Show your capability!???Very easy problem.

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Answer 1 · 2017-08-26T08:54:54

The moment of inertia of the rod about the perpendicular axis passing through its mid point
$I_(rod)=1/12xxM_(rod)xxL^2$ ,
where mass of the rod $M_(rod)=0.75kg$
and length of the rod $L=0.4m$
So $I_(rod)=1/12xx0.75xx0.4^2=0.01kgm^2$
Now the moment of inertia of each ring of mass $m=1kg$ at initial position $r_1=0.1m$ from center of rotation $O$
$I_1=mr_1^2=1xx0.1^2=0.01kgm^2$
And the moment of inertia of each ring of mass $m=1kg$ at final position $r_2=0.2m$ from center of rotation $O$
$I_2=mr_2^2=1xx0.2^2=0.04kgm^2$
Given that the initial angular velocity of the system for initial position of the ring be $omega_1=30$ rad/s
Let the angular velocity of the system in final position of the ring be $omega_2$
Hence applying conservation of angular momentum we can write
$(I_(rod)+2I_2)omega_2=(I_(rod)+2I_1)omega_1$

$=>omega_2=((I_(rod)+2I_1)omega_1)/(I_(rod)+2I_2)$

$=>omega_2=((0.01+2xx0.01)xx30)/(0.01+2xx0.4)$ rad/s

$=>omega_2=10$ rad/s

Answer 2 · 2017-08-26T11:08:07

See below.

The kinetic energy is conserved so

$E_1=1/2J_1 omega_1^2=E_2=1/2J_2omega_2^2$

Here

$J_i =1/2(J_r+2r_i^2m)$ with

$r_1 = 0.10$
$r_2=0.20$
$omega_1 = 30$
$m = 1$ and

$J_r=(m_r L^2)/12 = (0.75 xx 0.40^2)/12$

so

$omega_2 = (sqrt(J_1/J_2))omega_1 = (sqrt((J_r+2r_1^2m)/(J_r+2r_2^2m)))omega_1 = 17.3205$ [rad/sec]