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There is a circle of rope that rotates at a uniform angular velocity $ω$. What is the formula for calculating the tension of the rope section? Without gravity, the density of the rope is $ρ$, the radius of the rope circle is $R$, and the section radius of the rope is $r$.

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Consider an infinitesimally small section of the of the string $d\theta$. The following diagram illustrates this:

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The tension is of the same magnitude throughout the rope, and it acts perpendicular to the vector from the center of the string to the point of action.

From this diagram, you can tell that only the x-components to the left matter, since the y-components of the tensions cancel out. The x-components of the two tension vectors must be equal to the force required for centripetal acceleration.

$$2T\text{ sin}(d\theta/2) = Td\theta = (dm)\omega^2R$$

The small bit of mass can be found as follows:

$$dm = \rho dV = \rho A dx = \rho \pi r^2 (r d\theta) = \rho \pi r^3 d\theta$$

Then, cancelling out the $d\theta$ on both sides of the equation:

$$\boxed{T = \pi \rho r^3 \omega^2 R}$$

Note: This solution assumes $r \ll R$.

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  • $\begingroup$ When r / R is about what value is r < < R? Is the unit of T Newton? $\endgroup$
    – enbin
    Jan 16 '20 at 5:10
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    $\begingroup$ Usually $r$ is considered much less than $R$ if $r/R$ is around 0.1 or less. T is in newtons: dimensionally, you notice that $\rho$ is $kg/m^3$, while $r^3$ is $m^3$, and $\omega^2$ is $1/s^2$, and $R$ is in $m$, which results in a total of $kg*m*s^{-2}$ which is Newtons. $\endgroup$
    – wheelix
    Jan 17 '20 at 4:51
  • $\begingroup$ physics.stackexchange.com/q/524670/176092, In the video of this problem, the rope rises into the sky maybe because the weight of the rope at the left end does not provide enough tension t. $\endgroup$
    – enbin
    Jan 17 '20 at 6:52
  • $\begingroup$ What if the angular velocity is increasing? $\endgroup$
    – enbin
    Jan 18 '20 at 21:31

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