. We can use conservation of energy to obtain an expression for $\dot\theta$.

Using the work-energy theorem: \[ W = \Delta T + \Delta V \]

The rod is sliding along frictionless surfaces, so there are no non-conservative forces present, and constraint forces do no work (See work done by a constraint force)

Therefore: \[\begin{aligned} T_1 + V_1 &= T_2 + V_2 \\ 0 + mgh_{C, 1} &= \frac{1}{2}mv_C^2 + \frac{1}{2}I_{C, \hat{k}}\omega^2 + mgh_{C, 2} \end{aligned}\]

From before: \[\begin{aligned} h_{C, 1} = y_{C, 1} = \ell\cos\phi \qquad h_{C, 2} = y_{C, 2} = \ell\cos\theta \end{aligned}\]

And the velocity of the center of mass at $\theta$: \[\begin{aligned} \vec{v}_C &= \dot{x}_{C,2}\hat{\imath} + \dot{y}_{C, 2}\hat{\jmath} \\ &= \ell\dot\theta\cos\theta \hat{\imath} - \ell\dot\theta\sin\theta \hat{\jmath} \end{aligned}\]

So $v_C$ is simply $l\dot\theta$.

Plugging into our energy conservation, where $I_{C, \hat{k}} = \frac{1}{12}m\ell^2$: \[\begin{aligned} mg\ell\cos\phi &= \frac{1}{2}m(\ell\dot\theta)^2 + \frac{1}{2}\left(\frac{1}{12}m\ell^2 \right) \dot\theta^2 + mg\ell\cos\theta \\ mg\ell\cos\phi &= \frac{1}{2}m\ell^2 \dot\theta^2 + \frac{1}{24}m\ell^2 \dot\theta^2 + mg\ell\cos\theta \\ mg\ell\cos\phi &= \frac{13}{24}m\ell^2 \dot\theta^2 + mg\ell\cos\theta \end{aligned}\]

Solving for $\dot\theta^2$: \[ \dot\theta^2 = \frac{24g}{13\ell}\left(\cos\phi - \cos\theta\right) \]

Differentiating to obtain $\ddot\theta$: \[\begin{aligned} 2\dot\theta\ddot\theta &= \frac{24g}{13\ell}\sin\theta\dot\theta \\ \ddot\theta & = \frac{12g}{13\ell}\sin\theta \end{aligned}\]

Plugging our results into $F_Q$: \[\begin{aligned} F_Q &= -\frac{m\ell\frac{24g}{13\ell}\left(\cos\phi - \cos\theta\right)}{2}\sin\theta + \frac{m\ell\frac{12g}{13\ell}\sin\theta}{2}\cos\theta \\ &= -\frac{12}{13}mg\sin\theta \left(\cos\phi - \cos\theta\right) + \frac{6}{13}mg\cos\theta\sin\theta \\ \end{aligned}\]

Setting $F_Q = 0$: \[\begin{aligned} \frac{12}{13}mg\sin\theta \left(\cos\phi - \cos\theta\right) &= \frac{6}{13}mg\cos\theta\sin\theta \\ 2\left(\cos\phi - \cos\theta\right) &= \cos\theta \\ 2\cos\phi - 2\cos\theta - \cos\theta &= 0 \\ 2\cos\phi - 3\cos\theta &= 0 \\ \end{aligned}\]

Solving for $\theta$ yields $\theta = \arccos\left(\frac{2}{3}\cos\phi\right)$, or when the rod has fallen 2/3 of the original height. Note if the rod was unconstrained, the constraint we used still holds until the rod has fallen 2/3 of the original height. We can use rigid body relations to find the velocity of point $Q$ once the rod loses contact with the wall, and it will have both vertical and horizontal components of velocity, as expected.

We added one more assumption to the ones made in #rkc-xcp. Namely, the rope can only slip. This simply means that the tension is the same at any point in the rope.

We will begin by drawing a free body diagram and applying Newton's second law on three bodies: $m_1$, $m_2$ and pulley $A$. We will neglect pulley $B$ as it is fixed to the wall, and the results obtained from it are trivial.

The free body diagram on pulley $A$ is as follows:

Applying Newton's second law: \[ 2T - T_2 = m_A a_A \]

Since the pulley is massless, that means $T_2 = 2T$.

The free body diagram on $m_1$ is as follows:

Applying Newton's second law: \[ 2T - m_1 g = m_1 a_1 \]

Finally, the free body diagram on $m_2$ is as follows:

Applying Newton's second law: \[ T - m_2 g = m_2 a_2 \]

We have 3 unknowns, 2 equations. We apply our constraint equation from #rkc-xcp to relate the accelerations of $m_1$ and $m_2$: \[ 2a_1 + a_2 = 0 \]

Solving for $T$ from Newton's second law on $m_2$: \[ T = m_2 a_2 + m_2 g \]

Plugging it into Newton's second law on $m_1$: \[ 2m_2 a_2 + 2m_2 g - m_1 g = m_1 a_1 \]

Since $a_2 = -2a_1$: \[\begin{aligned} -4m_2 a_1 + 2m_2 g - m_1g &= m_1 a_1 \\ g\left(2m_2 - m_1 \right) &= a_1\left(m_1 + 4m_2\right) \end{aligned}\]

Solving for $a_1$: \[ a_1 = \frac{g\left(2m_2 - m_1 \right)}{m_1 + 4m_2} \]

Plugging that result into Newton's second law on $m_1$ and solving for $T$: \[ T = \frac{m_1\frac{g\left(2m_2 - m_1 \right)}{m_1 + 4m_2} + m_1g}{2} \]

Simplifying a little by multiplying the top and bottom by $m_1 + 4m_2$: \[ T = \frac{m_1g(2m_2 - m_1) + m_1g(m_1 + 4m_2)}{2(m_1 + 4m_2)} \]

We obtain our final result for the tension in the rope $T$: \[ T = \frac{3m_1m_2g}{m_1 + 4m_2} \]