Answer
$3.75m/s$
Work Step by Step
We can find the required initial speed as follows:
$L_i=L_f$
$\implies mvr=(I+mr^2)\omega_f$
This can be rearranged as:
$v=\frac{(I+mr^2)\omega_f}{mr}$
We plug in the known values to obtain:
$v=\frac{512Kg.m^2+(34.0)(2.31m)^2}{(34.0Kg)(2.31m)}(0.425rad/s)$
$v=3.75m/s$