[Q1] Find the velocity of point B, $\vec{v}_B$.

The motion of point B is governed by two constraints:

1. Fixed length of AB: Since point A is fixed and the length $l_{AB} = a$ is constant, point B must move in a circle of radius $a$ around A. This implies that the velocity vector $\vec{v}_B$ is perpendicular to the rope segment AB.
2. Inextensible rope over pulley: The free end of the rope is pulled with a constant speed $v$. This means the length of the rope segment BC is decreasing at a rate of $v$. The rate of change of length $l_{BC}$ is the projection of $\vec{v}_B$ onto the direction of BC. Therefore, the component of $\vec{v}_B$ along the direction from B to C is $v$.

Let $v_B$ be the magnitude of the velocity $\vec{v}_B$. Let's determine the angle between the vector $\vec{v}_B$ and the segment BC.

• The direction of $\vec{v}_B$ is perpendicular to AB.
• The angle between the segments AB and BC is $\alpha + \beta$.
• The angle $\phi$ between $\vec{v}_B$ and BC is therefore $|\frac{\pi}{2} - (\alpha+\beta)|$.

The projection of $\vec{v}_B$ onto BC is $v_B \cos\phi = v$.

$$v_B \cos\left(\frac{\pi}{2} - (\alpha+\beta)\right) = v$$ $$v_B \sin(\alpha+\beta) = v$$

Solving for the magnitude $v_B$:

$$v_B = \frac{v}{\sin(\alpha+\beta)}$$

The velocity vector $\vec{v}_B$ has this magnitude and a direction perpendicular to AB, pointing towards the line AC.

[Q2] Find the acceleration of point B along the direction of AB, $a_{B, AB}$.

Since point B moves in a circular path of radius $a$ around the fixed point A, it has a radial (centripetal) acceleration directed towards the center A. The magnitude of this acceleration is given by:

$$a_{rad} = \frac{v_B^2}{a}$$

The question asks for the acceleration component along the direction of AB (from A to B). This is opposite to the direction of the centripetal acceleration (from B to A).

$$a_{B, AB} = -a_{rad} = -\frac{v_B^2}{a}$$

Substituting the expression for $v_B$ from Q1:

$$a_{B, AB} = -\frac{1}{a} \left( \frac{v}{\sin(\alpha+\beta)} \right)^2 = -\frac{v^2}{a \sin^2(\alpha+\beta)}$$

[Q3] Find the acceleration of point B along the direction of BC, $a_{B, BC}$.

We consider the motion of B relative to the fixed point C. The acceleration component along BC is the radial component of acceleration in a polar coordinate system centered at C. The general expression for radial acceleration is $a_r = \ddot{r} - r\dot{\theta}^2$.

• The radial distance is $r = l_{BC}$. Its rate of change is $\dot{l}_{BC} = -v$ (since the rope is pulled in). As $v$ is constant, the second derivative is $\ddot{l}_{BC} = 0$.
• The angular velocity is $\dot{\theta} = \omega_{BC}$. The term $-r\dot{\theta}^2$ is the centripetal acceleration, directed towards C.

The acceleration component along the direction CB (towards C) is $\ddot{l}_{BC} - l_{BC}\omega_{BC}^2 = 0 - l_{BC}\omega_{BC}^2$. The component along BC (away from C) is the negative of this:

$$a_{B, BC} = l_{BC}\omega_{BC}^2$$

The angular velocity $\omega_{BC}$ is related to the component of B's velocity perpendicular to BC, $v_{B, \perp BC}$: $\omega_{BC} = \frac{v_{B, \perp BC}}{l_{BC}}$. Substituting this gives:

$$a_{B, BC} = l_{BC} \left( \frac{v_{B, \perp BC}}{l_{BC}} \right)^2 = \frac{v_{B, \perp BC}^2}{l_{BC}}$$

Now, we find $v_{B, \perp BC}$ and $l_{BC}$:

• $v_{B, \perp BC}$ is the component of $\vec{v}_B$ perpendicular to BC.

$v_{B, \perp BC} = v_B \sin\phi = v_B \sin\left(\frac{\pi}{2} - (\alpha+\beta)\right) = v_B \cos(\alpha+\beta)$.

Substituting $v_B = \frac{v}{\sin(\alpha+\beta)}$:

$$v_{B, \perp BC} = \frac{v \cos(\alpha+\beta)}{\sin(\alpha+\beta)} = v \cot(\alpha+\beta)$$

• $l_{BC}$ is found using the Law of Sines in triangle ABC:

$\frac{l_{BC}}{\sin\alpha} = \frac{a}{\sin\beta} \implies l_{BC} = \frac{a \sin\alpha}{\sin\beta}$.

Finally, we substitute these into the expression for $a_{B, BC}$:

$$a_{B, BC} = \frac{(v \cot(\alpha+\beta))^2}{a \sin\alpha / \sin\beta} = \frac{v^2 \cot^2(\alpha+\beta) \sin\beta}{a \sin\alpha}$$ $$a_{B, BC} = \frac{v^2 \sin\beta \cos^2(\alpha+\beta)}{a \sin\alpha \sin^2(\alpha+\beta)}$$