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Suppose a horizontal rod (ignore gravity) of mass $m$ and length $L$ is rotating about one of its ends with constant angular velocity $ω$. Then, the tension must decrease as we move away from the axis. Same goes for the centripetal force. Then on an element $dx$ at $x$ distance from the axis the forces must be $T$ and $T-dT$. But if we wish to write the centrifugal force, how would it vary? Shouldn't it increase as we move away from the axis as there is more and more material to pull away? And in that case, shouldn't the tension also increase? But tension is a real force, and therefore can't depend on frame of reference.

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  • $\begingroup$ Only the tension is physically there! The tension is the physically real thing that provides the centripetal acceleration. Centrifugal effect only appears if you are in the rotating frame of reference. It will equal the centripetal acceleration, but opposite in direction. You should try to understand the thing without centrifugal, first. $\endgroup$ May 2, 2023 at 5:41
  • $\begingroup$ I understand that the centripetal force is provided by a real force, but why does the variation appear to be opposite from a rotating frame? $\endgroup$ May 2, 2023 at 6:03
  • $\begingroup$ The proper way to learn about centrifugal effect is to mathematically derive it. The whole thing is a stupid trick, and we should not explain anything in terms of it as much as we can. $\endgroup$ May 2, 2023 at 6:09

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You are missing the point that if you are rotating with the rod, the rod is not accelerating (moving) relative to you.

The force causing the centripetal acceleration still exists so to make Newton's law work a fictitious (centrifugal) force, of the same magnitude and in the opposite direction, is added so that the net force on any element of the rod is zero.

Thus, in the rotating frame, the tension and the centrifugal force are always equal in magnitude.

Consider an element of the rod mass $\Delta m$ at a distance $r$ from the axis of rotation, in an inertial frame $F=ma\Rightarrow T(r)-T(r+\Delta r) = \Delta m \,r \,\omega^2$ and in the rotating frame $F=ma\Rightarrow T(r)-T(r+\Delta r) -\Delta m \,r \,\omega^2= \Delta m \,0=0$.


The last bit of my answer has caused some debate so let me explain what I did.

A centripetal force produces an acceleration $-\omega^2\,\vec r = -\omega^2\,r\,\hat r$ where $\vec r=r\,\hat r$ is measured from the axis of rotation.
Consider an element of the rod between $\vec r$ and $\vec r+d\vec r$ with the forces acting on the element due to the tension in the rod being $\vec T(\vec r)=T(\vec r)\,\hat r$ and $\vec T(\vec r+d\vec r)=T(\vec r+d\vec r)\,\hat r$ respectively.

Using Newton's second law,

$\vec F = m \vec a \Rightarrow T(\vec r+d\vec r)\,\hat r - T(\vec r)\,\hat r= dm\,(-\omega^2\,r\,\hat r)\Rightarrow T(\vec r) - T(\vec r+d\vec r)=dm\,\omega^2\,r$,

where the mass of the element is $dm$.

This is the equation I used to illustrate switching the $dm\,\omega^2\,r$ term from the right for an inertial frame to the left for a non-inertial frame.

As to what $dT$ is?

$T(\vec r+d\vec r)\,\hat r - T(\vec r)\,\hat r= dT\,\hat r \Rightarrow T(\vec r+d\vec r) - T(\vec r)= dT$.

If $dm = \rho \,dr$ where $\rho$ is the mass per unit length of the rod, $dT=-\rho \,\omega^2\, r\,dr$.

Knowing that $T(r=L) = 0$ where $L$ is the length of the rod and integrating the expression yields,

$T(r) = -\frac {\rho \,\omega^2}{2}(L^2-r^2)$.

Thus the component of the tension is always negative, which means that the tension at $r$ is in the $(-\hat r)$ direction, ie towards the axis of rotation.

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  • $\begingroup$ Am I right in saying that $T(r+\Delta r)$ would be $T-dT$ or should it be $T+dT$? $\endgroup$ May 2, 2023 at 7:13
  • $\begingroup$ The tension force is weaker as it goes out to the tips. $\endgroup$ May 2, 2023 at 8:17
  • $\begingroup$ @naturallyInconsistent Using $T-dT$ gives negative value for tension. See this. $\endgroup$ May 2, 2023 at 8:41
  • $\begingroup$ @Farcher why do you take $−\hat r$ ? Is it because $\hat r$ points outward and $\vec a$ inward? Doesn't this imply that tension points outward? $\endgroup$ May 2, 2023 at 12:16
  • $\begingroup$ @Farcher why do we get a negative quantity? $\endgroup$ May 2, 2023 at 14:31
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The centrifugal force

$$ dF_c=dm\,\omega^2\,x $$

where x is the distance from the rotations axes . if the rotation object is a uniform the area $~A(x)=A~$. thus the mass element is: $$dm=\rho\,A\,dx$$

if you "cut" the rod at x you obtain

$$T(x+dx)-T(x)=dT=dm\,\omega^2\,x=\rho\,A\,\omega^2\,x\,dx$$

integrate

$$T(x)=\frac 12\,\rho\,A\,\omega^2\,x^2+c$$

with

$$T(L)=0\quad,\Rightarrow c=-\frac 12\,\rho\,A\,\omega^2\,L^2\quad\Rightarrow\\ T(x)=\frac 12\,\rho\,A\,\omega^2\,\left(x^2-L^2\right)$$

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