1. System Analysis

We have three identical cylinders (A, B, C) initially rotating with $\omega_0$ in the same direction (let’s say Clockwise, CW). When brought into contact, friction acts at the interfaces (A-B and B-C) until slipping stops.

No-Slip Condition: For two cylinders to roll without slipping, their tangential velocities at the contact point must be equal. Since the contact points are on opposite sides, this requires their angular velocities to be equal in magnitude but opposite in direction.

Let the final angular velocity magnitude be $\omega_f$.
Since A and C are symmetric, $\omega_A = \omega_C = \omega_f$ (CW).
For B to roll with A and C, it must reverse direction: $\omega_B = \omega_f$ (Counter-Clockwise, CCW).

2. Impulse Equations

Let $J$ be the magnitude of the linear impulse exerted by friction at each interface during the entire process. We define Clockwise (CW) as positive.

Cylinder A:
Friction opposes the CW rotation. Torque is negative (CCW).
Equation: $I(\omega_f – \omega_0) = -J R$

Cylinder C:
By symmetry with A: $I(\omega_f – \omega_0) = -J R$

Cylinder B:
Initial $\omega_0$ (CW). Final $-\omega_f$ (CCW).
At left interface (with A): A moves down, B moves up. Friction on B is Down (CW torque).
At right interface (with C): C moves up, B moves down. Friction on B is Up (CW torque? No, wait).
Let’s check torques on B again.
Left contact: Force on B is Down. Torque is $\vec{r} \times \vec{F} = (-R)\hat{i} \times (-F)\hat{j} = +RF \hat{k}$ (CCW).
Right contact: Force on B is Up. Torque is $\vec{r} \times \vec{F} = (+R)\hat{i} \times (+F)\hat{j} = +RF \hat{k}$ (CCW).
Both torques on B are CCW (negative in our convention).
Equation: $I(-\omega_f – \omega_0) = -J R – J R = -2JR$

3. Solving for Final Angular Velocity

We have:

1. $I(\omega_0 – \omega_f) = JR$
2. $I(\omega_0 + \omega_f) = 2JR$

Substitute (1) into (2):

$$I(\omega_0 + \omega_f) = 2 \cdot [I(\omega_0 – \omega_f)]$$ $$\omega_0 + \omega_f = 2\omega_0 – 2\omega_f$$ $$3\omega_f = \omega_0$$ $$\omega_f = \frac{\omega_0}{3}$$

4. Conclusion

(a) The final angular velocities are:

$$\omega_A = \frac{\omega_0}{3} \text{ (CW)}, \quad \omega_B = \frac{\omega_0}{3} \text{ (CCW)}, \quad \omega_C = \frac{\omega_0}{3} \text{ (CW)}$$

(b) No, you cannot use the principle of conservation of angular momentum for the *entire system* as a simple sum $\sum I \omega = \text{const}$ because the axles exert external forces to keep the cylinders in place. However, a specific linear combination of angular momenta ($L_B – L_A – L_C$) is conserved due to the cancellation of internal impulses.