The rod is massless (inertia-less). The only inertia in the system comes from the beads. Since the system is isolated (no external torque mentioned maintaining speed), Angular Momentum is conserved.

For a system with zero Moment of Inertia of the constraint (rod), conservation of angular momentum leads to the particles moving in a straight line tangent to the initial path (Newton’s 1st Law). The rod simply rotates to follow them without exerting a tangential force.

Conclusion (b): The path of the bead in an inertial frame is a straight line.

The bead moves with constant linear velocity \( v \) equal to its initial tangential velocity. \[ v = r_0 \omega_0 \] Given: \( 2l \) is rod length, \( 1.2l \) is cord length. Initial radius \( r_0 = 1.2l / 2 = 0.6l \). Initial \( \omega_0 = 1/3 \) rad/s. \[ v = (0.6l)(1/3) = 0.2l \text{ m/s} \]

The bead leaves when its distance from the center is \( r_f = l \). The path is a straight line tangent to the start circle. Distance traveled \( d = \sqrt{r_f^2 – r_0^2} \). \[ d = \sqrt{l^2 – (0.6l)^2} = \sqrt{0.64l^2} = 0.8l \]

Time taken: \[ t = \frac{d}{v} = \frac{0.8l}{0.2l} = 4 \text{ s} \]