Question

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):

Answer 1

Draw a diagram of given problem.

Calculate ratio of distance covered by rotational as well as translational motion.
According to diagram, angular momentum is conserved about $A$
$L_1=\frac{2}{5}m{R}^2{\omega }_0$
$L_2=\frac{2}{5}m{R}^2\omega +m{R}^2\omega$
$=\frac{7}{5}m{R}^2\omega$
$L_1=L_2$

Therefore,
$\omega =\frac{2}{7}{\omega }_0$
${\omega }_0>\omega$
$V>V_0$

As sphere rotates and as well as translates, it means sphere accelerates forward and rotation decelerates. Therefore, calculating angular momentum

$\omega ={\omega }_0-\alpha t$

$\alpha =\frac{{\omega }_0-\omega }{t}$

${\omega }^2={\omega }_0^2-2a\theta (\text{Using}{v}^2=u^2+2as)$

$\theta =\frac{({\omega }_0-\omega )(\omega +{\omega }_0)}{2\alpha }=\frac{\alpha t(\omega +\frac{7}{2}\omega )}{2\alpha }=\frac{9}{4}\omega t$

$R\theta =\frac{9}{4}\omega Rt$

$V=V_0+at(V_0=0)$

$a=\omega R/t$

$V^2=V_0^2+2aS$

$S=\frac{V\cdot V}{2a}=\frac{\left(\omega R\right)(V\cdot t)}{2\cdot \omega R}=\frac{Vt}{2}$

Hence, ratio of distance is $9/2$ i.e., 4.5.