As airplane propeller has a moment of inertia (about its center) equal to 58 kg

Question

As airplane propeller has a moment of inertia (about its center) equal to 58 kg m^2. It spins at an initial rate of 3.30rpm. The airplane's engine then uniformly accelerates the propeller up to a find rate of 1450 rpm. The entire acceleration takes 30 s. Show your work throughout. What is the propeller's angular acceleration? If the propeller measure 1.4 m from center to tip, what is the total linear distance traveled by one tip of the propeller during the acceleration? Suppose that friction A air drag always opposes ihe spinning propeller with a constant torque of 75 N m. regardless of the propeller's speed. (In real life, the force of friction & air drag is NOT actually at all angular speeds.) In that case, what must he the engine's torque acting on the propeller to create the acceleration described here?

Explanation / Answer

A) Angular accel. = 2(1450 - 330)/(60)(30)= 3.9 rad/s²

B) angular displ. = avg angular velocity x time
avg angular velocity = 330 + 1120/2 = 890 rpm
avg angular veloicty = 890 x 2/60 = 93.2 rad/s
angular displ. = (93.2)(30) = 2796 rad

linear displacement equivalent to angluar displacement of 2796 rad
S = (2796)(1.4) = 3914.4 m

C) Net torque = I = (58)(3.9) = 226.2 N-m
If frictional torque = 75 N-m
then engine torque = 75 + 226.2 = 301.2 N-m

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