Question

A rod of mass m and length fits into a hollow tube of same length and mass. The tube is rotated with an initial angular velocity ω₀ and the rod slips through the rough hollow surface. The angular velocity of the rod as it slips out of the tube is

• A. $\frac{\omega_{0}}{2}$
• B. $\frac{\omega_{0}}{4}$
• C. $\frac{\omega_{0}}{16}$
• D. $\frac{\omega_{0}}{7}$

Answer 1

Answer: (B)

$\frac{\omega_{0}}{4}$

Answer 2

There is no external torque acting on the system. Hence angular momentum remains conserved.

I₁ω₁ = I₂ω₂

I₁ = 2M²/3; I₂ = [2M(2)²/3] ⇒ ω₂ = ω_o/4