A wheel of radius `R`, mass `m` and moment of inertia `I` is pulled along a horizontal surface by application of force `F` to as rope unwinding from the axel of radius, `r` as shown in figure. Friction is sufficient for pure rolling of the wheel.

a. What is the linear acceleration of the wheel?
b. Calculate the frictional force that acts on the wheel.

Explanation

The equations of the motion of the wheel are
`F=f=ma`……….i
`Fxxr+fxxR=Ialpha`…….ii
in case of pure rolling `a=Ralpha`………iii

`Fxxr+(F-ma)R=I/Ra`
or `a=(F(r+R)R)/((I+mR^(2)))`
We can find acceleration of centre of wheel by another approach also. The point of contact will be instantaneous centre of rotation (`ICR`), and we apply torque equation about `ICR`.
`F(R+r)=(I+mR^(2))alphaimplies alpha=(F(R+r))/((I+mR^(2)))`
Acceleration of centre of mass `a=alphaR=(F(R+r)R)/((I+mR^(2)))`
and the frictional force is
`f=F-ma=F[1-((r+R)R)/((I/m+R^(2)))]=(F[(I/m)-Rr])/([(I/m)+R^(2)])`
For `I/m=Rr` Frictional force is zero.
For `I/mgtRr,` frictional force positive i.e, it acts in backward direction.
For `I/mltRr,` frictional force is negative, i.e. it acts in forward direction.