A spool of thread of mass `m` and moment of inertia `I` is attached to the block of mass `M`. Along a horizontal line passing through the centre of the spool, a constant force `F` is applied as shown. The surface under the block is smooth while beneath the spool is rough.
Show that if `M=(I)/(r(R-r))`, no frictional force will act on the spool. Now suppose `M=(2I)/(r(R-r))`, find in which direction will the frictional force act? Also find in this case the rate of increase of length `'ab'` of the thread at time `t` after the application of force. (No slipping occurs.)

Explanation

For translational motion
`F-T-f=ma_(c.m.)`……(`1`)
For rotational motion
`(Tr+fR)=Ialpha`……(`2`)
As per given condition if frictional force is zero then,
`rArralpha=(a_(c.m.))/(R )`
Hence, `Tr+fR=I(a_(c.m.))/(R )`
`rArrT=(Ia_(c.m.))/(Rr)-f.(R )/(r )`....(`3`)
`rArr T=Ma'`
Since string `ab` remains tangent, hence acceleration of point `a` and that of point `b` is same.
`rArr a'=(a_(c.m.))/(R )(R-r)`
`rArrT=M(a_(c.m.))/(R )(R-r)`......(`4`)
From Eqs. (`3`) and (`4`) , we get
`M(a_(c.m.))/(R)(R-r)=I(a_(c.m.))/(R )(R-r)=I(a_(c.m.))/(R.r)-f.(R )/(r )`.....(`5`)
Putting `f=0` in Eq. (`5`)
`M=(I)/(r(R-r))`
Putting the value of `M=(2I)/(r(R-r))` in Eq.(`5`), we get
`f=(-Ia_(c.m.))/(R^(2))`.......(`6`)
Negative sign indicates that friction is acting in the opposite direction of our assumption i.e. in forward direction.
From Eqs. (`1`), (`3`) and (`6`), we get
`F-(2Ia_(c.m.))/(Rr)+(Ia_(c.m.))/(R^(2))=ma_(c.m.)`
`rArr a_(c.m.)=(F)/((m+(2I)/(Rr)-(I)/(R^(2))))rArralpha=(F)/(R(m+(2I)/(Rr)+(I)/(R^(2))))`
`:. T=ma'`
`(d^(2)(ab))/(dt^(2))=alpha(R-r)-a'=(F(R-r))/(R(m+(2I)/(Rr)-(I)/(R^(2))))-(2I_(cm))/(mRr)`
`=(F(R-r))/(R(m+(2I)/(Rr)-(I)/(R^(2))))-(2I)/(mRr)(F)/((m+(2I)/(Rr)-(I)/(R^(2))))`
`=(F)/(R(m+(2I)/(Rr)-(I)/(R^(2))))[R-r-(2I)/(mr)]`
`:. ` Rate of increase in length `ab=(=v_(ab))=(F)/(R(m+(2I)/(Rr)-(I)/(R^(2))))[R-r-(2I)/(mr)]`