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?
Best Answer
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.