Answer:
(a) I = 87.766 kg.m²
(b) m = 1611.864 kg
(c) ω = 4.7 rad/s
Explanation:
Given;
radius of the wheel, r = 0.330 m
applied force, F = 250 N
angular acceleration, α = 0.940 rad/s²
<u>Part (a)</u> the moment of inertia of the wheel
τ = F x r = Iα
I = (F x r) / α
where;
I is the moment of inertia
I = (250 x 0.33) / 0.94
I = 87.766 kg.m²
<u>Part (b)</u> the mass of the wheel
I = ¹/₂mr²
2I = mr²
m = (2I) / r²
where;
m is the mass of the wheel
m = (2 x 87.766) / (0.33²)
m = 1611.864 kg
<u>Part (c) </u>angular velocity after 5.00 s
ω = ω₀ + αt
where;
ω is the angular velocity after 5.00 s
ω₀ is the initial angular velocity = 0
ω = 0 + (0.94 x 5)
ω = 0 + 4.7
ω = 4.7 rad/s