full question: A helicopter rotor blade can be considered a long thin rod. Part A: If each of the three rotor helicopter blades is 3.65 m long and has a mass of 110 kg, calculate the moment of inertia of the three rotor blades about the axis of rotation?
Part B:How much torque must the motor apply to bring the blades from rest up to a speed of 5.8 rev/s in 7.8 s?
I got part A: 1470 kg*m^2
how do I go about solving part B?
Part B:How much torque must the motor apply to bring the blades from rest up to a speed of 5.8 rev/s in 7.8 s?
I got part A: 1470 kg*m^2
how do I go about solving part B?
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I'm not sure you calculated the moment of inertia quite right because I came up with a slightly different number (but they're only off by about 5 kg*m^2, so it might just be a rounding error).
To answer part B, first you figure out the angular acceleration using α=Δω/t (where α is the angular acceleration, Δω is the change in angular velocity in, and t is the amount of time), then once you have that you use τ=I*α (where τ is the torque, and I is the moment of inertia, and α must be in radians/sec^2). Δω and t are given and you've already calculated I, so the rest should be a piece of cake. Make sure you have your angular acceleration in units of radians/sec^2 before you use the torque equation or you'll get the wrong answer!
To answer part B, first you figure out the angular acceleration using α=Δω/t (where α is the angular acceleration, Δω is the change in angular velocity in, and t is the amount of time), then once you have that you use τ=I*α (where τ is the torque, and I is the moment of inertia, and α must be in radians/sec^2). Δω and t are given and you've already calculated I, so the rest should be a piece of cake. Make sure you have your angular acceleration in units of radians/sec^2 before you use the torque equation or you'll get the wrong answer!