[Q1] Tension in the string at the moment of release

We analyze the forces and accelerations at the instant the system is released from rest ($t=0$). The string AB is initially vertical.

1. Kinematic Constraint Let $\vec{r}_1$ and $\vec{r}_2$ be the position vectors of masses $m_1$ and $m_2$. The string has a fixed length $L$, so $|\vec{r}_2 - \vec{r}_1|^2 = L^2$. Differentiating with respect to time twice and evaluating at $t=0$ (where velocities are zero) gives the constraint on the initial accelerations $\vec{a}_1$ and $\vec{a}_2$:

$$(\vec{a}_2 - \vec{a}_1) \cdot (\vec{r}_2 - \vec{r}_1) = 0$$

This means the relative acceleration $(\vec{a}_2 - \vec{a}_1)$ is perpendicular to the string. Equivalently, the acceleration components along the string must be equal: $\vec{a}_1 \cdot \hat{u} = \vec{a}_2 \cdot \hat{u}$, where $\hat{u}$ is the unit vector along the string.

2. Equations of Motion Let's use a standard coordinate system with x-axis horizontal and y-axis vertical (upwards).

• The wire is at an angle $\alpha$ with the horizontal. The acceleration of $m_1$ is $\vec{a}_1$, directed along the wire.

  $$\vec{a}_1 = a_1(\cos\alpha, -\sin\alpha)$$

• The string is vertical, so $\hat{u}=(0, 1)$ (pointing from $m_2$ to $m_1$). The tension on $m_2$ is $\vec{T} = (0, T)$. The gravitational force on $m_2$ is $\vec{W}_2 = (0, -m_2g)$.

• The acceleration of $m_2$ is $\vec{a}_2 = (a_{2x}, a_{2y})$.

• Applying the kinematic constraint:

  $$a_1(-\sin\alpha) = a_{2y}$$

• Applying Newton's second law to $m_2$:

  • x-direction: $0 = m_2 a_{2x} \implies a_{2x}=0$.
  • y-direction: $T - m_2g = m_2 a_{2y} = m_2(-a_1 \sin\alpha)$.
  $$T = m_2g - m_2 a_1 \sin\alpha \quad (1)$$

• Now consider $m_1$. Its acceleration $a_1$ is along the wire. We project the forces on $m_1$ onto the wire's direction.

  • Gravitational force component along the wire: $m_1g \sin\alpha$.
  • Tension force on $m_1$ is directed vertically downwards: $\vec{T}_1 = (0, -T)$. Its component along the wire is $-T\sin\alpha$.
  • Newton's second law for $m_1$ along the wire:
  $$m_1g \sin\alpha - T\sin\alpha = m_1 a_1 \quad (2)$$

  Wait, let's re-check the tension component. The string is vertical, pulling $m_1$ downwards. The wire goes down at angle $\alpha$. So the vertical pull has a component along the wire. The component of a vertical vector $(0, -T)$ on the wire direction vector $(\cos\alpha, -\sin\alpha)$ is $(0,-T)\cdot(\cos\alpha, -\sin\alpha) = T\sin\alpha$. So the equation for $m_1$ is:

  $$m_1g \sin\alpha + T\sin\alpha = m_1 a_1 \quad (2')$$

3. Solving for Tension T From (1), we express $a_1$: $a_1 = \frac{m_2g - T}{m_2 \sin\alpha}$. Substitute this into (2'):

$$m_1g \sin\alpha + T\sin\alpha = m_1 \left( \frac{m_2g - T}{m_2 \sin\alpha} \right)$$ $$m_1 m_2 g \sin^2\alpha + T m_2 \sin^2\alpha = m_1 m_2 g - m_1 T$$ $$T(m_1 + m_2 \sin^2\alpha) = m_1 m_2 g (1 - \sin^2\alpha)$$ $$T = \frac{m_1 m_2 g \cos^2\alpha}{m_1 + m_2 \sin^2\alpha}$$

[Q2] Angle for no swinging

The condition "the string does not swing" implies that the string maintains a constant angle with the vertical. This means both masses move together as a single system with a common acceleration $\vec{a}$ directed down the wire.

1. System Acceleration Consider the two masses as a single system of mass $M = m_1 + m_2$. The net external force component along the wire is the sum of the gravitational force components.

$$F_{\text{net, along wire}} = (m_1g + m_2g)\sin\alpha$$

Applying Newton's second law to the system:

$$(m_1 + m_2)g \sin\alpha = (m_1 + m_2)a$$

The common acceleration of the system down the wire is:

$$a = g \sin\alpha$$

2. Equilibrium in an Accelerated Frame Let's analyze the forces on mass $m_2$ in the non-inertial frame of reference that accelerates with the system ($\vec{a}$ down the wire). In this frame, $m_2$ is in equilibrium. The forces acting on it are:

1. Gravity: $\vec{W}_2 = m_2\vec{g}$ (vertically down)
2. Tension: $\vec{T}$ (along the string)
3. Inertial (fictitious) force: $\vec{F}_{in} = -m_2\vec{a}$ (directed up along the wire)

For equilibrium, the vector sum of these forces is zero: $\vec{T} + \vec{W}_2 + \vec{F}_{in} = 0$. This means the tension $\vec{T}$ must balance the vector sum of gravity and the inertial force: $\vec{T} = -(\vec{W}_2 + \vec{F}_{in})$.

Let's decompose the gravitational force $\vec{W}_2$ into components parallel and perpendicular to the wire:

• Component parallel to the wire (downwards): $m_2g \sin\alpha$.
• Component perpendicular to the wire (into the wire): $m_2g \cos\alpha$.

The inertial force $\vec{F}_{in}$ has magnitude $m_2a = m_2(g\sin\alpha)$ and is directed up along the wire. This force exactly cancels the parallel component of gravity on $m_2$.

Therefore, the tension $\vec{T}$ only needs to balance the perpendicular component of gravity. This means the string must be oriented perpendicular to the wire, pointing from $m_2$ towards the wire.

3. Angle with the Vertical If the string is perpendicular to the wire, we can find its angle $\beta$ with the vertical. The wire is at an angle $\alpha$ with the horizontal. A line perpendicular to the wire is at an angle of $90^\circ$ to the wire. Thus, the angle of this perpendicular line with the vertical is $\alpha$.

So, the string must make an angle $\beta = \alpha$ with the vertical.