Answer
The ratio of its translational kinetic energy to its rotational kinetic energy about the central axis parallel to its length is $~~1$
Work Step by Step
We can write an expression for the rotational inertia of a thin-walled pipe:
$I = Mr^2$
We can find an expression for the rotational kinetic energy of the pipe when it has a speed $v$:
$K_{rot} = \frac{1}{2}I\omega^2$
$K_{rot} = \frac{1}{2}(Mr^2)(\frac{v}{r})^2$
$K_{rot} = \frac{1}{2}Mv^2$
The translational kinetic energy of the pipe when it has a speed $v$ is $\frac{1}{2}Mv^2$
We can see that the ratio of its translational kinetic energy to its rotational kinetic energy about the central axis parallel to its length is $~~1$.
