A constant horizontal force of `10 N` is applied at the centre of a wheel of mass `10 kg` and radius `0.30 m`. The wheel rolls without slipping on the horizontal surface, and the acceleration of the centre of mass is `0.60 m//s^(2)`.
a. What are the magnitude and direction of the frictional force on the wheel?
b. What is the rotational inertia of the wheel about an axis through its centre of mass and perpendicular to the plane of the wheel?

Explanation

a. The direction of the frictional force will be in backwark direction of motion of the wheel.
`10-f=ma`
Substituting `m=10kg` and `a=0.60m//s^(2)` in the above equation, we get
`F_(r)=4N`
b. Equation of motion for rotation of wheel
`f_(r)xxR=Ialpha`
which gives
`I=f_(r)xxR/alpha(4xx0.30)/((0.60/0.30))`
or `I=0.6kgm^(2)`