Determine the angular acceleration of the motion:
A flywheel of mass 15 kg and radius 20 cm is wound by a rope that carries a weight A of mass 5 kg at its free end as illustrated in Figure. Determine the angular acceleration of the motion, supposing gravitational acceleration = 10 m/sec^{2} for purpose of simplified calculations, and letting the following two cases:
(i) If friction at the bearing of flywheel-shaft is zero.
(ii) If frictional couple developed at the bearing of shaft is 2 N-m.
Solution
While block A travels a distance s downward, this has certain linear velocity V and acceleration a. The equivalent angular velocity ω and angular acceleration α of the flywheel of radius (r = 0.2 m) are specified by
V = r ω = 0.2 ω
a = r α = 0.2 α
s = r θ
Here, θ is the angular displacement of the flywheel. Likewise to equation in linear motion given by
V ^{2} = 2 a s
The equation in angular velocity is
ω^{2} = 2 α θ
By using principle of conservation of energy letting datum level as A′ which is the final position of A, we have
W_{A} × s = W V ^{2 } /2g + I_{m } ω^{2} /2+ C × θ
(i) While frictional couple C = 0;
5 × 10 × (0.2 θ) = 5 (0.2)^{2}/2 ω^{2} + 15 (0.2)^{2} /2 × ω^{2}/2 + 0
10 ×θ = ω^{2} × (0.2) ^{2} /2[5 + 7.5]
10 ×θ = α θ × (0.2)^{2 }× 12.5
∴ α = 20 rad / sec ^{2.}
(ii) 50 × 0.2 θ = α θ (0.2)^{2} × 12.5 + 2 × θ
∴ α = 16 rad / sec^{2.}