If the pulling force `F` is removed from a body rolling on a rough surface, the friction force also disappears, because the point of contact has no more tendency to slip with respect to the surface. The linear and angular speeds of the body remain constant and the body keeps on rolling.
As body is rolling, the friction will be of static nature and `Q`-unknown (both in magnitude and direction) .Let friction act in the backward direction.
Equation of motion `F-f=ma_(c)`....i
here we can apply torque equation about centre of mass as well as about point of contact with ground (instantenous centre of rotation)
About centre of mass : `Fx-fR-I_(c)alpha` ..........ii
About point of contact `O:F(R+r)+I_(c)alpha`
`implies F(R+r)=(I_(c)+MR^(2))alpha`..........iii
Taking torque equation about `O`, we cann directly calculate the value of `alpha`
Hence `alpha=(F(R+r))/((I_(c)+MR^(2)))`........iv
As the body is rolling `a_(c)=alphaR=(F(R+r)R)/((I_(c)+MR^(2)))`
or `a_(c)=F/M[((1+r/R))/((1+I_(C))/(MR^2))]` .............v
Substituting value of `a_(c)` in eqn i we get
force of friction `f=F[(I_(C)-MrR)/(I_(c)+MR^(2))]`...........vi
a. If force acts through `C` figure. i.e, `r=0`
From eqn v we get
`a_(C)=F/M[1/(1+I_(C)/(MR^(2)))]`
From eqn vi we get
`f=F[1/(1+(MR^2)/I_(C))]`
From the values of `a_(c)` and , it is clear that
body moves forward
friction acts backward
b. Force acts above `C` i.e, `r` is positive
From eqn v and vi it is clear that
the body moves forward
rotation about the centre of mass is clockwise
the friction force may act forward or backward
Backward if `rlt(I_(c))/(MR)`
Forward if `rgt(I_(c))/(MR)`
No friction force acts when `r=(I_(C))/(MR)`
c. Force acts below `C` i.e, `r` is negative
Frictioin force acts backward
`r` is negative
the body moves forward.
Rotation about the centre of mass is clockwise.
Friction force acts backward `(fltF)`.
