Answer:
α = F/(k×m×r)
Explanation:
When the wheel is pulled to turn in a counterclockwise direction, the wheel will have a moment of inertia given by Iw = k×m×r²
Where k = the radius of gyration of the wheel which is a dimensionless quantity less than one.
m = the mass of the wheel
r = the radius of the wheel
First and foremost, we relate the torque (τ) about the axle of the wheel to the force (F) applied on the wheel and we have that τ = r × F
We then relate the torque on the wheel to the angular acceleration (α), we have that τ = Iw × α, where Iw is the moment of inertia of the wheel as explained above
Substituting for torque τ and moment of inertia I into the above equation we have that
r × F = k×m×r² × α
solving for α we have that
α = r × F /(k×m×r²)
Therefore
α = F/(k×m×r)
