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10.6: Sample problems and solutions
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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