Q.The ratio of the acceleration for a solid sphere (mass $m$ and radius $R$) rolling down an incline of angle $\theta$ without slipping, and slipping down the incline without rolling, is,

For a round object of mass $m$ and radius $R$ rolling without slipping down an incline of angle $\theta$, two forces act along the slope: the driving component of gravity, $mg\sin\theta$, and an opposing static frictional force $f$ (the perpendicular component $mg\cos\theta$ is simply balanced by the normal force $N$).

Setting up the two equations of motion. For translation along the incline: $mg\sin\theta-f=ma$. For rotation about the object's own center (about which $mg\sin\theta$ produces no torque, since it passes through the center; only friction does, via $\tau=Rf$): $Rf=I\alpha$. Using $a=R\alpha$ and $I=mK^2$ (radius of gyration $K$), the friction force can be eliminated between the two equations, giving

Speed at the bottom. Using $v^2=u^2+2as$ with $u=0$ and incline length $s=h/\sin\theta$ (for vertical drop $h$):

Time to reach the bottom. Using $v=u+at$ with $u=0$:

$t=\sqrt{\dfrac{2h\!\left(1+K^2/R^2\right)}{g\sin^2\theta}}.$ …