A wheel of radius `R`, mass `m` with an axle of radius `r` is placed on a horizontal surface. Its moment of inertia is `I=mR^(2)`. Unwinding a rope from its axel a force `F` is applied to pull it along a horizontal surface. The friction is sufficient enough for its pure rolling `(angletheta=0^(@))`

Find the condition for which frictional force acts in backward direction

Options

A. `(I//m)gtRr`
B. `(2I//m)gtRr`
C. `((Isqrt(2))/m)gtRr`
D. `(I/(msqrt(2)))gtRr`

Answer

A

Explanation

`F-f ma, F(r)+f(R)=Ialpha`
`a=Ralpha`
Solving `a=(F(r+R)R)/((R+mR^(2)))` and `f=F-ma`
`=(F[(I//m)-Rr])/([(I//m)+r^(2)])`
`f ` is positive for `(I//m)gtRr` for frictional force acts in backwards direction.