Answer:
Mass moment of inertia of the flywheel = 140.17 kg.m²
Step-by-step explanation:
Consider the moment equation about the center of the flywheel
∑ M(g) = ∑ [m(g)(effective)]
-m(a) g r + τ = -I’ α – m(A) a(A) r
m(a)gr - τ = I’ a(A)/r + m(A) a(A) r, where m(a)g is weight of the block, r is the radius of the flywheel, α is the angular acceleration, I’α is the coupled, τ is the torsional moment of flywheel, and m(A) a(A) is force acting on block A.
Substitute 15 kg for m(A), 9.81,/s² for g and 0.6m for r
15 x 9.81 x 0.6 – τ = -I’(a(A)/0.6) + 15 x a(A) x 0.6
88.29 – τ = 1.667I’a(A) + 9a(A), equation 1
For the first case, a block of 15 kg weight falls 3 m from rest in 4.6 s.
Determine the acceleration of the block
S = ut + ½ a(A)t²
Substitute 3 m for s, 0 for u and 4.6 s for t
3 = 0 x 4.6 + ½ a(A) x (4.6)²
a(A) = 0.2836m/s²
Substitute in equation 1
88.29 – τ = 1.667I’ x 0.2836 + 9 x 0.2836
τ = 85.7376 - 0.4728I’
For the second case, a block of 30 kg weight falls 3 m in 3.1 s
Substitute 30 kg for m(A), 9.81 m/s² for g, and 0.6 m for r in equation 1
30 x 9.81 x 0.6 – τ = I(a(A)/0.6) + 30 x a(A) x 0.6
176.58 – τ = 1.667I’a(A) + 18a(A), equation 2
Determine acceleration of block
S = ut + ½ a(A)t²
Substitute 3 m for s, 0 for I and 3.1 s for t
3 = 0 x 3.1 + ½ a(A) (3.1)²
a(A) = 0.6243m/s²
Substitute 0.6243m/s² for a(A) in equation 2
176.58 – τ = 1.677I’ x 0.6243 + 18 x 0.6243
τ = 165.3426 - 1.0407I’
Solve both equations for I’
165.3426 – 1.0407I’ = 85.7376 – 0.4728I’
0.56791I’ = 79.605
I’ = 140.1718582 = 140.17 kg.m²