= Solution
For <rolling without slipping> through height $h$, <conservation of energy> gives
$$
mgh=\tfrac12mv^2+\tfrac12I(v/R)^2,\qquad
\boxed{v^2=\frac{2gh}{1+I/(mR^2)}.}
$$
A uniform solid sphere has $I/(mR^2)=2/5$, a solid cylinder $1/2$, and a pipe $(1+r_{\rm in}^2/R^2)/2>1/2$. Thus \b[the ball is fastest, then the solid cylinder, then the pipe]. A smaller <moment of inertia> leaves more <energy> in translation. This assumes release from rest through the same height, adequate <static friction> and negligible rolling resistance; a thin pipe has $v^2=gh$.
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