Answer
a) $\omega d(\frac{1}{2}-\frac{I}{2md^2})$
b) $\omega d$
c) $\omega d(2+\frac{I}{md^2})$
Work Step by Step
a) We find using conservation of angular momentum that:
$mvd+I\omega=(md^2+I)(\frac{1}{2}\omega)$
$mvd=\frac{1}{2}md^2\omega-\frac{1}{2}I\omega$
$v=\frac{1}{2}\omega d-\frac{I\omega}{2md}$
$v=\omega d(\frac{1}{2}-\frac{I}{2md^2})$
b) $mvd+I\omega=(md^2+I)(\omega)$
$mvd=md^2\omega$
$v= \omega d$
c) We use a similar process to find:
$mvd+I\omega=(md^2+I)(2\omega)$
$mvd=2md^2\omega +I\omega$
$v=2\omega d+\frac{I\omega}{md}$
$v= \omega d(2+\frac{I}{md^2})$
