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    • https://phys.libretexts.org/Courses/Berea_College/Introductory_Physics%3A_Berea_College/10%3A_Linear_Momentum_and_the_Center_of_Mass/10.06%3A_Sample_problems_and_solutions
      The two are related by: \[\begin{aligned} ds = Rd\theta\end{aligned}\] The mass element, \(dm\), can then be expressed in terms of the mass per unit length of the wire and the length, \(Rd\theta\), of...The two are related by: \[\begin{aligned} ds = Rd\theta\end{aligned}\] The mass element, \(dm\), can then be expressed in terms of the mass per unit length of the wire and the length, \(Rd\theta\), of the mass element: \[\begin{aligned} dm = \lambda ds = \lambda Rd\theta\end{aligned}\] We also need to express the \(y\) position of the mass element using \(\theta\): \[\begin{aligned} y = R\sin\theta\end{aligned}\] Now that we have expressed \(dm\) and \(y\) in terms of \(\theta\), we can determi…

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