Question

A spool with a thread wound on it, of mass $m$, rests on a rough horizontal surface. Its moment Of inertia about its own axis is $\gamma m\omega R^2$,where $R$ is the outer radius of spool and $\gamma$ is a constant. Radius of the wound thread is equal to $r$. The spool is pulled without sliding by the thread with a constant force $F$ directed at an angle $\alpha$. If the acceleration component along x axis is given by $a_x=\frac{F(cos\alpha -\frac{r}{R})}{(b+\gamma )m}$, find the value of $b$.

Answer 1

$Fcos\alpha -F_f=m{a}_x$....(1)
$F_{f}R-Fr=I\alpha =\gamma m{R}^2\alpha$ or $F_f-F\frac{r}{R}=\gamma m{a}_x$......(2)
Adding (1) and (2), we get $a_x=\frac{F(cos\alpha -\frac{r}{R})}{(1+\gamma )m}$
(Note for acceleration to be produced as $cos\alpha >r/R$)