Two masses 400 g and 350 g are suspended from the ends of a light string passing over a heavy pulley of radius 2 cm. When released from rest the heavier mass is observed to fall 81 cm in 9 s. The rotational inertia of the pulley i $s _______$ kg·m².
(g = 9.8 m/s²)

Question

Two masses 400 g and 350 g are suspended from the ends of a light string passing over a heavy pulley of radius 2 cm. When released from rest the heavier mass is observed to fall 81 cm in 9 s. The rotational inertia of the pulley i $s _______$ kg·m².

(g = 9.8 m/s²)

G1 Physics · Accepted Answer

The heavier mass falls from rest through s = 81 cm = 0.81 m in t = 9 s. Using s = ½at², we get a = 2s/t² = 2×0.81/81 = 0.02 m/s². Let m₁ = 0.400 kg, m₂ = 0.350 kg, R = 0.020 m. For an Atwood machine with a pulley of moment of inertia I, a = (m₁-m₂)g/ [m₁+m₂+I/R²]. Therefore, 0.02 = (0.400-0.350)(9.8)/ [0.400+0.350+I/(0.020)²]. The numerator is 0.49. Hence the effective denominator is 0.49/0.02 = 24.5. Therefore, 0.75 + I/R² = 24.5. Thus, I/R² = 23.75. Hence, I = 23.75(0.020)² = 23.75×4×10⁻⁴ = 9…

Two masses 400 g and 350 g are suspended from the ends of a light string passing over a heavy pulley of radius 2 cm. When released from rest the heavier mass is observed to fall 81 cm in 9 s. The rotational inertia of the pulley i s _______ kg·m². (g = 9.8 m/s²)