Finding Centripetal Force in a Closed-Loop String

AI Thread Summary
To find the centripetal force in a closed-loop string, the mass at a point can be expressed using the differential mass element dm = Mds/2πR, where M is the total mass and R is the radius. The centripetal force on that element is calculated as dF = dm ω²r, with ω being the angular speed and r the distance from the axis of rotation. Concerns arise about integrating dF, as the formula appears to work for finite masses but seems problematic for continuous distributions. However, integrating over ds with a constant mass allows for the calculation of force at each point rather than the total force. The discussion emphasizes the need to adapt the approach for continuous mass distributions while acknowledging the validity of the derived formulas.
dragonlorder
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Homework Statement


I never thought I would have this kind of elementary problem
consider a string closed-loop spinning around an axis, and its shape is a circle, I wanted to find the centripital force at each point. (uniform density is assumed). I have problem expressing the mass


Homework Equations





The Attempt at a Solution


I apply F=ma , and F=m f(angle), the acceleration is obviously a function of the angle. But how do I write the mass at a point of a continuous string. The formula F=m f(angle) works for 1000, or 10000 mass points spinning around an axis, since the mass is given for each mass point, but what about a string, its continuous. But if I write dm (differential form), it would distort the formula...
 
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If you have a continuous string, you can pick a point on the loop and consider a piece of arc length ds at that point. Then if the mass of the loop is M and its radius R, the mass of your piece is

dm = Mds/2πR

Once you have this, the centripetal force on that element is

dF = dm ω2r

where ω is he instantaneous angular speed and r the distance of your element of mass dm to the axis of rotation.

Is this what you had in mind?
 
kuruman said:
If you have a continuous string, you can pick a point on the loop and consider a piece of arc length ds at that point. Then if the mass of the loop is M and its radius R, the mass of your piece is

dm = Mds/2πR

Once you have this, the centripetal force on that element is

dF = dm ω2r

where ω is he instantaneous angular speed and r the distance of your element of mass dm to the axis of rotation.

Is this what you had in mind?

yea, but I feel weird having to integrate dF. The formula works for finite masses, but as the number goes infinite, it fails. But the formula for continuous rope should have the form similar to that of finite. that's why I feel weird to integrate dF.
 
dragonlorder said:
yea, but I feel weird having to integrate dF.

Why do you want to integrate? The question, as you posted it, asks for the force at each point, not the total force.
The formula works for finite masses, but as the number goes infinite, it fails.

If you do integrate, the mass will be a constant; you will integrate over ds.
 
tms said:
Why do you want to integrate? The question, as you posted it, asks for the force at each point, not the total force.


If you do integrate, the mass will be a constant; you will integrate over ds.

thats exactly the problem I am having. how to modify that formula for the continuous situation
 
dragonlorder said:
thats exactly the problem I am having. how to modify that formula for the continuous situation

You have already been given the main part of the answer.
 
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