Question

 A certain type of propeller blade can be modeled as a thin uniform bar 2.60 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. You may want to review (Page).

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 Part A

 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?

Answer 1

here,

length of bar , l = 2.6 m

mass , m = 24 kg

the frictional torque , Tff = 5 N.m

applied force , F = 40 N

r = 1 m

let the angular acceleration be alpha

equating the torques

I * alpha = T

m * l^2 /12 * alpha = ( F * r - Tff)

24 * 2.6^2 /12 * alpha = ( 40 * 1 - 5)

solving for alpha

alpha = 2.59 rad/s^2

the maximum angular acceleration is 2.59 rad/s^2