✔ 最佳答案
moment of inertia of wheel:
I = 2*ML²/12 + M(L/2)² = ML²(1/6 + 1/4) = 5ML²/12 = (5/3)M(L/2)²
shortcut:
motive force F = M*g
equilvalent mass m = M + (5/3)M = (8/3)M
acceleration a = F / m = g / (8/3) = 3g/8 ◄ acceleration
tension T = M(g - a) = M(g - 3g/8) = 5Mg / 8 ◄ tension
long cut:
tension T = M(g - a)
and also T = torque / radius = τ / (L/2)
but τ = I*α = I*a/r = (5/3)M(L/2)² * a / (L/2) = 5MLa / 6, so
T = M(g - a) = (5MLa/6) / (L/2) → mass M cancels
g - a = 5a / 3
g = 8a/3
a = 3g/8 ◄
and T = M(g - 3g/8) = 5Mg / 8 ◄
as before.
Hope this helps!