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A certain type of propeller blade can be modeled as a thin uniform bar 2.50 m long and of mass 24.0 kg. The blade rotates on an axle that is perpendicular to it and through its center. However, the axle does have friction. . If the friction produces a torque of 5 N·m, what maximum angular acceleration can the blad have if a technician pulls down on the blade with a 40.0 N force at a point that is 1 m from the axle?

User Rweisse
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Final answer:

The maximum angular acceleration of the propeller blade is 2.8 rad/s² when a technician applies a 40.0 N force, 1 m from the axle, overcoming the frictional torque of 5 N·m.

Step-by-step explanation:

The question is asking to determine the maximum angular acceleration that a propeller blade can have under certain conditions. The blade has a length of 2.50 m with a mass of 24.0 kg and experiences a frictional torque of 5 N·m at its axle. Additionally, a technician applies a 40.0 N force at a point 1 m away from the axle.

To calculate the angular acceleration, we must first find the net torque (Τnet) acting on the blade, which is the sum of the torque from the technician's force and the frictional torque. The formula for torque (Τ) due to a force (F) is Τ = F × r, where r is the distance from the axis of rotation to where the force is applied. Subsequently, we can calculate the moment of inertia (I) of the blade with the formula I = (1/12) × m × L2 for a thin uniform bar, where m is the mass and L is the length of the bar. Finally, the angular acceleration (α) can be found using α = Τnet / I according to Newton's second law for rotation.

The technician's torque is calculated as Τtech = 40.0 N × 1 m = 40.0 N·m. The net torque is thus Τnet = Τtech - friction torque = 40.0 N·m - 5 N·m = 35.0 N·m.

The moment of inertia for the propeller blade is I = (1/12) × 24.0 kg × (2.50 m)2 = 12.5 kg·m2.

Utilizing the equation for angular acceleration, we find α = Τnet / I = 35.0 N·m / 12.5 kg·m2 = 2.8 rad/s2.

Therefore, the maximum angular acceleration the blade can have is 2.8 rad/s2.

User Loeschg
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