A solid sphere is set into rotation at an angular velocity and it is then placed on a rough horizontal surface. The ratio of distances covered by rotational and translational motions up to the start of the pure rolling is (Assume uniformly accelerated motion up to start of pure rolling):

Angular momentum about A is conserved.

$\begin{array}{l}{L}_1=\frac{2}{5}m{R}^2{\omega }_0\\ L_2=\frac{2}{5}m{R}^2\omega +m{R}^2\omega =\frac{7}{5}m{R}^2\omega \\ L_1=L_2\\ \omega =\frac{2}{7}{\omega }_0\\ {\omega }_0>\omega \end{array}$

$\text{ and }v>v_0$

Sphere accelerates forward and rotation decelerates.

$\begin{array}{l}\omega ={\omega }_0-\alpha t\\ \alpha =\frac{{\omega }_0-\omega }{t} .......\left(i\right)\\ {\omega }^2={\omega }_0^2-2\alpha \theta \\ \theta =\frac{\left({\omega }_0-\omega \right)\left(\omega +{\omega }_0\right)}{2\alpha }=\frac{\alpha t\cdot \left(\omega +\frac{7}{2}\omega \right)}{2\alpha }=\frac{9}{4}\omega t\\ R\theta =\frac{9}{4}\omega Rt ....\left(ii\right)\\ v=v_0+at\left(v_0=0\right)\\ a=\omega R/t\\ v^2=v_0^2+2aS\\ S=\frac{v\cdot v}{2a}=\frac{\left(\omega R\right)(v\cdot t)}{2\cdot \omega R}=\frac{vt}{2} .......\left(iii\right)\\ \text{ Ratio of distance }=9/2\end{array}$