Step 1: Force Analysis and Symmetry

Let the central pivot be $P$. Since the pivot is light (massless), the net force acting on it must be zero at all times. The forces acting on the pivot are the tensions from the three rods.

Let $\vec{T}_A, \vec{T}_B, \vec{T}_C$ be the tension forces exerted by the rods on the pivot. $$ \vec{T}_A + \vec{T}_B + \vec{T}_C = 0 $$ Since the rods are arranged symmetrically at angles of $120^\circ$, the only way three vectors at these angles can sum to zero is if their magnitudes are equal.

Thus, the tension in each rod is the same:

$$ T_A = T_B = T_C = T $$

Consequently, the force acting on each particle is a tension force $T$ directed radially inward along the respective rod.

Step 2: Dynamics of Particles

We apply Newton’s Second Law in the radial direction (towards the pivot) for each particle. Note that the rods are light, so they only transmit force along their length.

• Particle A ($m$): $$ T = m a_A \implies a_A = \frac{T}{m} $$
• Particle B ($2m$): $$ T = 2m a_B \implies a_B = \frac{T}{2m} $$
• Particle C ($3m$): $$ T = 3m a_{C,\text{net radial}} $$ Wait! $a_{C,\text{net radial}}$ is the net acceleration along the rod. Let’s denote the magnitude of the acceleration vector associated with the tension force simply as $a_C = \frac{T}{3m}$.

Step 3: Kinematic Constraints

The rods are rigid, so the length $l$ is constant. This imposes a constraint relating the acceleration of the particles to the acceleration of the pivot, $\vec{a}_P$.

For any particle connected to the pivot by a rod of length $l$, the component of the particle’s acceleration along the rod (towards the center) is given by:

$$ \vec{a}_{\text{particle}} \cdot \hat{n} = \vec{a}_P \cdot \hat{n} + \frac{v_{rel}^2}{l} $$

Where $\hat{n}$ is the unit vector pointing from the particle to the pivot.

• For A and B: They are momentarily at rest ($v=0$). Thus, their acceleration is purely due to the tension and the pivot’s movement. $$ a_A = \vec{a}_P \cdot \hat{n}_A $$ $$ a_B = \vec{a}_P \cdot \hat{n}_B $$
• For C: Particle C has a velocity $u$ perpendicular to the rod. This creates a centripetal acceleration term relative to the pivot. $$ a_C = \vec{a}_P \cdot \hat{n}_C + \frac{u^2}{l} $$ Here, $a_C$ represents the component of C’s acceleration provided by the tension force (since tension is the only radial force).

Step 4: Solving for Tension

We now substitute the dynamic expressions ($a = F/m$) into the kinematic constraints:

1. $\frac{T}{m} = \vec{a}_P \cdot \hat{n}_A$
2. $\frac{T}{2m} = \vec{a}_P \cdot \hat{n}_B$
3. $\frac{T}{3m} = \vec{a}_P \cdot \hat{n}_C + \frac{u^2}{l}$

To solve for $T$, we sum these three equations. Notice that the unit vectors $\hat{n}_A, \hat{n}_B, \hat{n}_C$ are arranged symmetrically at $120^\circ$. A key property of such vectors is:

$$ \hat{n}_A + \hat{n}_B + \hat{n}_C = 0 $$

Summing the three equations:

$$ \left( \frac{T}{m} + \frac{T}{2m} + \frac{T}{3m} \right) = \vec{a}_P \cdot (\hat{n}_A + \hat{n}_B + \hat{n}_C) + \frac{u^2}{l} $$ $$ T \left( 1 + \frac{1}{2} + \frac{1}{3} \right) = \vec{a}_P \cdot (0) + \frac{u^2}{l} $$ $$ T \left( \frac{6 + 3 + 2}{6} \right) = \frac{u^2}{l} $$ $$ \frac{11T}{6} = \frac{mu^2}{l} $$ $$ T = \frac{6mu^2}{11l} $$

Step 5: Final Accelerations

Now substitute $T$ back into the acceleration formulas for each mass:

• Acceleration of A: $$ |\vec{a}_A| = \frac{T}{m} = \frac{1}{m} \left( \frac{6mu^2}{11l} \right) = \frac{6u^2}{11l} $$
• Acceleration of B: $$ |\vec{a}_B| = \frac{T}{2m} = \frac{1}{2m} \left( \frac{6mu^2}{11l} \right) = \frac{3u^2}{11l} $$
• Acceleration of C: $$ |\vec{a}_C| = \frac{T}{3m} = \frac{1}{3m} \left( \frac{6mu^2}{11l} \right) = \frac{2u^2}{11l} $$