A uniform solid cylinder of mass `m` rests on two horizontal planks as shown in Fig. A thread is wound on the cylinder. The hanging end of the thread is pulled vertically up with a force `F`. What is the maximum magnitude of the force `F` which still does not bring about any sliding of the cylinder, if the coefficient of friction between the cylinder and the planks is equal to `mu`. What is the maximum acceleration of the axis of the cylinder?

Explanation

As the force `F` is applied the cylinder tends to rotate along anticlockwise direction with the point of contact `C` tending to move to the right. Hence, friction acts along the left (as shown in figure) If `alpha` is the angular acceleration of the cylinder then from `tau=Ialpha` we have

`r(F-r)=(mr^(2)//2)alpha`...i
If `a` is the linear acceleratioin of the cylinder's `CM`, then from `F =ma` we have
`f=ma`........ii
For pure rolling,
`a=alphar`............iii
Dividing Eq i by eqn ii we get
`((F-f))/f=(ralpha)/(2a)=1/2`
`impliesF=(3f/2)impliesF_("max")=(3/2)f_("max")`
`[f_("max")=muNimpliesmu(mg-F)]`
`implies F_("max")=(3mu)/2[m-=LF_("max")]=(3mumg)/(2+3mu)`
from eq ii
`a=f/m`
`implies a_("max")=f_("max")/m=2/(3m)F_("max")=2/(3m)((3mumg)/(2+3mu))=(2mug)/(2+3mu)`