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 is kg m2. (g = 9.8 m/s2)
- 8.3 10-3
- 4.75 10-3
- 1.86 10-2
- 9.5 10-3
Answer
(4) 9.5 10-3
Solution
Starting from rest, s = 12at2, so
0.81 = 12 a 81
a = 0.02 m/s2
For the heavier mass m1g - T1 = m1a, for the lighter mass T2 - m2g = m2a, and for the pulley (T1 - T2)r = I = I ar. Adding these,
(m1 - m2)g = (m1 + m2 + Ir2 )a
Ir2 = (m1 - m2)ga - (m1 + m2) = 0.05 9.80.02 - 0.75 = 24.5 - 0.75 = 23.75 kg
I = 23.75 (0.02)2 = 9.5 10-3 kg m2
Answer: (4)