Solution 1: Intersecting Rods

Problem 1: Velocity of Intersection Point P

Step 1: Coordinate Geometry of Intersection

Let the intersection point be $P$. Consider the triangle $\Delta ACP$ formed by the vertical axis and the two rods.

Using the Sine Rule in $\Delta ACP$:

$$ \frac{AP}{\sin \beta} = \frac{CP}{\sin \alpha} = \frac{AC}{\sin(\pi – (\beta – \alpha))} = \frac{h}{\sin(\beta – \alpha)} $$

This gives us the instantaneous lengths of the rod segments from the pivot to the intersection:

$$ l_1 = AP = \frac{h \sin \beta}{\sin(\beta – \alpha)}, \quad l_2 = CP = \frac{h \sin \alpha}{\sin(\beta – \alpha)} $$

Step 2: Kinematic Analysis

The velocity of the geometric point $P$ can be described relative to either rod. The velocity $\vec{v}_P$ is the vector sum of the rod’s rotational velocity at that point and the sliding velocity of the intersection along the rod.

For Rod AB: $\vec{v}_P = \vec{v}_{rot,1} + v_{\parallel} \hat{u}_1$ (where $\hat{u}_1$ is along AB)
For Rod CD: $\vec{v}_P = \vec{v}_{rot,2} + v_{slide,2} \hat{u}_2$ (where $\hat{u}_2$ is along CD)

We need to find $v_{\parallel}$ (component along AB). To eliminate the unknown sliding velocity along CD ($v_{slide,2}$), we take the component of the velocity perpendicular to Rod CD.

Step 3: Solving for Velocity

Equating the components along the normal to rod CD ($\hat{n}_2$):

$$ (\vec{v}_{rot,1} + v_{\parallel} \hat{u}_1) \cdot \hat{n}_2 = \vec{v}_{rot,2} \cdot \hat{n}_2 $$

Analyzing the dot products (where angle between rods is $\theta = \beta – \alpha$):

$\vec{v}_{rot,2}$ is purely parallel to $\hat{n}_2$, so its projection is just $l_2 \omega_2$.
$\vec{v}_{rot,1}$ makes an angle $\theta$ with $\hat{n}_2$, projection is $l_1 \omega_1 \cos \theta$.
$\vec{v}_{\parallel}$ makes an angle $90+\theta$ with $\hat{n}_2$, projection is $v_{\parallel} \sin \theta$.

Substitute angular velocities $\omega_1 = v_1/L$ and $\omega_2 = v_2/L$ (noting opposite rotation directions):

$$ v_{\parallel} \sin(\beta-\alpha) = -\left[ l_2 \frac{v_2}{L} + l_1 \frac{v_1}{L} \cos(\beta-\alpha) \right] $$

Step 4: Substitution and Simplification

Substitute $l_1$ and $l_2$ from Step 1:

$$ v_{\parallel} = -\frac{h}{L \sin^2(\beta-\alpha)} \left[ v_2 \sin \alpha + v_1 \sin \beta \cos(\beta-\alpha) \right] $$

Using the expansion $\sin \alpha = \sin(\beta – (\beta-\alpha)) = \sin\beta \cos(\beta-\alpha) – \cos\beta \sin(\beta-\alpha)$:

Final Answer: $$ v_{P \parallel AB} = \frac{h \left| v_2 \sin(\beta-\alpha) \cos\beta – (v_1+v_2)\sin\beta \cos(\beta-\alpha) \right|}{L \sin^2(\beta-\alpha)} $$