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A string is wrapped around a pulley with a radius of 2.0 cm and no appreciable friction in its axle. The pulley is initially not turning. A constant force of 50 N is applied to the string, which does not slip, causing the pulley to rotate and the string to unwind. If the string unwinds 1.2 m in 4.9 s, what is the moment of inertia of the pulley?

Respuesta :

Answer:

2

Explanation:

2

asifejaz20

Answer:

Moment of inertia = I = 0.2 kg.m^2

Explanation:

According to equation of motion:

s = ½ at^2

a = 2s/t^2

a = (2 × 1.2)/4.9^2 = 0.09996 m/s^2

The Linear and angular acceleration will be:

a =rα

α = a/r = 0.09996/0.02 = 4.9979 rad/s^2

Torque = T = rF = (0.02)(50) = 1 N

Now,

Moment of inertia = I = T/ α = 1/(4.9979) = 0.2 kg.m^2

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