Centrifugal Force

If you consider a distributed mass "m" [kg/m] the force measured at radius "r" will be:

F= ∫(m*dr)*r*ω² integral from r to R where R= radius at end of string.

The result (in classical mechanics) will be F(r)= m*ω²*(R²-r²)/2.

The string supports the sum of forces required to maintain the masses on the circular path since they have the trend to follow a straight trajectory. The string is loaded with a centrifugal force because on the masses must act a centripetal force for the circular path.

Tx nick name, just trying to understand what you did. With m specified in kg/m, and I calculate it for the last meter, the applicable mass is just m, right? Now if R is fairly large, it means that R² - r² ≈ R² and F(R-1) ≈ m*ω²*R²/2, which somehow does not look right. Should it not boil down to just F(R-1) ≈ m*ω²*R?

What am I missing?

-J

The units! "m" being in kg/m it is NOT a concentrated mass. When you compute for a concentrated mass the radius is the distance between rotation center and COG of considered mass. In the way you put it would be F(1)=m*1*(R-1/2)*ω².

When the mass is uniformly distributed the mass element one has to consider is m*dr this is the reason I put it between brackets. This mass element is at the distance "r" to the rotation center thus the element of force will be dF=(m*dr)*r*ω². If you want to know the force generated in the string by the mass between r and R you must integrate the elementary force as:

F=∫dF from r to R → F=∫(m*dr)*r*ω² = m*ω²*(R²-r²)/2

Now let us look at the units: [m]=F*L^-1*T^2/L; [ω]= T^-1 ; [R²-r²]=L^2

→[F]= F* L^-2*T^2*(T^-1)²*L^2 = F so that the equation is CORRECT!

If you want to introduce the influence of v/c it is possible → dm= dmo/K where K is the correction factor which affects the mass as function of the speed at the considered level. "dmo" is the "static" mass element.

I hope that with those explanations all is clear if not I am every time available for more. If needed you may redirect the one who put the question to me.

Tx again nick name, as Codey pointed out, I've made a silly mistake with my "Now if R is fairly large, it means that R² - r² ≈ R² and F(R-1) ≈ m*ω²*R²/2".

Your equation does in fact give F(R-1) ≈ m ω² R for my example, as expected.

I will now contemplate the relativistic implications further along the lines that you suggested...

-J

Hello Jorrie

I agree with Nick's result F(r)= m*ω²*(R²-r²)/2. If you put r=R, F=0, as expected. If r=0, F=m*ω²*R²/2.

Considering the last meter, r=R-1. I think where you're going wrong is saying R>>r, implying R²-r² ≈ R². If R is large and r=R-1, R is not >> r, they're fairly close. Putting r=R-1, F(R-1)=m*ω²*(R²-(R-1)²)/2=

m*ω²*(2R-1)/2. If R>>1, F≈m*ω²*2*R/2. Alternatively, by factoring, R²-r² = (R-r)*(R+r). R-r=1 and R+r=2R-1≈2R.

Also need to be careful with units. It looks as if a length dimension has disappeared but of course it hasn't or there'd be something wrong. Strictly speaking it should be F(R-1)=m*ω²*(2R*1m-1m²)/2.

Cheers......Codey