= Solution
Let $s$ increase down the ramp. On a plane, <static friction> $F$ acts uphill. Translation and rotation give
$$
m\ddot s=mg\sin\theta-F,\qquad
Fa=I\dot\omega,\qquad \dot\omega=\ddot s/a.
$$
With $I=2ma^2/5$, elimination yields \b[$\ddot s=(5/7)g\sin\theta$]. <Static friction> does no work at the stationary contact point.
For a <sphere rolling in a symmetric groove>, the distance from the centre to the instantaneous contact axis AB is $h=a\sin(\phi/2)$, so $v=h\omega$. The <kinetic energy> and work-energy equation are
$$
K=\frac12\left(m+\frac I{h^2}\right)\dot s^2,\qquad
\left(m+\frac I{h^2}\right)\ddot s=mg\sin\theta.
$$
Hence
$$
\boxed{\ddot s=\frac{5g\sin\theta}{5+2/\sin^2(\phi/2)}.}
$$
For the stated angles this is $5g/18$. Using $g=9.81\,\mathrm{m\,s^{-2}}$ and a one-second descent,
$$
\boxed{v_g=2.725\,\mathrm{m\,s^{-1}},\qquad
K_g=\frac12mv_g^2\left(1+\frac{2}{5\sin^2 45^\circ}\right)=33.4\,\mathrm J.}
$$
On the level floor the rolling <radius> becomes $a$, so an <energy-preserving change of rolling radius> gives $K_g=\tfrac7{10}mv_f^2$. Therefore
$$
v_f=\sqrt{\frac97}\,v_g=3.090\,\mathrm{m\,s^{-1}},\qquad
\boxed{t_{\rm pins}=\frac{20\,\mathrm m}{v_f}=6.47\,\mathrm s.}
$$
The floor <speed> exceeds the groove <speed> because less of the same total <kinetic energy> is rotational. Holding the translational <speed> fixed at the transition would violate the stipulated <energy> balance with the new no-slip constraint.
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