Alarge grinding wheel in the shape of a solid cylinder of radius 0.330 m is free to rotate on a frictionless, vertical axle. a constant tangential force of 260 n applied to its edge causes the wheel to have an angular acceleration of 0.810 rad/s2. (a) what is the moment of inertia of the wheel?

Refer to the diagram shown below.

Let I = the moment of inertia of the wheel.
α = 0.81 rad/s², the angular acceleration
r = 0.33 m, the radius of the weel
F = 260 N, the applied tangential force

The applied torque is
T = F*r
   = (260 N)*(0.33 m)
   = 85.8 N-m

By definition,
T = I*α

Therefore,
I = T/α
  = (85.8 N-m)/(0.81 rad/s²)
  = 105.93 kg-m²

105.93 kg-m²

Ithink the answer would be a 

The answer is periosteum