Roational Mechanics of Helicopter Blades

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Homework Help Overview

The discussion revolves around the rotational mechanics of helicopter blades, specifically focusing on calculating forces, torque, and work associated with the rotor blades during operation. The problem involves a uniform helicopter rotor blade with given dimensions and mass, and participants are tasked with determining the force on a bolt, the required torque to accelerate the rotor, and the work done during this process.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the use of formulas related to rotational motion, including force, torque, and work. There is discussion on the moment of inertia for different configurations of the rotor blade, with some questioning the assumptions made regarding the blade's mass distribution.

Discussion Status

Some participants have attempted calculations and shared their results, noting discrepancies with expected answers. There is an ongoing exploration of the correct moment of inertia for the blade's rotation, with suggestions to reconsider the axis of rotation. The discussion reflects a mix of attempts and clarifications without reaching a definitive consensus.

Contextual Notes

Participants mention confusion regarding the application of formulas and the impact of unit conversions. There is also a reference to external resources for moments of inertia, indicating a reliance on various sources for clarification.

Zyxer22
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Homework Statement



A uniform helicopter rotor blade is 8.82 m long, has a mass of 108 kg, and is attached to the rotor axle by a single bolt. (a) What is the magnitude of the force on the bolt from the axle when the rotor is turning at 302 rev/min? (Hint: For this calculation the blade can be considered to be a point mass at its center of mass. Why?) (b) Calculate the torque that must be applied to the rotor to bring it to full speed from rest in 7.14 s. Ignore air resistance. (The blade cannot be considered to be a point mass for this calculation. Why not? Assume the mass distribution of a uniform thin rod.) (c) How much work does the torque do on the blade for the blade to reach a speed of 302 rev/min?

Homework Equations



F=mv^{2}/r

F= mr\omega^{2}

Since it's a uniform rod;

I=\frac{1mL^{2}}{12}
T = I\alpha
\alpha = \omega/t
W=.5I\omega^{2}

The Attempt at a Solution



Using F= mr\omega^{2}
I changed 302rev/min to 31.625 rad/s
plugging in m= 108kg r = 4.41m

F= 4.76 * 10^5
-->This answer is wrong.


\alpha = \omega/t
\alpha ~ 4.429 rad/s^{2}

I=\frac{mL^{2}}{12}
T = I\alpha
T = 775.58 Nm
--> This answer is also wrong

W=.5I\omega^{2}
W=87530 J
--> Also wrong

Any help would be greatly appreciated, thanks :)
 
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This question is still getting the better of me if anyone could help.
 
Zyxer22 said:
Using F= mr\omega^{2}
I changed 302rev/min to 31.625 rad/s
plugging in m= 108kg r = 4.41m

F= 4.76 * 10^5
-->This answer is wrong.
I can't see any mistake, other than units of F have been omitted.

\alpha = \omega/t
\alpha ~ 4.429 rad/s^{2}

I=\frac{mL^{2}}{12}
T = I\alpha
T = 775.58 Nm
--> This answer is also wrong
I believe moment of inertia for a rod rotated about its end is I = (m L2) /3

http://en.wikipedia.org/wiki/List_of_moments_of_inertia"
 
Last edited by a moderator:
Right, rotated about it's end the moment of inertia is I = (m L2) /3
I was assuming that the rotational axis is in the center of the blade though, leading to my equation.
Using I = (m L2) /3 actually leads to correct answers. Thanks ^^
 
Scratch that... I get the correct answers according to the book, but using the numbers from online I get the wrong answers... makes no sense to me
 

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