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Imagine something lika a helicopter rotor and blade, where you could measure the centripetal force from the centre of axis to the very base of the blade. The blade is uniform in profile and all of the centripetal force due to the mass of the blade is exerted on the first section of the blade.

The equation for centripetal force is $Fc = mv^2/r$.
where m= mass kg, v= velocity in radians, r= radius in metres, so the force on any unit length of blade is greater near the centre of rotation than near the tip.

How can I calculate the force due to a blade extending from say 0.1 m to 2m from the centre of rotation and probably plot it against angular velocity. Say the mass per metre length is 2kg. My limited knowledge of maths suggest integration is required?

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You can safely assume that a helicopter blade is a thin long rod. Denote the length of the helicopter blade with $l$. Introduce coordinate $x$ running from the rotor axis along the blade of the rotor. Obviously $x$ goes from 0 to $l$.

Now consider a small piece of the blade of length $dx$. What is the mass of that piece of the blade. If the whole blade has mass $m$, mass per unit of length is $m/l$. So the mass of the small piece of the blade is:

$$dm=\frac ml dx$$

The centripetal force coming from this small piece of the rotor is:

$$dF=dm\frac{v^2}{x}$$

...where $v$ denotes the veocity of the blade part. It's not particularly suitable to work with so we'll use the fact that:

$$v=x\omega$$

...where $\omega$ denotes the angular velocity of the rotor. Now we have:

$$dF=\frac ml \omega^2xdx$$

The force $dF$ is oriented along the blade so you can safely summarize contributions to the cenripetal force coming from different blade parts. In particular, for a part of the rotor starting at $x=x_1$ and ending at $x=x^2$ the centripetal force is:

$$F=\int_{x_0}^{x_1}\frac ml\omega^2 xdx=\frac{m\omega^2}{2l}(x_1^2-x_0^2)$$

For the whole blade, $x_0=0$ and $x_1=l$, the centripetal force at the root of the blade is:

$$F=\frac{ml\omega^2}2$$

Note that $\omega$ is angular velocity of the rotor in radians per second. If you are, for example, given a number of rotations per minute $n$, you'll have to recalculate:

$$\omega=\frac{2\pi n}{60}=\frac{\pi n}{30}$$

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  • $\begingroup$ Thanks for that, I cant quite see how you get from the left to the right hand side of equation 5, $\endgroup$ – user36093 Jun 12 '19 at 8:38
  • $\begingroup$ It's polite to upvote/accept the answer if you find it useful. $\endgroup$ – Oldboy Jun 12 '19 at 10:34
  • $\begingroup$ I did but but the upvote doesnt show because I am new, apparently. $\endgroup$ – user36093 Jun 12 '19 at 15:53

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