A wheel rolls smoothly along a floor while being accelerating horizontally by a force `vecF_(n)` applied at its top. The wheel has radius R = 0.500 m, mass m = 1.60 kg, and acceleration magnitude `a = 2.00 m//s^(2)`. The frictional force on the wheel is in the forward direction and has magnitude f = 1.80 N. What is the rotational inertia of the wheel?

Explanation

The (linear) acceleration a of a rolling object is the acceleration of the center, which is the center of mass com. (2) Newton.s second law written for x components relates the forces on the wheel to the resulting acceler ation. (3) Newton.s second law written in angular form relates the torques on the wheel to the resulting angu lar acceleration. (4) The linear acceleration and the angular acceleration are related by the radius according to Eq. 11-6 (`a_("com") = alpha R`).
Calculations: Newton.s second law for motion (`F_("net, y")= ma_(x)`) along the x axis gives us
`F_(n)+ f= ma`
`F_(a)= ma-f= (1.60kg)(2.00m//s^(2))-1.80N= 1.40N`
Next, using Newton.s second law for angular motion (`tau_("net") = Ialpha`), with torques and the angular acceleration = measured around the center of the wheel, we have
`-RF+Rf= Ialpha`
On the left side of this equation we have the torque due to the applied force, at distance R from the wheel.s center and perpendicular to that distance. We insert a minus sign because were that torque to act alone, it would cause a clockwise rotation. We also have the torque due to the frictional force. However, this torque acting alone would cause a counterclockwise rotation, and so we use a plus sign. We can relate the angular acceleration to the lin ear acceleration of the center with Eq. 11-6 (`a_("com") = alphaR`). Substituting for `alpha` and solving for I, we find
`I= (R^(2)(-F+f))/(a)= ((0.5)^(2)(-1.40N+ 1.80N))/(2.00m//s^2)= 0.050kg.m^(2)`