Moment of inertie of rotor blaes of a helicopter

[SnowOwl18]

having some major issues with this unit, mainly since the homework was assigned and we haven't been lectured on it yet...so it's new to me. anyway, if anyone can help me out with this problem, that'd be awesome.

-----A helicopter rotor blade can be considered a long thin rod. If each of the three rotor helicopter blades is L = 3.85m long and has a mass of m = 158kg, calculate the moment of inertia of the three rotor blades about the axis of rotation.
How much torque must the motor apply to bring the blades up to a speed of 5.28 revolutions per second in 8.83s? --------

The inertia equation I found for this problem is I= 1/3 ml^2...but i wasn't sure if i calculate this for each blade or just once. And also, I don't know what the units are. The second part of the problem, T=Fl...so (3.85) (158)(9.8) = 5961.34 N*m...but how do i calculate bringing the bladed up to the given speed in a certain amount of time? Thanks for any help.

 

[Tide]

If the torque is constant then you should have no difficulty using

I \frac {d^{ 2} \theta}{dt^2} = \tau

to find the time required to bring the rotors up to the desired rotational speed.

 

[SnowOwl18]

k i'll try that for the second part of the problem, but i still need to solve the first part in order to do the second, i think. so my question still remains, do i need to use the equation i found for each of the blades? or do i just calculate it once? and what are the units? o and thanks for your help, tide. :)

 

[Tide]

You calculated the moments of inertia for each of the blades rotating about an axis perpendicular to their lengths so all you have to do is add them together to find the total moment. As for units, surely you can figure out the units if you know the units of length and mass.

 

[SnowOwl18]

ah yes, i should've realized that. thanks a bunch. but just one question...in the equation you gave, is d the length of each blade? and to get theta, i tried to take 5.28rps and multiplied by 2pi...

I did the equation, assuming d was the length as follows:

[(2342)(3.85^2)(33.175)] / [(3.85) (8.83^2)]...but my answer came out wrong. Sorry for all the questions btw, we just haven't been taught this chapter yet, but i need to begin the homework because it takes so long. Thanks

 

[Tide]

Well, no, d is not the length of a rotor. It stands for derivative which, apparently, you haven't covered yet.

Does this look more familiar to you?

I \omega = \tau \times t

 

[SnowOwl18]

oh thank you! yeah, I'm taking an algebra based physics course...i haven't learned calculus yet, only precalculus. ) Sorry that I didn't clarify that.

 

[CartoonKid]

what is the answer? Is it 8.8kNm for torque?

 

[Angie913]

I don't know if you got the answer, but I have the exact same problem on my CAP homework today. For the first part, the equation for the moment of inertia for a rod is (1/3)mr^2 so you use that get your answer and multiply by 3 for each blade. The second part you just find the angular velocity and use one of the equations for time ( like angular velocity final= angular velocity initial + angular acceleration x time ) and you can get your angular acceleration. Torque is Inertia times angular acceleration, so there you go.