Answer
(a) $1.04\space rev/s$
(b) $0.79\space rev/s$
Work Step by Step
Please see the attached image first.
In this case, there is no external torque. So its angular momentum is conserved.
We can write,
(a) $I_{1}\omega_{1}=I_{2}\omega_{2}$ ; Let's plug known values into this equation
$5.31\space kgm^{2}\times 0.95\times 2\pi\space s^{-1}+24.7\space m/s \times 146\times 10^{-3}kg\times1.23\space m=(5.31+0.146\times1.23^{2})kgm^{2}\space \omega_{2}$
$31.7\space kgm^{2}/s+4.4kgm^{2}/s=5.53\space kgm^{2}\space \omega_{2}$
$6.53\space rad/s=\omega_{2}$
$1.04\space rev/s=\omega_{2}$
(b) $I_{1}\omega_{1}=I_{2}\omega_{2}$ ; Let's plug known values into this equation
$5.31\space kgm^{2}\times 0.95\times 2\pi\space s^{-1}-24.7\space m/s \times 146\times 10^{-3}kg\times1.23\space m=(5.31+0.146\times1.23^{2})kgm^{2}\space \omega_{3}$
$31.7\space kgm^{2}/s-4.4kgm^{2}/s=5.53\space kgm^{2}\space \omega_{3}$
$4.94\space rad/s=\omega_{3}$
$0.79\space rev/s=\omega_{3}$
