Roational Mechanics of Helicopter Blades

AI Thread Summary
The discussion revolves around calculating the forces, torque, and work related to a helicopter rotor blade with specific dimensions and mass. The initial calculations for force using the formula F=mrω² were incorrect due to a misunderstanding of the blade's moment of inertia, which should be I=(mL²)/3 for a rod rotating about its end. The user also faced challenges in determining the torque and work done, initially arriving at incorrect values. After clarification, the correct moment of inertia was applied, leading to accurate results. The conversation highlights the importance of understanding the physical principles behind rotational mechanics in solving such problems.
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 ^^
 
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