A large commercial wind turbine is rotating at a maximum angular velocity of 22

Question

A large commercial wind turbine is rotating at a maximum angular velocity of 22
rpm. Unfortunately, the wind turbine requires immediate maintenance and must be
stopped by a process called furling which applies an opposing torque using a motor
within the turbine. If the magnitude of the perpendicular force applied to the blades
to produce the opposing torque, is equivalent to the force due to centripetal
acceleration of the spinning blades, what is the magnitude and direction of the
opposing torque that needs to be applied? The mass of the blade and hub of the
turbine is 22000 kg and the blades have a diameter of 40 m.

Explanation / Answer

Given Mass of the blade and hub of the turbine, m = 22000 kg Radius of the blade, r = 20 m Angular velocity of the turbine , = 22 rpm = 22 *(2 rad /60s) =2.30 rad /s Opposing torque, = r F sin90 Perpendicular force , F = centripetal force (m r 2)                                     = (22000 kg) (20 m ) (2.30 rad /s)^2                                  F = 2.327 *10^6 N Torque , = 20 m (2.327 *10^6 N)                  = 4.65 *10^7 N-m Direction of the opposing torque is clock wise                        

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