the moment of inertia of wheel= 1/2mr²
m= mass of wheel= 2.8 kg
r= radius
i= 1/2 (2.8)(0.35)²=0.1715 kgm²
(a) torque = F r
F= force=5 N
r= radius=0.35 m
τ= 5 (0.35)
τ= 1.75 Nm
(b) for angular acceleration, we use
τ= iα
τ= torque= 1.75 Nm
i= moment of inertia=0.1715 kgm²
so 1.75 =0.1715 α
α=10.2 rad/s²
(c) for angular displacement θ, we use the kinematic equation,
θ= ωo t + 1/2α t²
ωo= initial angular velocity=0
t= time=4 s
α= angular acceleration=10.2 rad/s²
so θ = 0 (4) + 1/2 (10.2)(4)²
θ=81.6 rad
