A 150-mm-diameter shaft with a mass of 20 kg is rotating at 900 rpm. A pulley mounted on the shaft has a mass moment of inertia of 0.15 kg ⋅ m2. If the shaft and the pulley coast to a stop due to a tangential frictional force of 8lb at the outer radius of the shaft, determine the time required.
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First, we need to convert the given units to SI units for consistency.
The diameter of the shaft is 150 mm, which is 0.15 m in SI units.
The rotational speed of the shaft is 900 rpm, which is 900 * 2π / 60 = 94.25 rad/s in SI units.
The tangential frictional force is 8 lb, which is 8 * 4.44822 = 35.58 N in SI units.
The moment of inertia of the system (shaft + pulley) is the sum of the moment of inertia of the shaft and the pulley. The moment of inertia of the shaft can be calculated using the formula for the moment of inertia of a cylinder, which is 0.5 * mass * radius^2.
The moment of inertia of the shaft is 0.5 * 20 kg * (0.15 m / 2)^2 = 0.28125 kg*m^2.
The total moment of inertia of the system is 0.28125 kg*m^2 + 0.15 kg*m^2 = 0.43125 kg*m^2.
The angular deceleration due to the frictional force can be calculated using the formula torque = moment of inertia * angular acceleration, or angular acceleration = torque / moment of inertia. The torque due to the frictional force is force * radius = 35.58 N * (0.15 m / 2) = 2.6685 N*m.
The angular deceleration is 2.6685 N*m / 0.43125 kg*m^2 = 6.18 rad/s^2.
The time required for the shaft and the pulley to coast to a stop can be calculated using the formula final angular velocity = initial angular velocity - angular acceleration * time, or time = (final angular velocity - initial angular velocity) / angular acceleration. Since the final angular velocity is 0, the time required is 94.25 rad/s / 6.18 rad/s^2 = 15.25 s.
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