Rotation Kinetic Energy of Two Helicopter Blades

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SUMMARY

The discussion focuses on calculating the total moment of inertia and rotational kinetic energy of two helicopter blades, each with a mass of 230 kg and a length of 6.7 m, rotating at an angular speed of 41 rad/s. The moment of inertia for a thin rod is determined using the formula I = (1/3) * m * L², resulting in a total moment of inertia of 2 * (1/3) * 230 kg * (6.7 m)². The rotational kinetic energy is then calculated using the formula KE = (1/2) * I * ω², where ω is the angular speed.

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A helicopter has two blades (see figure), each of which has a mass of 230 kg and can be approximated as a thin rod of length 6.7 m. The blades are rotating at an angular speed of 41 rad/s. (a) What is the total moment of inertia of the two blades about the axis of rotation? (a) Determine the rotational kinetic energy of the spinning blades.

i know i have to use Torque=I * angular acceleration
but i don't know how all that fits in?
 
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i know i have to use Torque=I * angular acceleration

Are you sure about that? The question is asking for the total moment of inertia, which you calculate based on the size and mass of the blades (well you look them up in your book). What's the equation for the moment of inertia of a thin rod?
 

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