Question

A horizontal F is applied on a spherical shell of radius R at distance R/2 above center as shown in figure. Shell is doing pure rolling motion and friction acting on it is $f=\frac{xF}{y}$ then, y-x is

Answer 1

$F-f=maF\frac{R}{2}+fR=\frac{2}{3}m{R}^2\theta a=R\theta$
$\therefore \frac{F}{2}+f=\frac{2}{3}ma..\left(1\right)F-f=ma...\left(2\right)onsolving\left(1\right)and\left(2\right)\frac{3}{2}F=\frac{5}{3}ma\Rightarrow ma=\frac{9F}{10}putting\left(3\right)valueinto\left(2\right)F-\frac{9F}{10}=ff=\frac{F}{10}$
given, $f=\frac{xF}{y}$
so $\frac{F}{10}=\frac{xF}{y}\Rightarrow y=10x$
Hence,
$y-x=10-1=9$