A hollow straight tube of length `l` and mass `m` can turn freely about its centre (fixed) on a smooth horizontal table. Another smooth uniform rod of same length and mass is fitted into the tube so that their centres coincide. They system is set in motion with an initial angular velocity `omega_0`. the angular velocity of the rod at an instant when the rod slips out of the tube is :

Options

A. `omega_0//3`
B. `omega_0//2`
C. `omega_0//4`
D. `omega_0//7`

Answer

D

Explanation

(d) Initial angular momentum about `O` is.
`L_i = 2((m l^2)/(12)) omega_0 = (m l^2)/(6) omega_0`
Final angular momentum about `O` is
`L_f = I_omega =[(m l^2)/(12) +((m l^2)/(12) + m l^2)] omega = (7)/(6) m l^2 omega`
Angular momentum will be conserved
`L_i = L_f` or `omega = (omega_0)/(7)`.
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