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# Centre of mass

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A stone is dropped at t = 0. A second stone, with twice the mass of the first, is dropped from the same point at 100ms.

(a) How far below the release point is the center of mass of the two stones at t=300? (Neither stone has yet reached the ground.)

(b) How fast is the center of mass of the two-stone system moving at that time?

Kreyn  Mar 23, 2017
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+1109
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Solution(s)

$$\text {(a) }\\ \text {Find coordinates (y) for masses } \displaystyle m_1(y_1) \text { and } m_2(y_2)\\ y_1 = 0.5(9.81)*(0.300)^2 = 0.441 \small \text{meters}\\ y_2 = 0.5(9.81)*(0.300 – 0.100)^2 = 0.196 \small \text{ meters}\\ \\ \small \text{ Formula for the center of mass (2 objects):}\\ \\ y_{com} = \dfrac{(m_1y_1 + m_2y_2)}{(m_1 + m_2)}\\ y_{com} = \dfrac{((1) (0.441) + (2)(0.196)}{(1) + (2)} = 0.278 \small \text{ meters} \leftarrow \small\color{green} \text{Answer for (a)}\\ \text {(b) }\\ \text{Velocity of center of mass}\\ v1=(g)(t)_1 = 9.8*0.300 = \small \text{ 2.94m/s }\\ v2=(g)(t)_2 = 9.8*0.200 = \small \text{ 1.96m /s }\\ v_{com} = \dfrac{((1)(2.94)+(2)(1.96))}{3} = 2.287 \small \text{ m/s} \leftarrow \small\color{green} \text{Answer for (b)}\\\\ \small \text{ }\\ \small \text{Theory & Formulas: Complements of Archimedes of Syracuse, Leonhard Euler, Sir Isaac Newton, Pappus of Alexandria,}\\ \small \text{Guido Ubaldi, Francesco Maurolico, Federico Commandino, Simon Stevin, Luca Valerio, Jean-Charles de la Faille, Paul Guldin,}\\ \small \text{John Wallis, Louis Carré, Pierre Varignon, and Alexis Clairaut.}\\ \small \text{ } \hspace{20em} \scriptsize \text {(Probably a few others, too. The COM concept is ancient) }\\ \small \text{Produced by Lancelot Link and company. }\\ \small \text{Directed by GingerAle. }\\ \small \text{Sponsored by Naus Corp. Quantum Pharmaceuticals. }\\ \small \text{ } \hspace{15em} \scriptsize \text { Making a better world by neutralizing the quantum dumbness of }\\ \small \text{ } \hspace{15em} \scriptsize \text { Blarney Masters and related dumb-dumbs }\\$$

GingerAle  Mar 24, 2017