800 mm D -x- A -800 mm
The 4-kg rod AB can slide freely inside the 6-kg tube. The rod was entirely within the tube (x = = 0) and released with no initial velocity relative to the tube when the
Given:
Mass of the rod AB is Mr=4kg
Mass of the tube CDis Ml=6kg
Angular velocity of the assembly ω1=5rad/sec
X=400mm=0.400m
Length of rod=length of tube=Lr=Lt=800mm=0.800 m
The moment of inertia of the rod is given by,
Ir=0.2133gm2
The moment of inertia of the tube is given by,
It=0.32kgm2
Using conservation of angular momentum,
Initial angular momentum(Li)=final angular momentum(Lf)
Li =Lf ….(1)
The Initial angular momentum of the whole body is given by,
Initially X=0m and ω1=5 rad /sec
So
Li=10.66 kg.m2/s
Similarly, we have to find out final angular momentum Lf.
At x=0.400 m
Equation(2) becomes
Plugging values of Li and Lf in equation (1),
10.66=4.05ω2
ω2=2.632 rad /sec
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