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A rod of total mass $M$ and length $x$ has its tip attached to a string of negligible mass, so it spins in a circular motion horizontally free from friction with constant tangential velocity.

I have observed that the tension in the string can found via the integral

$$\text{tension}=(V_{\text{tangent}})^2 \int_0^x\mathrm dm/\text{length}$$

where $\mathrm dm$ is a small element of mass located along the rod.

However, suppose the rod is still attached to the string in a similar manner and instead I am only given its total mass $M$ and the location of its center of mass; how would I find the tension of the string in this scenario given that the rod is not attached at the point of its center of mass?

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Your equation is wrong because $V_{tangent}$ varies along the rod. It should be inside the integral.

Finding the tension in the string by the 2nd method is much more sensible. All you need to do is apply the usual formula for centripetal force : $T=Mv^2/R$ where $R$ is the radius of the circle traced out by the CM of the rod.

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  • $\begingroup$ For the second scenario i said we know the center of mass on the rod, but the rod is attached to the string NOT on the point of its center of mass, but yeah I now realize Vtangent varies $\endgroup$
    – Prandals
    Commented Jan 23, 2017 at 20:12
  • $\begingroup$ Yes. $R$ is the distance from the CM of the rod to the centre of the circle. If the rod is uniform and the length of string is $L$ then $R=L+\frac12x$. $\endgroup$ Commented Jan 23, 2017 at 20:15
  • $\begingroup$ What if it wasnt uniform? Would knowing the center of mass be useless? $\endgroup$
    – Prandals
    Commented Jan 23, 2017 at 20:24
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    $\begingroup$ $T=Mv^2/R$ applies even if the rod is not uniform. Knowing that the rod is uniform helps in locating the CM. If the rod is not uniform but you are told (or can otherwise calculate) the position of the CM relative to the end of the rod, you can still use the same formula. ... The integral method and the CM method will give you the same answer for $T$. $\endgroup$ Commented Jan 23, 2017 at 20:28
  • $\begingroup$ Remember that V(r) in this case is proportional to r, since it is rotating as a solid. So that means that V^2(r)/r is also proportional to r. With that in mind you should fairly easily be able to show the indicated result with regard to the COM. $\endgroup$
    – user93146
    Commented Jun 21, 2018 at 15:28

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