Calculating Torque for Helicopter Blade Rotation

AI Thread Summary
To calculate the torque required for a helicopter blade to reach an angular speed of 320 rev/min in 6.70 seconds, the moment of inertia must be determined. The correct moment of inertia for a uniform rod about its end is 1/3 mL^2, which can be derived using the parallel axis theorem. The angular acceleration was calculated to be 5 rad/s², and using the formula T=Ia, the resulting torque needed is 7.6 x 10^3 N.m. It is crucial to use the correct moment of inertia to arrive at the accurate torque value. Proper application of these principles leads to the correct solution.
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Homework Statement



Consider a simple helicopter blade, attached to a bolt at its end. The bolt is made to rotate by a powerful motor, and the blade therefore rotates. The blade has a length L= 7.8 m and mass M= 100 kg. The blade is shaped approximately like a long thin, uniform rod.
What torque must the bolt exert on the blade in order to bring it from rest to full angular speed 320 rev/min in time t= 6.70 s

Homework Equations



Before, I worked out the force that ther bolt exerts on the blade when the blade is spinning at full speed to be 4.4 x 10^5 N. thus, force= 4.4 x 10^5 N. I thought that i could just find the angular acceleration: wf=wi + at and then find the moment of inertia and use T=Ia but i don't know what the moment of inertia would be. Would it be 1/12mL^2 or 1/3mL^2

The Attempt at a Solution


I found the angular acceleration to be 5/rad/s^2.. but i just don't know what the moment of inertia would be becoz i worked them out using the above equations and i keep getting the wrong answer. the right answer is 7.6 x 10^3 N.m. Is T=Ia the right forumla to be using or am i on the wrong track? Any help would be muchly appreciated! thankyou!
 
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9giddjl said:
… use T=Ia but i don't know what the moment of inertia would be. Would it be 1/12mL^2 or 1/3mL^2

Hi 9giddjl! :smile:

(try using the X2 tag just above the Reply box :wink:)

It's 1/12 mL2 about the centre of a rod,

and using the parallel axis theorem it's 1/12 mL2 + 1/4 mL2 = 1/3 mL2 about the end :wink:
 
i used 1/2 ML^2 for the moment of inertia and it yielded the correct answer! thankyou for ur help wise teacher!:)
 
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